{ "schema": "cie-topical-past-papers/answers-v1", "subject": "9701", "subject_name": "Chemistry", "generated": "2026-10-21", "coverage": { "question_records": 443, "available_mark_scheme_entries": 443, "unavailable_mark_scheme_entries": 0 }, "records": [ { "id": "9701-2021-m-42-q01", "question_id": "9701-2021-m-42-q01", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 8, "status": "available", "reason": null, "text": "1(a) Co2+ = [Ar] 3d7 (4s0) 1\nCo3+ = [Ar] 3d6 (4s0)\n\n1(b) M1/2: Any two of: 3\n• Co3+ is reduced Co2+\n• oxygen gas/O is evolved\n2\n• E of Co3+ greater than E of O\n2\nM3: no change (to [Co(edta)]–) / not feasible OWTTE\n\n1(c) Any two of VISUAL observations: 2\n• condensation on tube / steam evolved\n• brown fumes / brown gas evolved\n• O formed that relights a glowing splint\n2\n• (solid) dissolves / turns to liquid\n\n1(d) M1: cationic radius / ion size increases (down the group) 2\nM2: less polarisation / distortion of nitrate ion / anion / NO –\n3\nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p05.png" ] }, { "id": "9701-2021-m-42-q02", "question_id": "9701-2021-m-42-q02", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 13, "status": "available", "reason": null, "text": "2(a)(i) M1 the only number extracted: 762, 1560, 496 2\nM2 correct multiplier, other four numbers used and calculation to the answer\n–272 = +416 + ½(496) + 762 + 1560 –141 + 798 + ΔH\nlattice\n∴ ΔH = –3915 (kJ mol–1) ecf\nlattice\n\n2(a)(ii) 20 × [0.9(+2) + 0.1(+3)] –2x = 0 ∴ x = 21 1\n\n2(a)(iii) • FeO more exothermic/more negative 2\n• Fe2+ has smaller radius/higher charge density (also same charge)\n• greater attraction/ stronger ionic bonds (between Fe2+ and O2–)\nAll three for two marks\n© UCLES 2021 Page 5 of 13\n\n2(b)(i) • Fe2+ reduced to Fe OR oxid no. Fe +2 → 0 1\n• Fe2+ oxidised to Fe3+ (in Fe O ) OR oxid no. Fe +2 → +3\n3 4\nBOTH bullets required\n\n2(b)(ii) 2O2– → O + 4e– 1\n2\n\n2(b)(iii) M1: coulombs and correct use of ÷ 96500 3\nM2: correct use of 3 and 8\nM3: correct use of 55.8 and answer\nM1: Q = It = 50 × 6 × 602 OR 1.08 × 106 C\nAND no. of faraday = 1.08 × 106 ÷ 96500 OR 11.2 / 11.19 mol e–\nM2: Fe2+ + 2Fe3+ + 8e– → 3Fe\n∴ moles of Fe = 3 / 8 × M1 = 4.20 mol Fe ecf\nM3: mass of Fe = 55.8 × M2 = 234.2 g ecf 3sf min\n\n2(c)(i) Any one of: 1\nsmall size / compact, low mass, high voltage OWTTE\n\n2(c)(ii) Li from +1 to +1 1\nFe from +3 to +2\n\n2(c)(iii) LiC + FePO → LiFePO + 6C 1\n6 4 4\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p05.png", "../answer-assets/9701_m21_ms_42-p06.png" ] }, { "id": "9701-2021-m-42-q03", "question_id": "9701-2021-m-42-q03", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 18, "status": "available", "reason": null, "text": "3(a) 3 bonding-pair centres and one lone pair (on iodine) 1\n\n3(b) 3I + 6NaOH → NaIO + 5NaI + 3H O 1\n2 3 2\n© UCLES 2021 Page 6 of 13\n\n3(c)(i) M1: E⦵ for IO – / H O = –0.68 + 1.19 = +0.51 (∴ feasible) 3\ncell 3 2 2\nM2: E⦵ for H O / I = +1.77 – 1.19 = +0.58 (∴ feasible)\ncell 2 2 2\nM3: 5H O + I → 4H O + 2IO – + 2H+\n2 2 2 2 3\n\n3(c)(ii) 2H O → 2H O + O 1\n2 2 2 2\n\n3(d)(i) M1: first order w.r.t. H O AND change in conc. × 1.5 gives increase rate × 1.5 (expts 3 / 4) 3\n2 2\nM2: first order w.r.t. IO – AND change in conc. × 2 gives increase rate × 2 (as reaction first order w.r.t. H O ) (expts 1 / 3)\n3 2 2\nM3: zeroth order w.r.t. H+ AND change in conc. has no effect on rate (expts 1 / 3 / 4 and 2)\n\n3(d)(ii) rate = k[H O ][IO –] ecf 1\n2 2 3\n\n3(d)(iii) M1: k = 8.82 × 10–5 ÷ (0.150 × 0.140) = 4.20 × 10–3 min 2sf ecf 2\nM2: mol–1 dm3 s–1 ecf\n\n3(e)(i) K = [Pb2+][IO –]2 1\nsp 3\n\n3(e)(ii) M1: 3.69 × 10–13 = x(2x)2 OR x = ∛(3.69 × 10–13 ÷ 4) 2\nM2: = 4.5(2) × 10–5 (mol dm–3) min 2sf ecf\n\n3(f)(i) M1: ΔS = ½(192) + ½(205) + ½(261) + 2(70) – 42 2\nM2: (+)427 (J K–1 mol–1) ecf\n\n3(f)(ii) ΔG (always) negative because 1\n• ΔH < 0 / negative OR exothermic AND\n• ΔS > 0 / positive OR –TΔS < 0 for all T\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p06.png", "../answer-assets/9701_m21_ms_42-p07.png" ] }, { "id": "9701-2021-m-42-q04", "question_id": "9701-2021-m-42-q04", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 11, "status": "available", "reason": null, "text": "4(a) (element that forms one or more stable) ions with incomplete/ partially filled 3d-orbitals/d-subshell 1\n\n4(b)(i) 1\nO AND T\nh d\n© UCLES 2021 Page 7 of 13\n\n4(b)(ii) M1: lower energy level (in between axes) 2\nOR OR\nM2: higher energy level (on the axes)\nOR\n\n4(c)(i) Circles round both N atoms and all four O– 1\n\n4(c)(ii) M1: (d–d) energy gap / ΔE is different 2\nM2: different frequency / wavelength (of light) absorbed\n\n4(c)(iii) ligand exchange / substitution / displacement / replacement 1\n\n4(c)(iv) Fe ( edds )–  1\n \nK =\nstab Fe ( H O ) 3+edds4–\n 2 6  \n\n4(c)(v) [Fe(edta)]– is more stable as it has the higher K 1\nstab\n\n4(c)(vi) K (edta) 1.26×1025 1\nK = stab = =3.17 × 104 (31658) min 2sf\nc K (edds) 3.98×1020\nstab\n© UCLES 2021 Page 8 of 13", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p07.png", "../answer-assets/9701_m21_ms_42-p08.png" ] }, { "id": "9701-2021-m-42-q05", "question_id": "9701-2021-m-42-q05", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 12, "status": "available", "reason": null, "text": "5(a)(i) M1: (cisplatin) can bond / bind with DNA / (nitrogenous) base A, T, C, G etc. 2\nM2: which prevents replication (of the DNA / strand)\nOR prevents cell division / prevents mitosis\nOR prevents transcription (and formation of mRNA)\n\n5(a)(ii) square planar 1\n\n5(a)(iii) the distance between two coordinating oxygens is too small to bond trans 1\nOR atoms in a bidentate ligand can only bond 90° not 180°\n\n5(a)(iv) +4 1\n\n5(b)(i) 1\nL\n\n5(b)(ii) M1: heat / reflux with acidified / alkaline KMnO4 (then acidify) 2\nM2: PCl OR SOCl / (heat with) PCl\n5 2 3\n\n5(b)(iii) C H O + 2PCl → C H O Cl + 2POCl + 2HCl 1\n8 6 4 5 8 4 2 2 3\nOR C H O + 2SOCl → C H O Cl + 2SO + 2HCl\n8 6 4 2 8 4 2 2 2\nOR 3C H O + 2PCl → 3 C H O Cl + 2H PO\n8 6 4 3 8 4 2 2 3 3\n\n5(b)(iv) 3\nM1: curly arrow from inside hexagon to C of electrophile\nM2: correct intermediate\nM3: curly arrow from C—H bond AND formation/loss of H+\n© UCLES 2021 Page 9 of 13", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p09.png" ] }, { "id": "9701-2021-m-42-q06", "question_id": "9701-2021-m-42-q06", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Polymerisation", "topic_slug": "9701-topic-35-polymerisation", "marks": 10, "status": "available", "reason": null, "text": "6(a) CO and H O / in words 1\n2 2\n\n6(b)(i) 1\nif more than one unit drawn ALLOW one repeat unit identified\n\n6(b)(ii) 2\nM1: presence of an ester group from the diol and COOH OR presence of an ester group from the fumaric acid and OH\nM2: rest of repeat unit including ‘dangling’ bonds\n\n6(b)(iii) C—C bonds are non-polar / polyalkenes cannot be hydrolysed 1\nOR polyesters / they can be broken down by hydrolysis\n\n6(c) 2\n© UCLES 2021 Page 10 of 13\n\n6(d) 3\nM1: (can be in words or diagram) substrate shape is complementary to active site\nM2: (can be in words or diagram) the substrate bind / bonds / fits (into the active site)\nM3: (can be in words or diagram) products are released\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-35-polymerisation/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p10.png", "../answer-assets/9701_m21_ms_42-p11.png" ] }, { "id": "9701-2021-m-42-q07", "question_id": "9701-2021-m-42-q07", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 28, "status": "available", "reason": null, "text": "7(a)(i) 2\nOR\nM1: peptide link shown\nM2: rest of Pro–Gly correct\n\n7(a)(ii) condensation ALLOW substitution / addition–elimination 1\n\n7(a)(iii) there is no H attached to the N 1\n\n7(b)(i) (C H NHCO H +) NaOH → C H NHCO Na + H O 1\n4 7 2 4 7 2 2\n\n7(b)(ii) 1\nskeletal only\n© UCLES 2021 Page 11 of 13\n\n7(b)(iii) LiAlH 1\n4\n\n7(c)(i) CH (CO C H ) : ● (di)ester 2\n2 2 2 5 2\nCH =CHCN: ● alkene ● nitrile/cyanide\n2\nAll three correct for two marks\n\n7(c)(ii) addition 1\n\n7(c)(iii) H / Ni OR H / Pt OR H / Pd 1\n2 2 2\n\n7(c)(iv) condensation / (nucleophilic) substitution / elimination 1\n\n7(c)(v) ethanol / C H OH / CH CH OH 1\n2 5 3 2\n\n7(c)(vi) 3\nM1/2: All four correct:\n• lone pair on NH\n2\n• curly arrow from N: to C of C—Cl\n• correct dipole on C—Cl\n• curly arrow from C—Cl to Cl\nM3: intermediate =\nOR\n\n7(c)(vii) Asterisk on *CHCO H 1\n2\n© UCLES 2021 Page 12 of 13\n\n7(d) 9 1\n\n7(e)(i) M1: 2\nM2: proton / H+ transferred from carboxylic acid to amine\n\n7(e)(ii) 4\nM1: glutamic acid towards + end (from the diagram)\nM2: proline and alanine towards – end (from the diagram)\nM3: Glu moves towards positive (pole) as negatively charged / contains a COO–\nOR Pro/Ala move towards negative (pole) as positively charged / contains a NH + / contains a NH +\n2 3\nM4: Ala moves farther than Pro because of lower M / size (with positive charge) ORA\nr\n\n7(f) M1: initial amount of C H N O SNa = 3.50 / 204.1 OR 0.0171(48) mol 4\n4 9 2 4\nAND amount of HCl added = 0.200 × 50.0 / 1000 OR 0.0100 mol\nM2: equilibrium amount of C H N O SNa = 0.0171(48) – 0.0100 OR 0.0071(48) mol\n4 9 2 4\nAND equilibrium amount of ACES = 0.0100 mol ecf\nM3: K = 10–6.88 = 1.32 ×10–7 (mol dm–3)\na\n[H+] = (1.32 × 10–7)0.01 / 0.0071(48) = 1.86 × 10–7 OR 1.8465 × 10–7 ecf\nM4: pH = –log(1.86 × 10–7) = 6.73 3sf min ecf\n© UCLES 2021 Page 13 of 13", "source_pages": [ 11, 12, 13 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_42-p11.png", "../answer-assets/9701_m21_ms_42-p12.png", "../answer-assets/9701_m21_ms_42-p13.png" ] }, { "id": "9701-2021-m-52-q01", "question_id": "9701-2021-m-52-q01", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 10, "status": "available", "reason": null, "text": "1(a) 1 °C 1\n\n1(b) M1: diagram of labelled polystyrene cup in a beaker AND containing labelled aqueous copper(II) sulfate 2\nM2: diagram of free-standing labelled thermometer with bulb clearly within the solution\n\n1(c) M1: plots after 4 mins and curved line of best fit back from 10 mins to 5 min 2\nM1: extrapolation to 4 mins and correct subtraction to determine ΔT\n\n1(d) M1: energy released = –25.(0) × 4.18 × ΔT in 1(c) = (–2560.5) J 2\nM2: enthalpy change of reaction = (M1 / 1000) / 1.25 × 10–2 = –205 (kJ mol–1)\n\n1(e)(i) no change / effect 1\nAND zinc is (already) in excess\n\n1(e)(ii) (heat loss) increases 1\nAND increased energy output / temperature rise OR more exothermic change occurs\n\n1(e)(iii) no change / effect 1\nAND twice the energy output nullified / cancelled out by twice the volume of water to be heated\nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_52-p05.png" ] }, { "id": "9701-2021-m-52-q02", "question_id": "9701-2021-m-52-q02", "subject": "9701", "year": 2021, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 20, "status": "available", "reason": null, "text": "2(a) M1: (250 cm3) volumetric flask 4\nM2: dissolve the solid / acid salt (1.89 g) (in the beaker) (using distilled) water\nM3: transfer / add to a (250 cm3 volumetric) flask AND rinse (with distilled water)\nM4: top up to mark (with distilled water)\n\n2(b)(i) to ensure that (exactly) 25.0 cm3 of solution has been delivered into the flask 1\n\n2(b)(ii) measuring cylinder 1\n\n2(b)(iii) difficult to see the 0.00 cm3 line 1\n\n2(b)(iv) (two) titres / readings within 0.1(0) (cm3) of each other 1\n© UCLES 2021 Page 5 of 6\n\n2(b)(v) colourless to pale purple 1\n\n2(b)(vi) M1: mol MnO –(aq) = 0.0200 × 24.40 / 1000 = 4.88 × 10–4 (mol) 3\n4\nM2: mol of C O 2–(aq) (in 25.0 cm3) = M1 × 5 / 2 = 1.22 × 10–3 (mol)\n2 4\nM3: mol of C O 2–(aq) (in solution A) = M2 × 250 / 25 = 1.22 × 10–2 (mol)\n2 4\n\n2(c)(i) (aqueous) sodium hydroxide 1\n\n2(c)(ii) M1: mol H+(aq) in 250 cm3 = 6.10 × 10–2 / 4 = 1.525 × 10–2 (mol) 3\nM2: ratio C O 2–(aq) : H+(aq)\n2 4\n(1.22 × 10–2 / 1.22 × 10–2) : 1.525 × 10–2 / 1.22 × 10–2 = 1 : 1.25 and b = 1.25\nM3: 2 – 1.25 a = 0.75\n\n2(c)(iii) M1: M of hydrated K H C O •dH O = 1.89 / 1.22 × 10–2 = 154.9 2\nr 0.75 1.25 2 4 2\nOR M of anhydrous K H C O = [(0.75 × 39.1) + (1.25 × 1.0) + (2 × 12.0) + (4 × 16.0)] = 118.575\nr 0.75 1.25 2 4\nM2: mass of water = hydrated mass – anhydrous mass = 154.9 – 118.575 = 36.325\nAND 36.325 / 18 = 2\n\n2(d) M1: initial readings 2\nmass of (empty) crucible / g\nmass of crucible + crystals before heating / g\nmass of crucible + crystals after heating / g\nM2: second reading of mass of crucible + crystals after re-heating\n© UCLES 2021 Page 6 of 6", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-March/ms/9701_m21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-march/9701_m21_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m21_ms_52-p05.png", "../answer-assets/9701_m21_ms_52-p06.png" ] }, { "id": "9701-2021-mj-41-q01", "question_id": "9701-2021-mj-41-q01", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "1(a) M1: increases down the group 3\nM2: radius / size of (cat)ion / M2+ increases\nM3: less polarisation / distortion of anion / hydroxide ion / hydroxide group / OH– / OH\n\n1(b)(i) Ca(OH) (s) ⇌ Ca2+(aq) + 2OH–(aq) 1\n2\n\n1(b)(ii) M1: 2\nK = [Ca2+][OH–]2\nsp\nOR\nK = 4x3\nsp\nM2: x = 3√5.02 × 10–6 / 4 = 0.0108 / 0.011 / 1.08 × 10–2 / 1.1 × 10–2 (mol dm–3) min 2 sf\n\n1(b)(iii) less soluble / decreases due to the common ion effect 1\nOR\ndecreases as equilibrium in (b)(i) has shifted to the left\nOR\ndecreases as [OH–] increases causing [Ca2+][OH–](2) to exceed its K\nsp\n© UCLES 2021 Page 5 of 17", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p05.png" ] }, { "id": "9701-2021-mj-41-q02", "question_id": "9701-2021-mj-41-q02", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 15, "status": "available", "reason": null, "text": "2(a) M1: (complexes have two sets of) d orbital(s) of different energy / d-d splitting occurs 4\nOR d orbital(s) / d (sub)-shell splits\nOR (inferred from a movement of an electron) from a lower d to higher d orbital\nM2: electron(s) promoted / excited\nOR electron(s) moves to higher (d–)orbital\nOR electron(s) jumps up (to d–orbital) / jumps to higher (d–orbital)\nM3: wavelength / frequency / light / photon / hν absorbed\nOR radiation / energy from visible (region) absorbed\nM4: colour seen is complementary (to colour absorbed)\nOR wavelength / frequency / colour / light not absorbed is transmitted / reflected / seen\n© UCLES 2021 Page 6 of 17\n\n2(b)(i) 4\nM1: All charges\nM2: One octahedral with correct 3D\nM3: All formulae\nM4: R is cis, S is trans\n\n2(b)(ii) dipoles cancel 1\n\n2(c)(i) M1: (a species) that donates two lone pairs / forms two coordinate bonds / two dative bonds 2\nM2: to a metal atom / metal ion\n© UCLES 2021 Page 7 of 17\n\n2(c)(ii) 1\nstructure of the picolinate anion ligand\n\n2(c)(iii) (coordination number) six 1\nAND (geometry around Cr) octahedral\n\n2(d)(i) (NH ) Cr O +6 1\n4 2 2 7\nCr O +3\n2 3\n\n2(d)(ii) (NH ) Cr O → N + Cr O + 4H O 1\n4 2 2 7 2 2 3 2\nQuestion Answer Marks", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p06.png", "../answer-assets/9701_s21_ms_41-p07.png", "../answer-assets/9701_s21_ms_41-p08.png" ] }, { "id": "9701-2021-mj-41-q03", "question_id": "9701-2021-mj-41-q03", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 17, "status": "available", "reason": null, "text": "3(a)(i) (an element) forming stable ion / ions / compound(s) / oxidation state(s) 1\nAND with partially filled / incomplete\nAND d orbitals / d subshell / d shell\n\n3(a)(ii) (melting point) higher 1\nAND (density) higher\n\n3(b)(i) M1: emf / potential difference / difference in electrode potential between two half-cells / two electrodes (in a cell) 2\nM2:\n(all solutions being) 1 mol dm–3\nAND either 1 atm OR 298 K\n© UCLES 2021 Page 8 of 17\n\n3(b)(ii) salt bridge, voltmeter, Cu(s), Cu2+(aq), Pt(s), Fe2+ and Fe3+(aq) 3\ntwo for one mark, four for two marks, six for three marks\n\n3(c)(i) M1: 2I– + 2Fe3+ → I + 2Fe2+ 2\n2\nM2: S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+\n2 8 4\n\n3(c)(ii) M1: I / I– +0.54 V AND Fe3+ / Fe2+ + 0.77 V AND [Fe(CN) ]3– / [Fe(CN) ]4– +0.36 V 2\n2 6 6\nM2:\nEo of I / I– is more positive / greater than Eo of [Fe(CN) ]3– / [Fe(CN) ]4–\n2 6 6\nOR Eo = –0.18 V so no reaction occurs\ncell\nOR Eo of Fe3+ / Fe2+ is more positive / greater than Eo of I / I–\n2\nOR Eo = 0.23 V so reaction occurs [1]\ncell\n\n3(d)(i) S O 2– and tartrate ions are both negatively charged / both reactants same charge 1\n2 8\nAND\nso repel each other OR have a high E\na\n\n3(d)(ii) C H O 2– + 2H O ⇌ 2CO + 2HCO – + 6H+ + 6e– 1\n4 4 6 2 2 2\n© UCLES 2021 Page 9 of 17\n\n3(e)(i) 3\nreagent structure of organic product type of reaction\nan excess of LiAlH reduction\n4\nan excess of CH COCl condensation\n3\nM1: product with LiAlH\n4\nM2: product with CH COCl\n3\nM3: both types of reaction\n\n3(e)(ii) 1\nOR dianion of tartrate with two cations present\n© UCLES 2021 Page 10 of 17", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p08.png", "../answer-assets/9701_s21_ms_41-p09.png", "../answer-assets/9701_s21_ms_41-p10.png" ] }, { "id": "9701-2021-mj-41-q04", "question_id": "9701-2021-mj-41-q04", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) M1: blue solid / blue ppt 3\nM2:\n[Cu(H O) ]2+ + 2OH– → Cu(OH) + 6H O\n2 6 2 2\nOR [Cu(H O) ]2+ + 2OH– → Cu(OH) (H O) + 2H O\n2 6 2 2 4 2\nM3: precipitation / acid-base\n\n4(a)(ii) M1: dark blue solution / deep blue solution 3\nM2: [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O\n2 6 3 3 4 2 2 2\nM3: ligand exchange / substitution / displacement / replacement\n\n4(b) M1: 2\nX CuSO and Y Cu\n4\nM2: type of reaction = redox / disproportionation\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p11.png" ] }, { "id": "9701-2021-mj-41-q05", "question_id": "9701-2021-mj-41-q05", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "5(a) measure volume / amount of oxygen formed / mass lost / and time / against time / per unit time 1\nOR\nmeasure absorbance / transmission against time / per unit time\n\n5(b)(i) time taken for the concentration / mass / amount 1\nof a reactant to fall to half (its original value) / to halve\n\n5(b)(ii) t = 150 s 1\n1 / 2\nAND evidence on graph / paper of one half-life\n\n5(b)(iii) no change 1\n\n5(c)(i) M1: evidence on graph of tangent AND 4 to 5 × 10–4 2\nM2: mol dm–3 s–1\n© UCLES 2021 Page 11 of 17\n\n5(c)(ii) (c)(i) / 0.10 AND s–1 1\n\n5(d) M1: NO + O → NO + O 2\n2 3 3 2\nM2: NO + NO → N O\n2 3 2 5\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p11.png", "../answer-assets/9701_s21_ms_41-p12.png" ] }, { "id": "9701-2021-mj-41-q06", "question_id": "9701-2021-mj-41-q06", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 18, "status": "available", "reason": null, "text": "6(a) M1: ethanoic acid > butanoic acid > water > ethanol 4\nM2: a reason given in terms of an electron donating or an electron withdrawing group for one of:\nstrengthening of O–H bond OR weakening of O–H bond OR stability of anion\nTwo out of the three alternatives M3, M4 and M5:\nM3: ethanol: positive inductive effect / electron donating effect of ethyl / alkyl / R group\nM4: butanoic acid: positive inductive effect / electron donating effect of propyl / alkyl / R group\nM5: (either ethanoic or butanoic) acid: negative inductive effect of either C=O or carbonyl OR negative charge delocalised\nover COO–\n\n6(b)(i) 3\nreagents and conditions observed change\nTollen’s reagent, warm silver mirror\ntest 1 OR\nFehling’s solution, warm (brick) red ppt / solid\ntest 2 acidified MnO –, warm decolourises OR bubbles\n4\nM1 / M2: reagents and conditions × 2\nM3: observations both correct\n© UCLES 2021 Page 12 of 17\n\n6(b)(ii) 2\nnumber of peaks in number of peaks in\ncompound\nproton NMR carbon-13 NMR\nHCO H 2 1\n2\nHO CCO H 1 1\n2 2\nHO CCH CH CO H 2 2\n2 2 2 2\none mark for three, four or five correct\ntwo marks for six correct\n\n6(b)(iii) OH peak disappears AND proton / H exchanges with deuterium 1\n\n6(c)(i) G = HOCH CH CH CH OH 2\n2 2 2 2\nH = NCCH CH CH CH CN\n2 2 2 2\n\n6(c)(ii) M1: step 1 NaOH(aq) + heat 4\nM2: step 2 acidified KMnO + heat / acidified K Cr O + heat\n4 2 2 7\nM3: step 3 CN– / KCN / NaCN + heat\nM4: step 4 LiAlH ALLOW Na in ethanol or H + Ni / Pd / Pt\n4 2\n\n6(d) 2\nM1: correct displayed amide linkage\nM2: the rest of the repeat unit correct including trailing bonds\n© UCLES 2021 Page 13 of 17", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p12.png", "../answer-assets/9701_s21_ms_41-p13.png" ] }, { "id": "9701-2021-mj-41-q07", "question_id": "9701-2021-mj-41-q07", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 13, "status": "available", "reason": null, "text": "7(a)(i) 2\n\n7(a)(ii) M1: step 1 hot KMnO / MnO – 3\n4 4\nM2: step 2 conc. H SO and conc. HNO\n2 4 3\nM3: step 3 Sn and conc. HCl (heat)\n\n7(b)(i) 3\nM1 / M2: each structure\nM3: both displayed linkage\n\n7(b)(ii) 1\nmolecular formula number of structural isomers formed\nC H N O 4\n9 19 3 4\n© UCLES 2021 Page 14 of 17\n\n7(c)(i) 1\n\n7(c)(ii) 2\none mark for two correct\ntwo marks for three correct\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p14.png", "../answer-assets/9701_s21_ms_41-p15.png" ] }, { "id": "9701-2021-mj-41-q08", "question_id": "9701-2021-mj-41-q08", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 7, "status": "available", "reason": null, "text": "8(a)(i) M1: CH CO H and CH CO – 3\n3 2 3 2\nM2: due to buffering action / acting as a buffer solution\nM3: CH CO H reacts with NaOH / OH– (forming CH CO – and water)\n3 2 3 2\nOR OH- reacts with H+ and equilibrium CH CO H ⇌ CH CO – + H+ shifts to the right\n3 2 3 2\n\n8(a)(ii) identifying CH CO – is present (with water) at the equivalence point 1\n3 2\nOR CH CO – react with water forming OH–\n3 2\nOR titrating a weak acid with a strong base\n© UCLES 2021 Page 15 of 17\n\n8(b) M1: moles MnO – = 0.025 × 0.0201 = 5.025 × 10–4 3\n4\nmoles V2+ = 5.025 × 10–4 × 5 / 3 = 8.375 × 10–4\nM2: moles VO – = 8.375 × 10–4\n3\nmass of NH VO = 116.9 × 8.375 × 10–4 = 0.0979 g\n4 3\nM3: % Purity of NH VO = 100 × 0.0979 / 0.15 = 65.3 must be 3 sf\n4 3\n© UCLES 2021 Page 16 of 17", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p15.png", "../answer-assets/9701_s21_ms_41-p16.png" ] }, { "id": "9701-2021-mj-41-q09", "question_id": "9701-2021-mj-41-q09", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 6, "status": "available", "reason": null, "text": "9(a) A B 6\nBr\nO\nO\nHO C CH\n3\nHO C CH\n3\nBr\nD E\nC\nO O\nCHI\n3\nHO C OH HO C Cl\nF\nO\nHO C O\none mark for each structure\n© UCLES 2021 Page 17 of 17", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_41-p17.png" ] }, { "id": "9701-2021-mj-42-q01", "question_id": "9701-2021-mj-42-q01", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 15, "status": "available", "reason": null, "text": "1(a)(i) [1s2] 2s22p63s23p63d3 1\n\n1(a)(ii) (a molecule or ion formed by a central) metal atom/ion surrounded by / bonded to one or more ligands 1\n\n1(b) 5\n[Cr(H O) ]3+(aq) formula of chromium species formed type of reaction\n2 6\n+ NaOH(aq) Cr(OH) or Cr(OH) (H O) precipitation\n3 3 2 3\n+ H O (aq) Cr O 2– / CrO 2– redox / oxidation\n2 2 2 7 4\n+ excess NH (aq) Cr(NH ) 3+ ligand substitution\n3 3 6\nchromium species: one mark for each correct species\ntype of reaction: two correct for one mark and three correct for two marks\n\n1(c) M1: ΔE is different OR energy gap between d-orbitals is different 2\nM2: different frequency / wavelength is absorbed\nOR different energy / light in visible region is absorbed\n\n1(d)(i) ethanoate ions are bidentate whereas as H O are monodentate 1\n2\nOR ethanoate ions form two dative bonds whereas H O forms one (dative) bond\n2\nOR ethanoate ions donate two lone pairs whereas H O donates one (lone) pair\n2\n\n1(d)(ii) (coordination number) six AND (geometry around Cr) octahedral 1\n\n1(d)(iii) coordinate / (dative) covalent 1\n\n1(e)(i) 4Cr2+ + O + 4H+  4Cr3+ + 2H O OR 2Cr2+ + O + 2H+  2Cr3+ + H O 2\n2 2 2 2 2\nM1: correct species\nM2: balancing\n\n1(e)(ii) Eo = 1.23 – (–0.41) = (+)1.64 V OR Eo = 0.68 – (–0.41) = (+)1.09 V 1\ncell cell\nvalue linked to (e)(i)\n© UCLES 2021 Page 5 of 16", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p05.png" ] }, { "id": "9701-2021-mj-42-q02", "question_id": "9701-2021-mj-42-q02", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 12, "status": "available", "reason": null, "text": "2(a) M1: increases down the group 3\nM2: radius / size of cation / M2+ increases OR charge density of cation / M2+ decreases\nM3: less polarisation / less distortion of anion / NO – ion OR less weakening of NO bond\n3\n\n2(b)(i) Pb(NO )  PbO + 2NO + ½O 1\n3 2 2 2\n\n2(b)(ii) lead nitrate / Pb(NO ) would decompose more / easier 1\n3 2\nAND as Pb2+ is smaller / Pb2+ has larger charge density (so more polarising)\n\n2(c)(i) BaC O  BaO + CO + CO OR BaC O  BaO + 2CO + ½O 1\n2 4 2 2 4 2\n\n2(c)(ii) M1: [a] initial moles MnO – = 0.0200 × 0.050 = 1.00 × 10–3 4\n4\n[b] moles Fe2+ = 0.050 × 0.0304 = 1.52 × 10–3\nM2: [a] moles MnO – unreacted = 1.52 × 10–3 / 5 = 3.04 × 10–4\n4\n[b] moles MnO – reacted = 1.00 × 10–3 – 3.04 × 10–4 = 6.96 × 10–4\n4\nM3: moles C O 2- reacted = 6.96 × 10–4 × 5/2 = 1.74 × 10–3\n2 4\nM4: mass of BaC O = 225.3 × 1.74 × 10–3 = 0.392 g\n2 4\n% Purity of BaC O = 100 × 0.392/0.50 = 78.4\n2 4\n\n2(d) M1: [OH–] = 2 × 0.12 = 0.24 (mol dm–3) 2\n[H+] = 1 × 10–14/0.24 = 4.17 × 10–14 / pOH = –log(0.24) OR 0.62\nM2: pH = –log[H+] = 13.4 OR pH = 14 – 0.6 = 13.4\n© UCLES 2021 Page 6 of 16", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p06.png" ] }, { "id": "9701-2021-mj-42-q03", "question_id": "9701-2021-mj-42-q03", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 15, "status": "available", "reason": null, "text": "3(a)(i) M1: voltage of an electrode / a half-cell compared to / connected to (standard) hydrogen electrode / half-cell 2\nM2: (at concentration of) 1 mol / dm3 AND (pressure of) 1 atm / 101 kPa (or in Pa) AND 298 K / 25°C\n\n3(a)(ii) 1\n\n3(a)(iii) 3\nOR\nM1 / M2: two 3D isomers of [Ru(en) ]3+\n3\nM3: optical / enantiomerism\n\n3(b)(i) Eo = 1.07 – 0.80 = (+)0.27 V 1\ncell\nAND direction of electron flow = Ag+ / Ag to Br / Br–\n2\n\n3(b)(ii) M1: Eo 3rd box ticked 2\ncell\nM2:\n[Ag+] decreases AND so (Ag+ / Ag) equilibrium shifts to the left\nOR\n[Ag+] decreases AND E for (Ag+ / Ag) becomes less positive / more negative\n© UCLES 2021 Page 7 of 16\n\n3(c)(i) (a species) that uses / shares a lone pair of electrons to form a coordinate bond to a metal atom / ion 1\n\n3(c)(ii) K = [Ag(S O ) 3–] [Br–] / [S O 2–]2 1\nc 2 3 2 2 3\n\n3(c)(iii) M1: K = K × K = 15.7 2\nc stab sp\nM2: 1 OR none / no units\n\n3(d) M1: highest [Ag(CN) ]– [Ag(S O ) ]3– [Ag(NH ) ]+ lowest 2\n2 2 3 2 3 2\nM2: K [Ag(CN) ]– is highest / [Ag(CN) ]– is the most stable\nstab 2 2\nOR higher K forms the more stable complex\nstab\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p07.png", "../answer-assets/9701_s21_ms_42-p08.png" ] }, { "id": "9701-2021-mj-42-q04", "question_id": "9701-2021-mj-42-q04", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 15, "status": "available", "reason": null, "text": "4(a)(i) M1: energy change when 1 mole of a ionic compound is formed 2\nM2: from its gaseous ions under standard conditions\n\n4(a)(ii) ΔH = (–2099) + (2× – 378) – (–2824) 2\nsol\nΔH = –31 kJ mol–1\nsol\nM1: use of ×2 as only multiplier\nM2: correct signs and evaluation\n\n4(a)(iii) M1: Cu2+ is smaller OR Cu2+ has a higher charge density 2\nM2: Cu2+ attracts water molecules more / stronger\nOR (Cu2+) forms stronger ion-dipole forces to water molecules\n\n4(b)(i) anode: chlorine / Cl 1\n2\ncathode: hydrogen / H\n2\n\n4(b)(ii) M1: Q = 0.75 × 60 × 60 = 2700 C AND 96 500 or 193 000 used 2\nM2: [a] moles of Ca = 2700 / 193 000 = 0.0140\n[b] mass = 0.0140 × 40.1 = 0.56 g\n© UCLES 2021 Page 8 of 16\n\n4(c)(i) measure / degree of (dis)order / randomness (of a system) 1\nOR the number of possible arrangements of the particles and their energy (in a given system)\n\n4(c)(ii) 1\nΔS is negative ΔS is zero ΔS is positive\nsolid dissolving in water \nwater solidifying to ice \n\n4(c)(iii) 2\ntwo correct for 1 mark, three correct for two marks:\n• starting at +8.6 kJ / in positive region close to the y-axis\n• line passes through x-axis around 100°C\n• negative gradient straight / curve line through the x-axis (no clear positive inflexions)\n\n4(d) M1: ΔH negative / – , ΔS negative / – 2\nM2: as temperature increase, ΔG becomes (more) positive / less negative ora\nOR at low(er) T, (ΔH more negative than TΔS) so ΔG is negative\nOR at high(er) T, (ΔH less negative than TΔS) so ΔG is positive\n© UCLES 2021 Page 9 of 16", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p08.png", "../answer-assets/9701_s21_ms_42-p09.png" ] }, { "id": "9701-2021-mj-42-q05", "question_id": "9701-2021-mj-42-q05", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 13, "status": "available", "reason": null, "text": "5(a) M1: 4\nethylamine phenylamine 4-nitrophenylamine\nmost basic least basic\nM2 / 3 / 4: explanation two correct for one mark, three correct for two marks, four correct for three marks\n• (basicity linked to) lone pair / p orbital on N AND being able accept / donate to / coordinate to a proton / H+\n• ethyl / alkyl group is electron donating / has a positive inductive effect (so lone pair on N is more able to accept a\nproton)\n• (phenylamines are less basic than ethylamine as) p orbital / lone pair on N is delocalised (into the ring so less able to\naccept a proton)\n• (4-nitrophenylamine is less basic than phenylamine as) nitro / NO group is electron withdrawing (so lone pair on N is\n2\nless able to accept a proton)\n\n5(b)(i) 1\n\n5(b)(ii) M1: 2\nstep 1: HNO , ⩽ 10°C OR NaNO , HCl(aq), ⩽ 10°C\n2 2\nM2:\nstep 2: NaOH / alkaline AND 1-naphthol / α-naphthol / structure below\n\n5(c)(i) 4-bromo-2-nitrobenzoic acid OR 4-bromo-2-nitro(-1-)benzenecarboxylic acid 1\n© UCLES 2021 Page 10 of 16\n\n5(c)(ii) 2\n\n5(c)(iii) M1: step 1 conc. H SO and conc. HNO 3\n2 4 3\nM2: step 2 Br and AlBr\n2 3\nM3: step 3 hot (alkaline / acidified) MnO – / KMnO\n4 4\n© UCLES 2021 Page 11 of 16", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p10.png", "../answer-assets/9701_s21_ms_42-p11.png" ] }, { "id": "9701-2021-mj-42-q06", "question_id": "9701-2021-mj-42-q06", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 11, "status": "available", "reason": null, "text": "6(a) 4\ncorrect isomers and correct assignment to peaks: mark as •  •  •  • \n\n6(b)(i) RCl + AlCl  R+ + AlCl – 1\n3 4\nOR Cl(CH ) COOH + AlCl  +(CH ) COOH + AlCl –\n2 3 3 2 3 4\n\n6(b)(ii) 2\n© UCLES 2021 Page 12 of 16\n\n6(b)(iii) SOCl OR PCl 1\n2 5\nALLOW PCl AND heat\n3\n\n6(b)(iv) 3\nM1: arrow to R+ OR arrow to positive carbon of +(CH ) COOH\n2 3\nM2: correct structure of intermediate\nM3: arrow from C-H bond into the ring AND H+\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p12.png", "../answer-assets/9701_s21_ms_42-p13.png" ] }, { "id": "9701-2021-mj-42-q07", "question_id": "9701-2021-mj-42-q07", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "7(a)(i) the power to which a concentration of a reactant is raised in the rate equation / law 1\n\n7(a)(ii) M1: (using expt 1 and 3) as [ClO ] × 2.5 rate × 6.25 so 2nd order 2\n2\nM2: (using expt 1 and 2) as [OH–] × 4 rate × 4 so 1st order\n\n7(a)(iii) rate = k[ClO ]2[OH–] 1\n2\n\n7(a)(iv) M1: 2\nk = rate / [ClO ]2[OH–]\n2\nk = 7.20 × 10-4 / (0.02)2(0.03)\nk = 60\nM2: mol–2 dm6 min–1\n© UCLES 2021 Page 13 of 16\n\n7(b)(i) 1\nstructure of phenol: C H OH OR\n6 5\n\n7(b)(ii) tangent drawn correctly AND rate = 0.015 / 260 = 5.8 × 10–5 1\nALLOW values consistent with tangent drawn at 100 sec\n\n7(c) 1\nAND half-life decreases (1st box)\n© UCLES 2021 Page 14 of 16", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p13.png", "../answer-assets/9701_s21_ms_42-p14.png" ] }, { "id": "9701-2021-mj-42-q08", "question_id": "9701-2021-mj-42-q08", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 10, "status": "available", "reason": null, "text": "8(a) M1: 3\nacyl chlorides > alkyl chlorides > aryl chlorides\nfastest slowest\nM2 / 3: Any two from:\n• acyl chlorides carbon in C-Cl bond is more electron deficient since it is also attached to an oxygen atom OR C-Cl\nbond is weakest / weakened in acyl chlorides since it is also attached to an oxygen atom / two electronegative atoms\n• aryl chlorides (no hydrolysis) C-Cl bond is part of delocalised system / partially double bond character (so C-Cl bond is\nstronger) OR lone pair / p-orbital on Cl delocalises with π ring (so C-Cl bond is stronger)\n• alkyl chlorides C-Cl bond strengthened by electron donating effect / positive inductive effect of alkyl / R group (as\ncompared to acyl chlorides) OR carbon atom has a smaller δ+ and the C-Cl bond is stronger (than the C-Cl bond in\nCOCl) due to (the carbon) being only attached to one electronegative atom\n\n8(b) 3\nALLOW amine salt for the third structure – mono or di ion\n\n8(c)(i) 1\n© UCLES 2021 Page 15 of 16\n\n8(c)(ii) 3\nchemical splitting pattern number of 1H atoms\nenvironment of proton\nshift (δ) (words required) responsible for the peak\n0.95 alkane / CH doublet 6\n3\n1.90 alkane / CH multiplet 1\nALLOW alkyne\n2.20 R / alkyl / CH next to doublet 2\n2\nC=O / COOH\nmark as • •  • •  • • \n© UCLES 2021 Page 16 of 16", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_42.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_42-p15.png", "../answer-assets/9701_s21_ms_42-p16.png" ] }, { "id": "9701-2021-mj-43-q01", "question_id": "9701-2021-mj-43-q01", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "1(a) M1: increases down the group 3\nM2: radius / size of (cat)ion / M2+ increases\nM3: less polarisation / distortion of anion / hydroxide ion / hydroxide group / OH– / OH\n\n1(b)(i) Ca(OH) (s) ⇌ Ca2+(aq) + 2OH–(aq) 1\n2\n\n1(b)(ii) M1: 2\nK = [Ca2+][OH–]2\nsp\nOR\nK = 4x3\nsp\nM2: x = 3√5.02 × 10–6 / 4 = 0.0108 / 0.011 / 1.08 × 10–2 / 1.1 × 10–2 (mol dm–3) min 2 sf\n\n1(b)(iii) less soluble / decreases due to the common ion effect 1\nOR\ndecreases as equilibrium in (b)(i) has shifted to the left\nOR\ndecreases as [OH–] increases causing [Ca2+][OH–](2) to exceed its K\nsp\n© UCLES 2021 Page 5 of 17", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p05.png" ] }, { "id": "9701-2021-mj-43-q02", "question_id": "9701-2021-mj-43-q02", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 15, "status": "available", "reason": null, "text": "2(a) M1: (complexes have two sets of) d orbital(s) of different energy / d-d splitting occurs 4\nOR d orbital(s) / d (sub)-shell splits\nOR (inferred from a movement of an electron) from a lower d to higher d orbital\nM2: electron(s) promoted / excited\nOR electron(s) moves to higher (d–)orbital\nOR electron(s) jumps up (to d–orbital) / jumps to higher (d–orbital)\nM3: wavelength / frequency / light / photon / hν absorbed\nOR radiation / energy from visible (region) absorbed\nM4: colour seen is complementary (to colour absorbed)\nOR wavelength / frequency / colour / light not absorbed is transmitted / reflected / seen\n© UCLES 2021 Page 6 of 17\n\n2(b)(i) 4\nM1: All charges\nM2: One octahedral with correct 3D\nM3: All formulae\nM4: R is cis, S is trans\n\n2(b)(ii) dipoles cancel 1\n\n2(c)(i) M1: (a species) that donates two lone pairs / forms two coordinate bonds / two dative bonds 2\nM2: to a metal atom / metal ion\n© UCLES 2021 Page 7 of 17\n\n2(c)(ii) 1\nstructure of the picolinate anion ligand\n\n2(c)(iii) (coordination number) six 1\nAND (geometry around Cr) octahedral\n\n2(d)(i) (NH ) Cr O +6 1\n4 2 2 7\nCr O +3\n2 3\n\n2(d)(ii) (NH ) Cr O → N + Cr O + 4H O 1\n4 2 2 7 2 2 3 2\nQuestion Answer Marks", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p06.png", "../answer-assets/9701_s21_ms_43-p07.png", "../answer-assets/9701_s21_ms_43-p08.png" ] }, { "id": "9701-2021-mj-43-q03", "question_id": "9701-2021-mj-43-q03", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 17, "status": "available", "reason": null, "text": "3(a)(i) (an element) forming stable ion / ions / compound(s) / oxidation state(s) 1\nAND with partially filled / incomplete\nAND d orbitals / d subshell / d shell\n\n3(a)(ii) (melting point) higher 1\nAND (density) higher\n\n3(b)(i) M1: emf / potential difference / difference in electrode potential between two half-cells / two electrodes (in a cell) 2\nM2:\n(all solutions being) 1 mol dm–3\nAND either 1 atm OR 298 K\n© UCLES 2021 Page 8 of 17\n\n3(b)(ii) salt bridge, voltmeter, Cu(s), Cu2+(aq), Pt(s), Fe2+ and Fe3+(aq) 3\ntwo for one mark, four for two marks, six for three marks\n\n3(c)(i) M1: 2I– + 2Fe3+ → I + 2Fe2+ 2\n2\nM2: S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+\n2 8 4\n\n3(c)(ii) M1: I / I– +0.54 V AND Fe3+ / Fe2+ + 0.77 V AND [Fe(CN) ]3– / [Fe(CN) ]4– +0.36 V 2\n2 6 6\nM2:\nEo of I / I– is more positive / greater than Eo of [Fe(CN) ]3– / [Fe(CN) ]4–\n2 6 6\nOR Eo = –0.18 V so no reaction occurs\ncell\nOR Eo of Fe3+ / Fe2+ is more positive / greater than Eo of I / I–\n2\nOR Eo = 0.23 V so reaction occurs [1]\ncell\n\n3(d)(i) S O 2– and tartrate ions are both negatively charged / both reactants same charge 1\n2 8\nAND\nso repel each other OR have a high E\na\n\n3(d)(ii) C H O 2– + 2H O ⇌ 2CO + 2HCO – + 6H+ + 6e– 1\n4 4 6 2 2 2\n© UCLES 2021 Page 9 of 17\n\n3(e)(i) 3\nreagent structure of organic product type of reaction\nan excess of LiAlH reduction\n4\nan excess of CH COCl condensation\n3\nM1: product with LiAlH\n4\nM2: product with CH COCl\n3\nM3: both types of reaction\n\n3(e)(ii) 1\nOR dianion of tartrate with two cations present\n© UCLES 2021 Page 10 of 17", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p08.png", "../answer-assets/9701_s21_ms_43-p09.png", "../answer-assets/9701_s21_ms_43-p10.png" ] }, { "id": "9701-2021-mj-43-q04", "question_id": "9701-2021-mj-43-q04", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) M1: blue solid / blue ppt 3\nM2:\n[Cu(H O) ]2+ + 2OH– → Cu(OH) + 6H O\n2 6 2 2\nOR [Cu(H O) ]2+ + 2OH– → Cu(OH) (H O) + 2H O\n2 6 2 2 4 2\nM3: precipitation / acid-base\n\n4(a)(ii) M1: dark blue solution / deep blue solution 3\nM2: [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O\n2 6 3 3 4 2 2 2\nM3: ligand exchange / substitution / displacement / replacement\n\n4(b) M1: 2\nX CuSO and Y Cu\n4\nM2: type of reaction = redox / disproportionation\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p11.png" ] }, { "id": "9701-2021-mj-43-q05", "question_id": "9701-2021-mj-43-q05", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "5(a) measure volume / amount of oxygen formed / mass lost / and time / against time / per unit time 1\nOR\nmeasure absorbance / transmission against time / per unit time\n\n5(b)(i) time taken for the concentration / mass / amount 1\nof a reactant to fall to half (its original value) / to halve\n\n5(b)(ii) t = 150 s 1\n1 / 2\nAND evidence on graph / paper of one half-life\n\n5(b)(iii) no change 1\n\n5(c)(i) M1: evidence on graph of tangent AND 4 to 5 × 10–4 2\nM2: mol dm–3 s–1\n© UCLES 2021 Page 11 of 17\n\n5(c)(ii) (c)(i) / 0.10 AND s–1 1\n\n5(d) M1: NO + O → NO + O 2\n2 3 3 2\nM2: NO + NO → N O\n2 3 2 5\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p11.png", "../answer-assets/9701_s21_ms_43-p12.png" ] }, { "id": "9701-2021-mj-43-q06", "question_id": "9701-2021-mj-43-q06", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 18, "status": "available", "reason": null, "text": "6(a) M1: ethanoic acid > butanoic acid > water > ethanol 4\nM2: a reason given in terms of an electron donating or an electron withdrawing group for one of:\nstrengthening of O–H bond OR weakening of O–H bond OR stability of anion\nTwo out of the three alternatives M3, M4 and M5:\nM3: ethanol: positive inductive effect / electron donating effect of ethyl / alkyl / R group\nM4: butanoic acid: positive inductive effect / electron donating effect of propyl / alkyl / R group\nM5: (either ethanoic or butanoic) acid: negative inductive effect of either C=O or carbonyl OR negative charge delocalised\nover COO–\n\n6(b)(i) 3\nreagents and conditions observed change\nTollen’s reagent, warm silver mirror\ntest 1 OR\nFehling’s solution, warm (brick) red ppt / solid\ntest 2 acidified MnO –, warm decolourises OR bubbles\n4\nM1 / M2: reagents and conditions × 2\nM3: observations both correct\n© UCLES 2021 Page 12 of 17\n\n6(b)(ii) 2\nnumber of peaks in number of peaks in\ncompound\nproton NMR carbon-13 NMR\nHCO H 2 1\n2\nHO CCO H 1 1\n2 2\nHO CCH CH CO H 2 2\n2 2 2 2\none mark for three, four or five correct\ntwo marks for six correct\n\n6(b)(iii) OH peak disappears AND proton / H exchanges with deuterium 1\n\n6(c)(i) G = HOCH CH CH CH OH 2\n2 2 2 2\nH = NCCH CH CH CH CN\n2 2 2 2\n\n6(c)(ii) M1: step 1 NaOH(aq) + heat 4\nM2: step 2 acidified KMnO + heat / acidified K Cr O + heat\n4 2 2 7\nM3: step 3 CN– / KCN / NaCN + heat\nM4: step 4 LiAlH ALLOW Na in ethanol or H + Ni / Pd / Pt\n4 2\n\n6(d) 2\nM1: correct displayed amide linkage\nM2: the rest of the repeat unit correct including trailing bonds\n© UCLES 2021 Page 13 of 17", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p12.png", "../answer-assets/9701_s21_ms_43-p13.png" ] }, { "id": "9701-2021-mj-43-q07", "question_id": "9701-2021-mj-43-q07", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 13, "status": "available", "reason": null, "text": "7(a)(i) 2\n\n7(a)(ii) M1: step 1 hot KMnO / MnO – 3\n4 4\nM2: step 2 conc. H SO and conc. HNO\n2 4 3\nM3: step 3 Sn and conc. HCl (heat)\n\n7(b)(i) 3\nM1 / M2: each structure\nM3: both displayed linkage\n\n7(b)(ii) 1\nmolecular formula number of structural isomers formed\nC H N O 4\n9 19 3 4\n© UCLES 2021 Page 14 of 17\n\n7(c)(i) 1\n\n7(c)(ii) 2\none mark for two correct\ntwo marks for three correct\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p14.png", "../answer-assets/9701_s21_ms_43-p15.png" ] }, { "id": "9701-2021-mj-43-q08", "question_id": "9701-2021-mj-43-q08", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 7, "status": "available", "reason": null, "text": "8(a)(i) M1: CH CO H and CH CO – 3\n3 2 3 2\nM2: due to buffering action / acting as a buffer solution\nM3: CH CO H reacts with NaOH / OH– (forming CH CO – and water)\n3 2 3 2\nOR OH- reacts with H+ and equilibrium CH CO H ⇌ CH CO – + H+ shifts to the right\n3 2 3 2\n\n8(a)(ii) identifying CH CO – is present (with water) at the equivalence point 1\n3 2\nOR CH CO – react with water forming OH–\n3 2\nOR titrating a weak acid with a strong base\n© UCLES 2021 Page 15 of 17\n\n8(b) M1: moles MnO – = 0.025 × 0.0201 = 5.025 × 10–4 3\n4\nmoles V2+ = 5.025 × 10–4 × 5 / 3 = 8.375 × 10–4\nM2: moles VO – = 8.375 × 10–4\n3\nmass of NH VO = 116.9 × 8.375 × 10–4 = 0.0979 g\n4 3\nM3: % Purity of NH VO = 100 × 0.0979 / 0.15 = 65.3 must be 3 sf\n4 3\n© UCLES 2021 Page 16 of 17", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p15.png", "../answer-assets/9701_s21_ms_43-p16.png" ] }, { "id": "9701-2021-mj-43-q09", "question_id": "9701-2021-mj-43-q09", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 6, "status": "available", "reason": null, "text": "9(a) A B 6\nBr\nO\nO\nHO C CH\n3\nHO C CH\n3\nBr\nD E\nC\nO O\nCHI\n3\nHO C OH HO C Cl\nF\nO\nHO C O\none mark for each structure\n© UCLES 2021 Page 17 of 17", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_43-p17.png" ] }, { "id": "9701-2021-mj-51-q01", "question_id": "9701-2021-mj-51-q01", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a)(i) the metal oxide 1\n\n1(a)(ii) count bubbles in a set time 1\nOR measure volume (of oxygen) in a certain time\nOR measure time to produce a certain volume\n\n1(b) Any two from: 2\nthe volume of the hydrogen peroxide / solution\nthe concentration of the hydrogen peroxide\ntemperature\nparticle size\n\n1(c) correctly drawn diagram showing: 3\nM1: contents of flask\nM2: sealed apparatus\nM3: collection of the gas in a measuring cylinder (graduations assumed) (via a delivery tube into water) / gas syringe\n\n1(d)(i) volume of gas / oxygen AND time 1\n\n1(d)(ii) plot a (volume / time) graph AND measure gradient / produce a tangent at t = 0 or at the start 1\n\n1(e) repeat (the experiments / trials) (until consistent results) 1\n© UCLES 2021 Page 5 of 9\n\n1(f) measuring the decrease / change in (total) mass of the solution / mixture / reaction vessel (and contents) (and time) 1\nOR\nmeasuring the mass / weight of the mixture/reaction vessel (and contents) AND time\n\n1(g) M1: (filter/remove) dry the residue/metal oxide 2\nM2: weigh the residue / metal oxide to see if the mass is unchanged\nOR\nM1: re-use with new hydrogen peroxide / do the same reaction again with the same catalyst\nM2: and check if the gas volumes identical / rate of reaction the same\n\n1(h) to allow the oxygen / gas (that is formed) to be released / escape / diffuse out 1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_51-p05.png", "../answer-assets/9701_s21_ms_51-p06.png" ] }, { "id": "9701-2021-mj-51-q02", "question_id": "9701-2021-mj-51-q02", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a)(i) M KMnO = 158.0 1\nr 4\n500 × 0.02 / 1000 = 0.01\n0.01 × 158.0 = 1.58 g\n\n2(a)(ii) (2 × 0.005 / 1.58) × 100 = 0.63% 1\n\n2(a)(iii) rinse the solid off the weighing boat into the beaker 1\nOR\nweighing the mass directly into the beaker\nOR\n(re)-weigh the weighing boat after transferring the KMnO into the beaker\n4\n\n2(a)(iv) Any two from: 2\nstir / agitate / mix (to ensure that the solid has dissolved)\nrinse the beaker and transfer the washings\nshaking / inverting / homogenising of volumetric solution\n© UCLES 2021 Page 6 of 9\n\n2(b)(i) rinse / run through / wash the burette with some of the KMnO solution 1\n4\nOR run some of the KMnO solution from the burette to fill to the tip\n4\n\n2(b)(ii) add the KMnO dropwise (near the end point) 1\n4\n\n2(c)(i) the Fe2+(aq) / Fe3+(aq) mixture measured using a measuring cylinder 1\nAND\nleads to increased likelihood of non-concordant titres\n\n2(c)(ii) (measure the volume) using a (volumetric / 25 cm3) pipette 1\n© UCLES 2021 Page 7 of 9\n\n2(d)(i) Set A 2\nrough titration 1 titration 2 titration 3\nfinal volume / cm3 18.40 17.25 34.55 18.00\ninitial volume / cm3 0.65 0.15 17.25 0.95\ntitre / cm3 17.75 17.10 17.30 17.05\nmean titre = 17.1 (cm3)\nSet B\nrough titration 1 titration 2\nfinal volume / cm3 45.05 43.60 43.70\ninitial volume / cm3 0.20 0.15 0.10\ntitre / cm3 44.85 43.45 43.60\nmean titre = 43.5 (cm3)\nM1: correct titres (to 2 decimal places)\nM2: correct means\n\n2(d)(ii) M1: 2\nn MnO – (aq) = 43.5 × 0.02 / 1000 = 8.70 × 10–4\n4\nn Fe2+ (aq) = 8.70 × 10–4 × 5 = 4.35 × 10–3\nM2:\n100\nmass of Fe 2+ (aq) = 4.35 × 10–3 × 55.8 × = 0.971 g\n25\n© UCLES 2021 Page 8 of 9\n\n2(d)(iii) n MnO - (aq) = 17.1 × 0.02 / 1000 = 3.42 × 10–4 1\n4\nn Fe2+ (aq) = 3.42 × 10–4 × 5 = 1.71 × 10–3\n250\nmass of Fe 2+ (aq) = 1.71 × 10–3 × 55.8 × = 0.954 g\n25\n\n2(d)(iv) 250 250 1\n% Fe3+ = ((2(d)(ii) × ) – 2(d)(iii))/(2(d)(ii) × )) × 100%\n100 100\nOR\n(2(d)(ii) × 2.5 – 2(d)(iii)) / (2(d)(ii) × 2.5) × 100%\n\n2(d)(v) reduce / use (in step 6) more than 100 cm3 of solution (Fe2+ and Fe3+) 1\n© UCLES 2021 Page 9 of 9", "source_pages": [ 6, 7, 8, 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_51-p06.png", "../answer-assets/9701_s21_ms_51-p07.png", "../answer-assets/9701_s21_ms_51-p08.png", "../answer-assets/9701_s21_ms_51-p09.png" ] }, { "id": "9701-2021-mj-52-q01", "question_id": "9701-2021-mj-52-q01", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a) negative 1\nAND\nbond formation is happening (between MgSO and H O)\n4 2\n\n1(b)(i) 8.05(°C) 1\n\n1(b)(ii) to allow the temperature of the water to equilibrate with room temperature 1\n\n1(b)(iii) M1: (measure) 50 cm3 (distilled) water and transfer into a polystyrene / polyfoam / plastic cup 4\nM2: measure the temperature of the water at (30 second) intervals (until constant / before adding MgSO )\n4\nM3: MgSO added at 180 s / 3 minutes\n4\nM4: stir and measure the temperature of the water at (30 second) intervals (until at least 8 mins have elapsed since time 0)\n\n1(c)(i) 0.025 × 7 × 18 = 3.15 1\nAND 50.00 – 3.15 = 46.85 (cm3)\n\n1(c)(ii) M1: burette 2\nM2: (apparatus) measures to 0.05 (cm3)\n\n1(d) use a larger number of moles (of hydrated magnesium sulfate/solid) 1\nOR\nuse a smaller volume of distilled water\n© UCLES 2021 Page 5 of 9\n\n1(e)(i) M1: 2\nΔH = –2125 / 1000 × 0.025 = –85.0\nexp1\nand\nΔH = 477.5 / 1000 × 0.025 = 19.1\nexp2\nM2:\nΔH = ΔH – ΔH\nexp1 exp2\n–85.0 – (+)19.1 = –104.1 kJ mol–1\n\n1(e)(ii) less exothermic 1\nAND\nfewer moles reacted\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_52-p05.png", "../answer-assets/9701_s21_ms_52-p06.png" ] }, { "id": "9701-2021-mj-52-q02", "question_id": "9701-2021-mj-52-q02", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a) butanoic acid is flammable 1\n\n2(b)(i) 250.(0) cm3 1\n\n2(b)(ii) M1: dissolve the solid / sodium butanoate (55.0 g) (in the beaker) (using a small volume of) (distilled) water\nM2: transfer / add to a (250 cm3 volumetric) flask AND rinse (with distilled water)\nM3: top the 250 cm3 volumetric flask up to mark (with distilled water)\n© UCLES 2021 Page 6 of 9\n\n2(b)(iii) 2\nmoles of volume of moles of volume of volume of\nacid acid base base water\n0.050 25.0 0.005 2.5 22.5\n0.050 25.0 0.008 4.0 21.0\n0.050 25.0 0.010 5.0 20.0\n0.050 25.0 0.025 12.5 12.5\n0.030 15.0 0.050 25.0 10.0\n0.010 5.0 0.050 25.0 20.0\n0.006 3.0 0.050 25.0 22.0\n0.004 2.0 0.050 25.0 23.0\nM1: column 4\nM2: column 5\n© UCLES 2021 Page 7 of 9\n\n2(c)(i) 1\nratio of –log [acid]\nmoles of acid moles of base pH\n[acid] / [base] [base]\n0.050 0.005 10.00 –1.00 3.82\n0.050 0.008 6.25 –0.80 4.01\n0.050 0.010 5.00 –0.70 4.12\n0.050 0.025 2.00 –0.30 4.70\n0.030 0.050 0.60 (+)0.22 5.04\n0.010 0.050 0.20 (+)0.70 5.52\n0.006 0.050 0.12 (+)0.92 5.74\n0.004 0.050 0.08 (+)1.10 5.91\n\n2(c)(ii) M1: 8 points plotted correctly 2\nM2: straight line of best fit drawn which must extend to cover all data points\n© UCLES 2021 Page 8 of 9\n\n2(c)(iii) M1: Plot at –0.3, 4.70 2\nM2:\n(actual) moles of acid (in the buffer) is lower (than recorded in the table)\nOR\n(actual) moles of base (in the buffer) is higher (than recorded in the table)\nOR\nconcentration of acid (in buffer) lower (than they think for this reading)\nOR\nconcentration of base (in buffer) higher (than they think for this reading)\nOR\nvolume of acid used (to make the buffer) is lower (than recorded in table / 25 cm3)\nOR\nvolume of base used (to make the buffer) is higher (than recorded in table / 12.5 cm3)\n\n2(c)(iv) pK = 4.82 1\na\n\n2(c)(v) M1: K = 1.51 × 10–5 (must be 3 SF) 2\na\nM2: mol dm–3\n\n2(d) endothermic 1\nAND\nK increases (with increasing temperature)\na\n© UCLES 2021 Page 9 of 9", "source_pages": [ 6, 7, 8, 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_52-p06.png", "../answer-assets/9701_s21_ms_52-p07.png", "../answer-assets/9701_s21_ms_52-p08.png", "../answer-assets/9701_s21_ms_52-p09.png" ] }, { "id": "9701-2021-mj-53-q01", "question_id": "9701-2021-mj-53-q01", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a)(i) the metal oxide 1\n\n1(a)(ii) count bubbles in a set time 1\nOR measure volume (of oxygen) in a certain time\nOR measure time to produce a certain volume\n\n1(b) Any two from: 2\nthe volume of the hydrogen peroxide / solution\nthe concentration of the hydrogen peroxide\ntemperature\nparticle size\n\n1(c) correctly drawn diagram showing: 3\nM1: contents of flask\nM2: sealed apparatus\nM3: collection of the gas in a measuring cylinder (graduations assumed) (via a delivery tube into water) / gas syringe\n\n1(d)(i) volume of gas / oxygen AND time 1\n\n1(d)(ii) plot a (volume / time) graph AND measure gradient / produce a tangent at t = 0 or at the start 1\n\n1(e) repeat (the experiments / trials) (until consistent results) 1\n© UCLES 2021 Page 5 of 9\n\n1(f) measuring the decrease / change in (total) mass of the solution / mixture / reaction vessel (and contents) (and time) 1\nOR\nmeasuring the mass / weight of the mixture/reaction vessel (and contents) AND time\n\n1(g) M1: (filter/remove) dry the residue/metal oxide 2\nM2: weigh the residue / metal oxide to see if the mass is unchanged\nOR\nM1: re-use with new hydrogen peroxide / do the same reaction again with the same catalyst\nM2: and check if the gas volumes identical / rate of reaction the same\n\n1(h) to allow the oxygen / gas (that is formed) to be released / escape / diffuse out 1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_53-p05.png", "../answer-assets/9701_s21_ms_53-p06.png" ] }, { "id": "9701-2021-mj-53-q02", "question_id": "9701-2021-mj-53-q02", "subject": "9701", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a)(i) M KMnO = 158.0 1\nr 4\n500 × 0.02 / 1000 = 0.01\n0.01 × 158.0 = 1.58 g\n\n2(a)(ii) (2 × 0.005 / 1.58) × 100 = 0.63% 1\n\n2(a)(iii) rinse the solid off the weighing boat into the beaker 1\nOR\nweighing the mass directly into the beaker\nOR\n(re)-weigh the weighing boat after transferring the KMnO into the beaker\n4\n\n2(a)(iv) Any two from: 2\nstir / agitate / mix (to ensure that the solid has dissolved)\nrinse the beaker and transfer the washings\nshaking / inverting / homogenising of volumetric solution\n© UCLES 2021 Page 6 of 9\n\n2(b)(i) rinse / run through / wash the burette with some of the KMnO solution 1\n4\nOR run some of the KMnO solution from the burette to fill to the tip\n4\n\n2(b)(ii) add the KMnO dropwise (near the end point) 1\n4\n\n2(c)(i) the Fe2+(aq) / Fe3+(aq) mixture measured using a measuring cylinder 1\nAND\nleads to increased likelihood of non-concordant titres\n\n2(c)(ii) (measure the volume) using a (volumetric / 25 cm3) pipette 1\n© UCLES 2021 Page 7 of 9\n\n2(d)(i) Set A 2\nrough titration 1 titration 2 titration 3\nfinal volume / cm3 18.40 17.25 34.55 18.00\ninitial volume / cm3 0.65 0.15 17.25 0.95\ntitre / cm3 17.75 17.10 17.30 17.05\nmean titre = 17.1 (cm3)\nSet B\nrough titration 1 titration 2\nfinal volume / cm3 45.05 43.60 43.70\ninitial volume / cm3 0.20 0.15 0.10\ntitre / cm3 44.85 43.45 43.60\nmean titre = 43.5 (cm3)\nM1: correct titres (to 2 decimal places)\nM2: correct means\n\n2(d)(ii) M1: 2\nn MnO – (aq) = 43.5 × 0.02 / 1000 = 8.70 × 10–4\n4\nn Fe2+ (aq) = 8.70 × 10–4 × 5 = 4.35 × 10–3\nM2:\n100\nmass of Fe 2+ (aq) = 4.35 × 10–3 × 55.8 × = 0.971 g\n25\n© UCLES 2021 Page 8 of 9\n\n2(d)(iii) n MnO - (aq) = 17.1 × 0.02 / 1000 = 3.42 × 10–4 1\n4\nn Fe2+ (aq) = 3.42 × 10–4 × 5 = 1.71 × 10–3\n250\nmass of Fe 2+ (aq) = 1.71 × 10–3 × 55.8 × = 0.954 g\n25\n\n2(d)(iv) 250 250 1\n% Fe3+ = ((2(d)(ii) × ) – 2(d)(iii))/(2(d)(ii) × )) × 100%\n100 100\nOR\n(2(d)(ii) × 2.5 – 2(d)(iii)) / (2(d)(ii) × 2.5) × 100%\n\n2(d)(v) reduce / use (in step 6) more than 100 cm3 of solution (Fe2+ and Fe3+) 1\n© UCLES 2021 Page 9 of 9", "source_pages": [ 6, 7, 8, 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9701_s21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-may-june/9701_s21_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s21_ms_53-p06.png", "../answer-assets/9701_s21_ms_53-p07.png", "../answer-assets/9701_s21_ms_53-p08.png", "../answer-assets/9701_s21_ms_53-p09.png" ] }, { "id": "9701-2021-on-41-q01", "question_id": "9701-2021-on-41-q01", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 9, "status": "available", "reason": null, "text": "1(a) moles of H = 462 / 24 000 = 0.01925 [1] 2\n2\nmolecules of H = 0.019 × 6.02 × 1023\n2\n= 1.16 × 1022 (1.1×1022 / 1.2 × 1022) [1] min 2sf ecf M1\n\n1(b) number of electrons = 1.16 × 1022 × 2 1\n= 2.32 × 1022 [1] min 2sf ecf 1a\n\n1(c) Q = 2.32 ×1022 × 1.6 × 10–19 1\n= 3.71 × 103 [1] min 2sf ecf 1b\n\n1(d) x = 3.71 × 103 / (14 × 60) 1\n= 4.4 (A) [1] min 2sf ecf 1c\n\n1(e)(i) ΔS = 262 + 205 – 140 = (+) 327 (J K–1 mol–1) [1] 1\n\n1(e)(ii) ΔG = ΔH – TΔS OR use of Gibbs equation [1] 2\nΔG = 572 – (298 × 0.327)\n= (+)474.6 (kJ mol–1) [1] min 3sf ecf 1e(i)\n\n1(e)(iii) becomes more feasible / spontaneous 1\nas TΔS is more positive / –TΔS becomes more negative  \nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p05.png" ] }, { "id": "9701-2021-on-41-q02", "question_id": "9701-2021-on-41-q02", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i H +Cl ( CH ) CO − 1\nK =   2 3 2  [1]\na ( )\nCl CH CO H\n 2 3 2 \n\n2(a)(ii) pK = – logK OR K = 10–pKa [1] 1\na a a\n\n2(a)(iii) [H+] = 10–4.0 = 1 × 10–4 [1] 1\n© UCLES 2021 Page 5 of 16\n\n2(a)(iv) • [HCl] = 1 × 10–4 ecf 2(a)(iii) 2\n• K = 10–4.52 = 3.02 × 10–5\na\n• [Cl(CH ) CO H] = (1 × 10–4)2 / 3.02 × 10–5\n2 3 2\n= 3.3 × 10–4 ecf\n[ HCl ] 1×10 −4\n• = =0.302 min 2sf ecf\nCl ( CH ) CO H 3.3×10 −4\n 2 3 2 \n\n2(b) M1 [H+] = 10–5 OR 1 × 10–5 3\nK = [H+][A–] / [HA] OR pH = pKa + log [A–] / [HA]\na\n  [1]\nM2 moles of A– = (1.35 × 10–5)(5 / 74) / (1 × 10–5)\nmoles of A– = 0.0912 [1] ecf\nM3 mass of sodium propanoate = 0.0912 × 96 = 8.76 [1] min 2sf ecf\n\n2(c) all of the (sodium) propanoate (ion) has been 2\nprotonated / converted to (propanoic) acid / neutralised [1]\nH+ is in excess / H+ is 0.1 mol dm–3 (from the H SO ) [1]\n2 4\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p05.png", "../answer-assets/9701_w21_ms_41-p06.png" ] }, { "id": "9701-2021-on-41-q03", "question_id": "9701-2021-on-41-q03", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 12, "status": "available", "reason": null, "text": "3(a) • enthalpy/energy change 2\n• one mole of electrons gained\n• by one mole of atoms\n• gaseous (atoms)\n\n3(b) Ca+(g) → Ca2+(g) + e- [1] 1\n© UCLES 2021 Page 6 of 16\n\n3(c) M1: selecting correct data 951, 844, 142 only 3\nM2: evaluation to give 249 (ΔH )\natom\nOR 2(951) = BE – 2(142) + 2(844)\nM3: evaluation to 498 (2 × 249) ecf M2\n951 = ΔH –142 + 844\natom\nΔH = 249\natom\nBE = 498 (kJ mol–1) [3]\n\n3(d)(i) attraction between nucleus / protons / nuclear charge 1\nand electron [1]\n\n3(d)(ii) repulsion between 1– ion / electrons of O– 1\nand electron [1]\n\n3(e) M1: selecting correct data 951, 1933, 3517 only (ignore signs) 2\nM2: evaluation to give –633 (ΔH) ecf\nf\nΔH = 951 + 1933 – 3517 = –633 (kJ mol–1) [2]\nf\n\n3(f) ionic charge / charge density (of the ions) [1] 2\ngreater (attractive) force between the ions [1]\n© UCLES 2021 Page 7 of 16", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p06.png", "../answer-assets/9701_w21_ms_41-p07.png" ] }, { "id": "9701-2021-on-41-q04", "question_id": "9701-2021-on-41-q04", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 8, "status": "available", "reason": null, "text": "4(a) • barium carbonate / Ba / BaCO 2\n3\n• larger ionic radius OR smaller charge density of cation / M2+ [1]\n• anion / CO 2– / carbonate ion is less distorted / less polarised OR C-O / C=O less weakened [1]\n3\n\n4(b) • calcium oxide / calcium hydroxide 3\n• CaSO / calcium sulfate is more soluble OR BaSO is less soluble [1]\n4 4\n• ΔH and ΔH are less exothermic / more endothermic (for BaSO ) [1]\nlatt hyd 4\n• ΔH is dominant factor / ΔH change is greater OR ΔH changes less [1]\nhyd hyd latt\n\n4(c)(i) mass of CO = 0.02 × 44 = 0.88 g [1] 1\n2\n\n4(c)(ii) (writes correct equation, deduces 3 CO per mole) 2\n2\nmoles of propane = 0.02 / 3 OR 0.00667 OR 1 / 150 [1]\nmass of propane = 0.02 / 3 × 44 = 0.293 g [1] ecf M1 × 44 3sf needed\nQuestion Answer Marks", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p08.png" ] }, { "id": "9701-2021-on-41-q05", "question_id": "9701-2021-on-41-q05", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "5(a)(i) +2 [1] 1\n\n5(a)(ii) 2\n[1]\nbond angle labelled 109.5° [1]\n© UCLES 2021 Page 8 of 16\n\n5(b)(i) [Co(H O) ]2+ + 2OH– → Co(H O) (OH) + 2H O OR [Co(H O) ]2+ + 2OH– → Co(OH) + 6H O [1] 2\n2 6 2 4 2 2 2 6 2 2\nblue precipitate [1]\n\n5(b)(ii) [Co(H O) ]2+ + 6NH → [Co(NH ) ]2+ + 6H O [1] 2\n2 6 3 3 6 2\nyellow / brown / straw solution [1]\n\n5(b)(iii) [Co(H O) ]2+ + 4Cl – → [CoCl ]2– + 6H O [1] 2\n2 6 4 2\nblue solution [1]\n\n5(b)(iv) ligand exchange [1] 1\n\n5(c)(i) (a species) that donates two lone pairs to form two dative bonds [1] 2\nto a (transition) metal atom / metal ion [1]\n\n5(c)(ii) 2\nM1 one correct 3D structure – trans or cis [1]\nM2 a correct optical isomer of M1 [must 3D] [1]\n© UCLES 2021 Page 9 of 16", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p08.png", "../answer-assets/9701_w21_ms_41-p09.png" ] }, { "id": "9701-2021-on-41-q06", "question_id": "9701-2021-on-41-q06", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 10, "status": "available", "reason": null, "text": "6(a) 2Cu2+ + 4I– → 2CuI + I OR 2CuSO + 4NaI → 2CuI + I + 2Na SO [1] 1\n2 4 2 2 4\n\n6(b)(i) M1: d-d orbital splitting occurs [1] 4\nM2: electron(s) promoted / excited [1]\nM3: wavelength / frequency of light is absorbed [1]\nM4: colour seen is complementary\nOR wavelength / frequency of light not absorbed is seen [1]\n\n6(b)(ii) (for Cu+) 3d10 OR 3d subshell full [1] 1\n\n6(c) M1: (Cu2+ / Cu+) Eo = (+)0.15 V AND (I / I–) Eo = (+)0.54 V [1] 2\n2\nM2: No, since (Eo ) negative / –0.39 V\ncell\nOR No, since (I / I–) is more positive than (Cu2+ / Cu+)\n2\nOR No, I is more easily reduced OR No, I stronger oxidant ORA [1]\n2 2\n\n6(d) M1: Cu2+ / Cu+ E becomes more positive as equilibrium shifts to the right [1] 2\nM2: The new E for Cu2+ / Cu+ is more positive than 0.54 / Eo (I / I–) [1]\n2\n© UCLES 2021 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p10.png" ] }, { "id": "9701-2021-on-41-q07", "question_id": "9701-2021-on-41-q07", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "7(a) 6 [1] 1\n\n7(b) M1: trend phenylethanoic acid > phenol > ethanol [1] 4\nM2: why phenylethanoic acid is the strongest\n• negative inductive electron withdrawing effect of C=O\nwhich weakens O-H bond / stabilises anion [1]\nM3: why phenol is stronger than ethanol / weaker than phenylethanoic acid\n• oxygen lone pair is delocalised into the ring system\nwhich weakens O-H bond / stabilises anion [1]\nM4: why ethanol is the weakest\n• electron donating alkyl / ethyl group\nwhich strengthens O-H bond / destabilises anion [1]\n\n7(c)(i) CH Cl + AlCl → +CH + AlCl – [1] 1\n3 3 3 4\n\n7(c)(ii) 3\nM1: arrow to CH + (arrow must come from inside the hexagon) [1]\n3\nM2: correct structure of intermediate [1]\nM3: arrow from C-H bond into the ring AND H+ seen [1]\n\n7(d)(i) Br + UV light [1] 1\n2\n© UCLES 2021 Page 11 of 16\n\n7(d)(ii) 1\n[1]\n\n7(d)(iii) CHECK Q is correct 2\nstep 2 – KCN in ethanol + heat [1]\nstep 3 – HCl(aq) + heat/reflux/boil [1]\n\n7(d)(iv) 1\n[1]\nALLOW any viable organic by-product from this radical substitution reaction,\ne.g. C H CH CH C H\n6 5 2 2 6 5\n© UCLES 2021 Page 12 of 16", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p11.png", "../answer-assets/9701_w21_ms_41-p12.png" ] }, { "id": "9701-2021-on-41-q08", "question_id": "9701-2021-on-41-q08", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 5, "status": "available", "reason": null, "text": "8(a) bromine decolourised OR orange / brown to colourless [1] 2\nwhite precipitate [1]\n\n8(b) no change [1] 1\n\n8(c) 1\n[1]\n\n8(d) 2,4,6-tribromophenylamine [1] ECF 8(c) for a bromophenylamine 1\nQuestion Answer Marks", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p13.png" ] }, { "id": "9701-2021-on-41-q09", "question_id": "9701-2021-on-41-q09", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 12, "status": "available", "reason": null, "text": "9(a) PCl OR PCl OR SOCl [1] 1\n5 3 2\n\n9(b)(i) amide [1] 1\n\n9(b)(ii) HCl / hydrogen chloride OR C H NH Cl / ethylammonium chloride [1] 1\n2 5 3\n\n9(c)(i) LiAl H [1] 1\n4\n\n9(c)(ii) reduction [1] 1\n\n9(d)(i) relative abundance = 2 carbons × 1.1 × 0.62 1\n= 1.36 / 1.4 [1] min 2sf\n\n9(d)(ii) CH CH + / C H + [1] 2\n3 2 2 5\n© UCLES 2021 Page 13 of 16\n\n9(e)(i) a [1] 1\n\n9(e)(iii) d [1] 1\n\n9(e)(iii) b, c, f [1] 1\n\n9(e)(iv) f [1] 1\nQuestion Answer Marks", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p13.png", "../answer-assets/9701_w21_ms_41-p14.png" ] }, { "id": "9701-2021-on-41-q10", "question_id": "9701-2021-on-41-q10", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 10, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 6, "status": "available", "reason": null, "text": "10(a) 1\n[1]\n© UCLES 2021 Page 14 of 16\n\n10(b) A and B 3\nany two from: [all net single positive charge]\n[1] × 2\nC at pH 1.0 [2+ positive charge]\n[1]\n© UCLES 2021 Page 15 of 16\n\n10(c) M1: peptide link correct and displayed unit including C=O 2\nM2: everything else correct\nOR\n[2]\n© UCLES 2021 Page 16 of 16", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_41-p14.png", "../answer-assets/9701_w21_ms_41-p15.png", "../answer-assets/9701_w21_ms_41-p16.png" ] }, { "id": "9701-2021-on-42-q01", "question_id": "9701-2021-on-42-q01", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "1(a) • enthalpy / energy change / given out / evolved / released 2\n• one mole is formed / made [1]\n• of compound / solid / lattice / crystal\n• (from) gaseous ions [1]\n\n1(b)b S–(g) + e– → S2–(g) [1] 1\n\n1(c)c (555 + 200 – 532 = 223, 223 × 8 = 1784) 3\nM1 selecting correct data 555, 200, 532 only, (ignore signs and multipliers) [1]\nM2 evaluation to give +223 [1]\nM3 multiplying M2 by 8 and evaluation ans (+) 1784 [1]\n\n1(d) (1619 + 555 – 2612 = –438) 2\nM1 selecting correct data 1619 555 2612 only, (ignore signs and multipliers) [1]\nM2 evaluation to give –438 [1]\n\n1(e)(i) ionic radius / size / sum of ionic radii [1] 2\nionic charge / product of ionic charges [1]\n\n1(e)(ii) M1 (size tends to make ΔHo of radium sulfide) less exothermic since the ions are larger [1] 2\nlatt\nM2 (charge tends to make ΔHo of radium sulfide) less exothermic since the ions are more highly charged [1]\nlatt\n\n1(e)(iii) (ionic) charge (since) 1\nAND\nΔHo of radium sulfide is more exothermic [1]\nlatt\n© UCLES 2021 Page 5 of 12", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p05.png" ] }, { "id": "9701-2021-on-42-q02", "question_id": "9701-2021-on-42-q02", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 10, "status": "available", "reason": null, "text": "2(a) M1 reference to equilibrium not being established / not at equilibrium [1] 2\nM2 concentration in water too high or conc in octan-1-ol too low owtte [1]\n\n2(b)(i) 1.1 / 1.13 / 1.127 / 1.1272 [1] 1\n\n2(b)(ii) M1 6.760 = (0.762 – x) / x [1] 3\nM2 7.76x = 0.762\nx = 0.0982 g (correct value scores 2) [1]\nM3 0.0982 / 74 = 0.0013 / 0.00133 mol [1]\n\n2(c)(i) 1.0x10–9 [1] 2\nmol3 dm–9 [1]\n\n2(c)(ii) yellow precipitate / yellow solid / yellow crystals [1] 2\ncommon ion effect / K has been exceeded [1]\nsp\nQuestion Answer Marks", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p06.png" ] }, { "id": "9701-2021-on-42-q03", "question_id": "9701-2021-on-42-q03", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 9, "status": "available", "reason": null, "text": "3(a) 0.351 / 24 = 0.015 (mol) [1] 2\n0.015 × 6.02 × 1023 = 9.0 × 1021 / 8.8 × 1021 [1]\n\n3(b) 1.76 × 1022 / 1.8 × 1022 [1] 1\n\n3(c) 2817 / 2816 / 2820 / 2800 C [1] 1\n\n3(d) 15 / 15.1 / 15.05 / 15.15 minutes [1] 1\n\n3(e)(i) –20 [1] 1\n© UCLES 2021 Page 6 of 12\n\n3(e)(ii) states / uses correct Gibbs equation [1] 2\nanswer = 190 / 191 / 190.0 [1]\n\n3(e)(iii) Becomes less feasible / less spontaneous / 1\nAND\nbecause ΔS is negative / TΔS becomes more negative / –TΔS becomes more positive [1]\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p06.png", "../answer-assets/9701_w21_ms_42-p07.png" ] }, { "id": "9701-2021-on-42-q04", "question_id": "9701-2021-on-42-q04", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "4(a) strontium nitrate AND because of larger cationic radius [1] 2\nNO – / nitrate ion / anion is less distorted / polarised\n3\nOR\nN–O or N=O bond less weakened / distorted [1]\n\n4(b) M1 magnesium oxide requires more water to dissolve AND because Sr(OH) is more soluble [1] 3\n2\nM2 lattice and hydration energies less for Sr(OH) [1]\n2\nM3 lattice energy is dominant factor / change in lattice energy is greater [1]\n\n4(c)(i) 0.13 [1] 1\n\n4(c)(ii) 26.7 cm3 must be 3SF [1] 1\n© UCLES 2021 Page 7 of 12", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p07.png" ] }, { "id": "9701-2021-on-42-q05", "question_id": "9701-2021-on-42-q05", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 15, "status": "available", "reason": null, "text": "5(a)(i) donates one lone pair / forms one dative bond AND to metal atom / to metal ion [1] 1\n\n5(a)(ii) Mo(CO) 0 / no charge / neutral / zero 2\n6\nFe(CN) 3–\n6\nboth formulae [1]\nboth charges [1]\n\n5(a)(iii) either of these renditions, with a CN on each bond: 1\nAND\n180o bond angle between two opposite bonds [1]\n\n5(b)(i) [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O 1\n2 6 3 3 4 2 2 2\nOR\nCu2+ + 2H O + 4NH → [Cu(NH ) (H O) ]2+ [1]\n2 3 3 4 2 2\n\n5(b)(ii) M1 d-orbitals split (could be in diagram) [1] 4\nM2 electrons transition to higher orbital / electrons promoted or excited [1]\nM3 wavelength / frequency / light / photon / hf / hn absorbed [1]\nM4 colour seen / reflected / transmitted is complement of colour absorbed [1]\n© UCLES 2021 Page 8 of 12\n\n5(c)(i) [Cu(NH ) (H O) ]2+ + 4HCl → CuCl 2– + 2H O + 4NH + 2\n3 4 2 2 4 2 4\nOR\n[Cu(NH ) (H O) ]2+ + 4Cl – (+ 4H+) → CuCl 2– + 2H O + 4NH (+ 4H+)\n3 4 2 2 4 2 3\nM1 formula and charge of CuCl 2– [1]\n4\nM2 rest of equation [1]\n\n5(c)(ii) ligand exchange / substitution / replacement / displacement [1] 1\n\n5(c)(iii) tetrahedral [1] 1\n\n5(c)(iv) yellow [1] 1\n\n5(c)(v) d-orbital splitting changes / ΔE changes [1] 1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p08.png", "../answer-assets/9701_w21_ms_42-p09.png" ] }, { "id": "9701-2021-on-42-q06", "question_id": "9701-2021-on-42-q06", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 7, "status": "available", "reason": null, "text": "6(a) red or pink to yellow / straw / brown / orange [1] 1\n\n6(b) [[Co(NH ) ]2+] / [[Co(H O) ]2+][NH ]6 [1] 1\n3 6 2 6 3\n\n6(c) Co(NH ) 2+ is more stable (than Co(H O) 2+) [1] 1\n3 6 2 6\n\n6(d)(i) Eo = + 0.30 [1] 1\ncell\n\n6(d)(ii) 4[Co(NH ) ]2+ + O + 2H O ⇌ 4[Co(NH ) ]3+ + 4OH– [1] 1\n3 6 2 2 3 6\n\n6(d)(iii) M1 No, because 1.82 V and 1.23 V [1] 2\nM2 Eo = –0.59 / 1.23 lower than 1.82 / 1.82 greater than 1.2 [1]\ncell\n© UCLES 2021 Page 9 of 12", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p09.png" ] }, { "id": "9701-2021-on-42-q07", "question_id": "9701-2021-on-42-q07", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 10, "status": "available", "reason": null, "text": "7(a) C H O [1] 1\n4 3 2\n\n7(b) 2\nbenzene-1,3-dicarboxylic acid 5 [1]\nbenzene-1,4-dicarboxylic acid 3 [1]\n\n7(c)(i) benzoic acid [1] 1\n\n7(c)(ii) COOH directs 3 position [1] 1\n\n7(c)(iii) electrophilic substitution [1] 1\n\n7(c)(iv) M1 curly arrow from within hexagon towards CH C+=O [1] 3\n3\nM2 correct intermediate [1]\nM3 curly arrow from C–H bond into hexagon and correct product Q [1]\n\n7(c)(v) MnO – / KMnO / manganateVII / permanganate aq / H+ / acidified / OH– then acid / alkaline then 1\n4 4\nacid heat / boil / reflux / T>50o\nOR\nalkaline iodine followed by acidification [1]\nQuestion Answer Marks", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p10.png" ] }, { "id": "9701-2021-on-42-q08", "question_id": "9701-2021-on-42-q08", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 14, "status": "available", "reason": null, "text": "8(a) M1 one diagram correct [1] 2\nM2 both diagrams correct 3D and different [1]\n© UCLES 2021 Page 10 of 12\n\n8(b) alanine because glutamic acid would have more than two / three peaks / absorbns / proton environments [1] 3\nreason why alanine has a doublet given as one neighbouring proton [1]\nglutamic acid would have (two) triplet(s) OR a multiplet [1]\nreason why alanine has a quartet / quadruplet given as three neighbouring protons [1]\n\n8(c)(i) 147 [1] 1\n\n8(c)(ii) 88 = [H NCH(CH )CO H]+ [1] 2\n2 2 2\n131 = [CH(CH CH CO H)CO H]+ [1]\n2 2 2 2\n\n8(d)(i) use of buffer [1] 1\n\n8(d)(ii) • correct circuit including DC power supply 2\n• paper or gel labelled [1]\n• sample towards the middle of the paper / gel\nOR\non cathode side [1]\n\n8(d)(iii) anode / positive / + 1\nAND\nanode / positive / + [1]\n\n8(d)(iv) M1 ala is –1 and glu is –2 [1] 2\nM2 ala is lighter / has lower M [1]\nr\n© UCLES 2021 Page 11 of 12", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p10.png", "../answer-assets/9701_w21_ms_42-p11.png" ] }, { "id": "9701-2021-on-42-q09", "question_id": "9701-2021-on-42-q09", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "9(a) 6\norganic starting material reagent and conditions\n1-butyl halide, e.g. NH under pressure or [1] + [1]\n3\nCH CH CH CH Br heated in sealed tube\n3 2 2 2\nbutanenitrile H and Ni or Pt / LiAlH / [1] + [1]\n2 4\nCH CH CH CN Na + ethanol\n3 2 2\nbutanamide LiAlH / Na + ethanol [1] + [1]\n4\nCH CH CH CONH\n3 2 2 2\n\n9(b) M1 butylamine > ammonia > phenylamine [1] 4\nM2 basicity related to ability of lp to accept proton / H+ [1]\nM3 butylamine is stronger because of positive inductive effect of alkyl group / C H [1]\n4 9\nM4 phenylamine is weaker because lp on N is delocalised into ring [1]", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p12.png" ] }, { "id": "9701-2021-on-42-q10", "question_id": "9701-2021-on-42-q10", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 10, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 5, "status": "available", "reason": null, "text": "10(a) C H O– / C H O–Na+ / C H ONa [1] 4\n6 5 6 5 6 5\nC H O– / C H O–Na+ / C H ONa [1]\n6 5 6 5 6 5\nC H N C H O–\n6 5 2 6 4\nor\nC H N C H OH or\n6 5 2 6 4\n[1]\nC H OCOC H CO C H [1]\n6 5 6 4 2 6 5\n\n10(b) 2 and 4 [1] 1\n© UCLES 2021 Page 12 of 12", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_42.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_42-p12.png" ] }, { "id": "9701-2021-on-43-q01", "question_id": "9701-2021-on-43-q01", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 9, "status": "available", "reason": null, "text": "1(a) moles of H = 462 / 24 000 = 0.01925 [1] 2\n2\nmolecules of H = 0.019 × 6.02 × 1023\n2\n= 1.16 × 1022 (1.1×1022 / 1.2 × 1022) [1] min 2sf ecf M1\n\n1(b) number of electrons = 1.16 × 1022 × 2 1\n= 2.32 × 1022 [1] min 2sf ecf 1a\n\n1(c) Q = 2.32 ×1022 × 1.6 × 10–19 1\n= 3.71 × 103 [1] min 2sf ecf 1b\n\n1(d) x = 3.71 × 103 / (14 × 60) 1\n= 4.4 (A) [1] min 2sf ecf 1c\n\n1(e)(i) ΔS = 262 + 205 – 140 = (+) 327 (J K–1 mol–1) [1] 1\n\n1(e)(ii) ΔG = ΔH – TΔS OR use of Gibbs equation [1] 2\nΔG = 572 – (298 × 0.327)\n= (+)474.6 (kJ mol–1) [1] min 3sf ecf 1e(i)\n\n1(e)(iii) becomes more feasible / spontaneous 1\nas TΔS is more positive / –TΔS becomes more negative  \nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p05.png" ] }, { "id": "9701-2021-on-43-q02", "question_id": "9701-2021-on-43-q02", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i H +Cl ( CH ) CO − 1\nK =   2 3 2  [1]\na ( )\nCl CH CO H\n 2 3 2 \n\n2(a)(ii) pK = – logK OR K = 10–pKa [1] 1\na a a\n\n2(a)(iii) [H+] = 10–4.0 = 1 × 10–4 [1] 1\n© UCLES 2021 Page 5 of 16\n\n2(a)(iv) • [HCl] = 1 × 10–4 ecf 2(a)(iii) 2\n• K = 10–4.52 = 3.02 × 10–5\na\n• [Cl(CH ) CO H] = (1 × 10–4)2 / 3.02 × 10–5\n2 3 2\n= 3.3 × 10–4 ecf\n[ HCl ] 1×10 −4\n• = =0.302 min 2sf ecf\nCl ( CH ) CO H 3.3×10 −4\n 2 3 2 \n\n2(b) M1 [H+] = 10–5 OR 1 × 10–5 3\nK = [H+][A–] / [HA] OR pH = pKa + log [A–] / [HA]\na\n  [1]\nM2 moles of A– = (1.35 × 10–5)(5 / 74) / (1 × 10–5)\nmoles of A– = 0.0912 [1] ecf\nM3 mass of sodium propanoate = 0.0912 × 96 = 8.76 [1] min 2sf ecf\n\n2(c) all of the (sodium) propanoate (ion) has been 2\nprotonated / converted to (propanoic) acid / neutralised [1]\nH+ is in excess / H+ is 0.1 mol dm–3 (from the H SO ) [1]\n2 4\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p05.png", "../answer-assets/9701_w21_ms_43-p06.png" ] }, { "id": "9701-2021-on-43-q03", "question_id": "9701-2021-on-43-q03", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 12, "status": "available", "reason": null, "text": "3(a) • enthalpy/energy change 2\n• one mole of electrons gained\n• by one mole of atoms\n• gaseous (atoms)\n\n3(b) Ca+(g) → Ca2+(g) + e- [1] 1\n© UCLES 2021 Page 6 of 16\n\n3(c) M1: selecting correct data 951, 844, 142 only 3\nM2: evaluation to give 249 (ΔH )\natom\nOR 2(951) = BE – 2(142) + 2(844)\nM3: evaluation to 498 (2 × 249) ecf M2\n951 = ΔH –142 + 844\natom\nΔH = 249\natom\nBE = 498 (kJ mol–1) [3]\n\n3(d)(i) attraction between nucleus / protons / nuclear charge 1\nand electron [1]\n\n3(d)(ii) repulsion between 1– ion / electrons of O– 1\nand electron [1]\n\n3(e) M1: selecting correct data 951, 1933, 3517 only (ignore signs) 2\nM2: evaluation to give –633 (ΔH) ecf\nf\nΔH = 951 + 1933 – 3517 = –633 (kJ mol–1) [2]\nf\n\n3(f) ionic charge / charge density (of the ions) [1] 2\ngreater (attractive) force between the ions [1]\n© UCLES 2021 Page 7 of 16", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p06.png", "../answer-assets/9701_w21_ms_43-p07.png" ] }, { "id": "9701-2021-on-43-q04", "question_id": "9701-2021-on-43-q04", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 8, "status": "available", "reason": null, "text": "4(a) • barium carbonate / Ba / BaCO 2\n3\n• larger ionic radius OR smaller charge density of cation / M2+ [1]\n• anion / CO 2– / carbonate ion is less distorted / less polarised OR C-O / C=O less weakened [1]\n3\n\n4(b) • calcium oxide / calcium hydroxide 3\n• CaSO / calcium sulfate is more soluble OR BaSO is less soluble [1]\n4 4\n• ΔH and ΔH are less exothermic / more endothermic (for BaSO ) [1]\nlatt hyd 4\n• ΔH is dominant factor / ΔH change is greater OR ΔH changes less [1]\nhyd hyd latt\n\n4(c)(i) mass of CO = 0.02 × 44 = 0.88 g [1] 1\n2\n\n4(c)(ii) (writes correct equation, deduces 3 CO per mole) 2\n2\nmoles of propane = 0.02 / 3 OR 0.00667 OR 1 / 150 [1]\nmass of propane = 0.02 / 3 × 44 = 0.293 g [1] ecf M1 × 44 3sf needed\nQuestion Answer Marks", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p08.png" ] }, { "id": "9701-2021-on-43-q05", "question_id": "9701-2021-on-43-q05", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "5(a)(i) +2 [1] 1\n\n5(a)(ii) 2\n[1]\nbond angle labelled 109.5° [1]\n© UCLES 2021 Page 8 of 16\n\n5(b)(i) [Co(H O) ]2+ + 2OH– → Co(H O) (OH) + 2H O OR [Co(H O) ]2+ + 2OH– → Co(OH) + 6H O [1] 2\n2 6 2 4 2 2 2 6 2 2\nblue precipitate [1]\n\n5(b)(ii) [Co(H O) ]2+ + 6NH → [Co(NH ) ]2+ + 6H O [1] 2\n2 6 3 3 6 2\nyellow / brown / straw solution [1]\n\n5(b)(iii) [Co(H O) ]2+ + 4Cl – → [CoCl ]2– + 6H O [1] 2\n2 6 4 2\nblue solution [1]\n\n5(b)(iv) ligand exchange [1] 1\n\n5(c)(i) (a species) that donates two lone pairs to form two dative bonds [1] 2\nto a (transition) metal atom / metal ion [1]\n\n5(c)(ii) 2\nM1 one correct 3D structure – trans or cis [1]\nM2 a correct optical isomer of M1 [must 3D] [1]\n© UCLES 2021 Page 9 of 16", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p08.png", "../answer-assets/9701_w21_ms_43-p09.png" ] }, { "id": "9701-2021-on-43-q06", "question_id": "9701-2021-on-43-q06", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 10, "status": "available", "reason": null, "text": "6(a) 2Cu2+ + 4I– → 2CuI + I OR 2CuSO + 4NaI → 2CuI + I + 2Na SO [1] 1\n2 4 2 2 4\n\n6(b)(i) M1: d-d orbital splitting occurs [1] 4\nM2: electron(s) promoted / excited [1]\nM3: wavelength / frequency of light is absorbed [1]\nM4: colour seen is complementary\nOR wavelength / frequency of light not absorbed is seen [1]\n\n6(b)(ii) (for Cu+) 3d10 OR 3d subshell full [1] 1\n\n6(c) M1: (Cu2+ / Cu+) Eo = (+)0.15 V AND (I / I–) Eo = (+)0.54 V [1] 2\n2\nM2: No, since (Eo ) negative / –0.39 V\ncell\nOR No, since (I / I–) is more positive than (Cu2+ / Cu+)\n2\nOR No, I is more easily reduced OR No, I stronger oxidant ORA [1]\n2 2\n\n6(d) M1: Cu2+ / Cu+ E becomes more positive as equilibrium shifts to the right [1] 2\nM2: The new E for Cu2+ / Cu+ is more positive than 0.54 / Eo (I / I–) [1]\n2\n© UCLES 2021 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p10.png" ] }, { "id": "9701-2021-on-43-q07", "question_id": "9701-2021-on-43-q07", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "7(a) 6 [1] 1\n\n7(b) M1: trend phenylethanoic acid > phenol > ethanol [1] 4\nM2: why phenylethanoic acid is the strongest\n• negative inductive electron withdrawing effect of C=O\nwhich weakens O-H bond / stabilises anion [1]\nM3: why phenol is stronger than ethanol / weaker than phenylethanoic acid\n• oxygen lone pair is delocalised into the ring system\nwhich weakens O-H bond / stabilises anion [1]\nM4: why ethanol is the weakest\n• electron donating alkyl / ethyl group\nwhich strengthens O-H bond / destabilises anion [1]\n\n7(c)(i) CH Cl + AlCl → +CH + AlCl – [1] 1\n3 3 3 4\n\n7(c)(ii) 3\nM1: arrow to CH + (arrow must come from inside the hexagon) [1]\n3\nM2: correct structure of intermediate [1]\nM3: arrow from C-H bond into the ring AND H+ seen [1]\n\n7(d)(i) Br + UV light [1] 1\n2\n© UCLES 2021 Page 11 of 16\n\n7(d)(ii) 1\n[1]\n\n7(d)(iii) CHECK Q is correct 2\nstep 2 – KCN in ethanol + heat [1]\nstep 3 – HCl(aq) + heat/reflux/boil [1]\n\n7(d)(iv) 1\n[1]\nALLOW any viable organic by-product from this radical substitution reaction,\ne.g. C H CH CH C H\n6 5 2 2 6 5\n© UCLES 2021 Page 12 of 16", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p11.png", "../answer-assets/9701_w21_ms_43-p12.png" ] }, { "id": "9701-2021-on-43-q08", "question_id": "9701-2021-on-43-q08", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 5, "status": "available", "reason": null, "text": "8(a) bromine decolourised OR orange / brown to colourless [1] 2\nwhite precipitate [1]\n\n8(b) no change [1] 1\n\n8(c) 1\n[1]\n\n8(d) 2,4,6-tribromophenylamine [1] ECF 8(c) for a bromophenylamine 1\nQuestion Answer Marks", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p13.png" ] }, { "id": "9701-2021-on-43-q09", "question_id": "9701-2021-on-43-q09", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 12, "status": "available", "reason": null, "text": "9(a) PCl OR PCl OR SOCl [1] 1\n5 3 2\n\n9(b)(i) amide [1] 1\n\n9(b)(ii) HCl / hydrogen chloride OR C H NH Cl / ethylammonium chloride [1] 1\n2 5 3\n\n9(c)(i) LiAl H [1] 1\n4\n\n9(c)(ii) reduction [1] 1\n\n9(d)(i) relative abundance = 2 carbons × 1.1 × 0.62 1\n= 1.36 / 1.4 [1] min 2sf\n\n9(d)(ii) CH CH + / C H + [1] 2\n3 2 2 5\n© UCLES 2021 Page 13 of 16\n\n9(e)(i) a [1] 1\n\n9(e)(iii) d [1] 1\n\n9(e)(iii) b, c, f [1] 1\n\n9(e)(iv) f [1] 1\nQuestion Answer Marks", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p13.png", "../answer-assets/9701_w21_ms_43-p14.png" ] }, { "id": "9701-2021-on-43-q10", "question_id": "9701-2021-on-43-q10", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 10, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 6, "status": "available", "reason": null, "text": "10(a) 1\n[1]\n© UCLES 2021 Page 14 of 16\n\n10(b) A and B 3\nany two from: [all net single positive charge]\n[1] × 2\nC at pH 1.0 [2+ positive charge]\n[1]\n© UCLES 2021 Page 15 of 16\n\n10(c) M1: peptide link correct and displayed unit including C=O 2\nM2: everything else correct\nOR\n[2]\n© UCLES 2021 Page 16 of 16", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_43-p14.png", "../answer-assets/9701_w21_ms_43-p15.png", "../answer-assets/9701_w21_ms_43-p16.png" ] }, { "id": "9701-2021-on-51-q01", "question_id": "9701-2021-on-51-q01", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a) It is not possible to measure a temperature when heating. 1\n\n1(b)(i) Mass of KHCO = 2.503 g 1\n3\nAND\nMass of K CO = 3.455 g\n2 3\n\n1(b)(ii) 2 ×0.0005 ×100 1\n= = 0.04(00)%\n2.503\nCorrect working must be shown\n\n1(c)(i) M1 3.5 °C for KHCO and1.5 °C for K CO in results table 2\n3 2 3\nAND\nq = 50 × 4.18 × 3.5 = 731.5 (J)\nM2 ΔH = M1 / (0.0250 × 1000) = + 29.26 kJ mol–1\n1\n\n1(c)(ii) M1 q = K CO = 50 × 4.18 × 1.5 = 313.5 (J) 2\n2 3\nM1\nM2 ΔH = − = –12.54 kJ mol–1\n2 0.025 ×1000\n\n1(d) M1 2 × 1(c)(i) = 2 × 29.26 = 58.52 kJ mol–1 2\nM2 M1 – 1(c)(ii) = +58.52 – (–12.54) = +71.06 kJ mol–1\n\n1(e) Any two from: 2\n• Heat loss OR heat gain (to / from surroundings)\n• Amount of solid not exactly 0.0250 moles\n• Experiment not carried out under standard conditions.\n© UCLES 2021 Page 5 of 8\n\n1(f) Add a lid to the polystyrene cup. 1\n\n1(g) Burette. 1\n\n1(h) Wear chemically resistant gloves 1\nAND\nReagent(s) are irritant(s).\n© UCLES 2021 Page 6 of 8", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_51-p05.png", "../answer-assets/9701_w21_ms_51-p06.png" ] }, { "id": "9701-2021-on-51-q02", "question_id": "9701-2021-on-51-q02", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a) M1 Addition of solid and acid must not lead to initial escape of gas from sealed system or produce excess volume 3\nmeasurement leading to a false volume reading e.g. Adding acid to solid already in the flask or vice versa.\nM2 Allow any leakproof apparatus capable of delivering a sample of gas into one collection vessel over water.\nM3 A vertical inverted measuring cylinder or burette filled with water positioned to collect the gas (via delivery tube).\n\n2(b)(i) 1\nV – V / cm3\nfinal t\n93\n71\n56\n43\n32\n25\n18\n15\n14\n6\n3\n0\n93, 71, 56, 43, 32, 25, 18, 15, 14, 6, 3, 0\n© UCLES 2021 Page 7 of 8\n\n2(b)(ii) M1 Selects and labels appropriate linear scales for axes AND All points plotted correctly AND covering at least half of the 2\ngrid for each axis.\nM2 Smooth curve of best fit line drawn going through most points.\n\n2(b)(iii) The point at 240 s. 1\n\n2(b)(iv) The volume of gas may have been read before 240 . 1\n\n2(b)(v) M1 Suitable construction lines shown on the graph. 2\nM2: Two half-lives read correctly from graph\nAND\nmean half-life calculated using values read correctly from graph. (Likely to be approx. 78 s)\n\n2(b)(vi) 0.693 1\n=\naverage 𝑡 calculated in 2(b)(v)\n(cid:2869)/(cid:2870)\n\n2(c) The value of k would increase 1\nAND\nas t would decrease.\n½\n\n2(d)(i) M1 Table complete with four different sets of values that would produce different concentrations, total volume = 50 cm3 2\nAND\nAll volumes of 2.0 mol dm–3 acid must be 10 cm3 or greater (to ensure an excess).\nM2 All concentrations correctly calculated.\n\n2(d)(ii) time (taken for fizzing to stop) 1\n\n2(d)(iii) The experiment should be repeated to identify / eliminate anomalies. 1\n© UCLES 2021 Page 8 of 8", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_51-p07.png", "../answer-assets/9701_w21_ms_51-p08.png" ] }, { "id": "9701-2021-on-52-q01", "question_id": "9701-2021-on-52-q01", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 11, "status": "available", "reason": null, "text": "1(a) 20.1 °C AND 0.53 g 1\n\n1(b) 0.53 / 46 = 0.0115 (mol) 1\n\n1(c) M1 q = 100 × 4.18 × 20.1 = 8401.8 J 2\nM2 ΔH = 8401.8 / 0.0115 × 1000 = –731 (kJ mol–1)\nc\n\n1(d) (2 × 0.05) / 20.1 × 100 = 0.498% 1\nWorking must be shown\n\n1(e) (accuracy would) increase 1\nand\nthe percentage error (for temperature / calculated mass of ethanol) is reduced\n(accuracy would) decrease\nand\nMore heat loss (due to high temperature rise)\n\n1(f) M1 (ΔH ) would be less exothermic / less negative 2\nc\nM2 moles (of ethanol) ‘combusted’ / ‘used’ (will appear to) increase (due to evaporation)\n\n1(g)(i) M1 heat loss (to the surroundings) 2\nM2 incomplete combustion\n\n1(g)(ii) bond enthalpies vary according to compound / molecule (the bond is situated in) 1\nOR\naverage bond enthalpies assume ethanol / water / substances are gaseous\n© UCLES 2021 Page 5 of 7", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_52-p05.png" ] }, { "id": "9701-2021-on-52-q02", "question_id": "9701-2021-on-52-q02", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 19, "status": "available", "reason": null, "text": "2(a) M1 (substances) are flammable 2\nM2 to control the temperature (of the reaction)\n\n2(b) Wear chemically resistant gloves 1\n\n2(c) To allow the halogenoalkane and sodium hydroxide to mix / react 1\n\n2(d) To indicate (the time) when the sodium hydroxide has been used up. 1\n\n2(e)(i) 2.40 × 10–3 1\n4.20 × 10–3\n7.41 × 10–3\n\n2(e)(ii) (1/t / rate of reaction increases from chlorobutane to iodobutane) bond enthalpy of carbon halogen bond decreases from 1\nC–Cl to C–I\n\n2(f) The concentration of sodium hydroxide. 1\n\n2(g)(i) M1 12 points plotted correctly 2\nM2 Curved line of best fit to cover data points\n\n2(g)(ii) point at t = 480 s 1\n\n2(g)(iii) Step 4 was omitted 1\n\n2(g)(iv) M1 Suitable construction lines shown on the graph. 2\nM2 Two half-lives correctly determined.\n\n2(g)(v) Yes 1\nAND\nhalf-life is constant\n© UCLES 2021 Page 6 of 7\n\n2(h)(i) (10 cm3) Pipette 1\n\n2(h)(ii) Insufficient (volume of) reaction mixture 1\n\n2(h)(iii) To quench the reaction 1\n\n2(i) The electrical conductivity of the mixture would not change enough to be measured (in a chemistry laboratory) 1\n© UCLES 2021 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_52-p06.png", "../answer-assets/9701_w21_ms_52-p07.png" ] }, { "id": "9701-2021-on-53-q01", "question_id": "9701-2021-on-53-q01", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a) It is not possible to measure a temperature when heating. 1\n\n1(b)(i) Mass of KHCO = 2.503 g 1\n3\nAND\nMass of K CO = 3.455 g\n2 3\n\n1(b)(ii) 2 ×0.0005 ×100 1\n= = 0.04(00)%\n2.503\nCorrect working must be shown\n\n1(c)(i) M1 3.5 °C for KHCO and1.5 °C for K CO in results table 2\n3 2 3\nAND\nq = 50 × 4.18 × 3.5 = 731.5 (J)\nM2 ΔH = M1 / (0.0250 × 1000) = + 29.26 kJ mol–1\n1\n\n1(c)(ii) M1 q = K CO = 50 × 4.18 × 1.5 = 313.5 (J) 2\n2 3\nM1\nM2 ΔH = − = –12.54 kJ mol–1\n2 0.025 ×1000\n\n1(d) M1 2 × 1(c)(i) = 2 × 29.26 = 58.52 kJ mol–1 2\nM2 M1 – 1(c)(ii) = +58.52 – (–12.54) = +71.06 kJ mol–1\n\n1(e) Any two from: 2\n• Heat loss OR heat gain (to / from surroundings)\n• Amount of solid not exactly 0.0250 moles\n• Experiment not carried out under standard conditions.\n© UCLES 2021 Page 5 of 8\n\n1(f) Add a lid to the polystyrene cup. 1\n\n1(g) Burette. 1\n\n1(h) Wear chemically resistant gloves 1\nAND\nReagent(s) are irritant(s).\n© UCLES 2021 Page 6 of 8", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_53-p05.png", "../answer-assets/9701_w21_ms_53-p06.png" ] }, { "id": "9701-2021-on-53-q02", "question_id": "9701-2021-on-53-q02", "subject": "9701", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a) M1 Addition of solid and acid must not lead to initial escape of gas from sealed system or produce excess volume 3\nmeasurement leading to a false volume reading e.g. Adding acid to solid already in the flask or vice versa.\nM2 Allow any leakproof apparatus capable of delivering a sample of gas into one collection vessel over water.\nM3 A vertical inverted measuring cylinder or burette filled with water positioned to collect the gas (via delivery tube).\n\n2(b)(i) 1\nV – V / cm3\nfinal t\n93\n71\n56\n43\n32\n25\n18\n15\n14\n6\n3\n0\n93, 71, 56, 43, 32, 25, 18, 15, 14, 6, 3, 0\n© UCLES 2021 Page 7 of 8\n\n2(b)(ii) M1 Selects and labels appropriate linear scales for axes AND All points plotted correctly AND covering at least half of the 2\ngrid for each axis.\nM2 Smooth curve of best fit line drawn going through most points.\n\n2(b)(iii) The point at 240 s. 1\n\n2(b)(iv) The volume of gas may have been read before 240 . 1\n\n2(b)(v) M1 Suitable construction lines shown on the graph. 2\nM2: Two half-lives read correctly from graph\nAND\nmean half-life calculated using values read correctly from graph. (Likely to be approx. 78 s)\n\n2(b)(vi) 0.693 1\n=\naverage 𝑡 calculated in 2(b)(v)\n(cid:2869)/(cid:2870)\n\n2(c) The value of k would increase 1\nAND\nas t would decrease.\n½\n\n2(d)(i) M1 Table complete with four different sets of values that would produce different concentrations, total volume = 50 cm3 2\nAND\nAll volumes of 2.0 mol dm–3 acid must be 10 cm3 or greater (to ensure an excess).\nM2 All concentrations correctly calculated.\n\n2(d)(ii) time (taken for fizzing to stop) 1\n\n2(d)(iii) The experiment should be repeated to identify / eliminate anomalies. 1\n© UCLES 2021 Page 8 of 8", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9701_w21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2021-oct-nov/9701_w21_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w21_ms_53-p07.png", "../answer-assets/9701_w21_ms_53-p08.png" ] }, { "id": "9701-2022-m-42-q01", "question_id": "9701-2022-m-42-q01", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 19, "status": "available", "reason": null, "text": "1(a)(i) oxidising agent [1] 2\nH O + 2H+ + 2I– → 2H O + I [1]\n2 2 2 2\n\n1(a)(ii) M1: K (93.8) = [I (cyclohexane)] ÷ [I (aq)] 2\npc 2 2\n93.8 = (0.390 / 15) ÷ (x / 20)\nM2: mass of I (aq), x = 5.54 × 10–3 (g) ecf\n2\n\n1(a)(iii) • K would be lower 2\npc\n• hexan-2-one is more polar (than cyclohexane)\nOR hexan-2-one is polar AND cyclohexane is non-polar\n• I is (therefore) less soluble in hexan-2-one\n2\nAll three correct for two marks\n\n1(b)(i) enthalpy change when one mole of a solute AND dissolves in water to form a solution of infinite dilution 1\n\n1(b)(ii) –(–629) + (–322) + (–293) = (+)14 (kJ mol–1) 1\n\n1(b)(iii) (cationic) charge density decreases Li+ to K+ [1] 2\nso lattice energies become less negative / less exothermic AND because less attraction between ions [1]\n\n1(c)(i) M1: moles of thiosulfate = 0.02230 × 0.150 = 3.345 × 10–3 2\nM2: [Cu2+] = 2 × ½ × 3.345 × 10–3 ÷ 0.0250 = 0.134 (mol dm–3) ecf\n\n1(c)(ii) starch 1\n\n1(c)(iii) 1\n3d\nxy\n\n1(d) Reduction of Fe3+: 2Fe3+ + 2I– → 2Fe2+ + I [1] 2\n2\nRegeneration of Fe3+: 2Fe2+ + S O 2– → 2Fe3+ + 2SO 2– [1]\n2 8 4\n© UCLES 2022 Page 5 of 12\n\n1(e) M1: ([Hg2+]) = 1.00 × 10–7 ÷ 454.4 = 2.20 × 10–10 (mol dm–3) 3\nM2: K = [Hg2+][I–]2 = 4[Hg2+]3 = 4.26 × 10–29 ecf\nsp\nM3: units = mol3 dm–9 ecf\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_42-p05.png", "../answer-assets/9701_m22_ms_42-p06.png" ] }, { "id": "9701-2022-m-42-q02", "question_id": "9701-2022-m-42-q02", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 9, "status": "available", "reason": null, "text": "2(a)(i) 1 mol liquid and 2 mol gas formed from 3 mol solid OR two solid compounds converted to a liquid and a gas 1\n\n2(a)(ii) M1: (as T increases) TΔS becomes greater (than ΔH) 2\nOR (as T increases) TΔS becomes more positive\nM2: (as T increases) feasibility will increase as ∆G becomes more negative\n\n2(b)(i) M1: = 314 + 131 – (19 + 3 × 187) use of values and correct stoichiometry 2\nM2: = –135 (J K–1 mol–1)\n\n2(b)(ii) M1: ∆G = 0 ∴ T = ∆H / ∆S = +219.3 × 103 ÷ –(b)(i) 2\nM2: = 1624(.4) (K)\n\n2(c) M1: 1.00 g Si is 1/28.1 = 0.0356 mol 2\n∴ moles of e– needed = 4 × mol Si = 0.142 faraday (3 sf)\nM2: Q = It ∴ t = M1 × 96500 ÷ 6 = 2289 (s) ecf\nQuestion Answer Marks", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_42-p06.png" ] }, { "id": "9701-2022-m-42-q03", "question_id": "9701-2022-m-42-q03", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 19, "status": "available", "reason": null, "text": "3(a)(i) (d-block) element that forms one or more stable ions with incomplete d subshell / incomplete d orbitals 1\n© UCLES 2022 Page 6 of 12\n\n3(a)(ii) • variable oxidation states 1\n• behave as catalysts\n• form complex ions / complexes\n• form coloured compounds / ions\nAny two for one mark\n\n3(b)(i) Ti is in +4 oxidation state so no d electrons / d0 OR Ti in TiO2+ has no d electrons / d0 [1] 2\ncannot absorb photons / light in visible spectrum OR no wavelength / frequency absorbed in visible spectrum [1]\n\n3(b)(ii) (1s2) 2s2 2p6 3s2 3p6 3d1 (4s0) 1\n\n3(b)(iii) 2TiO2+ + 4H+ + Zn → 2Ti3+ + 2H O + Zn2+ 1\n2\n\n3(c)(i) M1: ∆G = – nE⦵ F AND n = 4 3\ncell\nM2: ∴ E⦵ = –436100 / –4(96500) = 1.13 V ecf\ncell\nM3: E⦵ = E⦵(O ,4H+|H O) – E⦵(TiO2+|Ti3+) = 1.23 – E⦵(TiO2+|Ti3+)\ncell 2 2\n∴ E⦵(TiO2+|Ti3+) = (+)0.1 (V) ecf\n\n3(c)(ii) Ti3+ empty / vacant d orbitals can form dative bonds / accept a lone pair from a ligand 1\nOR Ti3+ has vacant d-orbitals which are energetically accessible\n\n3(c)(iii) the E⦵ of the half-cell must be greater than +1.23 V / E⦵ of the O |H+ half-cell 1\n2\nas E⦵ < 0 and the reaction does not occur\ncell\n\n3(d)(i) the number of co-ordinate bonds being formed by the metal atom/ion 1\n\n3(d)(ii) TiO + 6HF → TiF 2– + 2H O + 2H+ 1\n2 6 2\nOR TiO + 6HF → TiF 2– + 2H O+\n2 6 3\n\n3(d)(iii) species with two lone pairs (of electrons) [1] 2\nthat form dative covalent / co-ordinate bonds to a central metal atom/ion [1]\n© UCLES 2022 Page 7 of 12\n\n3(d)(iv) mirror image of isomer I trans isomer 2\n\n3(d)(v) isomer I AND cis isomer drawn by candidate ecf [1] 2\ndipoles do not cancel / partial charges do not cancel [1]\nQuestion Answer Marks", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_42-p06.png", "../answer-assets/9701_m22_ms_42-p07.png", "../answer-assets/9701_m22_ms_42-p08.png" ] }, { "id": "9701-2022-m-42-q04", "question_id": "9701-2022-m-42-q04", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 18, "status": "available", "reason": null, "text": "4(a)(i) In F: (phenyl)amine AND carboxylic acid 2\nIn J: phenol AND ester\nAny two for one mark\nAll four for two marks\n\n4(a)(ii) 0 (zero) in F AND 2 (two) in J 1\n© UCLES 2022 Page 8 of 12\n\n4(b)(i) step 1 CH Cl AND AlCl [1] 3\n3 3\nstep 2 D = [1]\nstep 4 (hot) Sn AND concentrated AND HCl [1]\n\n4(b)(ii) 2\n[1]\nCOOH group is electron-withdrawing group and 3,5-/meta- directing [1]\n\n4(c)(i) C H O 1\n9 18\n\n4(c)(ii) hydrolysis [1] 2\nacid–base / neutralisation [1]\n\n4(d)(i) C H OH + Na → C H O(–)Na(+) + ½H 1\n6 5 6 5 2\n\n4(d)(ii) 1\n\n4(d)(iii) • (CO)O—H bond weaker / more easy to donate H+ in K 2\n• owing to negative inductive / electron withdrawing effect of C=O / COOH group\n• carboxylate anion stabilised / phenoxide anion is less stabilised\nAll three for two marks\n© UCLES 2022 Page 9 of 12\n\n4(e) p-orbital on oxygen overlaps with ring / π system OR lone pair of e– on oxygen is delocalised into the ring [1] 3\nelectron density in ring increases [1]\nattracts/polarises electrophile better [1]\nQuestion Answer Marks", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_42.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_42-p08.png", "../answer-assets/9701_m22_ms_42-p09.png", "../answer-assets/9701_m22_ms_42-p10.png" ] }, { "id": "9701-2022-m-42-q05", "question_id": "9701-2022-m-42-q05", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 21, "status": "available", "reason": null, "text": "5(a)(i) CH CHClCOOH + H O ⇌ CH CHClCOO– + H O+ 1\n3 2 3 3\nOR CH CHClCOOH ⇌ CH CHClCOO– + H+\n3 3\n\n5(a)(ii) M1: [H+] = 10–1.51 = 0.0309 (mol dm– 3) 2\nM2: K = 0.03092/0.60 = 1.592 × 10–3 ecf\na\npK = –log 1.592 × 10–3 = 2.80 ecf\na\n\n5(a)(iii) CH CHCl COOH is a stronger acid (than propanoic acid, owing to higher [H+] [1] 2\n3\nbecause electron-withdrawing effect of Cl (substituent) AND weakens O—H / carboxylate anion stabilised [1]\n\n5(b)(i) pH at which a molecule has no overall charge / is neutral OR pH at which it exists as a zwitterion / dipolar ion 1\n\n5(b)(ii) CH CH(N+H )COOH 1\n3 3\n\n5(b)(iii) 1\n\n5(b)(iv) 2\nmiddle amide group shown displayed [1]\nrest of structure correct [1]\n© UCLES 2022 Page 10 of 12\n\n5(b)(v) condensation 1\n\n5(b)(vi) C is biodegradable / easily hydrolysed 1\n\n5(c)(i) pair I = CH CHCl COOH (acid) and CH CHCl COO– (c.base) 1\n3 3\npair II = NH (base) and NH + (c.acid)\n3 4\n\n5(c)(ii) evidence on graph / paper of one half-life (use of data) / t = 500 s [1] 2\n1/2\nconstant half-life (= first order) [1]\n\n5(c)(iii) so that [NH ] is (effectively) constant AND doesn’t affect the rate / zero order 1\n3\n\n5(c)(iv) M1: when [NH ] increases ×1.5, rate increase ×1.5 AND first order (w.r.t. [NH ]) / rate is proportional to [NH ] 2\n3 3 3\nM2: S 2 DEP on M1\nN\n(because rate is first order w.r.t. to [2-chloropropanoate] and [NH ])\n3\n\n5(c)(v) greater proportion of particles have E ⩾ E [1] 2\nA\nfrequency of (effective) collisions increases AND rate increases\nOR rate of collisions increases AND rate increases [1]\n\n5(c)(vi) 1\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_42-p10.png", "../answer-assets/9701_m22_ms_42-p11.png" ] }, { "id": "9701-2022-m-42-q06", "question_id": "9701-2022-m-42-q06", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 14, "status": "available", "reason": null, "text": "6(a) HOCH COOH + 2SOCl → ClCH COCl + 2SO + 2HCl 1\n2 2 2 2\n\n6(b) to remove / neutralise excess H+ / acid produced 1\nOR to react with any acidic by-products / HCl / SO\n2\nOR to react with any unreacted W\n© UCLES 2022 Page 11 of 12\n\n6(c) 4\nM1: curly arrow from lone pair on :NH to carbonyl C(δ+)=O\n2\nM2: correct dipole on δ+C=Oδ– AND curly arrow from bond C=O to O(δ–)\nM3: correct structure of the intermediate (inc. charges)\nM4: curly arrow from lone pair on :O– to C=O AND curly arrow from C—Cl to Cl\n\n6(d) N / nitrogen can donate its lone pair / LP / pair of electrons 1\n\n6(e)(i) mobile = ethyl ethanoate 1\nstationary = SiO / silica or Al O / alumina\n2 2 3\n\n6(e)(ii) R = distance travelled by solute / substance / compound / component 1\nf\n÷ distance travelled by solvent (front)\n\n6(e)(iii) X is more attracted / more affinity / adsorbed more to the stationary phase 1\nOR lidocaine dissolves better in the solvent ORA\n\n6(f)(i) quartet AND triplet 1\n\n6(f)(ii) • δ 7.1 = attached to aromatic ring / H—Ar 2\n• δ 3.0 = alkyl next to C=O / —CH –C=O\n(2)\n• δ 2.3 = alkyl next to aromatic ring / H C–Ar\n(3)\nAll three correct for two marks\n\n6(f)(iii) 9 (nine) 1\n© UCLES 2022 Page 12 of 12", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_42-p11.png", "../answer-assets/9701_m22_ms_42-p12.png" ] }, { "id": "9701-2022-m-52-q01", "question_id": "9701-2022-m-52-q01", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 9, "status": "available", "reason": null, "text": "1(a) allow (crucible / contents) to cool 1\n\n1(b) to prevent spitting of crystals 1\n\n1(c) pipe-clay triangle 1\n\n1(d) heat to constant mass 1\n\n1(e)(i) M1: (mass H O) = 3.15 (g) AND (mass CuSO ) = 4.34 (g) 3\n2 4\nM2: n(H O) = 3.15 / 18 = 0.175\n2\nAND\nn(CuSO ) = 4.34 / 159.6 = 0.02719…\n4\nM3: 0.02719… / 0.02719… AND 0.175 / 0.02719… = 6.44\n\n1(e)(ii) If value of x in 1(e)(i) is above 5: 1\nCuSO must have decomposed / crystals were wet before heating / some crystals were lost when heated (e.g. during\n4\nspitting) / impurity that decomposes or evaporates\nIf value of x in 1(e)(i) is below 5:\ndid not heat it enough / not all water driven off / unreactive impurity\n\n1(f) [(2 × 0.005) / 13.60] × 100 = 0.0735 (%) 1\n© UCLES 2022 Page 5 of 7", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_52-p05.png" ] }, { "id": "9701-2022-m-52-q02", "question_id": "9701-2022-m-52-q02", "subject": "9701", "year": 2022, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 21, "status": "available", "reason": null, "text": "2(a)(i) solution of known concentration 1\n\n2(a)(ii) M1: measure 28.55 cm3 of CH COOH 4\n3\nM2: using burette / graduated pipette\nM3: (transfer/add CH COOH to) a 250 cm3 volumetric flask\n3\nM4: top the (volumetric) flask up to (250 cm3) mark with distilled water\n\n2(b) (pure) CH COOH is corrosive 1\n3\n\n2(c)(i) 2\n3 4\n4.47 0.0752\n6.32 0.110\n8.94 0.154\n12.6 0.213\n17.9 0.368\n25.3 0.428\nM1: column 3 all values to 3 SF\nM2: column 4 all values to 3 SF\n\n2(c)(ii) conductivity 1\n\n2(c)(iii) to calibrate the conductivity meter 1\n\n2(c)(iv) temperature 1\n\n2(d) M1: 6 points plotted correctly 2\nM2: line of best fit drawn\n© UCLES 2022 Page 6 of 7\n\n2(e) M1: 5th point circled at 17.9, 0.368 2\nM2: ethanoic acid was diluted incorrectly\n\n2(f)(i) M1: co-ordinates read and recorded correctly 2\nM2: gradient calculated to 3 SF\n\n2(f)(ii) M1: (gradient calculated in (f)(i) / 3.91)2 2\n(0.0169 / 3.91)2 = 1.87 × 10–5\nM2: mol dm–3\n\n2(g) M1: percentage error of experimental value = [(1.34 – 1.28) / 1.34] × 100 = 4.5% 2\nM2: valid and the calculated actual percentage error is less than the maximum percentage error\n© UCLES 2022 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-March/ms/9701_m22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-feb-march/9701_m22_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m22_ms_52-p06.png", "../answer-assets/9701_m22_ms_52-p07.png" ] }, { "id": "9701-2022-mj-41-q01", "question_id": "9701-2022-mj-41-q01", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 16, "status": "available", "reason": null, "text": "1(a) M1 H and H decrease / both become less exothermic / less negative 3\nlatt hyd\nM2 H decreases / changes less/becomes less exothermic by a smaller extent OR H decreases / changes\nlatt hyd\nmore / dominant factor\nM3 H becomes less exothermic / less negative\nsol\nOR H becomes (more) endothermic / (more) positive\nsol\nOR H = H – H expression AND reaction becomes less exothermic\nsol hyd latt\n\n1(b) Mg: fizzing 1\nBa: (fizzing and) white solid/ppt forms\n\n1(c) M1 solubility of BaSO 2\n4\n= √1.08  10–10 = 1.04  10–5 (mol dm–3)\nM2 = 1.04  10–5  233.4 / 10 = 2.43  10–4 (g per 100 cm3 of solution) min 2sf\n\n1(d)(i) –1473 = 180 + 503 + 965 + Ho – 2469 3\nf\n\nHo of SO 2–(g) = –652 kJ mol–1\nf 4\nM1 correct five values used [1]\nM2 only correct five values used [1]\nM3 correct signs and evaluation [1]\n\n1(d)(ii)  BaSO is more negative/bigger 2\n4\n as Ba2+ is smaller OR Ba2+ has a larger charge\n stronger force of attraction between the ions\nOne mark for two correct\nTwo marks for all three correct\n© UCLES 2022 Page 5 of 15\n\n1(e)(i) M1 Go = 0 so T = Ho / So 2\nr\nM2\nT = 132 / 0.616 = 214.3 K\nT = –58.7 °C min 2sf\n\n1(e)(ii) M1 So = (203 + (70  8) + (2  192)) – (427 – (2  95)) = +530 J K–1 mol–1 3\nM2 Go = Ho – TSo\nM3 Go = 133 – (298  0.530) = –24.9 kJ mol–1 ecf 1dp min\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p05.png", "../answer-assets/9701_s22_ms_41-p06.png" ] }, { "id": "9701-2022-mj-41-q02", "question_id": "9701-2022-mj-41-q02", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "2(a) forms one or more stable ions / compounds / oxidation states 1\nwith incomplete / partially filled (3)d-orbital(s) / d-shell / d-subshell\n\n2(b) 1\nOR\n\n2(c)(i) the catalyst and the reactants are in a different state / phase 1\n\n2(c)(ii) M1 adsorption of reactants to the surface of the catalyst 3\nM2 bonds in the reactants weaken (lowering the activation energy)\nM3 reaction occurs and the products are desorbed\n© UCLES 2022 Page 6 of 15\n\n2(d) M1 reaction 1: 2KMnO + H SO  Mn O + H O + K SO 2\n4 2 4 2 7 2 2 4\nOR 2KMnO + 2H SO  Mn O + H O + 2KHSO\n4 2 4 2 7 2 4\nM2 reaction 2: Mn O  2MnO + 1.5O\n2 7 2 2\n\n2(e)(i) M1 [Mn(H O) ]2+ + 2OH–  Mn(OH) + 6H O 2\n2 6 2 2\nM2 precipitation / acid-base / deprotonation\n\n2(e)(ii) M1 [Mn(H O) ]2+ + 4Cl–  [MnCl ]2– + 6H O 2\n2 6 4 2\nM2 ligand exchange / substitution / replacement / displacement\n\n2(e)(iii) MnO AND redox 1\n2\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p06.png", "../answer-assets/9701_s22_ms_41-p07.png" ] }, { "id": "9701-2022-mj-41-q03", "question_id": "9701-2022-mj-41-q03", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 15, "status": "available", "reason": null, "text": "3(a)(i) evidence of tangent drawn at t = 40 s and calculation of gradient 1\n= 0.000170 (mol dm–3 s–1) min 2sf\n\n3(a)(ii) M1 evidence of construction lines and calculation of two t 2\n1/2\nOR evidence of construction lines and times for halving of concentration\nM2 deduction: constant half-life / constant time between halving of concentration  1st order\n\n3(b) M1 two half-lives in 320 s so t = 160 s 2\n1 / 2\nM2 k = 0.693 / 160 = 0.00433 s–1 ecf min 2sf\n\n3(c)(i) voltage / EMF / potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard 1\nconditions\n\n3(c)(ii) ions move (from the salt bridge) to maintain charge balance / complete the circuit 1\n© UCLES 2022 Page 7 of 15\n\n3(c)(iii) 3\nPt, H (g), good delivery system, 298 K, 1 atm\n2\nPt, Cr O 2-(aq), H+(aq) / Cr3+(aq), 1 mol dm–3 seen once, voltmeter\n2 7\nthree for one mark, six for two marks, nine for three marks\n\n3(c)(iv) SHE labelled negative AND arrow in the external circuit moving away from this electrode 1\n\n3(d)(i) M1 Cr O 2– + 3CH CHO + 8H+ ⇌ 2Cr3+ + 3CH COOH + 4H O 2\n2 7 3 3 2\nALLOW Cr O 2– + 3CH CHO + 5H+ ⇌ 2Cr3+ + 3CH COO– + 4H O\n2 7 3 3 2\nM1 E o = +2.27 (V)\ncell\n\n3(d)(ii) M1 Go = -nFE o 2\ncell\nM2 Go = –4  96 500  2.01 = –775 860 J mol–1\nGo = –776 kJ mol–1 min 3sf\n© UCLES 2022 Page 8 of 15", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p07.png", "../answer-assets/9701_s22_ms_41-p08.png" ] }, { "id": "9701-2022-mj-41-q04", "question_id": "9701-2022-mj-41-q04", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "4(a)  five 3d orbitals in the isolated Fe2+ ion of same energy 2\n splitting two higher and three lower d orbitals\n energy of non-degenerate d orbitals in the complex more than degenerate d orbitals in isolated ion\ntwo for one mark, three for two marks\n\n4(b)(i) M1 (a species) that donates / uses two lone pairs 2\nM2 to form dative / coordinate bonds to a metal atom / metal ion / TM / TE / metal\n\n4(b)(ii) 3\nM1 one correct 3D diagram with three bipy ligands\nM2 both diagrams correct\nM3 optical isomerism\n\n4(c) [Fe(CN) ]3– AND equilibrium lies most to the left / lowest Eo value 1\n6\n\n4(d) M1 N sp2 AND C sp2 3\nM2  bonds are formed by end-on-end overlap orbitals between C-H / C-C / C-N\nM3  bonds are formed by sideways overlap of p orbitals between C-N / C-C\n© UCLES 2022 Page 9 of 15\n\n4(e) 3\nM1 first curly arrow to Cl+\nM2 correct intermediate\nM3 second curly arrow AND H+ formed / lost\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p09.png", "../answer-assets/9701_s22_ms_41-p10.png" ] }, { "id": "9701-2022-mj-41-q05", "question_id": "9701-2022-mj-41-q05", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 16, "status": "available", "reason": null, "text": "5(a) M1 benzoic acid > phenol > phenylmethanol 3\nM2 / M3 Any two of:\n in benzoic acid negative inductive effect of C=O AND O-H bond is weakened\nOR due to delocalisation of minus charge by C=O / 2O carboxylate ion is stabilised\n in phenol lone pair on oxygen is delocalised into the ring AND O-H bond is weakened\n in phenyl methanol positive inductive effect of CH group AND O-H bond is strengthened\n2\n\n5(b) 3\nbenzoic acid phenylmethanol phenol\nNa(s)   \nNaOH(aq)   \nNa CO (aq)   \n2 3\nThree correct for one mark, six correct for two marks, nine correct for three marks\n© UCLES 2022 Page 10 of 15\n\n5(c)(i) POCl and HCl AND SO and HCl 1\n3 2\n\n5(c)(ii) all the by-products / SO and HCl are gaseous OR no liquid by-products formed 1\n2\n\n5(d)(i) 4\nOn the left-hand side:\n lone pair on O\n correct arrow from O to C (of C=O)\n dipole on C=O\n correct arrow on C=O\nM1 / M2 Two correct for one mark, four correct for two marks\nOn the right-hand side:\nM3 correct intermediate\nM4 arrow from lone pair on O– to C-O bond AND arrow from C-Cl to Cl\n\n5(d)(ii) addition-elimination 1\n© UCLES 2022 Page 11 of 15\n\n5(e)(i) 2\nOR\n\n5(e)(ii) hydrolysis OR heating in dilute acid / alkali 1\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p10.png", "../answer-assets/9701_s22_ms_41-p11.png", "../answer-assets/9701_s22_ms_41-p12.png" ] }, { "id": "9701-2022-mj-41-q06", "question_id": "9701-2022-mj-41-q06", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 12, "status": "available", "reason": null, "text": "6(a) a mixture containing equal amounts of each optical isomer 1\n\n6(b) 1\ncarbon‑13 NMR proton NMR\nnumber of peaks in CDCl 4 5\n3\n\n6(c)(i) the pH at which an amino acid exists as a zwitterion 1\nOR\nthe pH at which an amino acid has no overall charge\n© UCLES 2022 Page 12 of 15\n\n6(c)(ii) 1\n\n6(d) 2\nM1 peptide bond displayed\nM2 rest of the structure correct + continuation bonds\n\n6(e) 3\nspot identity\nE Asn\nF Lys-Asn\nG Lys\nM1 table correctly completed\nM2 Lys and Lys-Asn are positively charged OR Asn is (nearly) uncharged\nM3 Lys-Asn has the highest M\nr\n\n6(f)(i) aluminum oxide / silica (on solid support) AND inert gas / named inert gas e.g. N 1\n2\n\n6(f)(ii) S AND R is the same as the unknown amino acid in both solvents 1\nf\n© UCLES 2022 Page 13 of 15\n\n6(g) mass of L = 58 / 44 * 5.52  10–2 = 7.28  10–2 g 1\nconc. of L = 7.28  10–2 / 116 = 6.27  10–4 (mol dm–3) min 2sf\nQuestion Answer Marks", "source_pages": [ 12, 13, 14 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p12.png", "../answer-assets/9701_s22_ms_41-p13.png", "../answer-assets/9701_s22_ms_41-p14.png" ] }, { "id": "9701-2022-mj-41-q07", "question_id": "9701-2022-mj-41-q07", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "7(a)(i) phenylamine AND amine AND ester 1\n\n7(a)(ii) sp carbons = 0, sp2 carbons = 7, sp3 carbons = 6 1\n\n7(b) 6 1\n\n7(c) lone pair on the N can accept a proton 1\n\n7(d)(i) 1\n\n7(d)(ii) step 1 M1 concentrated HNO and H SO 2\n3 2 4\nstep 2 M2 hot (alkaline) KMnO (followed by addition of H+)\n4\n\n7(e) step 4 M1 HOCH CH N(CH CH ) 3\n2 2 2 3 2\nstep 5 M2 Sn AND HCl\nM3 concentrated (HCl) AND heat / reflux\n\n7(f)(i) M1 ratio of the concentration of a solute in two solvents 2\nM2 at equilibrium\n© UCLES 2022 Page 14 of 15\n\n7(f)(ii) M1 K = [procaine] / [procaine] 2\npc oct water\n1.77 = (x / 50)/(0.5 – x / 75)\nM2 1.77 = 1.5x / 0.5 – x\n0.885 –1.77x = 1.5x\nx = 0.271 g min 2sf\n© UCLES 2022 Page 15 of 15", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_41-p14.png", "../answer-assets/9701_s22_ms_41-p15.png" ] }, { "id": "9701-2022-mj-42-q01", "question_id": "9701-2022-mj-42-q01", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 16, "status": "available", "reason": null, "text": "1(a) M1 H and H both become less exothermic / less negative 3\nlatt hyd\nM2 H changes less / becomes less exothermic by a smaller extent\nhyd\nOR H changes more / dominant factor / changes faster\nlatt\nM3 H becomes more exothermic / more negative\nsol\nOR H becomes less endothermic / less positive\nsol\n\n1(b)(i) M1 K = [Be2+][OH–]2 2\nsp\nM2 mol3 dm–9 ecf\n\n1(b)(ii) M1 solubility of Be(OH) = 2.40  10–6 / 43 = 5.58  10–8 (mol dm–3) ecf (b)(i) 2\n2\nM2 K = 4  (5.58  10–8)3 = 6.95  10–22 ecf min 2sf\nsp\n\n1(c)(i) as a molecule or ion formed by a central metal atom / metal ion dative bonded to / surrounded by one or more ligands 1\n\n1(c)(ii) 1\ntetrahedral and correct 3D structure\n\n1(d)(i) d orbitals that are energetically accessible 1\nOR empty / vacant d orbitals AND form dative bonds / accept a lone pair from a ligand\n© UCLES 2022 Page 5 of 16\n\n1(d)(ii) 2\ncoordination\ncomplex ion shape polarity\nnumber\ncis-[Pt(en)Cl ] square planar 4 polar\n2\n[Ag(NH ) ]+ linear 2 non-polar\n3 2\n[Fe(C O ) ]3- octahedral 6 non-polar\n2 4 3\nthree for one mark, six for two marks\n\n1(e)(i) the equilibrium constant for the formation of the complex ion in a solvent (from its constituent ions or molecules) 1\n\n1(e)(ii) M1 K = [Ni(en) 2+] / [Ni(H O) 2+][en]3 2\nstab 3 2 6\nM1 units = mol–3 dm9 ecf\n\n1(e)(iii) the equilibria lie to the right / product side OR equilibrium 1 lies more to the right 1\nAND\nen complex is more stable / [Ni(en) ]2+ is more stable\n3\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p05.png", "../answer-assets/9701_s22_ms_42-p06.png" ] }, { "id": "9701-2022-mj-42-q02", "question_id": "9701-2022-mj-42-q02", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 9, "status": "available", "reason": null, "text": "2(a) the d and s sub-shells / orbitals are close / similar in energy 1\n\n2(b) 1\n© UCLES 2022 Page 6 of 16\n\n2(c) M1 (in a complex / in the presence of ligands) the d-orbitals are split into two sets of orbitals 3\nM2 as an electron is excited / promoted\nM3 visible light is absorbed AND colour seen is complementary\n\n2(d)(i) Cu(OH) : (pale / light) blue 1\n2\n[Cu(H O) (NH ) ]2+: dark / deep blue\n2 2 3 4\n\n2(d)(ii) (addition of OH–) increases [OH–] AND shifts equilibrium to the right 1\n\n2(e) 2\nM1 one correct 3D diagram with the six correct ligands\nM2 both 3D diagrams correct\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p06.png", "../answer-assets/9701_s22_ms_42-p07.png" ] }, { "id": "9701-2022-mj-42-q03", "question_id": "9701-2022-mj-42-q03", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 9, "status": "available", "reason": null, "text": "3(a)(i)  enthalpy change / energy change 2\n one mole of electrons\n (gained by) one mole of gaseous atoms\ntwo for one mark, three for two marks\n\n3(a)(ii) (energy required to overcome) the repulsion between the electron and anion / negative ion 1\n© UCLES 2022 Page 7 of 16\n\n3(a)(iii)  less negative / less exothermic down the group 2\n greater the distance between the nucleus and (the shells of the) electrons\nOR atomic radii increases\nOR more shielding by inner shells\n the less attraction between nucleus\nand incoming electron (and the less energy released)\ntwo for one mark, three for two marks\n\n3(b) M1 use of correct seven numbers only in calculation / energy cycle 3\nM2 only 2  used correctly\nM3 correct signs and evaluation ecf\n–208 = 131 + 906 + 1733 + 62 + 151 + 2x – 2605\n2x = –586\nx = –293 kJ mol–1\n\n3(c) first box ticked 1\nAND Cd2+ larger / Cd2+ lower charge density\nAND less attraction between the ions / weaker ionic bonds\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p07.png", "../answer-assets/9701_s22_ms_42-p08.png" ] }, { "id": "9701-2022-mj-42-q04", "question_id": "9701-2022-mj-42-q04", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 14, "status": "available", "reason": null, "text": "4(a)(i) M1 all five points plotted correctly 2\nM2 best-fit straight line (ruler) with negative gradient drawn\n\n4(a)(ii) M1 gradient correctly calculated OR gradient working seen 2\nM2 gradient = –So\nSo evaluated correctly ecf So = (+)160 ± 10 (J K–1 mol–1)\n\n4(b)(i) 2HCO –  CO 2– + CO + H O 1\n3 3 2 2\n© UCLES 2022 Page 8 of 16\n\n4(b)(ii) M1 ionic radius of M+ / cationic radius increases 2\nOR charge density of ion / M+ decreases down Group 1\nM2 less distortion / polarisation of the anion / HCO –\n3\nOR CO bond / C-O / C=O less weakened\n\n4(c)(i) M1 solution which resists changes in pH when 2\nopposes / resists change in pH\nM2 when small amount of acid / H+ or alkali / base / OH– is added\n\n4(c)(ii) M1 (with acid) HCO – + H+ → H CO OR HCO – + H O+ → H CO + H O 2\n3 2 3 3 3 2 3 2\nM2 (with alkali) H CO + OH– → HCO – + H O\n2 3 3 2\n\n4(c)(iii) M1 K = 10–6.35 = 4.47  10–7 3\na\nM2 [H+] = 4.47  10–7 / 14.1 = 3.17  10–8 ecf\nM3 pH = –log [H+] = 7.5 ecf from a calculated [H+] min 2sf\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p08.png", "../answer-assets/9701_s22_ms_42-p09.png" ] }, { "id": "9701-2022-mj-42-q05", "question_id": "9701-2022-mj-42-q05", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 8, "status": "available", "reason": null, "text": "5(a) 3\nsubstance liberated substance liberated\nelectrolyte\nat the anode at the cathode\nPbBr (l) Br / bromine Pb / lead\n2 2\nconcentrated NaCl(aq) Cl / chlorine H / hydrogen\n2 2\nCu(NO ) (aq) O / oxygen (+ H O) Cu / copper\n3 2 2 2\ntwo for one mark, four for two marks, six for three marks\n© UCLES 2022 Page 9 of 16\n\n5(b)(i) F = Le OR F is directly proportional to L 1\n\n5(b)(ii) number of Cu2+ formed = 0.35 / 63.5 = 5.51  10–3 4\nQ = I  t = 0.60  30  60 = 1080 C\nnumber of electrons = 1080 / 1.610-19 = 6.75  1021 ecf\nnumber of Cu2+ ions = 6.75  1021 / 2 = 3.375  1021 ecf\nnumber of Cu2+ ions per mole (L) = 3.375  1021 / 5.51  10-3 = 6.12  1023 ecf min 2sf\nall five points for four marks\nALLOW valid alternate calculations of L\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p09.png", "../answer-assets/9701_s22_ms_42-p10.png" ] }, { "id": "9701-2022-mj-42-q06", "question_id": "9701-2022-mj-42-q06", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 16, "status": "available", "reason": null, "text": "6(a)(i) NO 2 NO 2 3\nH\n+H+\n+\nNO +\n2\nM1 first curly arrow from within hexagon to N of the NO +\n2\nM2 correct intermediate\nM3 second curly arrow from C-H bond into the ring AND H+ formed / lost\n\n6(a)(ii) HSO – + H+  H SO 1\n4 2 4\n© UCLES 2022 Page 10 of 16\n\n6(b) M1 benzoic acid M2 phenol 2\nOR\n\n6(c) M1 phenol > benzene > benzoic acid 4\nM2 / M3\nphenol:\n lone pair / p-orbital on oxygen\ndelocalises into the ring / overlaps with -delocalised ring\n accepts / attracts / polarises NO + / electrophiles better\n2\nbenzoic acid:\n COOH / C=O is an electron-withdrawing / positive inductive effect\ntwo for one mark, three for two marks\nM4\n(phenol-oxygen) increases electron density in the ring (as compared to benzene as a result of the OH group)\nOR\n(benzoic acid–COOH) decreases electron density in the ring (as compared to benzene as a result of the COOH group)\n© UCLES 2022 Page 11 of 16\n\n6(d)(i) 3\n\n6(d)(ii) step 1 M1 Sn and hydrochloric acid / HCl 3\nM2 concentrated (HCl) + heat / reflux (followed by NaOH)\nstep 2 M3 HNO and (HCl) ⩽10°C\n2\nOR NaNO , HCl and ⩽10°C\n2\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p10.png", "../answer-assets/9701_s22_ms_42-p11.png", "../answer-assets/9701_s22_ms_42-p12.png" ] }, { "id": "9701-2022-mj-42-q07", "question_id": "9701-2022-mj-42-q07", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 8, "status": "available", "reason": null, "text": "7(a) TMS reference OR standard OR to define d = 0 1\nD O solvent OR identification of O-H / N-H protons / group\n2\n© UCLES 2022 Page 12 of 16\n\n7(b)(i) 2\nnumber of peaks observed number of peaks observed in\nketone\nin proton NMR spectrum carbon-13 NMR spectrum\npentan-2-one 4 5\npentan-3-one 2 3\n3-methylbutanone 3 4\nthree for one mark, six for two marks\n\n7(b)(ii) M1 3-methylbutanone 2\nM2 3-methylbutanone AND pentan-2-one\n\n7(c)(i) six / 6 1\n\n7(c)(ii) C H O 1\n21 34 5\n\n7(c)(iii) a substance that is able to rotate the plane of polarised light in opposite directions 1\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p12.png", "../answer-assets/9701_s22_ms_42-p13.png" ] }, { "id": "9701-2022-mj-42-q08", "question_id": "9701-2022-mj-42-q08", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Polymerisation", "topic_slug": "9701-topic-35-polymerisation", "marks": 11, "status": "available", "reason": null, "text": "8(a) M1 chloroethanoic acid > ethanoic acid > phenol > ethanol 4\nM2 correct link of acidity once can be implied from M1\nweakens O—H / carboxylate anion stabilised\nM3 / M4 explanation linked to structure\n (ClCH CO H > ethanoic acid) due to electronegative / electron withdrawing / negative inductive effect of Cl\n2 2\n (ethanoic acid > phenol) due to electronegative / electron withdrawing / negative inductive effect of COOH / C=O\n (phenol > ethanol) due to lone pair of oxygen overlapping / delocalised into the ring\n (ethanol weakest) alkyl group is electron donating / positive inductive effect\ntwo for one mark, four for two marks\n© UCLES 2022 Page 13 of 16\n\n8(b)  oxidation 2\n (solution) decolourises OR purple  colourless / pale pink OR bubbles\n HOOCCOOH + [O]  2CO + H O\n2 2\nOR 5HOOCCOOH + 2MnO – + 6H+  10CO + 8H O + 2Mn2+\n4 2 2\ntwo for one mark, three for two marks\n\n8(c) 2\nM1\nM2\n\n8(d) 2\nM1 peptide linkage shown displayed with saturated C each side\nM2 rest of structure correct AND continuation bonds\n© UCLES 2022 Page 14 of 16\n\n8(e) addition polymers do not hydrolyse 1\nOR condensation polymers can hydrolyse", "source_pages": [ 13, 14, 15 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-35-polymerisation/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p13.png", "../answer-assets/9701_s22_ms_42-p14.png", "../answer-assets/9701_s22_ms_42-p15.png" ] }, { "id": "9701-2022-mj-42-q09", "question_id": "9701-2022-mj-42-q09", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 9, "status": "available", "reason": null, "text": "9(a) cyclohexylamine > ammonia > phenylamine 3\n (order of basicity) linked to ability of N to accept a proton / donate its lone pair (to a proton)\n alkyl / cyclo / hexyl group is electron donating group / positive inductive group (and increases electron density on N)\n lone pair / p-orbital from N in phenylamine delocalised / overlaps with ring (and decreases electron density on N)\ntwo for one mark, three for two marks\n\n9(b)(i) (nucleophilic) addition-elimination 1\n\n9(b)(ii) 4\nM1 / M2\n lone pair on N\n correct arrow from (lone pair) N to C (of C=O)\n dipole on C=O\n correct arrow on C=O\ntwo for one mark, four for two marks\nM3 correct intermediate\nM4 arrow from lone pair on O– to C-O bond AND arrow from C-Cl to Cl\n© UCLES 2022 Page 15 of 16\n\n9(b)(iii) O O 1\nC C\nH C N CH\n3 3\n© UCLES 2022 Page 16 of 16", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_42-p15.png", "../answer-assets/9701_s22_ms_42-p16.png" ] }, { "id": "9701-2022-mj-43-q01", "question_id": "9701-2022-mj-43-q01", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 16, "status": "available", "reason": null, "text": "1(a) M1 H and H decrease / both become less exothermic / less negative 3\nlatt hyd\nM2 H decreases / changes less/becomes less exothermic by a smaller extent OR H decreases / changes\nlatt hyd\nmore / dominant factor\nM3 H becomes less exothermic / less negative\nsol\nOR H becomes (more) endothermic / (more) positive\nsol\nOR H = H – H expression AND reaction becomes less exothermic\nsol hyd latt\n\n1(b) Mg: fizzing 1\nBa: (fizzing and) white solid/ppt forms\n\n1(c) M1 solubility of BaSO 2\n4\n= √1.08  10–10 = 1.04  10–5 (mol dm–3)\nM2 = 1.04  10–5  233.4 / 10 = 2.43  10–4 (g per 100 cm3 of solution) min 2sf\n\n1(d)(i) –1473 = 180 + 503 + 965 + Ho – 2469 3\nf\n\nHo of SO 2–(g) = –652 kJ mol–1\nf 4\nM1 correct five values used [1]\nM2 only correct five values used [1]\nM3 correct signs and evaluation [1]\n\n1(d)(ii)  BaSO is more negative/bigger 2\n4\n as Ba2+ is smaller OR Ba2+ has a larger charge\n stronger force of attraction between the ions\nOne mark for two correct\nTwo marks for all three correct\n© UCLES 2022 Page 5 of 15\n\n1(e)(i) M1 Go = 0 so T = Ho / So 2\nr\nM2\nT = 132 / 0.616 = 214.3 K\nT = –58.7 °C min 2sf\n\n1(e)(ii) M1 So = (203 + (70  8) + (2  192)) – (427 – (2  95)) = +530 J K–1 mol–1 3\nM2 Go = Ho – TSo\nM3 Go = 133 – (298  0.530) = –24.9 kJ mol–1 ecf 1dp min\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p05.png", "../answer-assets/9701_s22_ms_43-p06.png" ] }, { "id": "9701-2022-mj-43-q02", "question_id": "9701-2022-mj-43-q02", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "2(a) forms one or more stable ions / compounds / oxidation states 1\nwith incomplete / partially filled (3)d-orbital(s) / d-shell / d-subshell\n\n2(b) 1\nOR\n\n2(c)(i) the catalyst and the reactants are in a different state / phase 1\n\n2(c)(ii) M1 adsorption of reactants to the surface of the catalyst 3\nM2 bonds in the reactants weaken (lowering the activation energy)\nM3 reaction occurs and the products are desorbed\n© UCLES 2022 Page 6 of 15\n\n2(d) M1 reaction 1: 2KMnO + H SO  Mn O + H O + K SO 2\n4 2 4 2 7 2 2 4\nOR 2KMnO + 2H SO  Mn O + H O + 2KHSO\n4 2 4 2 7 2 4\nM2 reaction 2: Mn O  2MnO + 1.5O\n2 7 2 2\n\n2(e)(i) M1 [Mn(H O) ]2+ + 2OH–  Mn(OH) + 6H O 2\n2 6 2 2\nM2 precipitation / acid-base / deprotonation\n\n2(e)(ii) M1 [Mn(H O) ]2+ + 4Cl–  [MnCl ]2– + 6H O 2\n2 6 4 2\nM2 ligand exchange / substitution / replacement / displacement\n\n2(e)(iii) MnO AND redox 1\n2\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p06.png", "../answer-assets/9701_s22_ms_43-p07.png" ] }, { "id": "9701-2022-mj-43-q03", "question_id": "9701-2022-mj-43-q03", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 15, "status": "available", "reason": null, "text": "3(a)(i) evidence of tangent drawn at t = 40 s and calculation of gradient 1\n= 0.000170 (mol dm–3 s–1) min 2sf\n\n3(a)(ii) M1 evidence of construction lines and calculation of two t 2\n1/2\nOR evidence of construction lines and times for halving of concentration\nM2 deduction: constant half-life / constant time between halving of concentration  1st order\n\n3(b) M1 two half-lives in 320 s so t = 160 s 2\n1 / 2\nM2 k = 0.693 / 160 = 0.00433 s–1 ecf min 2sf\n\n3(c)(i) voltage / EMF / potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard 1\nconditions\n\n3(c)(ii) ions move (from the salt bridge) to maintain charge balance / complete the circuit 1\n© UCLES 2022 Page 7 of 15\n\n3(c)(iii) 3\nPt, H (g), good delivery system, 298 K, 1 atm\n2\nPt, Cr O 2-(aq), H+(aq) / Cr3+(aq), 1 mol dm–3 seen once, voltmeter\n2 7\nthree for one mark, six for two marks, nine for three marks\n\n3(c)(iv) SHE labelled negative AND arrow in the external circuit moving away from this electrode 1\n\n3(d)(i) M1 Cr O 2– + 3CH CHO + 8H+ ⇌ 2Cr3+ + 3CH COOH + 4H O 2\n2 7 3 3 2\nALLOW Cr O 2– + 3CH CHO + 5H+ ⇌ 2Cr3+ + 3CH COO– + 4H O\n2 7 3 3 2\nM1 E o = +2.27 (V)\ncell\n\n3(d)(ii) M1 Go = -nFE o 2\ncell\nM2 Go = –4  96 500  2.01 = –775 860 J mol–1\nGo = –776 kJ mol–1 min 3sf\n© UCLES 2022 Page 8 of 15", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p07.png", "../answer-assets/9701_s22_ms_43-p08.png" ] }, { "id": "9701-2022-mj-43-q04", "question_id": "9701-2022-mj-43-q04", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "4(a)  five 3d orbitals in the isolated Fe2+ ion of same energy 2\n splitting two higher and three lower d orbitals\n energy of non-degenerate d orbitals in the complex more than degenerate d orbitals in isolated ion\ntwo for one mark, three for two marks\n\n4(b)(i) M1 (a species) that donates / uses two lone pairs 2\nM2 to form dative / coordinate bonds to a metal atom / metal ion / TM / TE / metal\n\n4(b)(ii) 3\nM1 one correct 3D diagram with three bipy ligands\nM2 both diagrams correct\nM3 optical isomerism\n\n4(c) [Fe(CN) ]3– AND equilibrium lies most to the left / lowest Eo value 1\n6\n\n4(d) M1 N sp2 AND C sp2 3\nM2  bonds are formed by end-on-end overlap orbitals between C-H / C-C / C-N\nM3  bonds are formed by sideways overlap of p orbitals between C-N / C-C\n© UCLES 2022 Page 9 of 15\n\n4(e) 3\nM1 first curly arrow to Cl+\nM2 correct intermediate\nM3 second curly arrow AND H+ formed / lost\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p09.png", "../answer-assets/9701_s22_ms_43-p10.png" ] }, { "id": "9701-2022-mj-43-q05", "question_id": "9701-2022-mj-43-q05", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 16, "status": "available", "reason": null, "text": "5(a) M1 benzoic acid > phenol > phenylmethanol 3\nM2 / M3 Any two of:\n in benzoic acid negative inductive effect of C=O AND O-H bond is weakened\nOR due to delocalisation of minus charge by C=O / 2O carboxylate ion is stabilised\n in phenol lone pair on oxygen is delocalised into the ring AND O-H bond is weakened\n in phenyl methanol positive inductive effect of CH group AND O-H bond is strengthened\n2\n\n5(b) 3\nbenzoic acid phenylmethanol phenol\nNa(s)   \nNaOH(aq)   \nNa CO (aq)   \n2 3\nThree correct for one mark, six correct for two marks, nine correct for three marks\n© UCLES 2022 Page 10 of 15\n\n5(c)(i) POCl and HCl AND SO and HCl 1\n3 2\n\n5(c)(ii) all the by-products / SO and HCl are gaseous OR no liquid by-products formed 1\n2\n\n5(d)(i) 4\nOn the left-hand side:\n lone pair on O\n correct arrow from O to C (of C=O)\n dipole on C=O\n correct arrow on C=O\nM1 / M2 Two correct for one mark, four correct for two marks\nOn the right-hand side:\nM3 correct intermediate\nM4 arrow from lone pair on O– to C-O bond AND arrow from C-Cl to Cl\n\n5(d)(ii) addition-elimination 1\n© UCLES 2022 Page 11 of 15\n\n5(e)(i) 2\nOR\n\n5(e)(ii) hydrolysis OR heating in dilute acid / alkali 1\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p10.png", "../answer-assets/9701_s22_ms_43-p11.png", "../answer-assets/9701_s22_ms_43-p12.png" ] }, { "id": "9701-2022-mj-43-q06", "question_id": "9701-2022-mj-43-q06", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 12, "status": "available", "reason": null, "text": "6(a) a mixture containing equal amounts of each optical isomer 1\n\n6(b) 1\ncarbon‑13 NMR proton NMR\nnumber of peaks in CDCl 4 5\n3\n\n6(c)(i) the pH at which an amino acid exists as a zwitterion 1\nOR\nthe pH at which an amino acid has no overall charge\n© UCLES 2022 Page 12 of 15\n\n6(c)(ii) 1\n\n6(d) 2\nM1 peptide bond displayed\nM2 rest of the structure correct + continuation bonds\n\n6(e) 3\nspot identity\nE Asn\nF Lys-Asn\nG Lys\nM1 table correctly completed\nM2 Lys and Lys-Asn are positively charged OR Asn is (nearly) uncharged\nM3 Lys-Asn has the highest M\nr\n\n6(f)(i) aluminum oxide / silica (on solid support) AND inert gas / named inert gas e.g. N 1\n2\n\n6(f)(ii) S AND R is the same as the unknown amino acid in both solvents 1\nf\n© UCLES 2022 Page 13 of 15\n\n6(g) mass of L = 58 / 44 * 5.52  10–2 = 7.28  10–2 g 1\nconc. of L = 7.28  10–2 / 116 = 6.27  10–4 (mol dm–3) min 2sf\nQuestion Answer Marks", "source_pages": [ 12, 13, 14 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p12.png", "../answer-assets/9701_s22_ms_43-p13.png", "../answer-assets/9701_s22_ms_43-p14.png" ] }, { "id": "9701-2022-mj-43-q07", "question_id": "9701-2022-mj-43-q07", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "7(a)(i) phenylamine AND amine AND ester 1\n\n7(a)(ii) sp carbons = 0, sp2 carbons = 7, sp3 carbons = 6 1\n\n7(b) 6 1\n\n7(c) lone pair on the N can accept a proton 1\n\n7(d)(i) 1\n\n7(d)(ii) step 1 M1 concentrated HNO and H SO 2\n3 2 4\nstep 2 M2 hot (alkaline) KMnO (followed by addition of H+)\n4\n\n7(e) step 4 M1 HOCH CH N(CH CH ) 3\n2 2 2 3 2\nstep 5 M2 Sn AND HCl\nM3 concentrated (HCl) AND heat / reflux\n\n7(f)(i) M1 ratio of the concentration of a solute in two solvents 2\nM2 at equilibrium\n© UCLES 2022 Page 14 of 15\n\n7(f)(ii) M1 K = [procaine] / [procaine] 2\npc oct water\n1.77 = (x / 50)/(0.5 – x / 75)\nM2 1.77 = 1.5x / 0.5 – x\n0.885 –1.77x = 1.5x\nx = 0.271 g min 2sf\n© UCLES 2022 Page 15 of 15", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_43-p14.png", "../answer-assets/9701_s22_ms_43-p15.png" ] }, { "id": "9701-2022-mj-51-q01", "question_id": "9701-2022-mj-51-q01", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a) 250.0 1\nnumber of moles NaCl =  0.200 = 0.05(00)\n1000\nmass of sodium chloride = 0.05(00)  (23.0 + 35.5) = 2.925 g\n\n1(b) (measure mass of weighing boat containing sodium chloride before transfer, then) 1\nmeasure the mass of weighing boat (and residue) (after transfer).\n\n1(c) M1 2\nadd a (small) volume of distilled water (to the small beaker)\nAND\ndissolve the sodium chloride\nM2\ntransfer the solution and washings into a 250 cm3 volumetric flask\nAND\nmake up to the mark with distilled water\n© UCLES 2022 Page 5 of 10\n\n1(d)(i) M1 Extra step 2\nwash the precipitate / residue (with cold distilled water) a\nOR\nreheat (and reweigh) until mass is constant b\nOR\nrinse the beaker into filter (washings) c\nOR\nfilter solution again d\nM2 Explanation\nto remove (unreacted) sodium chloride and/or lead compound a\nOR\nto ensure all water has been removed from the solid b\nOR\nto ensure all solid/ppt has been transferred into the filter c\nOR\nto collect any precipitate that passed through the filter paper first time d\n\n1(d)(ii) 20.05 1\n100 = 1(.0%)\n10\ncorrect working must be shown\n\n1(d)(iii) repeat the experiment and/or compare with results from other students 1\nAND\nconsistent results are obtained\n© UCLES 2022 Page 6 of 10\n\n1(e)(i) Two appropriate straight lines of best fit drawn that intersect. 1\n\n1(e)(ii) M1 volume of intersection correctly read from graph plotted in (e)(i). 2\n(expected volumes: NaCl(aq) ≈ 33.25 cm3, lead compound(aq) ≈ 16.75 cm3)\nM2 molar ratio consistent with volumes given for M1.\n(expected ratio = 2 : 1)\n\n1(f) empirical formula consistent with molar ratio given in (e)(ii). 1\n(expected answer: PbCl )\n2\n© UCLES 2022 Page 7 of 10\n\n1(g) investigation 1 2\nrepeat measurement (of height of precipitate) until it is constant\nOR\nleave (for a longer time) until height of precipitate is constant (before measuring)\ninvestigation 2\nuse a smaller volume of lead compound (in each case)\nOR\nadd larger volumes of sodium chloride (until it is in excess)\nQuestion Answer Marks", "source_pages": [ 5, 6, 7, 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_51-p05.png", "../answer-assets/9701_s22_ms_51-p06.png", "../answer-assets/9701_s22_ms_51-p07.png", "../answer-assets/9701_s22_ms_51-p08.png" ] }, { "id": "9701-2022-mj-51-q02", "question_id": "9701-2022-mj-51-q02", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a)(i) thermostatically controlled water bath 1\n\n2(a)(ii) flammable substance(s) are used (in the experiment) 1\n\n2(b)(i) (50 cm3) burette 1\n\n2(b)(ii) 10(.00) cm3 (volumetric) pipette 1\n\n2(c) reduces the reaction rate (of hydrolysis reaction) 1\nOR\nquenches the reaction\n\n2(d) volume of sodium hydroxide (solution used in titration) 1\n© UCLES 2022 Page 8 of 10\n\n2(e)(i) 1\ntime / s titre, V / cm3 V – V / cm3\nt final t\n60 1.25 46.00\n300 7.75 39.50\n600 17.75 29.50\n900 20.00 27.25\n1200 24.25 23.00\n1500 28.40 18.85\n1800 31.15 16.10\n2700 38.00 9.25\nFinal 47.25 (0.00)\n\n2(e)(ii) concentration of the 2-bromo-2-methylpropane 1\n\n2(e)(iii) M1 all points plotted correctly 2\nM2 smooth curve of best fit line drawn passing close to all points except anomaly\n\n2(e)(iv) M1 selects the point most anomalous to the plotted line of best fit 2\n(point at 600, 29.5 expected)\nM2 reaction was not effectively quenched\n© UCLES 2022 Page 9 of 10\n\n2(e)(v) M1 3\ncorrectly placed construction lines shown on the graph to determine at least one half-life correctly\nOR\nboth co-ordinates from line of best fit correctly recorded in the form (x,y) for either first t OR second t\n½ ½\nM2 two sets of co-ordinates from line of best fit correctly recorded for first t AND second t in the form (x,y)\n½ ½\nM3 two half-lives correctly calculated from listed co-ordinates (expected value: 1159 s)\n\n2(e)(vi) first order 1\nAND\nhalf-lives are constant (within experimental error)\n© UCLES 2022 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_51-p08.png", "../answer-assets/9701_s22_ms_51-p09.png", "../answer-assets/9701_s22_ms_51-p10.png" ] }, { "id": "9701-2022-mj-52-q01", "question_id": "9701-2022-mj-52-q01", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 12, "status": "available", "reason": null, "text": "1(a) M1 sealed apparatus 2\nM2 collection of the gas in a labelled inverted (250 cm3) measuring cylinder (via a delivery tube into water)\n\n1(b)(i) remove oxide layer by using sandpaper 1\n\n1(b)(ii) mass (of Mg) 1\n\n1(c)(i) Method 1 2\nM1 mol HCl used = 0.50  30 / 1000 OR 0.015 mol\nM2 mass Mg needed = 0.015 / 2  24.3 = 0.18(225) g\nMethod 2\nM1 mol of Mg used = 0.16 / 24.3 OR 0.00658(436)\nM2 vol of HCl needed = 0.00658 …  2  1000 / .50 = 26.3 cm3\n\n1(c)(ii) acid is in excess 1\n\n1(d) to make sure the gas is at room temperature 1\n\n1(e)(i) 1  100 / 146 = 0.68(49..%) 1\n\n1(e)(ii) 146 / (0.16 / 24.3) = 22 174 cm3 1\n\n1(f) M1 weakness 2\nloss of hydrogen/gas before the flask is stoppered OR between steps 4 and 5\nM2 improvement\nensure reagents are separate within sealed flask before mixing\n© UCLES 2022 Page 5 of 7", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_52-p05.png" ] }, { "id": "9701-2022-mj-52-q02", "question_id": "9701-2022-mj-52-q02", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 18, "status": "available", "reason": null, "text": "2(a) temperature 1\n\n2(b) Any two from: 2\nvolume of H O (aq)\n2 2\nconcentration of H O (aq)\n2 2\nvolume of KI(aq) / catalyst\nconcentration of KI(aq) / catalyst\n\n2(c)(i) M1 50(.0) cm3 of H O 3\n2 2\nM2 using burette transfer / add to a (250 cm3) volumetric flask\nM3 top the (volumetric) flask up to (250 cm3) mark with distilled water\n\n2(c)(ii) use chemically resistant gloves 1\n\n2(d)(i) 2\ntemp 1 / T / K–1 initial rate / mol s–1 log initial rate\n293 0.00341 5.75  10–6 –5.24\n298 0.00336 7.94  10–6 –5.10\n303 0.00330 1.17  10-5 –4.93\n308 0.00325 1.45  10–5 –4.84\n312 0.00321 2.19  10–5 –4.66\n319 0.00313 3.72  10–5 –4.43\n325 0.00308 5.25  10–5 –4.28\n328 0.00305 6.31  10–5 –4.20\nM1 1/T column correct to three significant figures\nM2 log initial rate column correct to three significant figures\n© UCLES 2022 Page 6 of 7\n\n2(d)(ii) M1 8 points plotted correctly 2\nM2 line of best fit drawn\n\n2(d)(iii) M1 most anomalous point circled 2\nM2 the recorded temperature (reading) is higher than the actual temperature for that set of readings\n\n2(d)(iv) M1 co-ordinates read and recorded correctly 2\nM2 gradient calculated to three significant figures\n\n2(d)(v) M1 gradient = – E / 2.303R OR any correct rearrangement of this 2\na\nM2 Correct calculation of E\na\n\n2(e) no AND there is an anomalous point on the graph 1\n© UCLES 2022 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_52-p06.png", "../answer-assets/9701_s22_ms_52-p07.png" ] }, { "id": "9701-2022-mj-53-q01", "question_id": "9701-2022-mj-53-q01", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a) 250.0 1\nnumber of moles NaCl =  0.200 = 0.05(00)\n1000\nmass of sodium chloride = 0.05(00)  (23.0 + 35.5) = 2.925 g\n\n1(b) (measure mass of weighing boat containing sodium chloride before transfer, then) 1\nmeasure the mass of weighing boat (and residue) (after transfer).\n\n1(c) M1 2\nadd a (small) volume of distilled water (to the small beaker)\nAND\ndissolve the sodium chloride\nM2\ntransfer the solution and washings into a 250 cm3 volumetric flask\nAND\nmake up to the mark with distilled water\n© UCLES 2022 Page 5 of 10\n\n1(d)(i) M1 Extra step 2\nwash the precipitate / residue (with cold distilled water) a\nOR\nreheat (and reweigh) until mass is constant b\nOR\nrinse the beaker into filter (washings) c\nOR\nfilter solution again d\nM2 Explanation\nto remove (unreacted) sodium chloride and/or lead compound a\nOR\nto ensure all water has been removed from the solid b\nOR\nto ensure all solid/ppt has been transferred into the filter c\nOR\nto collect any precipitate that passed through the filter paper first time d\n\n1(d)(ii) 20.05 1\n100 = 1(.0%)\n10\ncorrect working must be shown\n\n1(d)(iii) repeat the experiment and/or compare with results from other students 1\nAND\nconsistent results are obtained\n© UCLES 2022 Page 6 of 10\n\n1(e)(i) Two appropriate straight lines of best fit drawn that intersect. 1\n\n1(e)(ii) M1 volume of intersection correctly read from graph plotted in (e)(i). 2\n(expected volumes: NaCl(aq) ≈ 33.25 cm3, lead compound(aq) ≈ 16.75 cm3)\nM2 molar ratio consistent with volumes given for M1.\n(expected ratio = 2 : 1)\n\n1(f) empirical formula consistent with molar ratio given in (e)(ii). 1\n(expected answer: PbCl )\n2\n© UCLES 2022 Page 7 of 10\n\n1(g) investigation 1 2\nrepeat measurement (of height of precipitate) until it is constant\nOR\nleave (for a longer time) until height of precipitate is constant (before measuring)\ninvestigation 2\nuse a smaller volume of lead compound (in each case)\nOR\nadd larger volumes of sodium chloride (until it is in excess)\nQuestion Answer Marks", "source_pages": [ 5, 6, 7, 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_53-p05.png", "../answer-assets/9701_s22_ms_53-p06.png", "../answer-assets/9701_s22_ms_53-p07.png", "../answer-assets/9701_s22_ms_53-p08.png" ] }, { "id": "9701-2022-mj-53-q02", "question_id": "9701-2022-mj-53-q02", "subject": "9701", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a)(i) thermostatically controlled water bath 1\n\n2(a)(ii) flammable substance(s) are used (in the experiment) 1\n\n2(b)(i) (50 cm3) burette 1\n\n2(b)(ii) 10(.00) cm3 (volumetric) pipette 1\n\n2(c) reduces the reaction rate (of hydrolysis reaction) 1\nOR\nquenches the reaction\n\n2(d) volume of sodium hydroxide (solution used in titration) 1\n© UCLES 2022 Page 8 of 10\n\n2(e)(i) 1\ntime / s titre, V / cm3 V – V / cm3\nt final t\n60 1.25 46.00\n300 7.75 39.50\n600 17.75 29.50\n900 20.00 27.25\n1200 24.25 23.00\n1500 28.40 18.85\n1800 31.15 16.10\n2700 38.00 9.25\nFinal 47.25 (0.00)\n\n2(e)(ii) concentration of the 2-bromo-2-methylpropane 1\n\n2(e)(iii) M1 all points plotted correctly 2\nM2 smooth curve of best fit line drawn passing close to all points except anomaly\n\n2(e)(iv) M1 selects the point most anomalous to the plotted line of best fit 2\n(point at 600, 29.5 expected)\nM2 reaction was not effectively quenched\n© UCLES 2022 Page 9 of 10\n\n2(e)(v) M1 3\ncorrectly placed construction lines shown on the graph to determine at least one half-life correctly\nOR\nboth co-ordinates from line of best fit correctly recorded in the form (x,y) for either first t OR second t\n½ ½\nM2 two sets of co-ordinates from line of best fit correctly recorded for first t AND second t in the form (x,y)\n½ ½\nM3 two half-lives correctly calculated from listed co-ordinates (expected value: 1159 s)\n\n2(e)(vi) first order 1\nAND\nhalf-lives are constant (within experimental error)\n© UCLES 2022 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9701_s22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-may-june/9701_s22_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s22_ms_53-p08.png", "../answer-assets/9701_s22_ms_53-p09.png", "../answer-assets/9701_s22_ms_53-p10.png" ] }, { "id": "9701-2022-on-41-q01", "question_id": "9701-2022-on-41-q01", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 11, "status": "available", "reason": null, "text": "1(a) 2\nM1 K+ (g) and Cl – (g)\nAND\nKCl (aq) OR K+ (aq) + Cl – (aq)\nM2 three correct directional arrows COND M1\n\n1(b) use of data –155, –2493 AND 2  –364 [1] 2\nH Mg2+ = –1920 (kJ mol–1) [1] min 3sf\nhyd\n\n1(c) • Mg2+ is smaller (than K+) 2\n• Mg2+ is greater charge (than K+)\n• greater attraction\nbetween Mg2+ and Cl – /between the ions (in MgCl )\n2\nOR stronger ionic bonds (in MgCl )\n2\n\n1(d)(i) enthalpy change when 1\none mole of gaseous atoms formed from the element (in its standard state at 298 K)\n\n1(d)(ii) enthalpy change when 1\nevery atom in one mole of gaseous atoms gains one electron\nOR one mole of gaseous atoms gains one mole of electrons\n\n1(e)(i) number of possible arrangements of particles and energy in a system 1\n© UCLES 2022 Page 5 of 16\n\n1(e)(ii) S is positive 1\nAND KCl(s) → K+(aq) + Cl –(aq) /\nionic lattice solid forms aqueous ions OWTTE [1]\nOR\nS is positive\nAND G is (therefore becomes) negative /\nTS is greater than H OWTTE [1]\nsol\n\n1(e)(iii) more soluble 1\nAND G is more negative at higher T /\nTS is more positive at higher T /\n–TS is more negative at higher ecf from (e)(ii) [sign S]\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p05.png", "../answer-assets/9701_w22_ms_41-p06.png" ] }, { "id": "9701-2022-on-41-q02", "question_id": "9701-2022-on-41-q02", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i) (homogeneous is in the) same phase / state as reactants 1\nAND (heterogeneous is in a) different phase / state to reactants\n\n2(a)(ii) 1 S O 2– + 2Fe2+ → 2Fe3+ + 2SO 2– [1] 2\n2 8 4\n2 2I– + 2Fe3+ → 2Fe2+ + I [1]\n2\n\n2(a)(iii) reactants are both anions / negatively charged 1\nAND so they repel each other OWTTE\n\n2(b)(i) rate = k[NO]2[O ] OR rate = 8.6  106 [NO]2[O ] 1\n2 2\n\n2(b)(ii) rate = 8.6  106  (7.2  10–4)2  1.9  10–3 1\nrate = 8.47  10–3 (mol dm–3 s–1) min 2sf\n\n2(c)(i) (reaction is) first order wrt cisplatin / overall 1\nOR rate is directly proportional to concentration of cisplatin\n\n2(c)(ii) 0.693 / 2.50  10–5 = (2.77  104 s) 1\nOR In 2 / 2.50  10–5 = (2.77  104 s)\n© UCLES 2022 Page 6 of 16\n\n2(c)(iii) initial concentration is 8.0  10–5 mol dm–3 2\n– five half-life periods have elapsed [1]\ntime = 5  27720 = 1.39  105 s [1] min 2 sf\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p06.png", "../answer-assets/9701_w22_ms_41-p07.png" ] }, { "id": "9701-2022-on-41-q03", "question_id": "9701-2022-on-41-q03", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 10, "status": "available", "reason": null, "text": "3(a)(i) (Ion (Sn2+/Sn4+) concentration) 1 mol dm–3 1\nAND 298 K (25 °C)\n\n3(a)(ii) • both half-cells have Pt or C electrode 3\n• Sn2+ / Sn4+ AND H+ solutions\n• feasible gas delivery system\n• H label\n2\n• V/voltmeter AND correct circuit AND salt bridge touching solution\n• salt bridge labelled\n\n3(a)(iii) no (reaction) AND 2\nboth Eo values (Sn2+ / Sn) –0.14 and (Cl / Cl –) +1.36 [1]\n2\nE is –1.5 V / E is negative\ncell cell\nOR Eo of Sn4+ / Sn2+ is more negative/smaller than Eo of Cl / Cl –[1]\n2\n\n3(a)(iv) Sn2+→Sn4+ and VO2+→V3+ [1] 2\nSn2++ 2VO2+ + 4H+ → Sn4+ + 2V3+ +2H O [1]\n2\n\n3(b) moles of Sn = 2.95 / 118.7 = 0.0249 moles 2\nmoles of Al (is 2 / 3 moles of Sn) = 0.0166 moles [1]\nmass of Al = 0.0166  27 = 0.447 / 0.448 g to 3sf [1] ecf\nQuestion Answer Marks\n© UCLES 2022 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p07.png" ] }, { "id": "9701-2022-on-41-q04", "question_id": "9701-2022-on-41-q04", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 9, "status": "available", "reason": null, "text": "4(a)(i) ratio of concentration of the solute in two solvents at equilibrium [1] 1\n\n4(a)(ii) 3.50 = (1.62 / 100) / (0.38 / x) [1] 2\nx= 82 (cm3) (82.0987654) [1] ecf M1 min 2sf\n\n4(b)(i) pH resists change when small amount of acid or base added 1\n\n4(b)(ii) CH CH CH COO– / salt of butanoic acid / sodium butanoate 1\n3 2 2\nAND NaOH\n\n4(c)(i) K = [Al 3+][OH–]3 1\nsp\n\n4(c)(ii) [OH-] = 3  2.47  10–9 OR 7.41  10–9 [1] 3\nK = [2.47  10–9][7.41  10–9]3\nsp\n= 1.01  10–33 min 2sf ecf from 1st mark [1]\nmol4 dm–12 [1]\n© UCLES 2022 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p08.png" ] }, { "id": "9701-2022-on-41-q05", "question_id": "9701-2022-on-41-q05", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "5(a) 1\n3d 3d 2\nxy z\n\n5(b)(i) • d orbital(s) of different energy / d-d splitting occurs 4\n• electron(s) promoted / excited\n• wavelength of visible light absorbed\nAND complementary colour seen\n• different energy gap / different E\nOR different frequency/wavelength of light is absorbed\n\n5(b)(ii) 1\n\n5(b)(iii) octahedral 1\nAND optical isomerism\n\n5(c) Fe2+ is smaller / has a smaller radius 2\nOR Fe2+ greater charge density [1]\npolarises/distorts the anion / CO (2)– more [1]\n3\n\n5(d)(i) colourless / (pale) green to pink / purple 1\n© UCLES 2022 Page 9 of 16\n\n5(d)(ii) moles MnO – = 0.0100  0.0350 OR 3.50  10–4 2\n4\nmoles Fe2+ = 5  3.50  10–4 OR 1.75  10–3 [1]\nmoles Fe2+ = 1.75  10–3  4 = 7.00  10–3\n% of Fe = (7.00  10–3  55.8) = 0.3906 g\n= 0.3906 / 2.62  100 = 14.9 % by mass [1] ecf M1 min 2sf\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p09.png", "../answer-assets/9701_w22_ms_41-p10.png" ] }, { "id": "9701-2022-on-41-q06", "question_id": "9701-2022-on-41-q06", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 11, "status": "available", "reason": null, "text": "6(a)(i) [Cu(H O) ]2+ + 2OH– → [Cu(OH) (H O) ] + 2H O OR 1\n2 6 2 2 4 2\nOR\n[Cu(H O) ]2+ + 2NaOH → [Cu(OH) (H O) ] + 2H O + 2Na+\n2 6 2 2 4 2\n\n6(a)(ii) [Cu(H O) ]2+ + 4Cl– → [CuCl ]2– + 6H O 1\n2 6 4 2\nOR\n[Cu(H O) ]2+ + 4HCl → [CuCl ]2– + 6H O + 4H+\n2 6 4 2\n\n6(b)(i) 1\ncis and trans drawn\n\n6(b)(ii) cis identified as polar AND trans identified as non-polar 1\nconditional on diagram\n\n6(c)(i) the equilibrium (constant) for the formation of the complex ion in a solvent from its constituent ions or molecules 1\n\n6(c)(ii) [Cu(NH ) (H O) ]2+ AND as K is large OWTTE 1\n3 4 2 2 stab\n\n6(c)(iii) K = [[Cu(NH ) (H O) ]2+] / [[Cu(H O) ]2+] [NH ]4 [1] 2\nstab 3 4 2 2 2 6 3\nmol–4 dm12 [1] ecf M1\n© UCLES 2022 Page 10 of 16\n\n6(c)(iv) [Cu(H O) ]2+] = (0.0074) ÷ (1.4  1013  0.574) 1\n2 6\n[Cu(H O) ]2+] = 5.01  10–15 min 2sf ecf 6(c)(iii)\n2 6\n\n6(d) [Ru(C H N ) Cl ]+ [1] 2\n12 8 2 2 2\n[Fe(C O ) ]3– [1]\n2 4 3\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p10.png", "../answer-assets/9701_w22_ms_41-p11.png" ] }, { "id": "9701-2022-on-41-q07", "question_id": "9701-2022-on-41-q07", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 17, "status": "available", "reason": null, "text": "7(a)(i) 1\n\n7(a)(ii) plane of polarised light will be rotated (in both isomers) [1] 2\nby same angle / equal amounts in opposite directions [1]\n\n7(b)(i) CH COCl AND HCl 1\n3\n© UCLES 2022 Page 11 of 16\n\n7(b)(ii) 3\nmethanol [1]\nester bond → primary alcohol\nOR amide → 2° amine AND benzene ring unchanged [1]\nrest of the structure of second compound is correct [1]\n\n7(b)(iii) Q < phenylamine < P [1] 3\nany three from:\nability of N to accept a proton OR donate its lone pair to a proton\nphenylamine\nlone pair of N delocalised into ring\nOR p-orbital on N overlaps with  cloud of ring (and decreases electron density on N)\ncompound P (2° amine)\nalkyl group has a positive inductive effect\n(and increases electron density on N)\ncompound Q (amide)\nlone pair of N (in amide) delocalised by C=O\nOR overlap of lone pair of N with C=O (and decreases electron density on N)\n\n7(c)(i) conc. HNO and H SO (25 °C < T ⩽ 60 °C) [1] 2\n3 2 4\nSn and conc. HCl and reflux (followed by NaOH(aq)) [1]\n© UCLES 2022 Page 12 of 16\n\n7(c)(ii) 2\nNH on ring gives diazonium ion [1]\n2\ndibromo compound with Br atoms 2,6 to amine group [1]\n\n7(d)(i) pH where the species is a zwitterion is the dominant form 1\nOR pH where the species is electrically neutral\n© UCLES 2022 Page 13 of 16\n\n7(d)(ii) 2\nrepeat unit identified by label / brackets / circle [1]\nthree monomers complete\nall amide/peptide bonds correct\nall CHCH C H / CHR groups correct\n2 6 5\ntrailing bonds or other convention(everything else correct) [1]\nQuestion Answer Marks", "source_pages": [ 11, 12, 13, 14 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p11.png", "../answer-assets/9701_w22_ms_41-p12.png", "../answer-assets/9701_w22_ms_41-p13.png", "../answer-assets/9701_w22_ms_41-p14.png" ] }, { "id": "9701-2022-on-41-q08", "question_id": "9701-2022-on-41-q08", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 8, "status": "available", "reason": null, "text": "8(a) 120° AND sp2 1\n\n8(b)(i) C H Cl + AlCl → CH CH + + AlCl – 1\n2 5 3 3 2 4\n© UCLES 2022 Page 14 of 16\n\n8(b)(ii) 3\ncurly arrow 1 AND curly arrow 2 [1]\ncorrect intermediate structure [1]\nC H C H AND H+ [1]\n6 5 2 5\n\n8(c)(i) (aqueous / alkaline) AgNO / silver nitrate 1\n3\n\n8(c)(ii) C H Cl + H O → C H OH + HCl 1\n2 5 2 2 5\n/ C H Cl + NaOH → C H OH + NaCl\n2 5 2 5\nAND Ag+ + Cl – → AgCl\nAND NO equation shown for C H Cl\n6 5\n\n8(c)(iii) lone pair / p-orbital from Cl overlaps with benzene ring 1\nAND\nstronger / partial double C-Cl bond OR difficult to break C-Cl bond\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p14.png", "../answer-assets/9701_w22_ms_41-p15.png" ] }, { "id": "9701-2022-on-41-q09", "question_id": "9701-2022-on-41-q09", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 12, "status": "available", "reason": null, "text": "9(a)(i) stationary: non-volatile (non-polar) liquid 1\nmobile: nitrogen / argon / any inert gas\n\n9(a)(ii) the time that the substance stays in the column 1\nOR time between injection and detection OWTTE\n© UCLES 2022 Page 15 of 16\n\n9(a)(iii) areas are 30, 100, 25 OR 30, 150, 25 (½  b  h) OR 100 / 155 [1] 2\n% of B = 100  (100 / 155) OR % of B = 100  150 / 205 =\n= 64.5 = 73.2 [1] ecf M1 areas\n\n9(b) 5 5 5 4 4 2\n\n9(c) 4 [1] 2\nsinglet, (two) triplet(s), multiplet (any order) [1]\n\n9(d)(i) D = CH CH CO CH CH [1] E = (CH ) CHCO CH 2\n3 2 2 2 3 3 2 2 3\n\n9(d)(ii) O-CH labelled F 1\n2\nAND three protons on neighbouring carbon / adjacent CH\n3\n\n9(d)(iii) both CH in isopropyl group labelled G 1\n3\nAND alkane / alkyl (protons)\n© UCLES 2022 Page 16 of 16", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_41.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_41-p15.png", "../answer-assets/9701_w22_ms_41-p16.png" ] }, { "id": "9701-2022-on-42-q01", "question_id": "9701-2022-on-42-q01", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 11, "status": "available", "reason": null, "text": "1(a) (energy change) when one mole of ionic solid is formed from gaseous ions 1\n\n1(b) (–2237 + 193 + 590 + 1150 + (2  121) – (2  364)) [1] 2\n= –790 [1]\n\n1(c) –342 and 1\nBr atom has larger radius\n\n1(d)(i) energy change when one mole dissolves in water [1] 2\nenergy change when one mole of gaseous ions dissolves in water [1]\n\n1(d)(ii) (–2237 – 83 + 1650) / 2 [1] 2\n= –335 [1]\n\n1(e)(i) negative and 1\nreduction in number of gas molecules\n\n1(e)(ii) TS becomes more negative [1] 2\nless feasible AND G becomes positive [1]\nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p05.png" ] }, { "id": "9701-2022-on-42-q02", "question_id": "9701-2022-on-42-q02", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i) rate = k[NO][O ] 1\n3\n\n2(a)(ii) 1.66  10–8 [1] 2\nmol dm–3 s–1 [1]\n\n2(a)(iii) not constant 1\nAND\noverall second order / not overall first order\n\n2(b)(i) graph is straight line clearly parallel to x-axis 1\n\n2(b)(ii) graph is straight line with negative gradient 1\n© UCLES 2022 Page 5 of 11\n\n2(b)(iii) adsorption of reactants onto catalyst surface [1] 3\nbonds in reactants weaken [1]\nreaction occurs followed by desorption of products [1]\n\n2(b)(iv) all active sites on catalyst surface are occupied 1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p05.png", "../answer-assets/9701_w22_ms_42-p06.png" ] }, { "id": "9701-2022-on-42-q03", "question_id": "9701-2022-on-42-q03", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 11, "status": "available", "reason": null, "text": "3(a)(i) the voltage produced by a half-cell compared with a standard hydrogen electrode [1] 2\n1 mol dm–3, 298 K, 1 atm [1]\n\n3(a)(ii) Mg wire and Pt wire [1] 3\nvoltmeter, salt bridge, complete circuit [1]\nsolutes Mg2+ and MnO –, Mn2+, H+ [1]\n4\n\n3(a)(iii) Mg is minus, Pt is plus 1\narrow points towards MnO – / Mn2+ half-cell\n4\n\n3(a)(iv) 3.90 V 1\n\n3(a)(v) 5Mg + 2MnO – +16H+ → 5Mg2+ + 2Mn2+ + 8H O 1\n4 2\n\n3(a)(vi) no change and 1\ndilution will make Mg2+ / Mg potential even more negative\n\n3(b) either: 2\n4.75  1022  2  1.60  10–19 = 15 200 C OR\n2  96 500  (4.75  1022) / 6.02  1023 = 15 228 C [1]\n15 200 / (15  60) = 16.9 A OR\n15 228 / (15  60) = 16.9 A [1]\n© UCLES 2022 Page 6 of 11", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p06.png" ] }, { "id": "9701-2022-on-42-q04", "question_id": "9701-2022-on-42-q04", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 9, "status": "available", "reason": null, "text": "4(a)(i) 27 x4 = 2  10–39 1\nx = 9.28  10–11 mol dm–3\n\n4(a)(ii) [Fe3+][0.02]3 = 2  10–39 [1] 2\n[Fe3+] = 2.5  10–34 [1]\n\n4(a)(iii) common ion (effect) 1\n\n4(b)(i) two species that differ by one H+ ion 1\n\n4(b)(ii) HBrO BrO– 1\nH O+ H O\n3 2\n\n4(b)(iii) [acid] = 4  10–2 2\n[H+]2 = 8.0  10–11 [1]\npH = 5.05 [1]\n\n4(b)(iv) 8.70 1\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p07.png" ] }, { "id": "9701-2022-on-42-q05", "question_id": "9701-2022-on-42-q05", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 16, "status": "available", "reason": null, "text": "5(a) a d-block element that forms one or more stable ions with incomplete d-orbitals 1\n\n5(b)(i) 3\ngeometry CN bond angle\n[Cu(H O) ]2+ octahedral 6 90 or 180\n2 6\n[CuCl ]2– tetrahedral 4 109 to 110\n4\n\n5(b)(ii) degenerate – of the same energy 1\nnon-degenerate – not of the same energy / different energy\n\n5(b)(iii) different frequency / wavelength / photon absorbed 1\n© UCLES 2022 Page 7 of 11\n\n5(c)(i) forms 2 dative / coordinate bonds 1\n\n5(c)(ii) one isomer correct with correct 3D [1] 2\nthree isomers different with correct 3D [1]\n\n5(c)(iii) e.g. 1 and 2 optical [1] 2\ne.g. 1 or 2 and 3 cis-trans / geometric [1]\n\n5(d) 0.0230 moles ethanedioate in whole 100 cm3 3\n0.00575 moles ethanedioate in 25 cm3 sample [1]\n0.00230 moles of MnO – needed [1]\n4\n230 cm3 [1]\n\n5(e) Cu2+ is smaller / has a smaller radius / than Ba2+ [1] 2\nCu2+ polarises / distorts the anion more [1]\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p07.png", "../answer-assets/9701_w22_ms_42-p08.png" ] }, { "id": "9701-2022-on-42-q06", "question_id": "9701-2022-on-42-q06", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "6(a)(i) 2\nblue or pink s / solid\nblue aq / aqueous\ntwo cells correct [1]\ntwo more cells correct [1]\n\n6(a)(ii) [Co(H O) ]2+ + 2OH– → Co(OH) + 6H O or 1\n2 6 2 2\n[Co(H O) ]2+ + 2OH– → Co(H O) (OH) + 2H O\n2 6 2 4 2 2\n\n6(a)(iii) [Co(H O) ]2+ + 4Cl– → [CoCl ]2– + 6H O 1\n2 6 4 2\n\n6(b)(i) the equilibrium constant for the formation of a complex ion in a solvent OR the equilibrium constant for the formation of a 1\ncomplex ion from its constituent ions\n© UCLES 2022 Page 8 of 11\n\n6(b)(ii) K = [[Co(NH ) ]2+] / [[Co(H O) ]2+][NH ]6 1\nstab 3 6 2 6 3\nor\n[[Co(NH ) ]2+] / [Co2+][NH ]6\n3 6 3\n\n6(b)(iii) mol–6 dm18 1\n\n6(b)(iv) 7.86  10–5 1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p08.png", "../answer-assets/9701_w22_ms_42-p09.png" ] }, { "id": "9701-2022-on-42-q07", "question_id": "9701-2022-on-42-q07", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 15, "status": "available", "reason": null, "text": "7(a) C > A > B [1] 3\nchlorine and C=O are electronegative / withdraw charge and this causes greatest weakening of O–H bond or greatest\nstabilisation of the anion [1]\n2nd oxygen / C=O is electronegative / withdraws charge and this weakens O–H bond or stabilises anion [1]\n\n7(b)(i) chloromethane 1\naluminium chloride\n\n7(b)(ii) CH Cl + AlCl → AlCl – + CH + 1\n3 3 4 3\n\n7(b)(iii) curly arrow from within benzene to CH + [1] 3\n3\npositively charged intermediate [1]\ncurly arrow from C–H bond into ring, C H CH , H+ [1]\n6 5 3\n\n7(b)(iv) hot alkaline KMnO [1] 2\n4\nC H CH + 3[O] → C H CO H + H O OR\n6 5 3 6 5 2 2\nC H CH + 3[O] + OH– → C H CO – + 2H O [1]\n6 5 3 6 5 2 2\n\n7(b)(v) 2 or 4-nitromethylbenzene and 3-nitrobenzoic acid 1\n\n7(c)(i) HNO , T between 0° and 10°C 1\n2\n\n7(c)(ii) warm / T>10°C and H O 1\n2\n© UCLES 2022 Page 9 of 11\n\n7(c)(iii) CH C H –N=N–C H (CH )OH [1] 2\n3 6 4 6 3 3\nT between 0° and 10°C and NaOH(aq) [1]\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p09.png", "../answer-assets/9701_w22_ms_42-p10.png" ] }, { "id": "9701-2022-on-42-q08", "question_id": "9701-2022-on-42-q08", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 10, "status": "available", "reason": null, "text": "8(a) C* marked on CH of T and nowhere else 1\n\n8(b)(i) R – C H CH COCH CH [1] 2\n6 5 2 2 3\n3.7 is C H CH COCH CH\n6 5 2 2 3\n2.5 is C H CH COCH CH\n6 5 2 2 3\n1.0 is C H CH COCH CH [1]\n6 5 2 2 3\n\n8(b)(ii) singlet 1\nand\nno H on neighbouring C\n\n8(c)(i) P and T 1\n\n8(c)(ii) P 1\n\n8(d) CDCl or CCl 1\n3 4\n\n8(e) no difference and 1\nno protons that exchange with D\n\n8(f) 6 5 [1] 2\n4 3 [1]\n© UCLES 2022 Page 10 of 11", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p10.png" ] }, { "id": "9701-2022-on-42-q09", "question_id": "9701-2022-on-42-q09", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "9(a) pH 7 +H N(CH ) CHNH COOH 2\n3 2 4 2\npH 9.47 +H N(CH ) CHNH COO– [1]\n3 2 4 2\npH 12 H N(CH ) CHNH COO– [1]\n2 2 4 2\n\n9(b) D ethanoyl chloride [1] 3\nE correct phenyl ester [1]\nF correct amide [1]\n\n9(c) Br substitutes at both 2 and 6 positions [1] 3\nboth phenol and COOH groups deprotonated [1]\nNO substitutes at one or both of 2 and 6 positions [1]\n2\n\n9(d)(i) distance travelled by amino acid divided by distance travelled by solvent [1] 1\n\n9(d)(ii) tyr is more soluble in the solvent used 1\nor\nlys is more attracted to the stationary phase used\n© UCLES 2022 Page 11 of 11", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_42-p11.png" ] }, { "id": "9701-2022-on-43-q01", "question_id": "9701-2022-on-43-q01", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 11, "status": "available", "reason": null, "text": "1(a) 2\nM1 K+ (g) and Cl – (g)\nAND\nKCl (aq) OR K+ (aq) + Cl – (aq)\nM2 three correct directional arrows COND M1\n\n1(b) use of data –155, –2493 AND 2  –364 [1] 2\nH Mg2+ = –1920 (kJ mol–1) [1] min 3sf\nhyd\n\n1(c) • Mg2+ is smaller (than K+) 2\n• Mg2+ is greater charge (than K+)\n• greater attraction\nbetween Mg2+ and Cl – /between the ions (in MgCl )\n2\nOR stronger ionic bonds (in MgCl )\n2\n\n1(d)(i) enthalpy change when 1\none mole of gaseous atoms formed from the element (in its standard state at 298 K)\n\n1(d)(ii) enthalpy change when 1\nevery atom in one mole of gaseous atoms gains one electron\nOR one mole of gaseous atoms gains one mole of electrons\n\n1(e)(i) number of possible arrangements of particles and energy in a system 1\n© UCLES 2022 Page 5 of 16\n\n1(e)(ii) S is positive 1\nAND KCl(s) → K+(aq) + Cl –(aq) /\nionic lattice solid forms aqueous ions OWTTE [1]\nOR\nS is positive\nAND G is (therefore becomes) negative /\nTS is greater than H OWTTE [1]\nsol\n\n1(e)(iii) more soluble 1\nAND G is more negative at higher T /\nTS is more positive at higher T /\n–TS is more negative at higher ecf from (e)(ii) [sign S]\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p05.png", "../answer-assets/9701_w22_ms_43-p06.png" ] }, { "id": "9701-2022-on-43-q02", "question_id": "9701-2022-on-43-q02", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i) (homogeneous is in the) same phase / state as reactants 1\nAND (heterogeneous is in a) different phase / state to reactants\n\n2(a)(ii) 1 S O 2– + 2Fe2+ → 2Fe3+ + 2SO 2– [1] 2\n2 8 4\n2 2I– + 2Fe3+ → 2Fe2+ + I [1]\n2\n\n2(a)(iii) reactants are both anions / negatively charged 1\nAND so they repel each other OWTTE\n\n2(b)(i) rate = k[NO]2[O ] OR rate = 8.6  106 [NO]2[O ] 1\n2 2\n\n2(b)(ii) rate = 8.6  106  (7.2  10–4)2  1.9  10–3 1\nrate = 8.47  10–3 (mol dm–3 s–1) min 2sf\n\n2(c)(i) (reaction is) first order wrt cisplatin / overall 1\nOR rate is directly proportional to concentration of cisplatin\n\n2(c)(ii) 0.693 / 2.50  10–5 = (2.77  104 s) 1\nOR In 2 / 2.50  10–5 = (2.77  104 s)\n© UCLES 2022 Page 6 of 16\n\n2(c)(iii) initial concentration is 8.0  10–5 mol dm–3 2\n– five half-life periods have elapsed [1]\ntime = 5  27720 = 1.39  105 s [1] min 2 sf\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p06.png", "../answer-assets/9701_w22_ms_43-p07.png" ] }, { "id": "9701-2022-on-43-q03", "question_id": "9701-2022-on-43-q03", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 10, "status": "available", "reason": null, "text": "3(a)(i) (Ion (Sn2+/Sn4+) concentration) 1 mol dm–3 1\nAND 298 K (25 °C)\n\n3(a)(ii) • both half-cells have Pt or C electrode 3\n• Sn2+ / Sn4+ AND H+ solutions\n• feasible gas delivery system\n• H label\n2\n• V/voltmeter AND correct circuit AND salt bridge touching solution\n• salt bridge labelled\n\n3(a)(iii) no (reaction) AND 2\nboth Eo values (Sn2+ / Sn) –0.14 and (Cl / Cl –) +1.36 [1]\n2\nE is –1.5 V / E is negative\ncell cell\nOR Eo of Sn4+ / Sn2+ is more negative/smaller than Eo of Cl / Cl –[1]\n2\n\n3(a)(iv) Sn2+→Sn4+ and VO2+→V3+ [1] 2\nSn2++ 2VO2+ + 4H+ → Sn4+ + 2V3+ +2H O [1]\n2\n\n3(b) moles of Sn = 2.95 / 118.7 = 0.0249 moles 2\nmoles of Al (is 2 / 3 moles of Sn) = 0.0166 moles [1]\nmass of Al = 0.0166  27 = 0.447 / 0.448 g to 3sf [1] ecf\nQuestion Answer Marks\n© UCLES 2022 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p07.png" ] }, { "id": "9701-2022-on-43-q04", "question_id": "9701-2022-on-43-q04", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 9, "status": "available", "reason": null, "text": "4(a)(i) ratio of concentration of the solute in two solvents at equilibrium [1] 1\n\n4(a)(ii) 3.50 = (1.62 / 100) / (0.38 / x) [1] 2\nx= 82 (cm3) (82.0987654) [1] ecf M1 min 2sf\n\n4(b)(i) pH resists change when small amount of acid or base added 1\n\n4(b)(ii) CH CH CH COO– / salt of butanoic acid / sodium butanoate 1\n3 2 2\nAND NaOH\n\n4(c)(i) K = [Al 3+][OH–]3 1\nsp\n\n4(c)(ii) [OH-] = 3  2.47  10–9 OR 7.41  10–9 [1] 3\nK = [2.47  10–9][7.41  10–9]3\nsp\n= 1.01  10–33 min 2sf ecf from 1st mark [1]\nmol4 dm–12 [1]\n© UCLES 2022 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p08.png" ] }, { "id": "9701-2022-on-43-q05", "question_id": "9701-2022-on-43-q05", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "5(a) 1\n3d 3d 2\nxy z\n\n5(b)(i) • d orbital(s) of different energy / d-d splitting occurs 4\n• electron(s) promoted / excited\n• wavelength of visible light absorbed\nAND complementary colour seen\n• different energy gap / different E\nOR different frequency/wavelength of light is absorbed\n\n5(b)(ii) 1\n\n5(b)(iii) octahedral 1\nAND optical isomerism\n\n5(c) Fe2+ is smaller / has a smaller radius 2\nOR Fe2+ greater charge density [1]\npolarises/distorts the anion / CO (2)– more [1]\n3\n\n5(d)(i) colourless / (pale) green to pink / purple 1\n© UCLES 2022 Page 9 of 16\n\n5(d)(ii) moles MnO – = 0.0100  0.0350 OR 3.50  10–4 2\n4\nmoles Fe2+ = 5  3.50  10–4 OR 1.75  10–3 [1]\nmoles Fe2+ = 1.75  10–3  4 = 7.00  10–3\n% of Fe = (7.00  10–3  55.8) = 0.3906 g\n= 0.3906 / 2.62  100 = 14.9 % by mass [1] ecf M1 min 2sf\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p09.png", "../answer-assets/9701_w22_ms_43-p10.png" ] }, { "id": "9701-2022-on-43-q06", "question_id": "9701-2022-on-43-q06", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 11, "status": "available", "reason": null, "text": "6(a)(i) [Cu(H O) ]2+ + 2OH– → [Cu(OH) (H O) ] + 2H O OR 1\n2 6 2 2 4 2\nOR\n[Cu(H O) ]2+ + 2NaOH → [Cu(OH) (H O) ] + 2H O + 2Na+\n2 6 2 2 4 2\n\n6(a)(ii) [Cu(H O) ]2+ + 4Cl– → [CuCl ]2– + 6H O 1\n2 6 4 2\nOR\n[Cu(H O) ]2+ + 4HCl → [CuCl ]2– + 6H O + 4H+\n2 6 4 2\n\n6(b)(i) 1\ncis and trans drawn\n\n6(b)(ii) cis identified as polar AND trans identified as non-polar 1\nconditional on diagram\n\n6(c)(i) the equilibrium (constant) for the formation of the complex ion in a solvent from its constituent ions or molecules 1\n\n6(c)(ii) [Cu(NH ) (H O) ]2+ AND as K is large OWTTE 1\n3 4 2 2 stab\n\n6(c)(iii) K = [[Cu(NH ) (H O) ]2+] / [[Cu(H O) ]2+] [NH ]4 [1] 2\nstab 3 4 2 2 2 6 3\nmol–4 dm12 [1] ecf M1\n© UCLES 2022 Page 10 of 16\n\n6(c)(iv) [Cu(H O) ]2+] = (0.0074) ÷ (1.4  1013  0.574) 1\n2 6\n[Cu(H O) ]2+] = 5.01  10–15 min 2sf ecf 6(c)(iii)\n2 6\n\n6(d) [Ru(C H N ) Cl ]+ [1] 2\n12 8 2 2 2\n[Fe(C O ) ]3– [1]\n2 4 3\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p10.png", "../answer-assets/9701_w22_ms_43-p11.png" ] }, { "id": "9701-2022-on-43-q07", "question_id": "9701-2022-on-43-q07", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 17, "status": "available", "reason": null, "text": "7(a)(i) 1\n\n7(a)(ii) plane of polarised light will be rotated (in both isomers) [1] 2\nby same angle / equal amounts in opposite directions [1]\n\n7(b)(i) CH COCl AND HCl 1\n3\n© UCLES 2022 Page 11 of 16\n\n7(b)(ii) 3\nmethanol [1]\nester bond → primary alcohol\nOR amide → 2° amine AND benzene ring unchanged [1]\nrest of the structure of second compound is correct [1]\n\n7(b)(iii) Q < phenylamine < P [1] 3\nany three from:\nability of N to accept a proton OR donate its lone pair to a proton\nphenylamine\nlone pair of N delocalised into ring\nOR p-orbital on N overlaps with  cloud of ring (and decreases electron density on N)\ncompound P (2° amine)\nalkyl group has a positive inductive effect\n(and increases electron density on N)\ncompound Q (amide)\nlone pair of N (in amide) delocalised by C=O\nOR overlap of lone pair of N with C=O (and decreases electron density on N)\n\n7(c)(i) conc. HNO and H SO (25 °C < T ⩽ 60 °C) [1] 2\n3 2 4\nSn and conc. HCl and reflux (followed by NaOH(aq)) [1]\n© UCLES 2022 Page 12 of 16\n\n7(c)(ii) 2\nNH on ring gives diazonium ion [1]\n2\ndibromo compound with Br atoms 2,6 to amine group [1]\n\n7(d)(i) pH where the species is a zwitterion is the dominant form 1\nOR pH where the species is electrically neutral\n© UCLES 2022 Page 13 of 16\n\n7(d)(ii) 2\nrepeat unit identified by label / brackets / circle [1]\nthree monomers complete\nall amide/peptide bonds correct\nall CHCH C H / CHR groups correct\n2 6 5\ntrailing bonds or other convention(everything else correct) [1]\nQuestion Answer Marks", "source_pages": [ 11, 12, 13, 14 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p11.png", "../answer-assets/9701_w22_ms_43-p12.png", "../answer-assets/9701_w22_ms_43-p13.png", "../answer-assets/9701_w22_ms_43-p14.png" ] }, { "id": "9701-2022-on-43-q08", "question_id": "9701-2022-on-43-q08", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 8, "status": "available", "reason": null, "text": "8(a) 120° AND sp2 1\n\n8(b)(i) C H Cl + AlCl → CH CH + + AlCl – 1\n2 5 3 3 2 4\n© UCLES 2022 Page 14 of 16\n\n8(b)(ii) 3\ncurly arrow 1 AND curly arrow 2 [1]\ncorrect intermediate structure [1]\nC H C H AND H+ [1]\n6 5 2 5\n\n8(c)(i) (aqueous / alkaline) AgNO / silver nitrate 1\n3\n\n8(c)(ii) C H Cl + H O → C H OH + HCl 1\n2 5 2 2 5\n/ C H Cl + NaOH → C H OH + NaCl\n2 5 2 5\nAND Ag+ + Cl – → AgCl\nAND NO equation shown for C H Cl\n6 5\n\n8(c)(iii) lone pair / p-orbital from Cl overlaps with benzene ring 1\nAND\nstronger / partial double C-Cl bond OR difficult to break C-Cl bond\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p14.png", "../answer-assets/9701_w22_ms_43-p15.png" ] }, { "id": "9701-2022-on-43-q09", "question_id": "9701-2022-on-43-q09", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 12, "status": "available", "reason": null, "text": "9(a)(i) stationary: non-volatile (non-polar) liquid 1\nmobile: nitrogen / argon / any inert gas\n\n9(a)(ii) the time that the substance stays in the column 1\nOR time between injection and detection OWTTE\n© UCLES 2022 Page 15 of 16\n\n9(a)(iii) areas are 30, 100, 25 OR 30, 150, 25 (½  b  h) OR 100 / 155 [1] 2\n% of B = 100  (100 / 155) OR % of B = 100  150 / 205 =\n= 64.5 = 73.2 [1] ecf M1 areas\n\n9(b) 5 5 5 4 4 2\n\n9(c) 4 [1] 2\nsinglet, (two) triplet(s), multiplet (any order) [1]\n\n9(d)(i) D = CH CH CO CH CH [1] E = (CH ) CHCO CH 2\n3 2 2 2 3 3 2 2 3\n\n9(d)(ii) O-CH labelled F 1\n2\nAND three protons on neighbouring carbon / adjacent CH\n3\n\n9(d)(iii) both CH in isopropyl group labelled G 1\n3\nAND alkane / alkyl (protons)\n© UCLES 2022 Page 16 of 16", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_43.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_43-p15.png", "../answer-assets/9701_w22_ms_43-p16.png" ] }, { "id": "9701-2022-on-51-q01", "question_id": "9701-2022-on-51-q01", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) (no. moles of AgCl formed = 3.63 / 143.4 1\n= 0.02531381)\nno. moles MgCl in the sample = 0.02531381 / 2\n2\n= 0.0126569\nanswer to at least 2SF\n\n1(a)(ii) M1: mass of MgCl = answer to (a)(i)  95.3 1\n2\n(= 1.2062 g)\nM2: % by mass = (M1 /1.52)  100 = (79.355%) 1\nanswer to at least 2SF\n\n1(b)(i) stir (when solid is mixed with water) 1\nOR\nincrease temperature (of water)\nOR\nincrease state of division of the (solid) mixture\n\n1(b)(ii) to remove magnesium nitrate / (excess) silver nitrate (from the precipitate before drying) 1\n\n1(b)(iii) (It is dried in an oven) to avoid (thermal) decomposition (of silver chloride / precipitate / solid) 1\n\n1(c)(i) ((2  0.005) / 1.52)  100 (= 0.66%) (0.6578947) 1\ncorrect working must be shown along with correct answer.\nanswer to at least 1SF\n\n1(c)(ii) use a larger mass of solid (mixture). 1\n\n1(c)(iii) (continue drying and) reweigh until mass remains constant 1\n© UCLES 2022 Page 5 of 8\n\n1(d) the (measured) mass / amount (of ‘AgCl ’ / solid / precipitate) would be greater (than the true value) 1\nAND\nthe % by mass (of MgCl in the sample) would be greater (than the true value)\n2\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_51-p05.png", "../answer-assets/9701_w22_ms_51-p06.png" ] }, { "id": "9701-2022-on-51-q02", "question_id": "9701-2022-on-51-q02", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "2(a) (wear) chemically resistant gloves. 1\n\n2(b)(i) M1 leakproof apparatus capable of delivering a sample of gas (for collection over water) 1\nM2: a suitable collection vessel filled with water correctly positioned in a water containing trough to collect the gas (via 1\ndelivery tube)\n\n2(b)(ii) do not collect the first gas to be delivered 1\n© UCLES 2022 Page 6 of 8\n\n2(c)(i) M1: all points plotted correctly 1\nM2: line of best fit drawn 1\n\n2(c)(ii) M1: two acceptable coordinates from LOBF expressed in the form (x,y) 1\nM2: gradient correctly calculated from points listed for M1 1\n(≈ 0.357 cm3 K–1)\nanswer correctly rounded to 3SF\n\n2(d)(i) ring drawn around point at 308 K and 88 cm3 1\n\n2(d)(ii) the (actual) temperature was higher than the recorded / measured temperature / 308K / 35 °C 1\n\n2(e)(i) (absolute) temperature (of the gas) 1\n\n2(e)(ii) more (repeated) measurements should be taken 1\nAND\nanomalies identified / eliminated\n\n2(f) gradient is smaller / line is less steep (for smaller volume of oxygen at start of experiment) 1\nAND\ngradient = nR/P\n© UCLES 2022 Page 7 of 8", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_51-p06.png", "../answer-assets/9701_w22_ms_51-p07.png" ] }, { "id": "9701-2022-on-51-q03", "question_id": "9701-2022-on-51-q03", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 3, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 7, "status": "available", "reason": null, "text": "3(a)(i) description of a (clock) method to measure the time taken for a particular amount of product (Br ) to be produced 1\n2\n\n3(a)(ii) X = 1, Y = 1 and Z = 2 2\nall 3 correct = ✓✓ 2 correct = ✓\n1 or 0 correct = ×\n\n3(b) burette 1\n\n3(c) (changing) the temperature (of the reagents) will affect the rate of reaction (so it must be kept constant) 1\n\n3(d) M1: measure (a volume of) 1 moldm-3 sulfuric acid using a burette into a volumetric flask 1\nAND\nmake up to the (calibration) mark with distilled water\nM2: (0.15  a) cm3 of 1 mol dm-3 sulfuric acid is measured to make up a cm3 (of the diluted solution) 1\n© UCLES 2022 Page 8 of 8", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_51-p08.png" ] }, { "id": "9701-2022-on-52-q01", "question_id": "9701-2022-on-52-q01", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a)(i) (reheat and) reweigh until mass remains constant 1\n\n1(a)(ii) increases the rate of reaction 1\nAnd\ndue to increase the surface area\n\n1(a)(iii) to ensure the reaction is complete 1\n\n1(b)(i) M1: 16.55 16.30 16.00 16.20 2\nM2: 16.25 cm3\n\n1(b)(ii) M1 nNaOH 2\n1.00  16.25 / 1000 = 0.01625\nM2 nHCl unreacted\n= 0.01625  250 / 25 = 0.1625\n\n1(b)(iii) M1 nHCl reacted 3\n[2.00  100 / 1000] – 0.1625 = 0.0375\nM2 nCaCO reacted\n3\n= 0.0375 / 2 = 0.01875\nand\nmass CaCO reacted\n3\n= 0.01875  100.1 = 1.876875g\nM3 %CaCO = [1.876875/2.136]  100\n3\n= 87.9 %\n\n1(c) (25cm3) volumetric pipette 1\n\n1(d) ((2 x 0.0005) / 2.136)  100 (= 0.0468%) 1\n\n1(e) the % by mass (of CaCO in eggshell sample) would be less (than the true value) 1\n3\nand\nthe (measured) mass (of eggshell) would be greater (than the true value)\n© UCLES 2022 Page 5 of 7\n\n1(f) volume of NaOH would be 162.5 cm3 1\n(which is too large for a 50 cm3 burette)\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_52-p05.png", "../answer-assets/9701_w22_ms_52-p06.png" ] }, { "id": "9701-2022-on-52-q02", "question_id": "9701-2022-on-52-q02", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a) M1 13 points plotted correctly 2\nM2 line of best fit drawn\n\n2(b)(i) (likely to be) point at 18 minutes. 1\n\n2(b)(ii) diameter measured too early 1\nor\nthe recorded / measured diameter is smaller than the actual diameter (for that reading\n\n2(c) M1: tangent drawn at t = 15 minutes 3\nM2 co-ordinates read and recorded correctly\nM3 ignore candidate’s response, award 1 mark to all candidates for M3\n\n2(d) average rate of diffusion = (3.6 / 30) 1\n= 0.12 cm minutes-1\n\n2(e) time 1\n\n2(f) repeat the experiment to show the points / results / readings are reproducible 1\n\n2(g)(i) Gradient comment 2\nM1: lower gradient for K Cr O (than KMnO )\n2 2 7 4\nRate of diffusion + Mr comment\nM2: diffusion would occur more slowly for K Cr O\n2 2 7\nand\nthe M is greater\nr\n\n2(g)(ii) concentration (of solution / s) 1\n© UCLES 2022 Page 6 of 7\n\n2(h)(i) (wear) chemically resistant gloves 1\n\n2(h)(ii) any two from: 2\nRate of diffusion\nsolids diffuse too slowly (through the gel)\nInsolubility\nsolid crystals may not dissolve into gel\nUniformity / size of crystals\nthe crystals are not of uniform size\n© UCLES 2022 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_52-p06.png", "../answer-assets/9701_w22_ms_52-p07.png" ] }, { "id": "9701-2022-on-53-q01", "question_id": "9701-2022-on-53-q01", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) (no. moles of AgCl formed = 3.63 / 143.4 1\n= 0.02531381)\nno. moles MgCl in the sample = 0.02531381 / 2\n2\n= 0.0126569\nanswer to at least 2SF\n\n1(a)(ii) M1: mass of MgCl = answer to (a)(i)  95.3 1\n2\n(= 1.2062 g)\nM2: % by mass = (M1 /1.52)  100 = (79.355%) 1\nanswer to at least 2SF\n\n1(b)(i) stir (when solid is mixed with water) 1\nOR\nincrease temperature (of water)\nOR\nincrease state of division of the (solid) mixture\n\n1(b)(ii) to remove magnesium nitrate / (excess) silver nitrate (from the precipitate before drying) 1\n\n1(b)(iii) (It is dried in an oven) to avoid (thermal) decomposition (of silver chloride / precipitate / solid) 1\n\n1(c)(i) ((2  0.005) / 1.52)  100 (= 0.66%) (0.6578947) 1\ncorrect working must be shown along with correct answer.\nanswer to at least 1SF\n\n1(c)(ii) use a larger mass of solid (mixture). 1\n\n1(c)(iii) (continue drying and) reweigh until mass remains constant 1\n© UCLES 2022 Page 5 of 8\n\n1(d) the (measured) mass / amount (of ‘AgCl ’ / solid / precipitate) would be greater (than the true value) 1\nAND\nthe % by mass (of MgCl in the sample) would be greater (than the true value)\n2\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_53-p05.png", "../answer-assets/9701_w22_ms_53-p06.png" ] }, { "id": "9701-2022-on-53-q02", "question_id": "9701-2022-on-53-q02", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "2(a) (wear) chemically resistant gloves. 1\n\n2(b)(i) M1 leakproof apparatus capable of delivering a sample of gas (for collection over water) 1\nM2: a suitable collection vessel filled with water correctly positioned in a water containing trough to collect the gas (via 1\ndelivery tube)\n\n2(b)(ii) do not collect the first gas to be delivered 1\n© UCLES 2022 Page 6 of 8\n\n2(c)(i) M1: all points plotted correctly 1\nM2: line of best fit drawn 1\n\n2(c)(ii) M1: two acceptable coordinates from LOBF expressed in the form (x,y) 1\nM2: gradient correctly calculated from points listed for M1 1\n(≈ 0.357 cm3 K–1)\nanswer correctly rounded to 3SF\n\n2(d)(i) ring drawn around point at 308 K and 88 cm3 1\n\n2(d)(ii) the (actual) temperature was higher than the recorded / measured temperature / 308K / 35 °C 1\n\n2(e)(i) (absolute) temperature (of the gas) 1\n\n2(e)(ii) more (repeated) measurements should be taken 1\nAND\nanomalies identified / eliminated\n\n2(f) gradient is smaller / line is less steep (for smaller volume of oxygen at start of experiment) 1\nAND\ngradient = nR/P\n© UCLES 2022 Page 7 of 8", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_53-p06.png", "../answer-assets/9701_w22_ms_53-p07.png" ] }, { "id": "9701-2022-on-53-q03", "question_id": "9701-2022-on-53-q03", "subject": "9701", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 3, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 7, "status": "available", "reason": null, "text": "3(a)(i) description of a (clock) method to measure the time taken for a particular amount of product (Br ) to be produced 1\n2\n\n3(a)(ii) X = 1, Y = 1 and Z = 2 2\nall 3 correct = ✓✓ 2 correct = ✓\n1 or 0 correct = ×\n\n3(b) burette 1\n\n3(c) (changing) the temperature (of the reagents) will affect the rate of reaction (so it must be kept constant) 1\n\n3(d) M1: measure (a volume of) 1 moldm-3 sulfuric acid using a burette into a volumetric flask 1\nAND\nmake up to the (calibration) mark with distilled water\nM2: (0.15  a) cm3 of 1 mol dm-3 sulfuric acid is measured to make up a cm3 (of the diluted solution) 1\n© UCLES 2022 Page 8 of 8", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9701_w22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2022-oct-nov/9701_w22_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w22_ms_53-p08.png" ] }, { "id": "9701-2023-m-42-q01", "question_id": "9701-2023-m-42-q01", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 1\n[Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂\n\n1(a)(ii) 3\none ● for each of the eight species (including state symbol) on the correct line\nAND one ● e–\nany three [1] any six [2] all nine [3]\n\n1(a)(iii) • EA becomes less negative/ less exothermic (down group / S to Te) 2\n• atomic radii increases OR outer shell gets farther from nucleus OR electron added at higher energy level\nOR more shielding (of outer shells)\n• less nuclear attraction\nOR less attraction for incoming/added electron\nany two [1] all three [2]\n\n1(a)(iv) M1: O2– (has same charge but) smaller (radius than S2–) ORA 1\nOR oxygen has a smaller ion (than S2–)\nM2: stronger ionic bond OR greater attraction between Zn2+ and O2– ORA 1\n© UCLES 2023 Page 5 of 17\n\n1(b)(i) S negative AND more moles / molecules of gaseous reactants ORA 1\nOR S negative AND moles / molecules of gas are reduced (in the reaction)\n\n1(b)(ii) S = 50.8 + 197.7 – 43.7 – 5.7 = (+)199.1 (J K–1 mol–1) 1\n\n1(b)(iii) G = H – TS ALLOW G = H – TS 1\n= +733 – (800 + 273)  0.218 = (+)499.086 (kJ mol–1) min 3sf 1\n\n1(c)(i) Zn(NO ) → ZnO + 2NO + ½O 1\n3 2 2 2\nOR 2Zn(NO ) → 2ZnO + 4NO + O\n3 2 2 2\n\n1(c)(ii) increases (in thermal stability down the group) 1\nAND (cat)ion(ic) radius / ion size increases (down the group)\nless polarisation / less distortion 1\nof anion/ of nitrate ion/NO – / NO 2–\n3 3\nOR less weakening of N—O bond\n\n1(c)(iii) Mg(NO ) only ALLOW Mg2+ / magnesium 1\n3 2\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_42-p05.png", "../answer-assets/9701_m23_ms_42-p06.png" ] }, { "id": "9701-2023-m-42-q02", "question_id": "9701-2023-m-42-q02", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 17, "status": "available", "reason": null, "text": "2(a) H PO 1\n3 2\n\n2(b)(i) electrode potential E would become more positive / less negative (than E⦵) 1\nlower [H PO –] AND shifts equilibrium to the right-hand side 1\n2 2\n\n2(b)(ii) +1.57 – 0.74 = (+)0.83 (V) 1\n© UCLES 2023 Page 6 of 17\n\n2(b)(iii) 3\n• Pt(s) / C / graphite\n• Cr(s)\n• H+(aq) / any strong acid\n• Cr3+(aq)\n• H (g)\n2\n• voltmeter / V\n• salt bridge labelled\n• conditions of 1 atm AND 1 mol dm–3 / 1 M / 1 mol / dm3\n• other liquid level (salt bridge) and wire to electrode\nany three [1] any six [2] all nine [3]\n\n2(b)(iv) Pt electrode positive AND flow of electrons anticlockwise (to the SHE) 1\n\n2(b)(v) H PO – + 3OH– + Ni2+ → HPO 2– + 2H O + Ni 1\n2 2 3 2\n\n2(c)(i) (6.4 / 24000) ÷ 60 = 4.44  10–6 (mol dm-3 s–1) min 2sf 1\n\n2(c)(ii) [H PO –] doubles / 2 from experiments 1 to 2 ORA 1\n2 2\nvolume of H produced doubles / 2 (∴ first order wrt [H PO –])\n2 2 2\n[H PO –]  3 and [OH–]  ½ from experiments 1 to 3 ORA 1\n2 2\nvolume of H produced falls to ¾ original\n2\n(if first order wrt [H PO –] then must be second order wrt [OH–])\n2 2\nALLOW input data into rate equation and show k is the same\nk =2.8  10–6 / k = 6.7  10–2 (1 / 15) / k = 4 for all experiments [2]\n© UCLES 2023 Page 7 of 17\n\n2(c)(iii) mol–2 dm6 s–1 1\n\n2(c)(iv) t = 0.693 / 8.25  10–5 = 8400 (s) OR t = In2 / 8.25  10–5 = 8401.8 (s) 1\n½ ½\n\n2(c)(v) (k ) increases (with temperature) 1\n1\n\n2(d) • reactants adsorb (to surface of catalyst) 2\n• bonds (in reactant) weaken\n• (reaction occurs and the) products are desorbed\nOR reaction occurs and substances are desorbed\nany two [1] all three [2]\nQuestion Answer Marks", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_42-p06.png", "../answer-assets/9701_m23_ms_42-p07.png", "../answer-assets/9701_m23_ms_42-p08.png" ] }, { "id": "9701-2023-m-42-q03", "question_id": "9701-2023-m-42-q03", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 9, "status": "available", "reason": null, "text": "3(a) (a d-block element forms one or more) 1\nstable ions with incomplete filled d-subshell\n\n3(b)(i) • they behave as catalysts 1\n• they form complex ions / complexes\n• they form coloured compounds / salts / ions\nany two\n\n3(b)(ii) the d and s sub-shells/orbitals 1\nare close/similar in energy\n\n3(c)(i) 0.02500  0.0300 OR 7.50  10–4 mol VO + 1\n2\nOR ½(28.15 + 28.10)/1000  0.0400 OR 1.13  10–3 mol C O 2–\n2 4\nUse of their values to show ratio of VO + : C O 2– = 1:1.5 1\n2 2 4\nALLOW any viable approach\n© UCLES 2023 Page 8 of 17\n\n3(c)(ii) 1\n3D structure\ncharge 1\n\n3(d)(i) VO + AND E⦵ of H O is largest / most positive value 1\n2 2 2\nOR VO + AND E⦵ is less positive than H O\n2 2 2\n\n3(d)(ii) +5 1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_42-p08.png", "../answer-assets/9701_m23_ms_42-p09.png" ] }, { "id": "9701-2023-m-42-q04", "question_id": "9701-2023-m-42-q04", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 27, "status": "available", "reason": null, "text": "4(a)(i) CH CONH + 4[H] → CH CH NH + H O 1\n3 2 3 2 2 2\n\n4(a)(ii) nucleophilic substitution 1\n(ammonia with) ethanol AND heat under pressure 1\nOR ethanol AND heat in a sealed tube\n\n4(a)(iii) (CH CH ) NH OR (CH CH ) N 1\n3 2 2 3 2 3\n© UCLES 2023 Page 9 of 17\n\n4(b) (least) phenylamine < ammonia < ethylamine (most) 1\nexplanation 3\n• (order of basicity) ability of base AND to accept a proton\nOR donate its lone pair (to a proton)\nphenylamine\n• lone pair / p-orbital from N delocalised / overlaps with (-)ring / benzene\nethylamine\n• alkyl / ethyl group is electron donating group / +I group\n• increases electron density on N (ethylamine) ORA\nany two [1] any three [2] all four [3]\n\n4(c) 1\nconcentrated HNO and H SO (and 25–60 °C) 1\n3 2 4\n(reduction with) Sn and concentrated HCl (heat) 1\n\n4(d)(i) 1\n\n4(d)(ii) nitrous acid / HNO 1\n2\nOR NaNO AND dilute HCl\n2\n© UCLES 2023 Page 10 of 17\n\n4(d)(iii) phenol 1\n\n4(d)(iv) NaOH / alkali 1\n\n4(d)(v) dyestuffs / dyes 1\n\n4(e)(i) species that uses one lone pair of electrons 1\nthat forms a single dative covalent bond to a central metal atom / ion 1\n\n4(e)(ii)\ncorrect 3D cis isomer 1\ncorrect 3D trans isomer 1\n\n4(e)(iii) 6 1\n\n4(f)(i) units = mol–4 dm12 1\n\n4(f)(ii) CdCl 2− 1\n 4 \n(K =)\nstab\n4\nCd2+Cl−\n  \n\n4(f)(iii) [CdCl 2–] = K  0.043  0.0724 = 7.28 / 7.3  10–4 (mol dm–3) min 2sf 1\n4 stab\n© UCLES 2023 Page 11 of 17\n\n4(f)(iv) CH NH (is basic so) reacts with water to produce OH– that reacts with Cd2+ 1\n3 2\nOR\nCH NH acts as a base and accepts a proton from Cd[H O) ]2+ / water\n3 2 2 6\nOR\n[Cd(H O) ]2+ + 4CH NH → [Cd(OH) (H O) ]2– + 4CH NH + OWTTE\n2 6 3 2 4 2 2 3 3\n\n4(f)(v) Cd(OH) 2– = most stable (2nd box) AND [Cd(H O) ]2+ = least stable (first box) 1\n4 2 6\nCd(OH) 2– has highest K (and all K values given > 1) 1\n4 stab stab\nQuestion Answer Marks", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_42-p09.png", "../answer-assets/9701_m23_ms_42-p10.png", "../answer-assets/9701_m23_ms_42-p11.png", "../answer-assets/9701_m23_ms_42-p12.png" ] }, { "id": "9701-2023-m-42-q05", "question_id": "9701-2023-m-42-q05", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 22, "status": "available", "reason": null, "text": "5(a) any three points from: 3\n• bond angle = 120°\nAND shape is hexagonal ring planar / trigonal planar\n• (carbons are) sp2 hybridised\n• contains delocalised electrons in the  bonds / system\n• (sp2 orbitals) overlap end-on-end/ head-on to form  bonds\n• a p orbital (from each carbon atom) overlaps sideways\n(with each other above and below the ring) forming  bonds\n\n5(b)(i) Cl + AlCl → Cl+ + AlCl – 1\n2 3 4\n© UCLES 2023 Page 12 of 17\n\n5(b)(ii) 3\n+ H+\nM1 curly arrow from inside hexagon of aromatic ring to Cl +\nCON going to a lone pair on Cl\nM2 intermediate\nM3 curly arrow from C—H bond to inside the ring AND H+\n\n5(c)(i) electrophilic substitution 1\n© UCLES 2023 Page 13 of 17\n\n5(c)(ii) 1\nOR\n4-substituted acyl / alkyl derivative\nALLOW\n3-substituted acyl / alkyl derivative\n© UCLES 2023 Page 14 of 17\n\n5(c)(iii) (most) acyl chloride > alkyl chloride > aryl chloride (least) 1\nany two from: 2\n• acyl chlorides: carbon (in C—Cl) is more electron deficient\nAND it is also attached to an oxygen atom / two electronegative atoms\nOR C—Cl bond is weakest / weakened in acyl chlorides\nAND it is also attached to an oxygen atom / two electronegative atoms\n• aryl chlorides (no hydrolysis) C—Cl bond is part of delocalised system / partially double bond character so C—Cl\nbond is stronger\nOR lone pair on Cl delocalises with  ring so C—Cl bond is stronger\n• alkyl chlorides carbon atom has a smaller +\nAND due to the carbon being only attached to one electronegative atom\nOR C—Cl bond strengthened\nAND by electron donating effect / positive inductive effect of alkyl / R group\n\n5(d) (CH ) CNH (in ethanol) [substitution] } in either order 1\n3 3 2\nLiAlH OR NaBH [reduction ] } in either order 1\n4 4\n1\n[substitution first] OR [reduction first]\n\n5(e)(i) rotate the plane of polarised light in the opposite direction 1\nOR stereoisomers / molecules that are non-superimposable mirror images\n\n5(e)(ii) need to separate the optical isomers to form the pure active isomer 1\nOR reduced / different biological activity of ‘other’ enantiomer\nOR lower yield of biologically active molecule / desired molecule\n\n5(e)(iii) chiral catalyst OR use of an enzyme 1\n© UCLES 2023 Page 15 of 17\n\n5(f)(i) ten / 10 1\n\n5(f)(ii) ( = 0.9–1.7) 3\n• 9H\n• singlet\n• –CH / alkane\n3\n( = 3.2–4.0) / ( = 2.3–3.0)\n• 1H\n• triplet\n• –CHO / alkyl next to electronegative atom\nOR Ar-CH / alkyl next to aromatic ring\n( = 6.0–9.0)\n• 4H\n• multiplet\n• H–Ar / attached to aromatic ring\nany three [1] any six [2] all nine [3]\nQuestion Answer Marks", "source_pages": [ 12, 13, 14, 15, 16 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_42.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_42-p12.png", "../answer-assets/9701_m23_ms_42-p13.png", "../answer-assets/9701_m23_ms_42-p14.png", "../answer-assets/9701_m23_ms_42-p15.png", "../answer-assets/9701_m23_ms_42-p16.png" ] }, { "id": "9701-2023-m-42-q06", "question_id": "9701-2023-m-42-q06", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 9, "status": "available", "reason": null, "text": "6(a)(i) SiO OR Al O OR silica OR alumina OWTTE 1\n2 2 3\n\n6(a)(ii) Cd2+ AND R of M (= 2/5) = 0.40 / same R as in the Table 6.1 1\nf f\n\n6(b) metal cations are less soluble in butan-1-ol (than in water) 1\nOR metal cations weaker ion-dipole forces with butan-1-ol\n\n6(c) H O AND SCN– 1\n2\n\n6(d)(i) C H O K + KOH → C H O K + H O OR C H O K + OH– → C H O K– + H O 1\n8 5 4 8 4 4 2 2 8 5 4 8 4 4 2\n© UCLES 2023 Page 16 of 17\n\n6(d)(ii) M1 initially: moles of KOH = 0.150  20.0 ÷ 1000 = 3.0  10–3 1\nAND moles of acid HA = 0.100  50.0÷1000 = 5.0  10–3\nM2 at equilibrium: moles of salt KA = 3.0  10–3 ecf 1\nAND moles of acid HA = 5.0  10–3 – moles of NaOH\n= 2.0  10–3 ecf\nM3 K = 10–5.40 = 3.98  10–6 1\na\n3.98  10–6 = [H+] [3.0  10–3] ÷ [2.0  10–3]\n[H+] = 2.65  10–6 ecf\nM4 pH = –log(2.65  10–6) = 5.58 / 5.6 ecf on a calculated and identified [H+] 1\nALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4\n© UCLES 2023 Page 17 of 17", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_42-p16.png", "../answer-assets/9701_m23_ms_42-p17.png" ] }, { "id": "9701-2023-m-52-q01", "question_id": "9701-2023-m-52-q01", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 15, "status": "available", "reason": null, "text": "1(a)(i) All titres to nearest 0.05 cm3 (10.50, 9.65, 9.90, 9.75) 1\n\n1(a)(ii) (9.65 + 9.75) ÷ 2 = 9.70 (cm3) 1\n\n1(b) A solution of known concentration 1\n\n1(c)(i) To allow to reach equilibrium 1\n\n1(c)(ii) To allow solid to settle 1\n\n1(d) Acid is in excess 1\n\n1(e) aqueous potassium manganate(VII) 1\n\n1(f) From colourless to pale pink 1\n\n1(g) Lower percentage error 1\n\n1(h)(i) M1 Mol MnO –(aq) 2\n4\n= 0.0200  21.10 / 1000 = 4.22  10–4 mol\nM2 Mol Fe2+(aq) = M1  5 = 2.11  10–3 mol\nAND\nConcentration Fe2+(aq)\n= mol Fe2+  1000 / 25 = 0.0844 mol dm–3\n\n1(h)(ii) Same as [Fe2+(aq)] 1\neqm\n\n1(h)(iii) 0.100 – 0.0844 = 0.0156 1\n\n1(h)(iv) M1 K = 0.0156 / (0.0844  0.0844) = 2.19 2\nc\nM2 dm3 mol–1\n© UCLES 2023 Page 5 of 8", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_52-p05.png" ] }, { "id": "9701-2023-m-52-q02", "question_id": "9701-2023-m-52-q02", "subject": "9701", "year": 2023, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) (not all) reactants may feature in the rate determining step 1\nOR\norder can only be determined experimentally\n\n2(b) M1 Dissolve the solid / potassium iodide (in the beaker) (using a small volume of) (distilled) water 3\nM2 Transfer / add to a (250 cm3 volumetric) flask\nAND\nRinse (with distilled water)\nM3 Top the 250 cm3 volumetric flask up to mark (with distilled water)\n\n2(c)(i) Repeat the experiment to show the results are reproducible. 1\n\n2(c)(ii) Volume of KI 1\n\n2(c)(iii) Step 4 1\n© UCLES 2023 Page 6 of 8\n\n2(d)(i) 2\nLog V Log(1 / t)\n0.699 –2.41\n1.00 –2.08\n1.10 –2.05\n1.18 -1.88\nX X\n1.30 –1.75\n1.40 –1.64\nM1 Log V Correct values to 3 SF\nM2 log(1 / t) Correct values to 3 SF\n\n2(d)(ii) M1 6 points plotted correctly 2\nM2 Line of best fit drawn\n\n2(d)(iii) Point at log V = 1.10 1\nand\n(recorded time is too high because) watch started too early or stopped too late\n\n2(d)(iv) M1 Co-ordinates read and recorded correctly 2\nM2 Gradient calculated to 2 decimal places\n© UCLES 2023 Page 7 of 8\n\n2(d)(v) percentage error of experimental value 1\n= (2(d)(iv) – 1.00) / 1.00]  100\n(assuming 2(d)(iv) is 1.10)\n= [(1.10 – 1.00) / 1.00]  100 = 10%\n(10% > maximum percentage error / 5.25% )\nAND\nother factors (beyond measurement caused the error)\n© UCLES 2023 Page 8 of 8", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2023-March/ms/9701_m23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-march/9701_m23_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m23_ms_52-p06.png", "../answer-assets/9701_m23_ms_52-p07.png", "../answer-assets/9701_m23_ms_52-p08.png" ] }, { "id": "9701-2023-mj-41-q01", "question_id": "9701-2023-mj-41-q01", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 19, "status": "available", "reason": null, "text": "1(a) M1 increases (down the group) 3\nM2 radius / size of (cat)ion / M2+ increases\nM3 less polarisation / distortion of anion / nitrate ion / NO – OR less weakening of N–O / N=O (bond)\n3\n\n1(b) Cu(NO )  CuO + 2NO + ½O 1\n3 2 2 2\n\n1(c) 4\ncopper-\nformula of copper-containing colour copper-\ncontaining\nspecies formed containing formed\nspecies\nA [Cu(H O) ]2+ (pale) blue\n2 6\nB Cu(H O) (OH) or Cu(OH) (pale) blue\n2 4 2 2\nC [Cu(NH ) (H O) ]2+ dark blue\n3 4 2 2\nD CuCl 2– yellow\n4\nTwo correct for one mark, four correct for two marks, six correct for three marks, eight correct for four marks.\n\n1(d)(i) M1 (a species) that donates more than two lone pairs 2\nM2 to form dative / coordinate bonds to a metal atom or ion\n\n1(d)(ii) six atoms circled, 2N and 4O from different CO – 1\n2\n\n1(d)(iii) the number of co-ordinate bonds being formed by the metal ion 1\n\n1(d)(iv) ligand exchange 1\n\n1(d)(v) [FeEDTA]– > [CrEDTA]– > [PbEDTA]2– 1\nhighest concn lowest concn AND K of [FeEDTA]– is highest\nstab\n© UCLES 2023 Page 6 of 17\n\n1(e) M1 moles of Cr3+ = 2.096  10–4 in 25.0 cm3) 3\nM2 moles of Cr3+ = 8.384  10–4 (in 100.0 cm3)\nmoles of Cr (SO ) nH O = 8.384  10–4 / 2 = 4.192  10–4\n2 4 3 2\nM3 M of Cr (SO ) nH O = 0.2550 / 4.192  10–4 = 608.3\nr 2 4 3 2\nn = (608.3 – 392.3) / 18 = 12\n\n1(f) M1 E is different 2\nM2 different frequency (of light) is absorbed\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p06.png", "../answer-assets/9701_s23_ms_41-p07.png" ] }, { "id": "9701-2023-mj-41-q02", "question_id": "9701-2023-mj-41-q02", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 11, "status": "available", "reason": null, "text": "2(a) geometrical / cis-trans AND optical 1\n\n2(b) square planar 1\n© UCLES 2023 Page 7 of 17\n\n2(c) 3\n1 2\n3 4\nTwo correct for one mark, three correct for two marks, four correct for three marks.\n© UCLES 2023 Page 8 of 17\n\n2(d)(i) 3\nM1 two correct curly arrows\nM2 correct dipole\nM3 correct structure of intermediate\n\n2(d)(ii) use an excess of ammonia OR limiting amount of oxirane 1\n\n2(d)(iii) 2\nM1\nM2 elimination / dehydration / condensation\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p07.png", "../answer-assets/9701_s23_ms_41-p08.png", "../answer-assets/9701_s23_ms_41-p09.png" ] }, { "id": "9701-2023-mj-41-q03", "question_id": "9701-2023-mj-41-q03", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "3(a)(i) the power to which the concentration of a reactant is raised in the rate equation 1\n© UCLES 2023 Page 9 of 17\n\n3(a)(ii) 1\nthe order of reaction with respect to [IO –] 1\n3\nthe order of reaction with respect to [H+] 2\nthe order of reaction with respect to [I–] 2\nthe overall order of reaction 5\nAll correct for one mark\n\n3(a)(iii) 1\n\n3(a)(iv) M1 k = rate / [I–]2[IO –][H+]2 = (4.20  10-2)/(0.0252  0.04x0.0152) = 7.47  106 2\n3\nM2 units = mol–4 dm12 min–1\n\n3(a)(v) 0.0709 = k  0.12  [H+]2  0.01252 1\n[H+] = 2.25  10-2\n\n3(a)(vi) x = 102 / 1 = 100 1\n\n3(b) M1 step 1 Fe3+ + I–  FeI2+ 3\nM2 step 2 FeI2+ + I–  Fe2+ + I – OR FeI2+ + I–  FeI + AND slowest step = step 2\n2 2\nM3 step 3 Fe3+ + I –  Fe2+ + I OR FeI + + Fe3+  2Fe2+ + I\n2 2 2 2\n© UCLES 2023 Page 10 of 17", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p09.png", "../answer-assets/9701_s23_ms_41-p10.png" ] }, { "id": "9701-2023-mj-41-q04", "question_id": "9701-2023-mj-41-q04", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 10, "status": "available", "reason": null, "text": "4(a) M1 bond angle = 120° AND carbons are sp2 3\nM2  bonds are formed by end-on-end / head on / head to head / linear overlap of orbitals\nM3  bonds are formed by sideways / lateral overlap of p orbitals\n\n4(b)(i) 2\n[1] [1]\n\n4(b)(ii) M1 step 1 (CH ) CHBr and FeBr / Al Br 3\n3 2 3 3\nM2 step 2 conc HNO and conc H SO\n3 2 4\nM3 step 3 Sn and conc HCl\n\n4(c) 3,4,5-trimethylphenylamine 1\n\n4(d) 1\ncompound number of peaks observed\nW 6\nZ 6\nBoth correct for one mark\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p11.png" ] }, { "id": "9701-2023-mj-41-q05", "question_id": "9701-2023-mj-41-q05", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "5(a)(i) palladium, platinum and rhodium / Pt, Pd, Rh 1\n\n5(a)(ii) a catalyst in a different state / phase to the reactants / substrate 1\n\n5(b)(i) measure / degree of disorder / randomness of a system 1\nOR the number of possible arrangements of the particles and the energy in a system\n© UCLES 2023 Page 11 of 17\n\n5(b)(ii) M1 So = (192.8) + 213.8 – 238.2 – 188.8 4\nSo = -20.4 (J K–1 mol–1)\nM2 Ho = (–45.9) + (–393.5) – (–101.7) – (–241.8)\nHo = -95.9 (kJ mol–1)\nM3 Go = Ho – TSo\nM4 Go = –95.9 – (298  –0.0204) = –89.8 (kJ mol–1)\n\n5(c) 4(NH ) CO + 6NO  7N + 8H O + 4CO 1\n2 2 2 2 2 2\n\n5(d) 1\n\n5(e)(i) pK = –log K AND pH = –log [H+] 1\na a\n\n5(e)(ii) M1 [H+] = √0.120  2.00  10–4 = 4.89(317)  10–3 2\nM2 pH = 2.31\n\n5(e)(iii) % ionisation = 4.89  10-3/0.12  100 = 4.1% 1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p11.png", "../answer-assets/9701_s23_ms_41-p12.png" ] }, { "id": "9701-2023-mj-41-q06", "question_id": "9701-2023-mj-41-q06", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Polymerisation", "topic_slug": "9701-topic-35-polymerisation", "marks": 10, "status": "available", "reason": null, "text": "6(a)(i) carboxylic acid, amine, alkene 2\nTwo correct for one mark, three correct for two marks\n© UCLES 2023 Page 12 of 17\n\n6(a)(ii) 2\noptical\nTwo correct for one mark, three correct for two marks.\n\n6(b)(i) 1\n\n6(b)(ii) hydrolysis 1\n\n6(c)(i) 1\n© UCLES 2023 Page 13 of 17\n\n6(c)(ii) 2\nM1 amide linkage displayed correctly\nM2 rest of the structure correct\n\n6(c)(iii) condensation polymers can be hydrolysed 1\nQuestion Answer Marks", "source_pages": [ 12, 13, 14 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-35-polymerisation/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p12.png", "../answer-assets/9701_s23_ms_41-p13.png", "../answer-assets/9701_s23_ms_41-p14.png" ] }, { "id": "9701-2023-mj-41-q07", "question_id": "9701-2023-mj-41-q07", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 18, "status": "available", "reason": null, "text": "7(a) M1 diethylamine > ethylamine > ethanamide 4\nexplanation\nM2 basicity linked to ability of lone pair on N to accept a proton / H+\nM3 electron donating ethyl group increases electron density on N / makes lone pair more available for donation\nM4 lone pair of electrons on N is delocalised into C=O group\n\n7(b)(i) M1 resists change in pH 2\nM2 when a small amount of acid or alkali is added\n\n7(b)(ii) M1 H NCH(CH )COOH + H+  H N+CH(CH )COOH 2\n2 3 3 3\nM2 H NCH(CH )COOH + OH–  H NCH(CH )COO– + H O\n2 3 2 3 2\n\n7(c)(i) 1\n© UCLES 2023 Page 14 of 17\n\n7(c)(ii) 2\nM1 displayed peptide bond (between two amino acids)\nM2 rest of the structure correct\n\n7(d) 3\nM1 relative positions of the spots drawn\nM2 Ala is a zwitterion / neutral / at its isoelectric point (at pH 6) OR ala-glu AND glu are negatively charged\nM3 glu has lower M OR ala-glu has higher M\nr r\n© UCLES 2023 Page 15 of 17\n\n7(e)(i) 3\nnumber of 1H atoms number of protons\nchemical splitting\nresponsible for the on the adjacent\nshift (δ) pattern\npeak carbon(s)\n1.4 doublet 3 1\n3.5 singlet 3 0\n4.0 quartet 1 3\nThree correct for one mark, six correct for two marks, nine correct for three marks.\n\n7(e)(ii) one extra peak for NH group seen in CDCl , AND H exchanged for D in D O 1\n2 3 2\nQuestion Answer Marks", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p14.png", "../answer-assets/9701_s23_ms_41-p15.png", "../answer-assets/9701_s23_ms_41-p16.png" ] }, { "id": "9701-2023-mj-41-q08", "question_id": "9701-2023-mj-41-q08", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 9, "status": "available", "reason": null, "text": "8(a) 1\nenergy change always always either negative\npositive negative or positive\nlattice energy \nenthalpy of hydration \nenthalpy of solution \nAll correct for one mark\n\n8(b) The energy / enthalpy change when 1 mole of gaseous ions is dissolved in water 1\n\n8(c)(i) M1 use of correct six numbers only 3\n682.8 178.2 590 1145 111.9 324.6\nM2 2 used correctly with Br (2  111.9 and 2  324.6)\nM3 correct signs and evaluation to give –2170.6 kJ mol–1\n© UCLES 2023 Page 16 of 17\n\n8(c)(ii) M1 use of correct three numbers only 2170.6 103.1 and 1579 2\nM2 correct signs & evaluation –347 kJ mol–1\n\n8(c)(iii) M1 Br– has a smaller ionic radius 2\nM2 Br– has stronger attractive forces with water molecules\n© UCLES 2023 Page 17 of 17", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_41-p16.png", "../answer-assets/9701_s23_ms_41-p17.png" ] }, { "id": "9701-2023-mj-42-q01", "question_id": "9701-2023-mj-42-q01", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 13, "status": "available", "reason": null, "text": "1(a)(i) 1\nARROW 1 starts from O– to the C–O bond\nAND\nARROW 2 starts at the C-O bond to other O– ion\n\n1(a)(ii) M1 increases (down the group) 3\nM2 (cat)ionic radius / ion size increases (down the group)\nOR charge density of M2+ decreases\nM3 less polarisation / distortion\nof anion / carbonate ion / CO (2)–\n3\n\n1(b)(i) M1 energy released when one mole of a ionic solid / compound is formed 2\nM2 from gas (phase) ion(s) / gaseous ion(s) (under standard conditions)\n\n1(b)(ii)  (H / it) becomes more positive/less negative 2\ndecomp\n(down the group)\n size /(ionic) radii of oxide ion is smaller (than carbonate ion) ORA\n so H of oxides becomes ORA\nlatt\nless exothermic faster OR less negative faster\nOR changes more OR changes faster\nAny two [1], all three [2]\n\n1(c)(i) 2MnO – + 6H+ + 5SO 2–  2Mn2+ + 3H O + 5SO 2– 1\n4 3 2 4\n© UCLES 2023 Page 6 of 19\n\n1(c)(ii) M1 M2 3\nany two bullets [1] or all four [2]\n moles MnO – = 0.025  22.40 / 1000 = 5.6  10–4\n4\n moles SO 2– = 5.6  10–4  5 / 2 = 1.4  10–3 (in 25 cm3) ecf from (c)(i) and bullet 1\n3\n moles SO 2– = 1.4  10–2 (in 250 cm3) ecf bullet 2\n3\n mass K SO = 1.4  10–2  158.3 = 2.2162 g ecf bullet 3\n2 3\nOR moles K SO (if 100% pure) = 3.40 ÷ 158.3 = 0.02148\n2 3\nM3\n% purity = 100  2.2162 / 3.40 = 65.2 / 65.3 % ecf min 2sf\nOR\n% purity = 100  0.014 / 0.02148 = 65.2 / 65.3 %\n\n1(d) S() tetrahedral 1\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p06.png", "../answer-assets/9701_s23_ms_42-p07.png" ] }, { "id": "9701-2023-mj-42-q02", "question_id": "9701-2023-mj-42-q02", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "2(a) Any two from: 1\n they have variable / multiple oxidation states OWTTE\n they behave as catalysts\n they form complex ions / complexes\n they form coloured compounds/ions\n\n2(b)(i) (is a molecule or ion formed by a central) metal atom / metal ion bonded / surrounded by (one or more) ligands 1\n© UCLES 2023 Page 7 of 19\n\n2(b)(ii) 4\nformula of colour / state of\nreagent added type of reaction\nspecies formed species formed\nbrown solution\nan excess of ligand\n[Co(NH ) ]2+ ALLOW yellow-\nNH (aq) 3 6 exchange\n3 brown solution\nan excess of ligand\n[CoCl ]2– blue solution\nconcentrated HCl 4 exchange\nCo(OH)\n2 blue ppt.\nNaOH(aq) OR precipitation\nALLOW pink ppt\nCo(OH) (H O)\n2 2 4\nformula and colour / state: any two [1], any four [2], all six [3]\ntype of reaction: all three correct [1]\n\n2(c)(i) M1 (a species) that donates / uses two lone pairs 2\nM2 to form a two dative / coordinate bond to a metal atom / metal ion\n© UCLES 2023 Page 8 of 19\n\n2(c)(ii) Any four from: 3\nAny two [1], any three [2], all four [3]\n\n2(c)(iii) oxidation state of cobalt (+)3 1\nAND type of stereoisomerism cis-trans OR geometric(al) OR optical\n© UCLES 2023 Page 9 of 19", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p07.png", "../answer-assets/9701_s23_ms_42-p08.png", "../answer-assets/9701_s23_ms_42-p09.png" ] }, { "id": "9701-2023-mj-42-q03", "question_id": "9701-2023-mj-42-q03", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 12, "status": "available", "reason": null, "text": "3(a) 1\nalways always can be either\nenergy change\npositive negative negative or positive\nbond energy \nenthalpy change of atomisation \nenthalpy change of formation \n\n3(b) M1 (enthalpy change) when one mole of gaseous atoms is produced IGNORE energy released 2\nM2 from its element(s) in its standard state / standard conditions / 298 K AND 1 atm\n\n3(c) M1 use of correct six numbers only –31 / 285 / 731 / –141 / 798 / 496 3\nM2 2  used correctly with Ag (2  285 (570) and 2  731 (1462)) AND 0.5 with O=O (496 (248))\nM3 correct signs and evaluation\n–31 = (2  285) + (2  731) + (–141) + (798) + x + (0.5  496)\nx = –2968 kJ mol–1\n\n3(d)  Ag Se Ag S Ag O OWTTE 2\n2 2 2\nleast exothermic most exothermic\n charge density of anion decreases down the group ORA\n/ radius/size of anion increases down the group\n/ Se2– largest radius / O2– smallest radius / O has smallest ionic radius\n less attraction between the ions / ionic bond gets weaker (with Ag Se) ORA\n2\nAny two [1], all three [2]\n\n3(e)(i) (K = ) [Ag+]2[SO 2–] [1] 1\nsp 3\n© UCLES 2023 Page 10 of 19\n\n3(e)(ii) x = 3√1.5  10–14 / 4 = 1.55  10–5 (mol dm-3) 1\n[Ag+] = 1.55  10–5  2 = 3.11  10–5 (mol dm-3) min 2sf ecf (e)(i)\n\n3(f)  feasibility / it increases as temperature increases ORA 2\n S is positive / S is >0 / entropy change is positive (and H is positive)\nOR –TS becomes more negative / TS becomes more positive\n as G becomes / is negative / G < 0\nAny two [1], all three [2]\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p10.png", "../answer-assets/9701_s23_ms_42-p11.png" ] }, { "id": "9701-2023-mj-42-q04", "question_id": "9701-2023-mj-42-q04", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "4(a)(i) the sum/total of the power / exponent to which a concentration of a reactant is raised in the rate equation / law 1\n\n4(a)(ii) M1 (expt 1 and 2) when [Fe3+]  3, rate  3 so first order w.r.t. Fe3+ 2\nM2\n(expt 1 and 3) when [Fe3+]  2, [I–]  2, rate  8 so second order w.r.t. I–\nOR\n(expt 2 and 3) when [Fe3+] ÷ 1.5, [I–]  2, rate  2.7 so second order w.r.t. I–\n\n4(a)(iii) (rate =) k[Fe3+][I–]2 [1] ecf (a)(ii) 1\n\n4(a)(iv) k = rate / [Fe3+][I–]2 = (2.64  10–4) / (0.04  0.022) 2\nk = 16.5 min 2sf ecf\nunits = mol-2 dm6 s–1 ecf\n\n4(a)(v) (k and rate of reaction) both increase 1\n© UCLES 2023 Page 11 of 19\n\n4(b)(i) step 1 as this has one mole of H O and one mole of I– 1\n2 2\nOR\nstep 1 as correct stoichiometry / coefficients of 1 H O and 1 I–\n2 2\nOR\nstep 1 as number of moles of each reactant are consistent with rate equation / their orders\n\n4(b)(ii) step 1 I oxid no –1  +1 AND O oxid no –1  –2 1\nOR\nstep 3 I oxid no –1  0 AND I oxid no +1  0\n\n4(b)(iii) intermediate AND formed (in step 2) and used up (in step 3) 1\nALLOW oxidising agent (in step 3) AND oxidises I– (to I )\n2\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p11.png", "../answer-assets/9701_s23_ms_42-p12.png" ] }, { "id": "9701-2023-mj-42-q05", "question_id": "9701-2023-mj-42-q05", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 14, "status": "available", "reason": null, "text": "5(a)(i) 3\nM1 M2 M3 BOTH\n\n5(a)(ii) M1 reaction 1: hydrogenation / reduction 2\nM2 mechanism 2: (free) radical substitution\n\n5(b) the substitution product is stabilised by delocalization of -electrons / by -electrons in the ring 1\nOR\nthe addition product is not stabilised by delocalisation (of -)electrons ALLOW addition product will remove -electron\ndelocalised system\n© UCLES 2023 Page 12 of 19\n\n5(c)(i) 3\nM1 curly arrow from inside the hexagon to S atom\nM2 structure of the intermediate\nM3 curly arrow from C-H bond into the ring AND formation/loss of H+\n\n5(c)(ii) 1\n\n5(d)(i) CH COOH + SOBr  CH COBr + HBr + SO 1\n3 2 3 2\n© UCLES 2023 Page 13 of 19\n\n5(d)(ii) M1 acyl bromides > acyl chlorides > alkyl chlorides 3\neasiest hardest\nM2 M3 Any two from:\nAcyl bromides easiest\n in acyl bromides C–Br is much weaker ORA AND due to less orbital overlap / Br having a larger atomic radii\nAcyl halides easier than alkyl chlorides\n carbon in C–Cl / C–Br / C–X bond is more +/electron deficient\nOR C–Cl / C–Br bond is weaker (than C-Cl in alkyl chlorides)\nAND attached to an oxygen atom / two electronegative atoms / electron withdrawing C=O group ORA\nalkyl chlorides hardest\n in alkyl chlorides C–Cl bond strengthened\nAND by positive inductive effect / electron donating effect of alkyl / R group\nQuestion Answer Marks", "source_pages": [ 12, 13, 14 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p12.png", "../answer-assets/9701_s23_ms_42-p13.png", "../answer-assets/9701_s23_ms_42-p14.png" ] }, { "id": "9701-2023-mj-42-q06", "question_id": "9701-2023-mj-42-q06", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 8, "status": "available", "reason": null, "text": "6(a)(i) 5 / five 1\n\n6(a)(ii) M1 benefit: 2\nhigher biological efficiency / activity (of the drug)\nOR less side effects\nOR smaller dose required (as drug more potent)\nM2 disadvantage:\nthe need to separate (a racemic mixture into a single stereoisomer)\nOR lower yield (of biologically active molecule/product)\nOR need a chiral catalyst / enzyme (in the synthesis so expensive)\n\n6(b)(i) carboxylic acid/carboxyl, ester, amide, amine 2\nAny two [1], all four [2]\n© UCLES 2023 Page 14 of 19\n\n6(b)(ii) 3\n3  [1] for each structure\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p14.png", "../answer-assets/9701_s23_ms_42-p15.png" ] }, { "id": "9701-2023-mj-42-q07", "question_id": "9701-2023-mj-42-q07", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 8, "status": "available", "reason": null, "text": "7(a) M1 p-orbital / lone pair from O atom is overlaps / delocalised into the ring 3\nM2 greater  electron density around the ring OR makes the ring more electron rich / positions 2,4,6 more electron rich\nM3 polarises electrophiles more easily\n\n7(b) iodobenzene AND as Br is more electronegative (than I / I+/I+ in the electrophile) 1\n\n7(c)(i) C H O / C H OH + Na  C H O–Na+ + ½H 1\n6 6 6 5 6 5 2\n\n7(c)(ii) 1\n© UCLES 2023 Page 15 of 19\n\n7(c)(iii) M1 reaction 1: aqueous HNO / dilute HNO 2\n3 3\nM2 reaction 2: alkaline/ NaOH AND C H N + (Cl–)\n6 5 2\nOR\nalkaline / NaOH AND benzene / phenyl diazonium ion / salt\nQuestion Answer Marks", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p15.png", "../answer-assets/9701_s23_ms_42-p16.png" ] }, { "id": "9701-2023-mj-42-q08", "question_id": "9701-2023-mj-42-q08", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Halogen compounds", "topic_slug": "9701-topic-31-halogen-compounds", "marks": 15, "status": "available", "reason": null, "text": "8(a)  chlorobenzene is less reactive than chloroethane OWTTE 2\n p-orbital / lone pair on Cl will overlap / delocalise into the ring\n due to partial double C-Cl bond OWTTE\nOR C-Cl bond strengthened (more)\nAny two [1], all three [2]\n\n8(b)(i) ethyl 3-chloropropanoate 1\n© UCLES 2023 Page 16 of 19\n\n8(b)(ii) W 4\nX\nO\nOH\nHO ONa\nY Z\nO O\nHO H N\n2\nOH O\nO\n4  [1] for each structure\n\n8(b)(iii) M1 step 1 NaCN / KCN in ethanol (heat) 2\nM2 step 2 H + Ni / H + Pt ALLOW LiAlH\n2 2 4\n\n8(c)(i) (because solvent / D / CDCl ) doesn’t give a signal / peak / absorption ORA 1\n3\n© UCLES 2023 Page 17 of 19\n\n8(c)(ii) 4\nchemical splitting number of 1H atoms\nenvironment of proton\nshift δ / ppm pattern responsible for the peak\n1.2 alkyl / alkane / (R-)CH triplet 3\n(3)\n2.8 alkyl next to C=O / triplet 2\nCH C=O\n(2)\n3.7 alkyl next to triplet 2\nelectronegative atom /\nCH -Cl\n(2)\n3.9 alkyl next to quartet / 2\nelectronegative atom / quadruplet\nCH -O\n(2)\nAny three [1], any six [2], any nine [3], all twelve [4]\n\n8(c)(iii) ( = 3.9) three H on neighbouring / adjacent C / it’s next to a CH 1\n3\n© UCLES 2023 Page 18 of 19", "source_pages": [ 16, 17, 18 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-31-halogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p16.png", "../answer-assets/9701_s23_ms_42-p17.png", "../answer-assets/9701_s23_ms_42-p18.png" ] }, { "id": "9701-2023-mj-42-q09", "question_id": "9701-2023-mj-42-q09", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 8, "status": "available", "reason": null, "text": "9(a) potential difference / voltage between the two half-cells / two electrodes (in a cell) under standard conditions 1\n\n9(b) 3\nsalt bridge voltmeter complete circuit [wires/salt bridge in/touching both solutions]\nCl Cl – good delivery system (no arrow required)\n2\nPt Pt Fe2+ and Fe3+\nAny three [1], any six [2], all nine [3]\n\n9(c) M1 Go ) = – nFE o OR –2  96500  0.59 2\ncell\n\nM2 Go = –2  96500  0.59 = –113870 J mol-1\nGo = –114 kJ mol-1 min 3sf ecf\n\n9(d)(i) E = 0.77 + (0.059 / 1) log (1/0.15) use of z = 1 1\nE = 0.82 V\n\n9(d)(ii) E = 0.59 + 0.77 – (answer to (d)(i)) = 0.54 V ecf 1\ncell\n© UCLES 2023 Page 19 of 19", "source_pages": [ 19 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_42-p19.png" ] }, { "id": "9701-2023-mj-43-q01", "question_id": "9701-2023-mj-43-q01", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 19, "status": "available", "reason": null, "text": "1(a) M1 increases (down the group) 3\nM2 radius / size of (cat)ion / M2+ increases\nM3 less polarisation / distortion of anion / nitrate ion / NO – OR less weakening of N–O / N=O (bond)\n3\n\n1(b) Cu(NO )  CuO + 2NO + ½O 1\n3 2 2 2\n\n1(c) 4\ncopper-\nformula of copper-containing colour copper-\ncontaining\nspecies formed containing formed\nspecies\nA [Cu(H O) ]2+ (pale) blue\n2 6\nB Cu(H O) (OH) or Cu(OH) (pale) blue\n2 4 2 2\nC [Cu(NH ) (H O) ]2+ dark blue\n3 4 2 2\nD CuCl 2– yellow\n4\nTwo correct for one mark, four correct for two marks, six correct for three marks, eight correct for four marks.\n\n1(d)(i) M1 (a species) that donates more than two lone pairs 2\nM2 to form dative / coordinate bonds to a metal atom or ion\n\n1(d)(ii) six atoms circled, 2N and 4O from different CO – 1\n2\n\n1(d)(iii) the number of co-ordinate bonds being formed by the metal ion 1\n\n1(d)(iv) ligand exchange 1\n\n1(d)(v) [FeEDTA]– > [CrEDTA]– > [PbEDTA]2– 1\nhighest concn lowest concn AND K of [FeEDTA]– is highest\nstab\n© UCLES 2023 Page 6 of 17\n\n1(e) M1 moles of Cr3+ = 2.096  10–4 in 25.0 cm3) 3\nM2 moles of Cr3+ = 8.384  10–4 (in 100.0 cm3)\nmoles of Cr (SO ) nH O = 8.384  10–4 / 2 = 4.192  10–4\n2 4 3 2\nM3 M of Cr (SO ) nH O = 0.2550 / 4.192  10–4 = 608.3\nr 2 4 3 2\nn = (608.3 – 392.3) / 18 = 12\n\n1(f) M1 E is different 2\nM2 different frequency (of light) is absorbed\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p06.png", "../answer-assets/9701_s23_ms_43-p07.png" ] }, { "id": "9701-2023-mj-43-q02", "question_id": "9701-2023-mj-43-q02", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 11, "status": "available", "reason": null, "text": "2(a) geometrical / cis-trans AND optical 1\n\n2(b) square planar 1\n© UCLES 2023 Page 7 of 17\n\n2(c) 3\n1 2\n3 4\nTwo correct for one mark, three correct for two marks, four correct for three marks.\n© UCLES 2023 Page 8 of 17\n\n2(d)(i) 3\nM1 two correct curly arrows\nM2 correct dipole\nM3 correct structure of intermediate\n\n2(d)(ii) use an excess of ammonia OR limiting amount of oxirane 1\n\n2(d)(iii) 2\nM1\nM2 elimination / dehydration / condensation\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p07.png", "../answer-assets/9701_s23_ms_43-p08.png", "../answer-assets/9701_s23_ms_43-p09.png" ] }, { "id": "9701-2023-mj-43-q03", "question_id": "9701-2023-mj-43-q03", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "3(a)(i) the power to which the concentration of a reactant is raised in the rate equation 1\n© UCLES 2023 Page 9 of 17\n\n3(a)(ii) 1\nthe order of reaction with respect to [IO –] 1\n3\nthe order of reaction with respect to [H+] 2\nthe order of reaction with respect to [I–] 2\nthe overall order of reaction 5\nAll correct for one mark\n\n3(a)(iii) 1\n\n3(a)(iv) M1 k = rate / [I–]2[IO –][H+]2 = (4.20  10-2)/(0.0252  0.04x0.0152) = 7.47  106 2\n3\nM2 units = mol–4 dm12 min–1\n\n3(a)(v) 0.0709 = k  0.12  [H+]2  0.01252 1\n[H+] = 2.25  10-2\n\n3(a)(vi) x = 102 / 1 = 100 1\n\n3(b) M1 step 1 Fe3+ + I–  FeI2+ 3\nM2 step 2 FeI2+ + I–  Fe2+ + I – OR FeI2+ + I–  FeI + AND slowest step = step 2\n2 2\nM3 step 3 Fe3+ + I –  Fe2+ + I OR FeI + + Fe3+  2Fe2+ + I\n2 2 2 2\n© UCLES 2023 Page 10 of 17", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p09.png", "../answer-assets/9701_s23_ms_43-p10.png" ] }, { "id": "9701-2023-mj-43-q04", "question_id": "9701-2023-mj-43-q04", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 10, "status": "available", "reason": null, "text": "4(a) M1 bond angle = 120° AND carbons are sp2 3\nM2  bonds are formed by end-on-end / head on / head to head / linear overlap of orbitals\nM3  bonds are formed by sideways / lateral overlap of p orbitals\n\n4(b)(i) 2\n[1] [1]\n\n4(b)(ii) M1 step 1 (CH ) CHBr and FeBr / Al Br 3\n3 2 3 3\nM2 step 2 conc HNO and conc H SO\n3 2 4\nM3 step 3 Sn and conc HCl\n\n4(c) 3,4,5-trimethylphenylamine 1\n\n4(d) 1\ncompound number of peaks observed\nW 6\nZ 6\nBoth correct for one mark\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p11.png" ] }, { "id": "9701-2023-mj-43-q05", "question_id": "9701-2023-mj-43-q05", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "5(a)(i) palladium, platinum and rhodium / Pt, Pd, Rh 1\n\n5(a)(ii) a catalyst in a different state / phase to the reactants / substrate 1\n\n5(b)(i) measure / degree of disorder / randomness of a system 1\nOR the number of possible arrangements of the particles and the energy in a system\n© UCLES 2023 Page 11 of 17\n\n5(b)(ii) M1 So = (192.8) + 213.8 – 238.2 – 188.8 4\nSo = -20.4 (J K–1 mol–1)\nM2 Ho = (–45.9) + (–393.5) – (–101.7) – (–241.8)\nHo = -95.9 (kJ mol–1)\nM3 Go = Ho – TSo\nM4 Go = –95.9 – (298  –0.0204) = –89.8 (kJ mol–1)\n\n5(c) 4(NH ) CO + 6NO  7N + 8H O + 4CO 1\n2 2 2 2 2 2\n\n5(d) 1\n\n5(e)(i) pK = –log K AND pH = –log [H+] 1\na a\n\n5(e)(ii) M1 [H+] = √0.120  2.00  10–4 = 4.89(317)  10–3 2\nM2 pH = 2.31\n\n5(e)(iii) % ionisation = 4.89  10-3/0.12  100 = 4.1% 1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p11.png", "../answer-assets/9701_s23_ms_43-p12.png" ] }, { "id": "9701-2023-mj-43-q06", "question_id": "9701-2023-mj-43-q06", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Polymerisation", "topic_slug": "9701-topic-35-polymerisation", "marks": 10, "status": "available", "reason": null, "text": "6(a)(i) carboxylic acid, amine, alkene 2\nTwo correct for one mark, three correct for two marks\n© UCLES 2023 Page 12 of 17\n\n6(a)(ii) 2\noptical\nTwo correct for one mark, three correct for two marks.\n\n6(b)(i) 1\n\n6(b)(ii) hydrolysis 1\n\n6(c)(i) 1\n© UCLES 2023 Page 13 of 17\n\n6(c)(ii) 2\nM1 amide linkage displayed correctly\nM2 rest of the structure correct\n\n6(c)(iii) condensation polymers can be hydrolysed 1\nQuestion Answer Marks", "source_pages": [ 12, 13, 14 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-35-polymerisation/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p12.png", "../answer-assets/9701_s23_ms_43-p13.png", "../answer-assets/9701_s23_ms_43-p14.png" ] }, { "id": "9701-2023-mj-43-q07", "question_id": "9701-2023-mj-43-q07", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 18, "status": "available", "reason": null, "text": "7(a) M1 diethylamine > ethylamine > ethanamide 4\nexplanation\nM2 basicity linked to ability of lone pair on N to accept a proton / H+\nM3 electron donating ethyl group increases electron density on N / makes lone pair more available for donation\nM4 lone pair of electrons on N is delocalised into C=O group\n\n7(b)(i) M1 resists change in pH 2\nM2 when a small amount of acid or alkali is added\n\n7(b)(ii) M1 H NCH(CH )COOH + H+  H N+CH(CH )COOH 2\n2 3 3 3\nM2 H NCH(CH )COOH + OH–  H NCH(CH )COO– + H O\n2 3 2 3 2\n\n7(c)(i) 1\n© UCLES 2023 Page 14 of 17\n\n7(c)(ii) 2\nM1 displayed peptide bond (between two amino acids)\nM2 rest of the structure correct\n\n7(d) 3\nM1 relative positions of the spots drawn\nM2 Ala is a zwitterion / neutral / at its isoelectric point (at pH 6) OR ala-glu AND glu are negatively charged\nM3 glu has lower M OR ala-glu has higher M\nr r\n© UCLES 2023 Page 15 of 17\n\n7(e)(i) 3\nnumber of 1H atoms number of protons\nchemical splitting\nresponsible for the on the adjacent\nshift (δ) pattern\npeak carbon(s)\n1.4 doublet 3 1\n3.5 singlet 3 0\n4.0 quartet 1 3\nThree correct for one mark, six correct for two marks, nine correct for three marks.\n\n7(e)(ii) one extra peak for NH group seen in CDCl , AND H exchanged for D in D O 1\n2 3 2\nQuestion Answer Marks", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p14.png", "../answer-assets/9701_s23_ms_43-p15.png", "../answer-assets/9701_s23_ms_43-p16.png" ] }, { "id": "9701-2023-mj-43-q08", "question_id": "9701-2023-mj-43-q08", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 9, "status": "available", "reason": null, "text": "8(a) 1\nenergy change always always either negative\npositive negative or positive\nlattice energy \nenthalpy of hydration \nenthalpy of solution \nAll correct for one mark\n\n8(b) The energy / enthalpy change when 1 mole of gaseous ions is dissolved in water 1\n\n8(c)(i) M1 use of correct six numbers only 3\n682.8 178.2 590 1145 111.9 324.6\nM2 2 used correctly with Br (2  111.9 and 2  324.6)\nM3 correct signs and evaluation to give –2170.6 kJ mol–1\n© UCLES 2023 Page 16 of 17\n\n8(c)(ii) M1 use of correct three numbers only 2170.6 103.1 and 1579 2\nM2 correct signs & evaluation –347 kJ mol–1\n\n8(c)(iii) M1 Br– has a smaller ionic radius 2\nM2 Br– has stronger attractive forces with water molecules\n© UCLES 2023 Page 17 of 17", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_43-p16.png", "../answer-assets/9701_s23_ms_43-p17.png" ] }, { "id": "9701-2023-mj-51-q01", "question_id": "9701-2023-mj-51-q01", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 9, "status": "available", "reason": null, "text": "1(a)(i) aqueous AND water has greater density (than ethoxyethane so it forms the lower layer) 1\n\n1(a)(ii) (10 cm3) (volumetric) pipette 1\n\n1(a)(iii) the acid transfers / dissolves into the water (before titration / reaction with sodium hydroxide) 1\n\n1(b)(i) 27.25 1\nn (NaOH) in titre = 0.5 (= 0.0136250)\n1000\nn(HOOCCH CH COOH(aq)) = 0.0136250 / 2 = 0.0068125\n2 2\n27.25   1000\n[HOOCCH 2 CH 2 COOH(aq)] =  0.5  2  = 0.6813 mol dm–3 (0.68125 mol dm–3)\n 1000   10\nat least 2 significant figures\n\n1(b)(ii) M1 [HOOCCH CH COOH(ethoxyethane)] = 0.1125 mol dm–3 2\n2 2\n(22.50  0.1 / 1000 ÷ 10 / 1000) / 2)\nM2 K = M1 / (b)(i)\npc\nExpected answer = 0.1125 / 0.68125 = 0.165(1376)\n(0.1614–0.1662)\nat least 2 significant figures\n\n1(b)(iii) it is not possible to pipette all (30 cm3) of the aqueous layer from the beaker (as some will always be left behind) 1\n\n1(b)(iv) use a larger volume (>30cm3) of water (in step 1) 1\n\n1(b)(v) K value would be smaller / decrease 1\npc\nAND\nbecause (equilibrium had not been attained and) more butanedioic acid remains in aqueous layer\n© UCLES 2023 Page 5 of 8", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_51-p05.png" ] }, { "id": "9701-2023-mj-51-q02", "question_id": "9701-2023-mj-51-q02", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 7, "status": "available", "reason": null, "text": "2(a) the application of the hydrolysed samples (to the paper) is more controlled 1\n\n2(b) to make the amino acids / spots visible 1\n\n2(c) lysine, tryptophan and leucine 2\ntwo correct for one mark, three correct for two marks\n\n2(d) two (of the three) amino acids (that compose the tripeptide) are the same 1\n\n2(e)(i) (the spots of two amino acids) have similar (not the same) R values 1\nf\n\n2(e)(ii) use a longer stationary phase / (chromatography) paper 1\nOR\nuse a different solvent / mobile phase\n© UCLES 2023 Page 6 of 8", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_51-p06.png" ] }, { "id": "9701-2023-mj-51-q03", "question_id": "9701-2023-mj-51-q03", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 3, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "3(a) M1 add (a small volume of) distilled water (to the 50 cm3 beaker) 3\nAND\ndissolve the (aromatic diazonium) compound / solid\nM2 transfer the solution and washings (using a funnel)\nAND\ninto a 100 cm3 volumetric flask\nM3 make up to the (calibration) mark / line with (distilled) water\nAND\nthen mix the solution (by inverting the flask)\n\n3(b)(i) thermostatically controlled water bath 1\n\n3(b)(ii) volume of (nitrogen) gas (produced) 1\n\n3(b)(iii) not possible to weigh apparatus (while temperature is maintained) in water bath 1\n\n3(c)(i) 1\ntime / min volume, V / cm3 V – V / cm3\nt final t\n0 0.0 57.2\n5 17.3 39.9\n9 27.0 30.2\n16 39.5 17.7\n21 42.6 14.6\n28 49.0 8.2\n36 52.8 4.4\nfinal 57.2 0.0\n© UCLES 2023 Page 7 of 8\n\n3(c)(ii) M1 all points plotted correctly 2\nM2 smooth curve of best fit line drawn passing close to all points except marked anomaly\n\n3(c)(iii) point most anomalous to the plotted line of best fit circled. 1\n\n3(c)(iv) (for anomalous point above the line of best fit) 1\nmeasurement (of volume) was made before the time recorded in the results table\n(for anomalous point below the line of best fit)\nmeasurement (of volume) was made after the time recorded in the results table\n\n3(c)(v) M1 two sets of co-ordinates from the line of best fit correctly recorded for first t AND second t in the form (x, y) 2\n½ ½\nM2 both half-lives correctly calculated from co-ordinates\n\n3(c)(vi) first order AND half-lives are constant (within experimental error) 1\n© UCLES 2023 Page 8 of 8", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_51-p07.png", "../answer-assets/9701_s23_ms_51-p08.png" ] }, { "id": "9701-2023-mj-52-q01", "question_id": "9701-2023-mj-52-q01", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 17, "status": "available", "reason": null, "text": "1(a)(i) 1.242 g / 1.08 = 1.15 cm3 1\n\n1(a)(ii) burette 1\n\n1(a)(iii) 1  100 / 1000  40 = 4.0(0) g 1\n\n1(b) M1 dissolve the solid / sodium hydroxide (in the beaker) (using a small volume of) (distilled) water 3\nM2 transfer / add to a (100 cm3 volumetric) flask AND rinse (with distilled water)\nM3 top the 100 cm3 volumetric flask up to mark (with distilled water)\n\n1(c) M1 mol C H COOCH = 1.242 / 136 = 0.00913 AND mol of NaOH = 0.100  (10 / 1000) = 0.001(00) 2\n6 5 3\nM2 mass of benzoic acid decreases AND NaOH no longer in excess / NaOH (now) the limiting (reagent) / not all\nC H COOCH reacts / statement that mol NaOH < mol ester\n6 5 3\n\n1(d)(i) allows heating to increase the rate of reaction without the loss of volatile substances 1\n\n1(d)(ii) (substances) are flammable 1\n\n1(d)(iii) to remove the anti-bumping granules (from the reaction mixture) 1\n\n1(e)(i) add (a little) (named) (acid–base) indicator AND the reaction mixture shows the colour for a (strong) acid in that indicator 1\n\n1(e)(ii) precipitate forms 1\n\n1(f) benzoic acid is less soluble in cold water 1\n\n1(g) dry the precipitate and reweigh until constant mass 1\n\n1(h)(i) 1.242  122 / 136 = 1.11 1\n\n1(h)(ii) 74.0(%) 1\nQuestion Answer Marks", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_52-p06.png" ] }, { "id": "9701-2023-mj-52-q02", "question_id": "9701-2023-mj-52-q02", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "2(a)(i) 1.995  10–7 (mol dm–3) or 2.00  10–7 (mol dm–3) 1\n© UCLES 2023 Page 6 of 7\n\n2(a)(ii) 3.98  10–14 (mol2 dm–6) 1\n\n2(a)(iii) as temperature increases K increases 1\nw\n\n2(b)(i) M1 1/T all correct to 3 significant figures AND standard form 2\nM2 log K all correct to 2 decimal places\nw\nT 1 / T K log K\nw w\n283 3.53  10–3 2.93  10–15 – 14.53\n293 3.41  10–3 6.81  10–15 – 14.17\n303 3.30  10–3 1.47  10–14 – 13.83\n313 3.19  10–3 2.92  10–14 – 13.53\n323 3.10  10–3 5.48  10–14 – 13.26\n333 3.00  10–3 1.09  10–13 – 12.96\n343 2.92  10–3 1.45  10–13 – 12.84\n\n2(b)(ii) M1 seven points plotted correctly 2\nM2 line of best fit drawn\n\n2(b)(iii) plot at 2.92  10 –3, –12.84 1\n\n2(b)(iv) pH was recorded at a lower temperature than 343 K 1\n\n2(b)(v) M1 co-ordinates read and recorded correctly 2\nM2 gradient calculated to three significant figures\n\n2(b)(vi) M1 gradient = –H / 2.303R OR H = –gradient  2.303  8.31 OR any correct rearrangement of this 2\nM2 correct calculation of H from 2(b)(v)\n© UCLES 2023 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_52-p06.png", "../answer-assets/9701_s23_ms_52-p07.png" ] }, { "id": "9701-2023-mj-53-q01", "question_id": "9701-2023-mj-53-q01", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 9, "status": "available", "reason": null, "text": "1(a)(i) aqueous AND water has greater density (than ethoxyethane so it forms the lower layer) 1\n\n1(a)(ii) (10 cm3) (volumetric) pipette 1\n\n1(a)(iii) the acid transfers / dissolves into the water (before titration / reaction with sodium hydroxide) 1\n\n1(b)(i) 27.25 1\nn (NaOH) in titre = 0.5 (= 0.0136250)\n1000\nn(HOOCCH CH COOH(aq)) = 0.0136250 / 2 = 0.0068125\n2 2\n27.25   1000\n[HOOCCH 2 CH 2 COOH(aq)] =  0.5  2  = 0.6813 mol dm–3 (0.68125 mol dm–3)\n 1000   10\nat least 2 significant figures\n\n1(b)(ii) M1 [HOOCCH CH COOH(ethoxyethane)] = 0.1125 mol dm–3 2\n2 2\n(22.50  0.1 / 1000 ÷ 10 / 1000) / 2)\nM2 K = M1 / (b)(i)\npc\nExpected answer = 0.1125 / 0.68125 = 0.165(1376)\n(0.1614–0.1662)\nat least 2 significant figures\n\n1(b)(iii) it is not possible to pipette all (30 cm3) of the aqueous layer from the beaker (as some will always be left behind) 1\n\n1(b)(iv) use a larger volume (>30cm3) of water (in step 1) 1\n\n1(b)(v) K value would be smaller / decrease 1\npc\nAND\nbecause (equilibrium had not been attained and) more butanedioic acid remains in aqueous layer\n© UCLES 2023 Page 5 of 8", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_53-p05.png" ] }, { "id": "9701-2023-mj-53-q02", "question_id": "9701-2023-mj-53-q02", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 7, "status": "available", "reason": null, "text": "2(a) the application of the hydrolysed samples (to the paper) is more controlled 1\n\n2(b) to make the amino acids / spots visible 1\n\n2(c) lysine, tryptophan and leucine 2\ntwo correct for one mark, three correct for two marks\n\n2(d) two (of the three) amino acids (that compose the tripeptide) are the same 1\n\n2(e)(i) (the spots of two amino acids) have similar (not the same) R values 1\nf\n\n2(e)(ii) use a longer stationary phase / (chromatography) paper 1\nOR\nuse a different solvent / mobile phase\n© UCLES 2023 Page 6 of 8", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_53-p06.png" ] }, { "id": "9701-2023-mj-53-q03", "question_id": "9701-2023-mj-53-q03", "subject": "9701", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 3, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "3(a) M1 add (a small volume of) distilled water (to the 50 cm3 beaker) 3\nAND\ndissolve the (aromatic diazonium) compound / solid\nM2 transfer the solution and washings (using a funnel)\nAND\ninto a 100 cm3 volumetric flask\nM3 make up to the (calibration) mark / line with (distilled) water\nAND\nthen mix the solution (by inverting the flask)\n\n3(b)(i) thermostatically controlled water bath 1\n\n3(b)(ii) volume of (nitrogen) gas (produced) 1\n\n3(b)(iii) not possible to weigh apparatus (while temperature is maintained) in water bath 1\n\n3(c)(i) 1\ntime / min volume, V / cm3 V – V / cm3\nt final t\n0 0.0 57.2\n5 17.3 39.9\n9 27.0 30.2\n16 39.5 17.7\n21 42.6 14.6\n28 49.0 8.2\n36 52.8 4.4\nfinal 57.2 0.0\n© UCLES 2023 Page 7 of 8\n\n3(c)(ii) M1 all points plotted correctly 2\nM2 smooth curve of best fit line drawn passing close to all points except marked anomaly\n\n3(c)(iii) point most anomalous to the plotted line of best fit circled. 1\n\n3(c)(iv) (for anomalous point above the line of best fit) 1\nmeasurement (of volume) was made before the time recorded in the results table\n(for anomalous point below the line of best fit)\nmeasurement (of volume) was made after the time recorded in the results table\n\n3(c)(v) M1 two sets of co-ordinates from the line of best fit correctly recorded for first t AND second t in the form (x, y) 2\n½ ½\nM2 both half-lives correctly calculated from co-ordinates\n\n3(c)(vi) first order AND half-lives are constant (within experimental error) 1\n© UCLES 2023 Page 8 of 8", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9701_s23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-may-june/9701_s23_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s23_ms_53-p07.png", "../answer-assets/9701_s23_ms_53-p08.png" ] }, { "id": "9701-2023-on-41-q01", "question_id": "9701-2023-on-41-q01", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "1(a)(i) M1:F + ClO → FClO + F [1] 2\n2 2 2\nM2:ClO + F → FClO [1]\n2 2\nOR\nM1: F + ClO → F ClO\n2 2 2 2\nM2: F ClO + ClO → 2FClO\n2 2 2 2\ntwo balanced equations MUST add to give the overall equation\n\n1(a)(ii) first step AND 1\nhas one mole / molecule of F and ClO [1]\n2 2\nOR same moles of reactants as orders in rate equation\n\n1(b)(i) rate = k [F ][Cl O ] / rate = 1.22 [F ][Cl O ] 1\n2 2 2 2\nAND second / 2 BOTH[1]\n\n1(b)(ii) rate = 1.22  2  10-3  2  10-3 = 4.88 x 10-6 [1] 1\nmin 2sfecf 1(b)(i)\n\n1(c)(i) k = (0.693 / 4) = 0.173 OR 1.73  10–1[1] 2\n1\nMUST BE 3SF\ns–1 [1]\n\n1(c)(ii) at 4 s = 0.001 1\nat 8 s = 0.0005\nat 12 s = 0.00025\nsmooth curve\nALL correct [1]\n\n1(c)(iii) tangent drawn at 0.00100 mol dm–3 1\nAND\ngradient dependent on their rate = Y / X in the range = 1.5–2.0 10–4 [1]\nIGNORE sign\n© UCLES 2023 Page 5 of 18", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p05.png" ] }, { "id": "9701-2023-on-41-q02", "question_id": "9701-2023-on-41-q02", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 14, "status": "available", "reason": null, "text": "2(a) (K =)[H+][OH–] OR (K =)[H3O+][OH–] [1] 1\nw w\n\n2(b)(i) M1: pH values 1.5 AND 12.5 [1] 2\nM2: conc of HCl = 10–1.5 = 0.0316 ecf [1]\nmin 2sf\n\n2(b)(ii) sodium chloride / NaCl AND 7 [1] 1\n\n2(b)(iii) (mixture Y) 1 to 3 AND (mixture Z) 11 to 13 [1] 1\n\n2(c)(i) H SO > CH CCl COOH > CH CH COOH [1] u / c 3\n2 4 3 2 3 2\nexplanation\n• H SO is fully dissociated / strong acid\n2 4\n• CH CH COOH / CH CCl COOH are partly dissociated / weak acids\n3 2 3 2\n• Cl / chlorine is electron-withdrawing / electronegative\n• alkyl group (in CH3CH2COOH) is electron donating\n• stabilises / destabilises anion OR weakens / strengthens O-H bond (linked correctly)\n• correct reference to release of / donation of / form H+ / proton\nany two [1] any four [2]\n\n2(c)(ii) • H O+ and H O 2\n3 2\n• H SO and HSO –\n2 4 4\n• HSO – and SO 2–\n4 4\nany one correct pair [1]\nall three correct pairs [2]\n\n2(d)(i) (the) ratio of the concentrations (of a solute between) 1\ntwo solvents / two liquids (at) equilibrium [1]\n\n2(d)(ii) mass = 5-x / x = 7.84 OR mass = x / 5-x = 7.84 1\nmass = 0.5656 g mass = 4.43 g\n[1] min 2sf\n© UCLES 2023 Page 6 of 18\n\n2(d)(iii) Any numbers in which volume of water is 7.84 times volume of hexane e.g. 78.4 cm3 water 10 cm3 hexane [1] u / c 1\ne.g mass = 0.566 g 177.4 cm3 water 22.6 cm3 hexane\nALLOW reverse ratio volume of hexane is 7.84 times volume of water when consistent with d(ii)\n\n2(d)(iv) Q is CH (CH ) OH 1\n3 2 4\nAND it is least polar\n/ contains a large non-polar hydrocarbon chain\n/ stronger id-id forces with hexane in Q OWTTE\n[1]\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p06.png", "../answer-assets/9701_w23_ms_41-p07.png" ] }, { "id": "9701-2023-on-41-q03", "question_id": "9701-2023-on-41-q03", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 17, "status": "available", "reason": null, "text": "3(a)(i) (number of possible) arrangements of particles / energy in a system 1\nOR\nmeasure / degree of disorder in / of a system [1]\n\n3(a)(ii) positive / + 1\nAND more gas molecules / particles in products\nOR more moles / molecules in products / RHS [1]\n\n3(b) H = (2  150) – (1  496) (= –196) [1] 1\nOR\nH = (2  150) + (4  460) – (1  496) – (4  460) (= –196)\n\n3(c) G = H –TS seen or used with correct signs[1] 3\n–238 = –196 – 298S\nS = 42 / 298\nS = (+)0.141 / 0.1409 (kJ K–1 mol–1)\nOR (+)141 / 140.9 (J K–1 mol–1) [1] ecf from G = H + TS\n141 = (2  70) + S(O2(g)) – (2  102)\nS, O2(g) = 205 / 204.94 (J K–1 mol–1) [1] ecf\n© UCLES 2023 Page 7 of 18\n\n3(d) iron(III) chloride / FeCl 3 1\nAND same state / phase as reactants / H O [1]\n2 2\n\n3(e)(i) hydrogen peroxide / H O AND +2.18 [1] 1\n2 2\n\n3(e)(ii) G = –nEo F [1] 2\ncell\nG = (–2  2.18  96500) = –420.7 (kJ mol–1) [1] ecf\n\n3(f)(i) Nernst: (E =) Eo + (0.059 / z)log[ox] / [red] [1] u / c 2\nOR (E =) Eo + (RT / zF)ln[ox] / [red]\nOR (E =) 1.82 + (0.059 / 1)log(0.02 / 2)\nE = 1.82 + (0.059 / 1)log(0.02 / 2) = (+)1.702 (V) [1] min 2sf\nOR\nE = 1.82 + [(8.314  298 / 1)  (96 500) ] ln(0.02 / 2) = (+)1.702\n\n3(f)(ii) H O + 2H+ + 2Co2+ → 2H O + 2Co3+ [1] 1\n2 2 2\nECF for reverse equation from (f)(i) if E > 1.77 V\n\n3(g)(i) enthalpy change when one mole of gaseous ions 1\nforms an aqueous solution / dissolves in water [1]\n© UCLES 2023 Page 8 of 18\n\n3(g)(ii) Substances with state symbols: 2\n• Al F3(s)\n• Al F3(aq) OR Al 3+(aq) + (3)F–(aq)\n• Al 3+(g) (3)F–(g)\n[1]\nchanges identified and three correct arrow directions (one given) [1]\n\n3(g)(iii) H = –4690 + (3  – 506) – (–209) = –5999 (kJ mol–1) 1\nlatt\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p07.png", "../answer-assets/9701_w23_ms_41-p08.png", "../answer-assets/9701_w23_ms_41-p09.png" ] }, { "id": "9701-2023-on-41-q04", "question_id": "9701-2023-on-41-q04", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "4(a)(i) • Co(OH) OR Co(OH) (H O) 2\n2 2 2 4\n• [Co(NH ) ]2+ OR [Co(NH ) ]3+\n3 6 3 6\n• [CoCl ]2–\n4\nany two [1] all three [2]\n\n4(a)(ii) pink to 1\nblue [1]\n\n4(b)(i) 2Ca(NO ) → 2CaO + 4NO + O 1\n3 2 2 2\nOR Ca(NO ) → CaO + 2NO + ½O [1]\n3 2 2 2\n© UCLES 2023 Page 9 of 18\n\n4(b)(ii) M1: Mg(NO ) below 480 °C, 3\n3 2\nBa(NO ) above 520 °C BOTH [1] u / c\n3 2\nM2: ionic radii of M2+ increases down the group\nOR radii of Ba2+ is greater (than Mg2+)\nOR charge density of M2+ decreases down the group\nOR charge density of Ba2+ is smaller (than Mg2+) [1] ORA u / c\nM3: (larger cations) polarise / distort anion / NO – less\n3\nOR\n(larger ions) weaken N-O / N=O (bond) less\n[1] ORA\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p09.png", "../answer-assets/9701_w23_ms_41-p10.png" ] }, { "id": "9701-2023-on-41-q05", "question_id": "9701-2023-on-41-q05", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "5(a) more than one (stable) oxidation state [1] 2\nempty / vacant (d) orbitals are energetically accessible\nOR empty / vacant (d) orbitals can form dative bonds with ligands\n[1]\n\n5(b)(i) oxid. no. = + 4 AND 1\ncoord. no. = 8 [1]\n© UCLES 2023 Page 10 of 18\n\n5(b)(ii) bond angle must go from bond to bond 2\nOR CN group to CN group\ndiagram\n• for Ag: NC—Ag—CN\n• angle = labelled 180°\n• for Cu:\n• angle = labelled 109–110°\nany two [1] all four [2]\n\n5(c)(i) K = [[Cu(CN) ]3–] / [Cu+] [CN–]4 [1] 1\nstab 4\nALLOW use of [[Cu(H O) ]+]\n2 6\n\n5(c)(ii) [Cu+] = 5.0  10–19 (mol dm–3) [1] min 1sf 1\nECF from reversed ratio 5(c)(i)\n\n5(d)(i) mol of I = 0.5  0.02  20.1 / 1000 = 2.01  10–4 [1] min 2sf 1\n2\n\n5(d)(ii) mol of Cu = 2.01  10–4  2  4 = 1.608 / 1.61  10–3 [1] ecf min 2sf 1\n\n5(d)(iii) % of Cu = 100  (1.60810–3  63.5) / 0.567 = 18.0 [1] ecf 1\n\n5(d)(iv) Cu2+ is d9 / d shell / sub-shell / orbitals is / are not full 2\nAND Cu+ is d10 / d shell / sub-shell / orbitals is / are full [1]\nd-d* transitions / d electron promotion\nnot possible / possible (linked correctly) [1]\n© UCLES 2023 Page 11 of 18", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p10.png", "../answer-assets/9701_w23_ms_41-p11.png" ] }, { "id": "9701-2023-on-41-q06", "question_id": "9701-2023-on-41-q06", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 6, "status": "available", "reason": null, "text": "6(a)(i) NH = monodentate 2\n3\nEDTA4– = polydentate / hexadentate\nCN– = monodentate\nC O 2– = bidentate\n2 4\nany two [1] all four [2]\n\n6(a)(ii) (ligand that) donates 3 lone pairs to central metal atom / ion 1\nOR\n(ligand that) forms 3 dative bonds to central metal atom / ion [1]\n\n6(a)(iii) ref to using the electrons / lone pair on (each) N / amine group [1] 1\n\n6(b)(i) trans = non-polar 1\ncis isomer 1 = polar\ncis isomer 2 = polar ALL [1]\n\n6(b)(ii) optical isomers / non-superimposable mirror images [1] 1\nALLOW enantiomers / they rotate polarised light differently\nQuestion Answer Marks", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p12.png" ] }, { "id": "9701-2023-on-41-q07", "question_id": "9701-2023-on-41-q07", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 8, "status": "available", "reason": null, "text": "7(a) C H N O S 2– OR C H N O S [1] 1\n16 10 2 7 2 16 10 2 7 2\n© UCLES 2023 Page 12 of 18\n\n7(b) 3\n[1]\n[1]\nECF from E\n[1]\n\n7(c) M1 step 1: HNO OR NaNO + HCl [1] 3\n2 2\nM2 step 1: T ⩽ 10 °C [1]\nM3 step 2: NaOH(aq) / alkaline conditions [1]\n\n7(d) 14 / fourteen [1] 1\n© UCLES 2023 Page 13 of 18", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p12.png", "../answer-assets/9701_w23_ms_41-p13.png" ] }, { "id": "9701-2023-on-41-q08", "question_id": "9701-2023-on-41-q08", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "8(a) phenol, amide AND alkene / C=C all three needed [1] 1\n\n8(b) 3\nM1: HBr [1] u / c\nM2: structure of the organic product[1]\nM3: correct balanced equation [1] ecf\n\n8(c) 2\nM1: ring reduced [1]\nM2: alkene reduced [1]\n\n8(d)(i) hot AND concentrated acidified AND MnO – / KMnO all [1] 1\n4 4\n© UCLES 2023 Page 14 of 18\n\n8(d)(ii) 3\n[1] for  values\n[1] for splitting\n[1] for explanations\n\n8(e)(i) 2\n[1]\n[1]\n\n8(e)(ii) hydrolysis AND neutralisation / acid-base[1] 1\n© UCLES 2023 Page 15 of 18\n\n8(f) 1\n[1]\nQuestion Answer Marks", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p14.png", "../answer-assets/9701_w23_ms_41-p15.png", "../answer-assets/9701_w23_ms_41-p16.png" ] }, { "id": "9701-2023-on-41-q09", "question_id": "9701-2023-on-41-q09", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 13, "status": "available", "reason": null, "text": "9(a)(i) 1\n[1]\n\n9(a)(ii) hot (alkaline) KMnO / MnO – [1] 2\n4 4\nSOCl OR PCl OR PCl + heat[1]\n2 5 3\n\n9(a)(iii) oxidation [1] 2\n(nucleophilic) substitution\n© UCLES 2023 Page 16 of 18\n\n9(b)(i)\n10\nand HCl\nThe red numbers refer to points 1 to 10 in the verbal MS that follows.\nbox one: 4\n• 1 dipole on C=O\n• 2 curly arrow from O of phenol to C of acyl chloride\n• 3 curly arrow from bond of C=O to O of acyl chloride\nbox two:\n• 4 curly arrow from O-H bond to what was O of phenol\n• 5 curly arrow from C-Cl bond to what was Cl of acyl chloride\n• 6 curly arrow from O to reform C=O\n• 7 + charge on what was O of phenol\n• 8 – charge on what was O of acyl chloride\n• 9 curly arrow start close to LP for bullet 3 OR 6 (from LP on O)\nbox three:\n• 10 HCl / H+Cl –\nany two [1] any five [2] any eight [3] all ten [4]\n\n9(b)(ii) (nucleophilic) addition-elimination [1] 1\n© UCLES 2023 Page 17 of 18\n\n9(c) chlorobenzene, chloroethane, benzoyl chloride [1] u / c 3\nexplanation linked to their order\n• correct link to: strengthening (ArCl / RCl) C-Cl bond\nOR weakening (RCOCl) C-Cl bond\nOR C-Cl bond has partially double bond character (ArCl)\nOR C-Cl is more difficult to break (linked correctly)\nchlorobenzene:\n• lone pair / p-orbital on Cl overlaps / delocalised / incorporated with ring\nbenzoyl chloride:\n• C of C-Cl has most electron deficient\nOR has an electronegative oxygen atom\n/ two electronegative atoms / electron withdrawing C=O group\nchloroethane\n• electron donating effect / positive inductive effect of alkyl / R group\nany two [1] any three [2]\n© UCLES 2023 Page 18 of 18", "source_pages": [ 16, 17, 18 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_41.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_41-p16.png", "../answer-assets/9701_w23_ms_41-p17.png", "../answer-assets/9701_w23_ms_41-p18.png" ] }, { "id": "9701-2023-on-42-q01", "question_id": "9701-2023-on-42-q01", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "1(a) 1 2\n0 [1]\n1\n2 [1]\n\n1(b)(i) The line should start at 1  10–5 and decrease to 0.773  10–5 / 7.73  10–6 and be straight for 10 s [1] 1\n\n1(b)(ii) its not possible AND the concentrations in the rate law are not known [1] 1\n\n1(c) 9.24  10–7 [1] 1\n\n1(d)(i) 630 [1] 1\n\n1(d)(ii) Three halvings taking equal times, these times to agree with their answer to di.[1] 1\n\n1(e)(i) CH COH=CH + I → CH C+(OH)CH I + I– [1] 1\n3 2 2 3 2\n\n1(e)(ii) step 1 AND has (both) substances in rate law.[1] 1\n\n1(e)(iii) CH C+OHCH and CH COCH OR 1\n3 3 3 3\nCH C+(OH)CH and CH C(OH)=CH OR\n3 3 3 2\nCH C+(OH)CH I and CH COCH I [1]\n3 2 3 2\nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p05.png" ] }, { "id": "9701-2023-on-42-q02", "question_id": "9701-2023-on-42-q02", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 11, "status": "available", "reason": null, "text": "2(a) resists pH change when small amount of acid or alkali is added [1] 1\n\n2(b)(i) C H COOH + NaOH → C H COO–Na+ + H O [1] 1\n6 5 6 5 2\n\n2(b)(ii) C H COO–Na+ + HNO → C H COOH + NaNO [1] 1\n6 5 3 6 5 3\n\n2(c) [H+] = 7.08  10–5 [1] 3\n[C H COOH] = 8.20  10–3 [1]\n6 5\n[C H COO–Na+] = 7.31  10–3 [1]\n6 5\n© UCLES 2023 Page 5 of 14\n\n2(d) neutralisation is exothermic [1] 2\nAll of the C H COOH has reacted AND excess KOH 1]\n6 5\n\n2(e)(i) If [X] = [Mg2+] = [Mg(C H COO–) ] 2\n6 5 2\n4X3 = 1.76  10–7 so X = 3.53  10–3 [1]\n0.940 g dm–3 [1]\n\n2(e)(ii) lower than AND common ion effect [1] 1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p05.png", "../answer-assets/9701_w23_ms_42-p06.png" ] }, { "id": "9701-2023-on-42-q03", "question_id": "9701-2023-on-42-q03", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 18, "status": "available", "reason": null, "text": "3(a)(i) hydrogen, delivery system, H+, platinum, [1] 1\n\n3(a)(ii) iron . . . . . hydrogen . . . . . iron [1] 1\n\n3(b)(i) (for specified V2+, V3+ or VO2+) Eo is more positive than / above –0.44 AND more negative than / below 0.77 V [1] 1\n\n3(b)(ii) V and VO + [1] 1\n2\n\n3(b)(iii) V + Fe2+ → V2+ + Fe OR 1\nVO + + 2H+ + Fe2+ → VO2+ + H O + Fe3+ [1]\n2 2\n\n3(c)(i) Nernst: E = 0.77 + (0.059 / z)log[ox] / [red] [1] 2\n0.947 [1]\n\n3(c)(ii) 2Fe3+ + Cl– + 2OH– → 2Fe2+ + ClO– + H O [1] 1\n2\n\n3(d) Eo = 1.33 V [1] 3\ncell\nGo = –nEo F [1]\ncell\n–257 [1]\n© UCLES 2023 Page 6 of 14\n\n3(e) 0.64  17  60 = 653 / 652.8 Coulombs [1] 3\n652.8 ÷ 1.6x10–19 = 4.08  1021 (number of electrons)\n4.08  1021 ÷ 2 = 2.04  1021 (number of atoms Fe) [1]\n0.185 ÷ 55.8 = 3.31  10–3 (number of moles Fe atoms\n2.04  1021 ÷ 3.31  10–3 = L = 6.153  1023 [1]\n\n3(f)(i) S = –179 [1] 3\nG = H –TS [1]\n–74.7 [1]\n\n3(f)(ii) less AND G becomes more positive [1] 1\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p06.png", "../answer-assets/9701_w23_ms_42-p07.png" ] }, { "id": "9701-2023-on-42-q04", "question_id": "9701-2023-on-42-q04", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "4(a) equilibrium constant for the formation of a complex in a solvent / 1\nfrom its constituents [1]\n\n4(b) +2 AND + 3 [1] 1\n\n4(c) Six [1] 1\n\n4(d) [Fe(EDTA)]– AND largest K [1] 1\nstab\n\n4(e) [[CuEDTA]2–] ÷ ([[Cu(H O) ]2+]  [EDTA2–]) = 6.311019 [1] 2\n2 6\n0.095 [1]\n\n4(f) different E OR different energy gap between d-orbitals [1] 2\nabsorption of different wavelength OR absorption of different frequency [1]\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p07.png" ] }, { "id": "9701-2023-on-42-q05", "question_id": "9701-2023-on-42-q05", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 6, "status": "available", "reason": null, "text": "5(a) MgC O → MgO + CO + CO [1] 1\n2 4 2\n© UCLES 2023 Page 7 of 14\n\n5(b) magnesium ethanedioate decomposes at lower T because Mg2+ has smaller radius than Ca2+ [1] 2\nso anion is more polarised by Mg2+ [1]\n\n5(c) (27.05 ÷ 1000)  0.02 = 5.41  10–5 moles MnO – [1] 3\n4\n5.41  10–5  5/2 = 1.3525  10–4 moles C O 2– [1]\n2 4\n0.00338 mol dm–3 [1]\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p07.png", "../answer-assets/9701_w23_ms_42-p08.png" ] }, { "id": "9701-2023-on-42-q06", "question_id": "9701-2023-on-42-q06", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 10, "status": "available", "reason": null, "text": "6(a)(i) donates one lp to metal atom or ion [1] 1\n\n6(a)(ii) metal atom or ion bonded to one or more ligands [1] 1\n\n6(a)(iii) has vacant d-orbitals which are energetically accessible [1] 1\n\n6(b)(i) octahedral 1\nsquare planar\noctahedral [1]\n\n6(b(ii) [Au(dien)(H O) Cl]2+ + 2Cl– → [Au(dien)Cl ] + 2H O [1] 1\n2 2 3 2\nOR [Au(dien)Cl ] + 2H O → [Au(dien)(H O) Cl]2+ + 2Cl–\n3 2 2 2\n\n6(b)(iii)\n[1]\n© UCLES 2023 Page 8 of 14\n\n6(b)(iv) 1\n[1]\n\n6(b)(v) [Rh(en) Cl ]+ AND 1\n2 2\noptical AND cis / trans OR geometric [1]\n\n6(b)(vi) 2\n[1]\ncis/trans OR geometric [1]\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p08.png", "../answer-assets/9701_w23_ms_42-p09.png" ] }, { "id": "9701-2023-on-42-q07", "question_id": "9701-2023-on-42-q07", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 11, "status": "available", "reason": null, "text": "7(a) CH Cl + AlCl [1] 2\n3 3\nhot alkaline) MnO – [1]\n4\n© UCLES 2023 Page 9 of 14\n\n7(b) 4, 110-160, 2\n1, 25-50 either order [1]\n4, 110-160,\n1, 160-185 either order [1]\n\n7(c)(i) Mark J and K together. 4\neither or J or K [1]\nJ or K [1]\nfor the ring substituted product – AlCl [1]\n3\nfor the side-chain substituted product – uv [1]\n© UCLES 2023 Page 10 of 14\n\n7(c)(ii) 3\nL [1]\nM [1]\nPCl + heat OR SOCl OR PCl [1]\n3 2 5", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p09.png", "../answer-assets/9701_w23_ms_42-p10.png", "../answer-assets/9701_w23_ms_42-p11.png" ] }, { "id": "9701-2023-on-42-q08", "question_id": "9701-2023-on-42-q08", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 12, "status": "available", "reason": null, "text": "8(a)(i) 2\ndiagram has trailing bonds, two or more monomer residues AND either repeat unit correct OR ester linkage correct [1]\ndiagram fully correct [1]\n© UCLES 2023 Page 11 of 14\n\n8(a)(ii) condensation AND ester [1] 1\n\n8(a)(iii) it is biodegradable because it can hydrolyse [1] 1\n\n8(b)(i) d values: 3\n12, 4.4 [1]\n5.0, 1.6 [1]\nsplitting patterns:\nsinglet, quartet / quadruplet, singlet, doublet [1]\n\n8(b)(ii) TMS / tetramethylsilane [1] 1\n\n8(b)(iii) CDCl will not give an absorption / peak OR 1\n3\nCHCl will give an absorption / peak [1]\n3\n\n8(c)(i) time between injection and detection [1] 1\n\n8(c)(ii) an unreactive gas AND a non-polar liquid [1] 1\n\n8(c)(iii) area of peak divided by total area of all peaks  100% [1] 1\n© UCLES 2023 Page 12 of 14", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p11.png", "../answer-assets/9701_w23_ms_42-p12.png" ] }, { "id": "9701-2023-on-42-q09", "question_id": "9701-2023-on-42-q09", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 14, "status": "available", "reason": null, "text": "9(a) chloroethane and ethylamine [1] 4\nheat, pressure, ethanol [1]\nN-ethyl ethanamide CH CONHC H [1]\n3 2 5\nLiAlH [1]\n4\nOther syntheses that work also got credit /\n\n9(b) phenylamine, ammonia, diethylamine [1] 3\n• Availability of lone pair on N to receive H+\n• phenylamine – LP, delocalised / overlaps into ring [1]\n• diethylamine – ethyl are electron donating [1]\nTwo bullet points for one mark, three bullet points for two marks.\n\n9(c) 2\nQ or [1]\nN or N=N circled AND dye / colouring[1]\n2\n© UCLES 2023 Page 13 of 14\n\n9(d)(i)\nbox one box two box three\n11\nHCl\nbox one: 4\n1 dipole on C=O\n2 LP on N of amine [1]\n3 CA from N of amine to C of acyl chloride\n4 CA from = to O of acyl chloride [1]\nbox two:\n5 + on what was N of amine and nowhere else\n6 – on what was O of acyl chloride and nowhere else\n7 LP on what was O of acyl chloride [1]\n8 CA from N-H bond to what was N of amine\n9 CA from C-Cl bond to what was Cl of acyl chloride\n10 CA from (LP on) O to reform C=O [1]\nbox three:\n11 HCl\n\n9(d)(ii) Addition-elimination 1\n© UCLES 2023 Page 14 of 14", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_42-p13.png", "../answer-assets/9701_w23_ms_42-p14.png" ] }, { "id": "9701-2023-on-43-q01", "question_id": "9701-2023-on-43-q01", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "1(a)(i) M1:F + ClO → FClO + F [1] 2\n2 2 2\nM2:ClO + F → FClO [1]\n2 2\nOR\nM1: F + ClO → F ClO\n2 2 2 2\nM2: F ClO + ClO → 2FClO\n2 2 2 2\ntwo balanced equations MUST add to give the overall equation\n\n1(a)(ii) first step AND 1\nhas one mole / molecule of F and ClO [1]\n2 2\nOR same moles of reactants as orders in rate equation\n\n1(b)(i) rate = k [F ][Cl O ] / rate = 1.22 [F ][Cl O ] 1\n2 2 2 2\nAND second / 2 BOTH[1]\n\n1(b)(ii) rate = 1.22  2  10-3  2  10-3 = 4.88 x 10-6 [1] 1\nmin 2sfecf 1(b)(i)\n\n1(c)(i) k = (0.693 / 4) = 0.173 OR 1.73  10–1[1] 2\n1\nMUST BE 3SF\ns–1 [1]\n\n1(c)(ii) at 4 s = 0.001 1\nat 8 s = 0.0005\nat 12 s = 0.00025\nsmooth curve\nALL correct [1]\n\n1(c)(iii) tangent drawn at 0.00100 mol dm–3 1\nAND\ngradient dependent on their rate = Y / X in the range = 1.5–2.0 10–4 [1]\nIGNORE sign\n© UCLES 2023 Page 5 of 18", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p05.png" ] }, { "id": "9701-2023-on-43-q02", "question_id": "9701-2023-on-43-q02", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 14, "status": "available", "reason": null, "text": "2(a) (K =)[H+][OH–] OR (K =)[H3O+][OH–] [1] 1\nw w\n\n2(b)(i) M1: pH values 1.5 AND 12.5 [1] 2\nM2: conc of HCl = 10–1.5 = 0.0316 ecf [1]\nmin 2sf\n\n2(b)(ii) sodium chloride / NaCl AND 7 [1] 1\n\n2(b)(iii) (mixture Y) 1 to 3 AND (mixture Z) 11 to 13 [1] 1\n\n2(c)(i) H SO > CH CCl COOH > CH CH COOH [1] u / c 3\n2 4 3 2 3 2\nexplanation\n• H SO is fully dissociated / strong acid\n2 4\n• CH CH COOH / CH CCl COOH are partly dissociated / weak acids\n3 2 3 2\n• Cl / chlorine is electron-withdrawing / electronegative\n• alkyl group (in CH3CH2COOH) is electron donating\n• stabilises / destabilises anion OR weakens / strengthens O-H bond (linked correctly)\n• correct reference to release of / donation of / form H+ / proton\nany two [1] any four [2]\n\n2(c)(ii) • H O+ and H O 2\n3 2\n• H SO and HSO –\n2 4 4\n• HSO – and SO 2–\n4 4\nany one correct pair [1]\nall three correct pairs [2]\n\n2(d)(i) (the) ratio of the concentrations (of a solute between) 1\ntwo solvents / two liquids (at) equilibrium [1]\n\n2(d)(ii) mass = 5-x / x = 7.84 OR mass = x / 5-x = 7.84 1\nmass = 0.5656 g mass = 4.43 g\n[1] min 2sf\n© UCLES 2023 Page 6 of 18\n\n2(d)(iii) Any numbers in which volume of water is 7.84 times volume of hexane e.g. 78.4 cm3 water 10 cm3 hexane [1] u / c 1\ne.g mass = 0.566 g 177.4 cm3 water 22.6 cm3 hexane\nALLOW reverse ratio volume of hexane is 7.84 times volume of water when consistent with d(ii)\n\n2(d)(iv) Q is CH (CH ) OH 1\n3 2 4\nAND it is least polar\n/ contains a large non-polar hydrocarbon chain\n/ stronger id-id forces with hexane in Q OWTTE\n[1]\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p06.png", "../answer-assets/9701_w23_ms_43-p07.png" ] }, { "id": "9701-2023-on-43-q03", "question_id": "9701-2023-on-43-q03", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 17, "status": "available", "reason": null, "text": "3(a)(i) (number of possible) arrangements of particles / energy in a system 1\nOR\nmeasure / degree of disorder in / of a system [1]\n\n3(a)(ii) positive / + 1\nAND more gas molecules / particles in products\nOR more moles / molecules in products / RHS [1]\n\n3(b) H = (2  150) – (1  496) (= –196) [1] 1\nOR\nH = (2  150) + (4  460) – (1  496) – (4  460) (= –196)\n\n3(c) G = H –TS seen or used with correct signs[1] 3\n–238 = –196 – 298S\nS = 42 / 298\nS = (+)0.141 / 0.1409 (kJ K–1 mol–1)\nOR (+)141 / 140.9 (J K–1 mol–1) [1] ecf from G = H + TS\n141 = (2  70) + S(O2(g)) – (2  102)\nS, O2(g) = 205 / 204.94 (J K–1 mol–1) [1] ecf\n© UCLES 2023 Page 7 of 18\n\n3(d) iron(III) chloride / FeCl 3 1\nAND same state / phase as reactants / H O [1]\n2 2\n\n3(e)(i) hydrogen peroxide / H O AND +2.18 [1] 1\n2 2\n\n3(e)(ii) G = –nEo F [1] 2\ncell\nG = (–2  2.18  96500) = –420.7 (kJ mol–1) [1] ecf\n\n3(f)(i) Nernst: (E =) Eo + (0.059 / z)log[ox] / [red] [1] u / c 2\nOR (E =) Eo + (RT / zF)ln[ox] / [red]\nOR (E =) 1.82 + (0.059 / 1)log(0.02 / 2)\nE = 1.82 + (0.059 / 1)log(0.02 / 2) = (+)1.702 (V) [1] min 2sf\nOR\nE = 1.82 + [(8.314  298 / 1)  (96 500) ] ln(0.02 / 2) = (+)1.702\n\n3(f)(ii) H O + 2H+ + 2Co2+ → 2H O + 2Co3+ [1] 1\n2 2 2\nECF for reverse equation from (f)(i) if E > 1.77 V\n\n3(g)(i) enthalpy change when one mole of gaseous ions 1\nforms an aqueous solution / dissolves in water [1]\n© UCLES 2023 Page 8 of 18\n\n3(g)(ii) Substances with state symbols: 2\n• Al F3(s)\n• Al F3(aq) OR Al 3+(aq) + (3)F–(aq)\n• Al 3+(g) (3)F–(g)\n[1]\nchanges identified and three correct arrow directions (one given) [1]\n\n3(g)(iii) H = –4690 + (3  – 506) – (–209) = –5999 (kJ mol–1) 1\nlatt\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p07.png", "../answer-assets/9701_w23_ms_43-p08.png", "../answer-assets/9701_w23_ms_43-p09.png" ] }, { "id": "9701-2023-on-43-q04", "question_id": "9701-2023-on-43-q04", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "4(a)(i) • Co(OH) OR Co(OH) (H O) 2\n2 2 2 4\n• [Co(NH ) ]2+ OR [Co(NH ) ]3+\n3 6 3 6\n• [CoCl ]2–\n4\nany two [1] all three [2]\n\n4(a)(ii) pink to 1\nblue [1]\n\n4(b)(i) 2Ca(NO ) → 2CaO + 4NO + O 1\n3 2 2 2\nOR Ca(NO ) → CaO + 2NO + ½O [1]\n3 2 2 2\n© UCLES 2023 Page 9 of 18\n\n4(b)(ii) M1: Mg(NO ) below 480 °C, 3\n3 2\nBa(NO ) above 520 °C BOTH [1] u / c\n3 2\nM2: ionic radii of M2+ increases down the group\nOR radii of Ba2+ is greater (than Mg2+)\nOR charge density of M2+ decreases down the group\nOR charge density of Ba2+ is smaller (than Mg2+) [1] ORA u / c\nM3: (larger cations) polarise / distort anion / NO – less\n3\nOR\n(larger ions) weaken N-O / N=O (bond) less\n[1] ORA\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p09.png", "../answer-assets/9701_w23_ms_43-p10.png" ] }, { "id": "9701-2023-on-43-q05", "question_id": "9701-2023-on-43-q05", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "5(a) more than one (stable) oxidation state [1] 2\nempty / vacant (d) orbitals are energetically accessible\nOR empty / vacant (d) orbitals can form dative bonds with ligands\n[1]\n\n5(b)(i) oxid. no. = + 4 AND 1\ncoord. no. = 8 [1]\n© UCLES 2023 Page 10 of 18\n\n5(b)(ii) bond angle must go from bond to bond 2\nOR CN group to CN group\ndiagram\n• for Ag: NC—Ag—CN\n• angle = labelled 180°\n• for Cu:\n• angle = labelled 109–110°\nany two [1] all four [2]\n\n5(c)(i) K = [[Cu(CN) ]3–] / [Cu+] [CN–]4 [1] 1\nstab 4\nALLOW use of [[Cu(H O) ]+]\n2 6\n\n5(c)(ii) [Cu+] = 5.0  10–19 (mol dm–3) [1] min 1sf 1\nECF from reversed ratio 5(c)(i)\n\n5(d)(i) mol of I = 0.5  0.02  20.1 / 1000 = 2.01  10–4 [1] min 2sf 1\n2\n\n5(d)(ii) mol of Cu = 2.01  10–4  2  4 = 1.608 / 1.61  10–3 [1] ecf min 2sf 1\n\n5(d)(iii) % of Cu = 100  (1.60810–3  63.5) / 0.567 = 18.0 [1] ecf 1\n\n5(d)(iv) Cu2+ is d9 / d shell / sub-shell / orbitals is / are not full 2\nAND Cu+ is d10 / d shell / sub-shell / orbitals is / are full [1]\nd-d* transitions / d electron promotion\nnot possible / possible (linked correctly) [1]\n© UCLES 2023 Page 11 of 18", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p10.png", "../answer-assets/9701_w23_ms_43-p11.png" ] }, { "id": "9701-2023-on-43-q06", "question_id": "9701-2023-on-43-q06", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 6, "status": "available", "reason": null, "text": "6(a)(i) NH = monodentate 2\n3\nEDTA4– = polydentate / hexadentate\nCN– = monodentate\nC O 2– = bidentate\n2 4\nany two [1] all four [2]\n\n6(a)(ii) (ligand that) donates 3 lone pairs to central metal atom / ion 1\nOR\n(ligand that) forms 3 dative bonds to central metal atom / ion [1]\n\n6(a)(iii) ref to using the electrons / lone pair on (each) N / amine group [1] 1\n\n6(b)(i) trans = non-polar 1\ncis isomer 1 = polar\ncis isomer 2 = polar ALL [1]\n\n6(b)(ii) optical isomers / non-superimposable mirror images [1] 1\nALLOW enantiomers / they rotate polarised light differently\nQuestion Answer Marks", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p12.png" ] }, { "id": "9701-2023-on-43-q07", "question_id": "9701-2023-on-43-q07", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 8, "status": "available", "reason": null, "text": "7(a) C H N O S 2– OR C H N O S [1] 1\n16 10 2 7 2 16 10 2 7 2\n© UCLES 2023 Page 12 of 18\n\n7(b) 3\n[1]\n[1]\nECF from E\n[1]\n\n7(c) M1 step 1: HNO OR NaNO + HCl [1] 3\n2 2\nM2 step 1: T ⩽ 10 °C [1]\nM3 step 2: NaOH(aq) / alkaline conditions [1]\n\n7(d) 14 / fourteen [1] 1\n© UCLES 2023 Page 13 of 18", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p12.png", "../answer-assets/9701_w23_ms_43-p13.png" ] }, { "id": "9701-2023-on-43-q08", "question_id": "9701-2023-on-43-q08", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "8(a) phenol, amide AND alkene / C=C all three needed [1] 1\n\n8(b) 3\nM1: HBr [1] u / c\nM2: structure of the organic product[1]\nM3: correct balanced equation [1] ecf\n\n8(c) 2\nM1: ring reduced [1]\nM2: alkene reduced [1]\n\n8(d)(i) hot AND concentrated acidified AND MnO – / KMnO all [1] 1\n4 4\n© UCLES 2023 Page 14 of 18\n\n8(d)(ii) 3\n[1] for  values\n[1] for splitting\n[1] for explanations\n\n8(e)(i) 2\n[1]\n[1]\n\n8(e)(ii) hydrolysis AND neutralisation / acid-base[1] 1\n© UCLES 2023 Page 15 of 18\n\n8(f) 1\n[1]\nQuestion Answer Marks", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p14.png", "../answer-assets/9701_w23_ms_43-p15.png", "../answer-assets/9701_w23_ms_43-p16.png" ] }, { "id": "9701-2023-on-43-q09", "question_id": "9701-2023-on-43-q09", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 13, "status": "available", "reason": null, "text": "9(a)(i) 1\n[1]\n\n9(a)(ii) hot (alkaline) KMnO / MnO – [1] 2\n4 4\nSOCl OR PCl OR PCl + heat[1]\n2 5 3\n\n9(a)(iii) oxidation [1] 2\n(nucleophilic) substitution\n© UCLES 2023 Page 16 of 18\n\n9(b)(i)\n10\nand HCl\nThe red numbers refer to points 1 to 10 in the verbal MS that follows.\nbox one: 4\n• 1 dipole on C=O\n• 2 curly arrow from O of phenol to C of acyl chloride\n• 3 curly arrow from bond of C=O to O of acyl chloride\nbox two:\n• 4 curly arrow from O-H bond to what was O of phenol\n• 5 curly arrow from C-Cl bond to what was Cl of acyl chloride\n• 6 curly arrow from O to reform C=O\n• 7 + charge on what was O of phenol\n• 8 – charge on what was O of acyl chloride\n• 9 curly arrow start close to LP for bullet 3 OR 6 (from LP on O)\nbox three:\n• 10 HCl / H+Cl –\nany two [1] any five [2] any eight [3] all ten [4]\n\n9(b)(ii) (nucleophilic) addition-elimination [1] 1\n© UCLES 2023 Page 17 of 18\n\n9(c) chlorobenzene, chloroethane, benzoyl chloride [1] u / c 3\nexplanation linked to their order\n• correct link to: strengthening (ArCl / RCl) C-Cl bond\nOR weakening (RCOCl) C-Cl bond\nOR C-Cl bond has partially double bond character (ArCl)\nOR C-Cl is more difficult to break (linked correctly)\nchlorobenzene:\n• lone pair / p-orbital on Cl overlaps / delocalised / incorporated with ring\nbenzoyl chloride:\n• C of C-Cl has most electron deficient\nOR has an electronegative oxygen atom\n/ two electronegative atoms / electron withdrawing C=O group\nchloroethane\n• electron donating effect / positive inductive effect of alkyl / R group\nany two [1] any three [2]\n© UCLES 2023 Page 18 of 18", "source_pages": [ 16, 17, 18 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_43.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_43-p16.png", "../answer-assets/9701_w23_ms_43-p17.png", "../answer-assets/9701_w23_ms_43-p18.png" ] }, { "id": "9701-2023-on-51-q01", "question_id": "9701-2023-on-51-q01", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 12, "status": "available", "reason": null, "text": "1(a)(i) M1: 27.30, 27.00, 26.70, 26.80 2\nM2: correct averaging of two (or more) titres within 0.10 cm3 of each other.\n(Expected answer = (26.70 + 26.80) / 2 = 26.75)\n\n1(a)(ii) M1: n(Na S O (aq)) = (a(i) / 1000)  0.001 3\n2 2 3\n(Expected answer = 26.75 / 1000 x 0.001 = 2.675  10-5)\nM2: n(O (aq)) = M1 / 4\n2\n(Expected answer = 2.675  10-5 / 4 = 6.6875 x 10-6)\nM3: concentration of oxygen = M2  1000 / 25 = M2 x 40\n(Expected answer = 6.6875  10-6  1000 / 25\n= 2.675  10-4 mol dm-3)\nAnswer to at least 3SF\n\n1(b) (25.0 cm3) volumetric pipette 1\n\n1(c) to avoid any oxygen from (trapped) air (in the flask) reacting with / oxidising Mn(OH) 1\n2\n\n1(d)(i) M1: All points plotted correctly. 2\nM2: smooth curve LOBF drawn.\n\n1(d)(ii) Correct value read from candidate’s graph. 1\nValue given to at least 3SF.\n(Expected value – 2.57  10-4 moldm-3)\n© UCLES 2023 Page 8 of 11\n\n1(d)(iii) M1 Point most anomalous to the plotted line of best fit circled. 2\nM2:\nFor anomalous point below the LOBF\nThe (actual) temperature > recorded / measured temperature (in the table).\nFor anomalous point above the LOBF.\nThe (actual) temperature < recorded / measured temperature (in the table).\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_51-p08.png", "../answer-assets/9701_w23_ms_51-p09.png" ] }, { "id": "9701-2023-on-51-q02", "question_id": "9701-2023-on-51-q02", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 18, "status": "available", "reason": null, "text": "2(a) M1: Leakproof reaction vessel capable of delivering a sample of gas. 3\nM2: Suitable method of collection and measurement of the gas produced. (eg. over water or gas syringe).\nM3: method for monitoring & maintaining temperature (thermostatically controlled water bath).\n\n2(b)(i) M1: A tangent line which is… 2\n• a straight line\n• starts at / passes through 0,0.\n• keeps to the left of the curve of the original line.\n• covers at least 25 cm3 (5 big squares) on y-axis.\n• passes between 4.0 and 5.5 minutes (inc.) at 25 cm3\nM2: correct calculation of the gradient of tangent line drawn.\n\n2(b)(ii) The total volume of gas produced divided by the time taken is the mean rate (rather than the initial rate). 1\n\n2(b)(iii) It (the line / curve) would be below / lower (than the one on the graph) (starting at 0,0). 1\n© UCLES 2023 Page 9 of 11\n\n2(b)(iv) A second curve which: 1\n• starts at 0,0 (± 1 small square)\n• is less steep than curve for 0.0750 moldm-3. (before any plateau)\n• produces 27.5 cm3 gas in total by 45 minutes. (± half a small square).\n• Extends from 0 – 45 minutes.\n\n2(c)(i) 1\nconcentration rate\n0.5 0.000714 7.14 10-4\n0.4 0.000455 4.55 10-4\n0.3 0.000313 3.13 10-4\n0.2 0.000234 2.34 10-4\n0.1 0.000123 1.23 10-4\n\n2(c)(ii) explanation using numbers from the table that whatever change is made to the concentration causes the same change in 1\nthe rate (within experimental error).\n\n2(c)(iii) time (taken to collect 0.0150dm3 of N ) 1\n2\n\n2(c)(iv) M1: Correct volume, 40(.0) cm3, (of 0.5(00) moldm-3 ) C H N Cl(aq) is used / measured. 3\n6 5 2\nM2: measure the volume from M1 C H N Cl using a (50 cm3) burette into a 100 cm3 volumetric flask.\n6 5 2\nM3: make up to the (calibration) mark with distilled water.\nAND\nthen mix the solution (by inverting the flask)\n\n2(c)(v) to make sure the concentration does not change (from 0.200 moldm-3) (by making sure the C H N Cl does not decompose) 1\n6 5 2\n© UCLES 2023 Page 10 of 11\n\n2(d)(i) Two acceptable reasons for low yield. 2\n1 Some C H N Cl had decomposed / reacted with water (before reacting with phenylamine) OR temperature rose above\n6 5 2\n5 oC (in step 1) OR phenylamine used was impure.\n2 Some dye / solid is not recovered / is lost through / left in filter OR some of the dye dissolves in the water / is aqueous (in\nstep 2).\n\n2(d)(ii) (Residual) water causes solid to have a different / inaccurate (lower) melting point (to the dry solid). 1\nOR\nTo ensure all water has been driven off to increase accuracy in measurement of melting point\n© UCLES 2023 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_51-p09.png", "../answer-assets/9701_w23_ms_51-p10.png", "../answer-assets/9701_w23_ms_51-p11.png" ] }, { "id": "9701-2023-on-52-q01", "question_id": "9701-2023-on-52-q01", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "1(a)(i) 9 points plotted correctly 2\nTwo straight lines of best fit which correctly intersect\n\n1(a)(ii) Temperature change = 1\nintersection point temperature (read from graph) – 18.8\n\n1(a)(iii) intersection point volume (read from graph) 1\n\n1(a)(iv) n(HCl) = 1.00  (25.00/1000) = 0.0250 1\n[NaOH] = 0.0250  (1000/1(a)(iii)) 1\n\n1(a)(v) The indicator: 1\nchanges colour rapidly / instantly\nOR\ndetects end-point / neutralisation rapidly / instantly\n\n1(b) (25 cm3) Volumetric pipette 1\n\n1(c) ((2  0.1)/2.5)  100 = 8(.00) (%) 1\n\n1(d) mcΔT = 1\n(25.00 + 22.10)  4.18 x 6.0\n= 1181.268 (J)\n= q/n(NaOH) (÷1000) 1\n= 1181.268 / (1.00  22.10/1000) ÷ 1000\n= – 53.5 (kJ mol–1)\n\n1(e) Heat loss (to the surroundings) 1\n\n1(f) Ethanoic acid is a weak acid 2\nEnergy is required for dissociation of ethanoic acid\n© UCLES 2023 Page 5 of 7", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_52-p05.png" ] }, { "id": "9701-2023-on-52-q02", "question_id": "9701-2023-on-52-q02", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "2(a) Contents of flask 3\nSealed apparatus\nCollection of the gas in a labelled suitable container\n\n2(b) 12.5(0) cm3 (of 2.00 mol dm–3) HCl 3\nUsing burette transfer and add to a volumetric flask\nTop the (volumetric) flask up to 250 cm3 mark with distilled water\n\n2(b) 12.5(0) cm3 (of 2.00 mol dm–3) HCl\nUsing burette transfer and add to a volumetric flask\nTop the (volumetric) flask up to 250 cm3 mark with distilled water\n\n2(c)(i) (The more concentrated) HCl is corrosive 1\nOR\n(The more concentrated) HCl is an irritant\n\n2(c)(ii) 1\n2.00 15.82 0.0640\n3.00 10.41 0.0961\n4.00 7.81 0.128\n5.00 6.25 0.160\n6.00 5.24 0.191\n\n2(c)(iii) Straight line going through the origin with a positive gradient 1\n\n2(c)(iv) Straight line labelled B passing through the origin but with steeper gradient than line A 1\n© UCLES 2023 Page 6 of 7\n\n2(c)(v) Rate = k [HCl] 1\n\n2(d)(i) (at higher concentrations) line curves (upwards) 2\nRate (1/t) increases as temperature increases\n\n2(d)(ii) Use a water bath 1\n\n2(e)(i) (remove oxide layer) by using sandpaper 1\n\n2(e)(ii) (zinc oxide + acid) doesn’t produce hydrogen (at the start) 1\n© UCLES 2023 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_52-p06.png", "../answer-assets/9701_w23_ms_52-p07.png" ] }, { "id": "9701-2023-on-53-q01", "question_id": "9701-2023-on-53-q01", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 12, "status": "available", "reason": null, "text": "1(a)(i) M1: 27.30, 27.00, 26.70, 26.80 2\nM2: correct averaging of two (or more) titres within 0.10 cm3 of each other.\n(Expected answer = (26.70 + 26.80) / 2 = 26.75)\n\n1(a)(ii) M1: n(Na S O (aq)) = (a(i) / 1000)  0.001 3\n2 2 3\n(Expected answer = 26.75 / 1000 x 0.001 = 2.675  10-5)\nM2: n(O (aq)) = M1 / 4\n2\n(Expected answer = 2.675  10-5 / 4 = 6.6875 x 10-6)\nM3: concentration of oxygen = M2  1000 / 25 = M2 x 40\n(Expected answer = 6.6875  10-6  1000 / 25\n= 2.675  10-4 mol dm-3)\nAnswer to at least 3SF\n\n1(b) (25.0 cm3) volumetric pipette 1\n\n1(c) to avoid any oxygen from (trapped) air (in the flask) reacting with / oxidising Mn(OH) 1\n2\n\n1(d)(i) M1: All points plotted correctly. 2\nM2: smooth curve LOBF drawn.\n\n1(d)(ii) Correct value read from candidate’s graph. 1\nValue given to at least 3SF.\n(Expected value – 2.57  10-4 moldm-3)\n© UCLES 2023 Page 8 of 11\n\n1(d)(iii) M1 Point most anomalous to the plotted line of best fit circled. 2\nM2:\nFor anomalous point below the LOBF\nThe (actual) temperature > recorded / measured temperature (in the table).\nFor anomalous point above the LOBF.\nThe (actual) temperature < recorded / measured temperature (in the table).\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_53-p08.png", "../answer-assets/9701_w23_ms_53-p09.png" ] }, { "id": "9701-2023-on-53-q02", "question_id": "9701-2023-on-53-q02", "subject": "9701", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 18, "status": "available", "reason": null, "text": "2(a) M1: Leakproof reaction vessel capable of delivering a sample of gas. 3\nM2: Suitable method of collection and measurement of the gas produced. (eg. over water or gas syringe).\nM3: method for monitoring & maintaining temperature (thermostatically controlled water bath).\n\n2(b)(i) M1: A tangent line which is… 2\n• a straight line\n• starts at / passes through 0,0.\n• keeps to the left of the curve of the original line.\n• covers at least 25 cm3 (5 big squares) on y-axis.\n• passes between 4.0 and 5.5 minutes (inc.) at 25 cm3\nM2: correct calculation of the gradient of tangent line drawn.\n\n2(b)(ii) The total volume of gas produced divided by the time taken is the mean rate (rather than the initial rate). 1\n\n2(b)(iii) It (the line / curve) would be below / lower (than the one on the graph) (starting at 0,0). 1\n© UCLES 2023 Page 9 of 11\n\n2(b)(iv) A second curve which: 1\n• starts at 0,0 (± 1 small square)\n• is less steep than curve for 0.0750 moldm-3. (before any plateau)\n• produces 27.5 cm3 gas in total by 45 minutes. (± half a small square).\n• Extends from 0 – 45 minutes.\n\n2(c)(i) 1\nconcentration rate\n0.5 0.000714 7.14 10-4\n0.4 0.000455 4.55 10-4\n0.3 0.000313 3.13 10-4\n0.2 0.000234 2.34 10-4\n0.1 0.000123 1.23 10-4\n\n2(c)(ii) explanation using numbers from the table that whatever change is made to the concentration causes the same change in 1\nthe rate (within experimental error).\n\n2(c)(iii) time (taken to collect 0.0150dm3 of N ) 1\n2\n\n2(c)(iv) M1: Correct volume, 40(.0) cm3, (of 0.5(00) moldm-3 ) C H N Cl(aq) is used / measured. 3\n6 5 2\nM2: measure the volume from M1 C H N Cl using a (50 cm3) burette into a 100 cm3 volumetric flask.\n6 5 2\nM3: make up to the (calibration) mark with distilled water.\nAND\nthen mix the solution (by inverting the flask)\n\n2(c)(v) to make sure the concentration does not change (from 0.200 moldm-3) (by making sure the C H N Cl does not decompose) 1\n6 5 2\n© UCLES 2023 Page 10 of 11\n\n2(d)(i) Two acceptable reasons for low yield. 2\n1 Some C H N Cl had decomposed / reacted with water (before reacting with phenylamine) OR temperature rose above\n6 5 2\n5 oC (in step 1) OR phenylamine used was impure.\n2 Some dye / solid is not recovered / is lost through / left in filter OR some of the dye dissolves in the water / is aqueous (in\nstep 2).\n\n2(d)(ii) (Residual) water causes solid to have a different / inaccurate (lower) melting point (to the dry solid). 1\nOR\nTo ensure all water has been driven off to increase accuracy in measurement of melting point\n© UCLES 2023 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9701_w23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2023-oct-nov/9701_w23_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w23_ms_53-p09.png", "../answer-assets/9701_w23_ms_53-p10.png", "../answer-assets/9701_w23_ms_53-p11.png" ] }, { "id": "9701-2024-m-42-q01", "question_id": "9701-2024-m-42-q01", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 26, "status": "available", "reason": null, "text": "1(a)(i) (enthalpy change when) one mole of a substance / solute 1\nAND dissolves in water / turns into an aqueous solution\n(to form a solution of infinite dilution)\n\n1(a)(ii) there is a (large) increase in entropy OR S is positive OR TS is positive 1\nso G is negative / TS outweighs H 1\n\n1(b)(i) anionic charge density decreases (down the group / Cl– to I–) 1\n(so hydration enthalpies become less negative / less exothermic because) 1\nless attraction of ion to water / the dipole–ion force weakens\n\n1(b)(ii) the difference between H and H remains roughly constant 1\nlatt hyd\nOR H and H become less exothermic by a similar amount\nlatt hyd\n\n1(b)(iii) H (K+(g)) = –629 + 21.0 –(–293) = –315 kJ mol–1 1\nhyd\n\n1(b)(iv) solubility of PbI = 7.1  10–9 = x·(2x)2 ∴ x = ∛(¼  7.1  10–9) 1\n2\nsolubility of PbI = 1.21  10–3 mol dm–3 min 2sf 1\n2\n\n1(b)(v) • Pb2+ has a greater charge 2\n• Pb2+ is smaller (than K+) OR (Pb2+) smaller ionic radius\n• greater attraction between Pb2+ and I–\nOR ionic bond between Pb2+ and I– is stronger\nOR lattice energy is more exothermic / more negative\nmark as  ✓ ✓\n\n1(c)(i) S = 2(155.5) + 2(116.1) – 4(106.3) – 2(213.6) – 205.2 1\nS = –514.4 (J K–1 mol–1) min 3sf ECF 1\n© Cambridge University Press & Assessment 2024 Page 5 of 13\n\n1(c)(ii) G = H – TS AND use of 298 K for T OR clear working to represent this 1\nG = –203.4 – 298(–514.4 / 1000)\nG = –50.1 OR –50.2 (kJ mol–1) OR G = –50108.8 J min 3sf ECF from (c)(i) 1\n(negative so spontaneous)\n\n1(c)(iii) increases (in thermal stability down the group) 1\nAND (cat)ionic radius / ion size increases (down the group)\nless polarisation of anion / C—O bond / distortion of carbonate ion / CO 2– 1\n3\nOR C–O is less weakened / stronger (down the group)\n\n1(d)(i) 2H+(aq) + 2e– → H (g) state symbols required 1\n2\n\n1(d)(ii) 21.35 1\nM1 moles of S 2 O 3 2– = × 0.100 = 2.135  10–3\n1000\nM2 calc of Q = 2.135  10–3  0.5  2  96500 = 206.02 / 205.68 ECF 1\nM3 = answer I = 206 / (8  60) = 0.428 OR 0.429 A min 2sf ECF 1\n\n1(e)(i) reaction 1 concentrated HNO AND concentrated H SO (concentrated seen once) 1\n3 2 4\nreaction 2 Sn AND (concentrated) HCl 1\n\n1(e)(ii) step 1 NaNO\n2\n+ HCl → HNO\n2\n+ NaCl 1\nstep 2 C H NH + HNO + H+ → C H N + + 2H O 1\n6 5 2 2 6 5 2 2\n\n1(e)(iii) nucleophilic (aromatic) substitution 1\n© Cambridge University Press & Assessment 2024 Page 6 of 13", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_42-p05.png", "../answer-assets/9701_m24_ms_42-p06.png" ] }, { "id": "9701-2024-m-42-q02", "question_id": "9701-2024-m-42-q02", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 13, "status": "available", "reason": null, "text": "2(a)(i) ● conjugate base of acid I = OH– 1\n● acid II = HNO ● conjugate base of acid II = NO –\n2 2\n\n2(a)(ii) lone pair on the N can be donated to a proton /H+ 1\nOR lone pair on the N can accept / gain a proton /H+\nOR lone pair on the N can form a dative bond to a proton/H+\n\n2(a)(iii) H O + CH COOH → H O+ + CH COO– 1\n2 3 3 3\n\n2(b)(i) (K ) = [H+][OH–] ALLOW (K ) = K  K 1\nw w a b\n\n2(b)(ii) endothermic 1\nAND equilibrium position moves right OR as water dissociates more\nOR K increases with temperature\nw\n\n2(b)(iii) M1 (K increases with temperature so) pH of neutral solution decreases 1\nw\nOR (from graph) K = 1.50  10–14 = [H+]2\nw\n∴ neutral pH = –½ log (1.50  10–14) = 6.91\nOR [H+] = 10–7.00\n∴ [OH–] = 1.50  10–14 / 10–7.00 = 1.50  10–7\nM2 pH 7 is therefore above neutral pH / is alkaline 1\nOR [OH–] > [H+] (so alkaline)\n\n2(c)(i) H\n2\nO(l) particles / molecules has more randomness / disorder 1\nOR\nH\n2\nO(l) has more ways to arrange particles / energy (than in solid)\n\n2(c)(ii) H O(g) particles / molecules has much more randomness / disorder 1\n2\nOR\nH O(g) has many more ways to arrange particles / energy (than in liquid)\n2\n\n2(c)(iii) +6030 / (70.1 – 48.0) = 272.85 / 272.9 / 273 K (which is 0 °C) 1\n\n2(d)(i) (+)1.62 V 1\n© Cambridge University Press & Assessment 2024 Page 7 of 13\n\n2(d)(ii) [OH–] = 10–14/10–11 = 1  10–3 1\nE = +0.40 – ½  0.059  log (10–3)2 = +0.40 – 0.059(–3) = +0.577 (V) min 2sf 1\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_42-p07.png", "../answer-assets/9701_m24_ms_42-p08.png" ] }, { "id": "9701-2024-m-42-q03", "question_id": "9701-2024-m-42-q03", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "3(a)(i) energies / energy levels of the 3d and the 4s (sub-shells / orbitals) are similar 1\nOR the difference between the 3d and the 4s is small\n\n3(a)(ii) Fe = [Ar] 3d6 4s2 AND 1\nFe3+ = [Ar] 3d5\n\n3(b)(i) (a molecule or ion formed by a central) metal atom / ion 1\nsurrounded / bonded by one or more ligands\n\n3(b)(ii) 6 / six 1\n\n3(b)(iii) bond angle 106–108° 1\nAND lone pair from O is donated (in a bond)\nOR one more of O’s electron pairs is now a bond/bonding pair (so repels less)\n\n3(b)(iv) M1 (degenerate) d orbitals split (into two energy levels) 1\nOR (degenerate) d orbitals become non-degenerate\nM2 as an electron moves up / to a higher energy level 1\nM3 absorption of light energy / energy in the visible region 1\nAND colour seen is complementary (to colour absorbed)\n\n3(b)(v) (d–d) energy gap / E /  is different 1\noct\ndifferent frequency / wavelength (of light) absorbed / transmitted / reflected 1\n\n3(c)(i) [Fe(H O) ]3+ + H PO → [Fe(H O) (H PO )]2+ + H O+ 1\n2 6 3 4 2 5 2 4 3\n© Cambridge University Press & Assessment 2024 Page 8 of 13\n\n3(c)(ii) [[Fe(H O) (SCN)]2+] 1\nK = 2 5\nstab\n[[Fe(H O) ]3+][SCN–]\n2 6\nmol–1 dm3 u/c 1\n\n3(c)(iii) 1.30  102 / 59.0 = 2.2(03) min 2sf 1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_42-p08.png", "../answer-assets/9701_m24_ms_42-p09.png" ] }, { "id": "9701-2024-m-42-q04", "question_id": "9701-2024-m-42-q04", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "4(a)(i) X [Ru(H\n2\nO)\n5\nCl ]2+ 1\nY [Ru(H\n2\nO)\n4\nCl2 ]+ ECF [1] if reversed 1\n\n4(a)(ii) 2\n\n4(b)(i) (pyrazine has) a lone pair on each N (atoms) / two lone pairs (on the N’s) 1\nAND which can be donated / form a coordinate / dative bond (with Ru)\n\n4(b)(ii) • shape is (hexagonal ring) planar / (trigonal) planar / 120o 2\n• carbons and nitrogens are sp2hybridised\n• a p orbital (from each atom) overlaps sideways/laterally\n(with each other above and below the ring forming  bonds)\nmark as  ✓ ✓\n© Cambridge University Press & Assessment 2024 Page 9 of 13\n\n4(b)(iii) 1/one 1\nall the carbon atoms are equivalent / in same environment 1\nOR pyrazine is a symmetrical molecule u/c\n\n4(b)(iv) +2 / 2+ AND +3 / 3+ OR +4 / 4+ AND +1 / 1+ 1\n\n4(c)(i) SOCl2 OR PCl3 OR PCl5 1\n\n4(c)(ii)\nM1 correct ester linkage in the middle of the polymer 1\n(displayed C=O bond attached to benzene ring)\nM2 rest of structure 1\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_42-p09.png", "../answer-assets/9701_m24_ms_42-p10.png" ] }, { "id": "9701-2024-m-42-q05", "question_id": "9701-2024-m-42-q05", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 17, "status": "available", "reason": null, "text": "5(a)(i) AlCl3 + CH\n3\nCl → AlCl4 – + CH\n3\n+ 1\n© Cambridge University Press & Assessment 2024 Page 10 of 13\n\n5(a)(ii) 1\n+ H+\ncurly arrow from inside hexagon to electrophile (CH +)\n3\nintermediate 1\ncurly arrow from C—H bond into the hexagon AND H+ formed 1\n\n5(a)(iii) C\n6\nH\n4\n(Cl)CH\n3\n+ 2[O] → C\n6\nH\n4\n(Cl)CHO + H\n2\nO 1\n\n5(a)(iv) step 3 HCN AND KCN (cat) OR KCN AND H\n2\nSO\n4\n/HCl 1\nstep 4 H\n2\nSO\n4\n(aq) OR HCl (aq) 1\n\n5(a)(v) addition–elimination / condensation 1\n\n5(a)(vi) (a strong base) it would hydrolyse the ester 1\n\n5(a)(vii) (a substance able to) rotate the plane of plane-polarised light 1\n\n5(a)(viii) any two of: 2\n• reduced / different biological activity of ‘other’ enantiomer ORA\n• avoids need to separate the optical isomers to form the pure active isomer\n• lower dosage required OR (drug is) more potent\n• higher yield (of biologically-active molecule)\n• no / less (harmful) side effects OR other isomer can have side effects\n\n5(b)(i) M1 [HA] = (75  10–3 / 180) ÷ 0.100 = 4.17  10–3 OR 1 / 240 (mol dm–3) 1\nM2 [H+] = (K  [HA])½= (10–3.49  4.17  10–3)½ = 1.16  10–3 (mol dm–3) 1\na\nM3 pH = –log [H+] = 2.93 to 2.94 min 2sf 1\n© Cambridge University Press & Assessment 2024 Page 11 of 13\n\n5(b)(ii) 1\n1\nCH COOH\n3\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_42-p10.png", "../answer-assets/9701_m24_ms_42-p11.png", "../answer-assets/9701_m24_ms_42-p12.png" ] }, { "id": "9701-2024-m-42-q06", "question_id": "9701-2024-m-42-q06", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 17, "status": "available", "reason": null, "text": "6(a)(i) pH at which a molecule has 1\nno overall charge / is neutral/ is a Zwitterion / charges cancel out\n\n6(a)(ii) 1\n\n6(b)(i) ethanol AND heat in a sealed tube 1\nOR ethanol AND high pressure\n\n6(b)(ii) C\n6\nH\n5\nCOCl / benzoyl chloride / benzoyl anhydride 1\n\n6(b)(iii) 2\n• carboxylic acid to primary alcohol COOH to CH OH\n2\n• amide to amine CONH to CH NH\n2\n• rest of the molecule is correct (carbons and benzene ring)\nmark as  ✓ ✓\n© Cambridge University Press & Assessment 2024 Page 12 of 13\n\n6(b)(iv) 2\n\n6(c)(i) M1 nitrogen lone pair in amides is delocalised with C=O 1\nM2 lone pair less available for donation / to accept H+ 1\nOR less electron density on N / NH so less able to accept H+\n2\n\n6(c)(ii) chloroethanoic acid is a stronger acid (than ethanoic acid) 1\nbecause electron-withdrawing (–I / inductive) effect of Cl substituent 1\nAND weakens O—H / carboxylate anion stabilised\n\n6(d)(i) 4\nproton a b c d e\n 9.0–13.0 6.0–9.0 2.3–3.0 3.2–4.0 1.0–5.0\nsplitting singlet multiplet triplet triplet singlet\nany three [1] any five [2] any seven [3] all nine [4]\n\n6(d)(ii) 1\nproton a b c d e\npresent in D 2 O ✓ ✓ ✓\n© Cambridge University Press & Assessment 2024 Page 13 of 13", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_42-p12.png", "../answer-assets/9701_m24_ms_42-p13.png" ] }, { "id": "9701-2024-m-52-q01", "question_id": "9701-2024-m-52-q01", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 18, "status": "available", "reason": null, "text": "1(a) M1 Order of weighing 2\n• 1 boat + AgNO is weighed\n3\n• 2 (AgNO transferred)\n3\n• 3 empty boat reweighed\nM2 table + units\n/ g\nMass of boat + AgNO before\n3\ntransfer\nMass of boat after transfer\nMass of AgNO transferred\n3\n\n1(b) M1 Dissolve the solid / silver nitrate (in the beaker) (using a small volume of) (distilled) water 3\nM2 Transfer / add to a (250 cm3 volumetric) flask\nand\nRinse (with distilled water)\nM3 Top the 250 cm3 volumetric flask up to mark (with distilled water)\n\n1(c) Sunlight decomposes silver nitrate 1\n\n1(d) Filter 1\n\n1(e)(i) (10 cm3) (volumetric) pipette 1\n\n1(e)(ii) Dropping pipette 1\n(as 1 cm3, not 1.00 cm3 is asked for)\n\n1(f) Wear chemically resistant gloves 1\n\n1(g) Solution X 1\n© Cambridge University Press & Assessment 2024 Page 5 of 8\n\n1(h)(i) 23.40; 22.35; 22.60; 22.45 1\n\n1(h)(ii) (22.35 + 22.45) ÷ 2 = 22.40 (cm3) 1\n\n1(h)(iii) M1 Mol AgNO used = 10.62 / 169.9 = 0.0625… (mol) 3\n3\nM2 Mol Ag+(aq) reacting with Cl–(aq)\n= M1  22.40/250\n= 5.60  10–3 (mol)\nM3 Conc Cl –(aq) = M2  1000 / 10\n= 0.560 (mol dm–3)\n\n1(h)(iv) [(0.05  2) / 22.60]  100 = 0.442% 1\nWorking must be shown\n\n1(i) Ag+ ions react with other substances in the sample of sea water 1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_52-p05.png", "../answer-assets/9701_m24_ms_52-p06.png" ] }, { "id": "9701-2024-m-52-q02", "question_id": "9701-2024-m-52-q02", "subject": "9701", "year": 2024, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 12, "status": "available", "reason": null, "text": "2(a) Step 3 1\n\n2(b) M1 Lowers the temperature (of reaction mixture) 2\nM2 Decreases concentration (in reaction mixture)\n© Cambridge University Press & Assessment 2024 Page 6 of 8\n\n2(c)(i) M1 Column 4 M2 Column 5 2\n1 2 3 4 5\nV t (V ∞ – V t )\n/ cm3 / cm3\n1 1 12.60 1.20 13.30\n2 13 17.70 6.30 8.20\n3 20 19.90 8.50 6.00\n4 30 22.10 10.70 3.80\n5 40\n6 50 24.90 13.50 1.00\n7 60 25.90 14.50 0.00\n\n2(c)(ii) Time 1\n\n2(c)(iii) Temperature (of flask B) 1\n\n2(c)(iv) Yes 1\nand\nThe value of the titre is accurate and corresponds to correct time\n\n2(c)(v) M1 6 points plotted correctly 2\nM2 Line of best fit drawn\n© Cambridge University Press & Assessment 2024 Page 7 of 8\n\n2(c)(vi) 25.90 – (V – V) at 40 minutes from the graph 1\n∞\n\n2(c)(vii) No and there is an anomalous point on the graph 1\nor\nYes and all the points lie on/close to the line of best fit\n© Cambridge University Press & Assessment 2024 Page 8 of 8", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2024-March/ms/9701_m24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-march/9701_m24_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m24_ms_52-p06.png", "../answer-assets/9701_m24_ms_52-p07.png", "../answer-assets/9701_m24_ms_52-p08.png" ] }, { "id": "9701-2024-mj-41-q01", "question_id": "9701-2024-mj-41-q01", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 12, "status": "available", "reason": null, "text": "1(a)(i) M1 strontium > calcium > magnesium 4\nM2H and H both become less exothermic / less negative\nlatt hyd\nM3H changes more\nlatt\nM4H becomes more exothermic / more negative\nsol\n\n1(a)(ii) pH increases as the concentration of [OH–] ions increases 1\n\n1(b) M1 [H+] = 10–12.2 OR 6.31  10–13 4\nM2 [OH–] = 0.0158 mol dm–3 ecf\nM3 [Ba(OH) ] = 0.01585 ÷ 2 = 7.92 ecf\n2\nM4 mol of Ba(OH) = 0.00792 ÷ 4 = 1.98  10–3\n2\nmass Ba(OH) = 0.339 g ecf\n2\n\n1(c)(i) K = [Fe2+][OH–]2 1\nsp\n\n1(c)(ii) M1 K = 4  (5.85  10–6)3 = 8.01  10–16 2\nsp\nM2 mol3 dm–9\nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p05.png" ] }, { "id": "9701-2024-mj-41-q02", "question_id": "9701-2024-mj-41-q02", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "2(a)(i) element which forms one or more stable ion with incomplete d orbitals 1\n\n2(a)(ii) they have vacant d orbitals that are energetically accessible 1\n\n2(b)(i) orbitals of the same energy 1\n\n2(b)(ii) 1\n© Cambridge University Press & Assessment 2024 Page 5 of 18\n\n2(c) M1 2AgNO + 2NaOH → Ag O + 2NaNO + H O 2\n3 2 3 2\nOR 2Ag+ + 2OH– → Ag O + H O\n2 2\nM2 Ag O + 4NH + H O → 2[Ag(NH ) ]OH\n2 3 2 3 2\nOR Ag+ + 2NH + OH– → [Ag(NH ) ]OH\n3 3 2\n\n2(d) M1 bond angle H–N–Ag is 109.5° 2\nM2 shape is linear AND bond angle for N–Ag–N is 180°\n\n2(e) M1 Ag O + H O + 2e– → 2Ag + 2OH– 2\n2 2\nM2 Zn + 2OH– → Zn(OH) + 2e–\n2\n\n2(f) For example: 2\nM1 presence of dps ligand bonded to two Ru\nM2 the rest of the structure correct\n© Cambridge University Press & Assessment 2024 Page 6 of 18", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p05.png", "../answer-assets/9701_s24_ms_41-p06.png" ] }, { "id": "9701-2024-mj-41-q03", "question_id": "9701-2024-mj-41-q03", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "3(a) Li C O H O → Li CO + H O + CO 1\n2 2 4 2 2 3 2\n\n3(b) M1 CaC O AND Ca2+has a smaller ionic radius / Ca2+has a higher charge density 2\n2 4\nM2 anion / C O 2– becomes more polarised / distorted\n2 4\n\n3(c) M1 d orbital(s) of different energy / d-d splitting occurs 3\nM2 electron(s) promoted / excited\nM3 light is absorbed AND colour seen is complementary\n\n3(d) 2K Fe(C O ) → 2K Fe(C O ) + K C O + 2CO 1\n3 2 4 3 2 2 4 2 2 2 4 2\n\n3(e) 2\n\n3(f) M1 HC O – + H+ → H C O 2\n2 4 2 2 4\nM2 HC O – + OH– → C O 2–+ H O\n2 4 2 4 2\n\n3(g) M1 2C O 2– + O (g) + 2H O(l) → 4CO (g) + 4OH–(aq) 2\n2 4 2 2 2\nM2 E = 0.40 – (–0.59) = 0.99 V\ncell\n© Cambridge University Press & Assessment 2024 Page 7 of 18", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p07.png" ] }, { "id": "9701-2024-mj-41-q04", "question_id": "9701-2024-mj-41-q04", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "4(a) M1 voltage of a half-cell compared to SHE 2\nM2 1 mol dm–3 AND 101 kPa AND 298 K\n\n4(b)(i) 3\nM1 Pt Ag Ag+\nM2 gas delivery system H H+\n2\nM3 voltmeter salt bridge wiring Pt to V to Ag\n4b(ii) E value would be more negative AND shifts Ag+ (+ e–) ⇋ Ag to the left 1\n\n4(c) enthalpy change when one mole of a solute dissolves in water 1\n\n4(d)(i) 2\nM1 two arrows in blue\nM2 correct species in red with all state symbols\n© Cambridge University Press & Assessment 2024 Page 8 of 18\n\n4(d)(ii) H = –811.6 (kJ mol–1) 1\nlatt\n\n4(e) M1 Mg(NO ) (s), NaNO (s), RbNO (s) 3\n3 2 3 3\nM2 Mg2+ has a higher charge than Na+ OR Rb+\nAND Na+ has a smaller radius than Rb+\nM3 correct statement relating magnitude of LE to attraction between ions OR strength of ionic bonds\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p08.png", "../answer-assets/9701_s24_ms_41-p09.png" ] }, { "id": "9701-2024-mj-41-q05", "question_id": "9701-2024-mj-41-q05", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 8, "status": "available", "reason": null, "text": "5(a)(i) tangent drawn at t = 0 AND gradient of tangent calculated at t = 0 1\nanswer between 0.016 to 0.040\n\n5(a)(ii) [I–] stays constant / [I–] does not change 1\nOR the overall order is one under these conditions\n\n5(b) M1 equation 1: 2Fe2+ + S O 2– → 2Fe3+ + 2SO 2– 2\n2 8 4\nM2 equation 2: 2Fe3+ + 2I– → 2Fe2+ + I\n2\n\n5(c) the rate constant and rate of reaction 1 will both increase 1\n\n5(d) k = 0.693 ÷ t 1\n½\nt = 0.693 ÷ 0.0158 = 43.9 s\n½\n\n5(e) NO + Br → NOBr OR NO + Br → NOBr + Br 2\n2 2 2\nNOBr + NO → 2NOBr OR Br + NO → NOBr\n2\nM1 two equations adding up to the overall equation\nM2 step 1 has one NO and one Br only and is identified as slow step\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 18", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p09.png" ] }, { "id": "9701-2024-mj-41-q06", "question_id": "9701-2024-mj-41-q06", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 15, "status": "available", "reason": null, "text": "6(a)(i) ratio of the concentrations of solute in two solvents 1\n\n6(a)(ii) M1 10.5 = (y ÷ 25) ÷ ((0.74–y) ÷ 40) 2\nM2 y = 0.642 g\n\n6(b) 2\nsp sp2 sp3\nKekulé benzene 0 6 0\nDewar benzene 0 4 2\nLadenburg benzene 0 0 6\nM1 row 1 and 2 correct\nM2 row 3 correct\n\n6(c) M1 120° AND hexagonal/trigonal planar 2\nM2 C-C has -bonds and -bonds AND C-H have -bonds only\n\n6(d) bond strain OR ring strain 1\n\n6(e) 1\npeaks\nDewar benzene 2\nLadenburg benzene 1\ndelocalised benzene 1\n\n6(f)(i) BrCH CH CH CH Br + FeBr → BrCH CH CH CH + + FeBr – 1\n2 2 2 2 3 2 2 2 2 4\n© Cambridge University Press & Assessment 2024 Page 10 of 18\n\n6(f)(ii) 3\nM1 curly arrow from inside the hexagon towards C+\nM2 structure of the intermediate\nM3 curly arrow from C–H bond into the ring AND H+\n© Cambridge University Press & Assessment 2024 Page 11 of 18\n\n6(f)(iii) M1 Y 2\nM2\nZ\nO\nOR\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p10.png", "../answer-assets/9701_s24_ms_41-p11.png", "../answer-assets/9701_s24_ms_41-p12.png" ] }, { "id": "9701-2024-mj-41-q07", "question_id": "9701-2024-mj-41-q07", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 12, "status": "available", "reason": null, "text": "7(a) methyl pentanoate 1\n\n7(b)(i) time between injection and detection 1\n\n7(b)(ii) 29.5% 1\n© Cambridge University Press & Assessment 2024 Page 12 of 18\n\n7(c)(i) 2\nester number of peaks seen number of peaks seen\nin proton (1H) NMR in carbon-13 NMR\nB 5 6\nC 4 5\nM1 row 1 correct\nM2 row 2 correct\n\n7(c)(ii) A and B 1\n© Cambridge University Press & Assessment 2024 Page 13 of 18\n\n7(d) 6\none mark for each structure: F, G, H, J, K and L\n© Cambridge University Press & Assessment 2024 Page 14 of 18\n\n7(d) six alternatives for F\n© Cambridge University Press & Assessment 2024 Page 15 of 18", "source_pages": [ 12, 13, 14, 15 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p12.png", "../answer-assets/9701_s24_ms_41-p13.png", "../answer-assets/9701_s24_ms_41-p14.png", "../answer-assets/9701_s24_ms_41-p15.png" ] }, { "id": "9701-2024-mj-41-q08", "question_id": "9701-2024-mj-41-q08", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 8, "status": "available", "reason": null, "text": "8(a)(i) two 1\n\n8(a)(ii) amide 2\namine\nester\ncarboxylic acid / carboxyl\nany two [1] all four [2]\n\n8(b)(i) M1 hydrolysis 2\nM2 acid–base\n\n8(b)(ii) 3\none mark for each structure\n© Cambridge University Press & Assessment 2024 Page 16 of 18", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p16.png" ] }, { "id": "9701-2024-mj-41-q09", "question_id": "9701-2024-mj-41-q09", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 7, "status": "available", "reason": null, "text": "9(a)(i) C H OH + Na → C H ONa + ½H 1\n6 5 6 5 2\n\n9(a)(ii) 1\nboth structures for one mark\n\n9(b) 3\nM1 correct dipole AND curly arrow from inside the hexagon to the carbon atom\nM2 structure of the intermediate\nM3 curly arrow from C-H bond into the ring AND H+\n© Cambridge University Press & Assessment 2024 Page 17 of 18\n\n9(c)(i) 1\nAll three curly arrows for one mark\n\n9(c)(ii) 1\n© Cambridge University Press & Assessment 2024 Page 18 of 18", "source_pages": [ 17, 18 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_41.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_41-p17.png", "../answer-assets/9701_s24_ms_41-p18.png" ] }, { "id": "9701-2024-mj-42-q01", "question_id": "9701-2024-mj-42-q01", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "1(a) M1 magnesium > calcium > strontium 4\nM2 H and H both become less exothermic / less negative\nlatt hyd\nM3 H changes less OR H is dominant factor\nlatt hyd\nM4 H becomes less exothermic / less negative / more positive / more endothermic\nsol\n\n1(b) M1H / energy change when 1 mole of an ionic solid / compound is formed 2\nM2 from gaseous ions (under standard conditions)\n\n1(c) M1 as ionic radii increases AND H less exothermic 2\nlatt\nM2 as ionic charge increases AND H increases/more exothermic\nlatt\n\n1(d)(i) 3\nany two [1] any three [2] all four [3]\n© Cambridge University Press & Assessment 2024 Page 5 of 15\n\n1(d)(ii) M1 selection of ONLY six correct values (–381, 89, 419, 279, –200, 640) AND use of  2 as only multiplier with K 2\nM2 correct evaluation of data used ecf\n–381 = (89  2) + (419  2) + 279 + (–200) + 640 + Ho\nlatt\nHo = –2116 (kJ mol–1)\nlatt\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p05.png", "../answer-assets/9701_s24_ms_42-p06.png" ] }, { "id": "9701-2024-mj-42-q02", "question_id": "9701-2024-mj-42-q02", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 9, "status": "available", "reason": null, "text": "2(a)(i) 2LiNO → Li O + 2NO + ½O 1\n3 2 2 2\n\n2(a)(ii) M1 radius / size of cation / M+ increases OR charge density of ion decreases 2\nM2 less polarisation / distortion of anion / nitrate ion / NO – / less weakening of NO bond\n3\n\n2(b) M1 M2 any two for one mark or all four for two marks: 3\n mol total MnO – = 0.125  0.0500 = 6.25  10–3\n4\n mol Fe2+ = 0.0400  0.0225 = 9.00  10–4\n mol unreacted MnO – = 9.00  10–4 ÷ 5 = 1.80  10–4 ecf\n4\n mol reacted MnO – = 6.25  10–3 - 1.80  10–4 = 6.07  10–3 ecf\n4\nM3\nmol NO – = 2.5  6.07  10–3 = 1.5175  10–2\n2\nconc NaNO = 4  1.5175  10–2 = 6.07-6.08  10–2 mol dm–3 ecf min 2sf\n2\n\n2(c)(i) 3MnO 2– + 4H+ → 2MnO – + MnO + 2H O 2\n4 4 2 2\nM1 MnO 2– as a reactant and MnO – + MnO products identified\n4 4 2\nM2 correct equation\n\n2(c)(ii) (E ) decreases / becomes less positive AND as [H+] decreases AND equilibrium shifts to the left 1\ncell\nOR in alkali the E = 0.60 – 0.56 = 0.04 V (working required)\ncell\n© Cambridge University Press & Assessment 2024 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p06.png" ] }, { "id": "9701-2024-mj-42-q03", "question_id": "9701-2024-mj-42-q03", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 5, "status": "available", "reason": null, "text": "3(a) M1 So = (213.8 + 2  248.2) – (237.8 + 3  205.2) 3\nSo = –143.2 (J K–1 mol–1)\nM2 Ho = (–393.5 + 2  –296.8) – (116.7)\nHo = –1103.8 (kJ mol–1)\nM3 Go = Ho – TSo\nGo = –1103.8 – (298  –0.1432) = –1061.1 to –1061.4 (kJ mol–1) ecf min 3sf\n\n3(b) M1 Go = Ho – TSo AND Go = 0 2\nOR T = Ho / So [1]\n\nM2 T = 261.6 ÷ 0.3655 = 715.7 / 716 / 715 K min 3sf\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p07.png" ] }, { "id": "9701-2024-mj-42-q04", "question_id": "9701-2024-mj-42-q04", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "4(a)(i) the (3)d and (4)s (sub-shells/orbitals) are close/similar in energy 1\n\n4(a)(ii) 1\n© Cambridge University Press & Assessment 2024 Page 7 of 15\n\n4(b)  precipitation solution A: e.g. NaOH / OH– 3\n observations: (pale) blue ppt. / solid\n product: Cu(OH) OR [Cu(OH) (H O) ] ecf from A\n2 2 2 4\n ligand substitution solution B: e.g. HCl / Cl –, NH\n3\n observations: dark/deep blue solution (with NH ) OR yellow solution (with Cl –)\n3\n product: [Cu(NH ) (H O) ]2+ OR [CuCl ]2–\n3 4 2 2 4\nany two [1] any four [2] all six [3]\n\n4(c) M1 (Ag+) d-subshell is full / complete OR d10 OR d-orbitals are full 2\nM2 no d-d(*) transition OR no d electrons promoted/excited\n\n4(d) M1 species with two lone pairs (of electrons) 2\nM2 that form dative (covalent) / co-ordinate bond(s) to a (central) metal atom / ion\n\n4(e)(i) 3\n\n4(e)(ii) optical AND cis-trans / geometical 1\n\n4(e)(iii) trans isomer / trans isomer correctly identified AND dipoles cancel / partial charges cancel / ion dipoles cancel 1\n© Cambridge University Press & Assessment 2024 Page 8 of 15", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p07.png", "../answer-assets/9701_s24_ms_42-p08.png" ] }, { "id": "9701-2024-mj-42-q05", "question_id": "9701-2024-mj-42-q05", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "5(a)(i) M1 (using exp 1 and 2) [NO]  3, rate  9 so 2nd order to NO 3\nM2 (using exp 1 and 3) [NO]  2, [Cl ]  4, rate  16 so 1st order to Cl\n2 2\nOR (using exp 2 and 3) [NO]  2 / 3, [Cl ]  4, rate  1.8 so 1st order to Cl\n2 2\nM3 (rate =) k [NO]2 [Cl ] ecf\n2\n\n5(a)(ii) M1 rate = k [NO]2[Cl ] 2\n2\nk = (3.68  10–2) ÷ (0.0252  0.015) = 3925.33 ecf (a)(i) min 2sf\nM2 dm6 mol–2 min–1 ecf (a)(i)\n\n5(b) 2\nAny two [1] all three [2]\n\n5(c) M1 O + NO → NO + O ALLOW O + NO → NO 2\n3 2 3 2 3 2 5\nM2 NO + NO → N O ALLOW NO + NO → N O + O\n3 2 2 5 5 2 2 5 2\nQuestion Answer Marks", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p09.png" ] }, { "id": "9701-2024-mj-42-q06", "question_id": "9701-2024-mj-42-q06", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 13, "status": "available", "reason": null, "text": "6(a)(i) species / molecules / pair that differ (by the presence or absence) of a H+ ion / proton 1\n\n6(a)(ii) HCOOH + H 2 O ⇌ H 3 O+ + HCOO– 1\n\n6(b)(i) [H+] [CH CH CH COO] 1\nK  3 2 2\na [CH CH CH COOH]\n3 2 2\n© Cambridge University Press & Assessment 2024 Page 9 of 15\n\n6(b)(ii) M1 K = 10–4.82 = 1.51  10–5 OR [H+] = 10–3.25 = 5.62  10–4 2\na\nM2 [HA] = (10–3.25)2 ÷ 1.51  10–5 = 0.021 / 0.0209 / 0.02089 (mol dm–3) ecf min 2sf\n\n6(c)(i) M1 a solution that resists / opposes / minimises changes in pH 2\nM2 when small amounts of acid / H+ and base / alkali / OH– are added to it\n\n6(c)(ii) Method 1 3\nM1 [H+] = 10–5.70 OR 2.00  10–6 OR 1.99526  10–6\nM2 moles CH COONa = 0.400  0.125 = 5.00  10–2\n3\nmoles CH COOH = (5.00  10–2  2.00  10–6) ÷ 1.78  10–5 OR 5.60  10–3\n3 eqm\nM3 [CH COOH] = 5.60  10–3  1000 / 600\n3 initial\n[CH COOH] = 0.00933–0.00936 (mol dm–3) ecf min 2sf\n3 initial\nMethod 2\nM1 pK = –log(1.78  10–5) OR 4.75\na\nM2 pH = pK + log[B–] / [HA]\na\n5.7 = 4.75 + log (0.05 / (0.6x)\n0.95 = log (0.05 / (0.6x)\nM3 100.95 = (0.05 / (0.6x)\n5.348x = 0.05\nx = 0.00935-0.00937 (mol dm–3) ecf min 2sf\n\n6(d)(i) HCOOH → CO + 2H+ + 2e– 1\n2\n\n6(d)(ii) M1 Q = 3.75  40  60 OR 9000 C AND use of 96500 OR 1.6  10–19  6.02  1023 2\nM2 moles of oxygen = 9000 ÷ 386000 = 0.0233\nvolume of O = 0.0233  24000 = 559.6 / 559.8 / 560 cm3 ecf min 2sf\n2\nALLOW [use of Q ÷ (1.6  10–19  6.02  1023)] = 560.6 cm3\n© Cambridge University Press & Assessment 2024 Page 10 of 15", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p09.png", "../answer-assets/9701_s24_ms_42-p10.png" ] }, { "id": "9701-2024-mj-42-q07", "question_id": "9701-2024-mj-42-q07", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 9, "status": "available", "reason": null, "text": "7(a)(i) 2-nitrobenzoic acid OR 2-nitrobenzenecarboxylic acid 1\n\n7(a)(ii) hot / reflux / heat AND (alkaline / acidified / neutral) MnO – / KMnO 1\n4 4\n\n7(a)(iii) COOH / carboxyl group is electron-withdrawing / electronegative AND 3- and 5- / meta- directing 1\n\n7(b)(i) 3\n\n7(b)(ii) M1 step 1: Fe / Sn, conc. HCl 3\nM2 step 2: HNO OR NaNO AND HCl\n2 2\nM3 step 1: heat / reflux / hot AND step 2: ⩽10 °C\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p11.png" ] }, { "id": "9701-2024-mj-42-q08", "question_id": "9701-2024-mj-42-q08", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 16, "status": "available", "reason": null, "text": "8(a) M1 benzylamine > ammonia > phenylamine 3\nM1 M2 Any two for one mark, all three for two marks:\n (basicity linked to) p orbital of N / lone pair on N AND being able to bond / accept / donate to / coordinate to a proton / H+\n (benzylamine) R / alkyl / CH group AND is electron donating / positive inductive effect / has + I effect\n2\n (phenylamine) p orbital of N / lone pair on N is overlaps / incorporated / delocalised in the ring / -bond system\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n8(b)(i) M1 white ppt. 2\nM2 structure\n\n8(b)(ii) M1 lone pair / p-orbital / electrons on the nitrogen / NH AND overlap / delocalised / incorporated AND with the (–electrons) 2\n2\nring /  system\nM2 increasing its electron density (of the ring) OR it can polarise the electrophile / Br better\n2\n\n8(c) lone pair / p-orbital on N is delocalised AND into C=O group / across the two electronegative O and N 1\n© Cambridge University Press & Assessment 2024 Page 12 of 15\n\n8(d) 4\nM1 / M2 any two for one mark, all four for two marks:\n lone pair on N\n correct arrow from lone pair N to C (of C=O)\n correct dipole on C=O\n correct arrow from the C=O bond to O atom\nM3 correct intermediate\nM4 arrow from O(–) to C–O bond AND arrow from C–Cl bond to Cl\nALLOW arrow from anywhere on O– to C–O bond\n\n8(e)(i) pH at which a molecule has no overall charge / is neutral / exist as a zwitterion 1\n\n8(e)(ii) C H CH CH(NH )COO– 1\n6 5 2 2\n\n8(f) 2\nOR\nM1 correct peptide bond displayed (there must be a saturated carbon attached to either side of the peptide group)\nM2 rest of the dipeptide correct\n© Cambridge University Press & Assessment 2024 Page 13 of 15", "source_pages": [ 11, 12, 13 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p11.png", "../answer-assets/9701_s24_ms_42-p12.png", "../answer-assets/9701_s24_ms_42-p13.png" ] }, { "id": "9701-2024-mj-42-q09", "question_id": "9701-2024-mj-42-q09", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 12, "status": "available", "reason": null, "text": "9(a) due to the electron-withdrawing effect/electronegative / –I effect of chlorine 1\nAND stabilising the anion / carboxylate ion OR weakening the O-H bond\n\n9(b)(i) (COOH) + 2SOCl → (COCl) + 2HCl + 2SO 1\n2 2 2 2\n\n9(b)(ii) 2\n\n9(b)(iii) 2\nM1 one correct repeat unit (within their structure)\nM2 the rest of the structure correct\necf on one incorrect monomer used\n[If structure partly or fully skeletal: continuation bonds must be dashed / wavy / different or have through brackets]\n\n9(c)(i) CHI / triiodomethane 1\n3\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n9(c)(ii) 4\nchemical shift splitting number of 1H atoms number of protons on\n() pattern responsible for the peak adjacent carbon atoms\n1.15 triplet 3 2\n2.25 singlet 3 0\n3.60 singlet 2 0\n3.95 quartet / quad 2 3\nruplet\nmark as any three [1] any six [2] any nine [3] all twelve [4]\n\n9(c)(iii) 1\nOR\n© Cambridge University Press & Assessment 2024 Page 15 of 15", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_42-p14.png", "../answer-assets/9701_s24_ms_42-p15.png" ] }, { "id": "9701-2024-mj-43-q01", "question_id": "9701-2024-mj-43-q01", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 12, "status": "available", "reason": null, "text": "1(a)(i) M1 strontium > calcium > magnesium 4\nM2H and H both become less exothermic / less negative\nlatt hyd\nM3H changes more\nlatt\nM4H becomes more exothermic / more negative\nsol\n\n1(a)(ii) pH increases as the concentration of [OH–] ions increases 1\n\n1(b) M1 [H+] = 10–12.2 OR 6.31  10–13 4\nM2 [OH–] = 0.0158 mol dm–3 ecf\nM3 [Ba(OH) ] = 0.01585 ÷ 2 = 7.92 ecf\n2\nM4 mol of Ba(OH) = 0.00792 ÷ 4 = 1.98  10–3\n2\nmass Ba(OH) = 0.339 g ecf\n2\n\n1(c)(i) K = [Fe2+][OH–]2 1\nsp\n\n1(c)(ii) M1 K = 4  (5.85  10–6)3 = 8.01  10–16 2\nsp\nM2 mol3 dm–9\nQuestion Answer Marks", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p05.png" ] }, { "id": "9701-2024-mj-43-q02", "question_id": "9701-2024-mj-43-q02", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "2(a)(i) element which forms one or more stable ion with incomplete d orbitals 1\n\n2(a)(ii) they have vacant d orbitals that are energetically accessible 1\n\n2(b)(i) orbitals of the same energy 1\n\n2(b)(ii) 1\n© Cambridge University Press & Assessment 2024 Page 5 of 18\n\n2(c) M1 2AgNO + 2NaOH → Ag O + 2NaNO + H O 2\n3 2 3 2\nOR 2Ag+ + 2OH– → Ag O + H O\n2 2\nM2 Ag O + 4NH + H O → 2[Ag(NH ) ]OH\n2 3 2 3 2\nOR Ag+ + 2NH + OH– → [Ag(NH ) ]OH\n3 3 2\n\n2(d) M1 bond angle H–N–Ag is 109.5° 2\nM2 shape is linear AND bond angle for N–Ag–N is 180°\n\n2(e) M1 Ag O + H O + 2e– → 2Ag + 2OH– 2\n2 2\nM2 Zn + 2OH– → Zn(OH) + 2e–\n2\n\n2(f) For example: 2\nM1 presence of dps ligand bonded to two Ru\nM2 the rest of the structure correct\n© Cambridge University Press & Assessment 2024 Page 6 of 18", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p05.png", "../answer-assets/9701_s24_ms_43-p06.png" ] }, { "id": "9701-2024-mj-43-q03", "question_id": "9701-2024-mj-43-q03", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "3(a) Li C O H O → Li CO + H O + CO 1\n2 2 4 2 2 3 2\n\n3(b) M1 CaC O AND Ca2+has a smaller ionic radius / Ca2+has a higher charge density 2\n2 4\nM2 anion / C O 2– becomes more polarised / distorted\n2 4\n\n3(c) M1 d orbital(s) of different energy / d-d splitting occurs 3\nM2 electron(s) promoted / excited\nM3 light is absorbed AND colour seen is complementary\n\n3(d) 2K Fe(C O ) → 2K Fe(C O ) + K C O + 2CO 1\n3 2 4 3 2 2 4 2 2 2 4 2\n\n3(e) 2\n\n3(f) M1 HC O – + H+ → H C O 2\n2 4 2 2 4\nM2 HC O – + OH– → C O 2–+ H O\n2 4 2 4 2\n\n3(g) M1 2C O 2– + O (g) + 2H O(l) → 4CO (g) + 4OH–(aq) 2\n2 4 2 2 2\nM2 E = 0.40 – (–0.59) = 0.99 V\ncell\n© Cambridge University Press & Assessment 2024 Page 7 of 18", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p07.png" ] }, { "id": "9701-2024-mj-43-q04", "question_id": "9701-2024-mj-43-q04", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "4(a) M1 voltage of a half-cell compared to SHE 2\nM2 1 mol dm–3 AND 101 kPa AND 298 K\n\n4(b)(i) 3\nM1 Pt Ag Ag+\nM2 gas delivery system H H+\n2\nM3 voltmeter salt bridge wiring Pt to V to Ag\n4b(ii) E value would be more negative AND shifts Ag+ (+ e–) ⇋ Ag to the left 1\n\n4(c) enthalpy change when one mole of a solute dissolves in water 1\n\n4(d)(i) 2\nM1 two arrows in blue\nM2 correct species in red with all state symbols\n© Cambridge University Press & Assessment 2024 Page 8 of 18\n\n4(d)(ii) H = –811.6 (kJ mol–1) 1\nlatt\n\n4(e) M1 Mg(NO ) (s), NaNO (s), RbNO (s) 3\n3 2 3 3\nM2 Mg2+ has a higher charge than Na+ OR Rb+\nAND Na+ has a smaller radius than Rb+\nM3 correct statement relating magnitude of LE to attraction between ions OR strength of ionic bonds\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p08.png", "../answer-assets/9701_s24_ms_43-p09.png" ] }, { "id": "9701-2024-mj-43-q05", "question_id": "9701-2024-mj-43-q05", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 8, "status": "available", "reason": null, "text": "5(a)(i) tangent drawn at t = 0 AND gradient of tangent calculated at t = 0 1\nanswer between 0.016 to 0.040\n\n5(a)(ii) [I–] stays constant / [I–] does not change 1\nOR the overall order is one under these conditions\n\n5(b) M1 equation 1: 2Fe2+ + S O 2– → 2Fe3+ + 2SO 2– 2\n2 8 4\nM2 equation 2: 2Fe3+ + 2I– → 2Fe2+ + I\n2\n\n5(c) the rate constant and rate of reaction 1 will both increase 1\n\n5(d) k = 0.693 ÷ t 1\n½\nt = 0.693 ÷ 0.0158 = 43.9 s\n½\n\n5(e) NO + Br → NOBr OR NO + Br → NOBr + Br 2\n2 2 2\nNOBr + NO → 2NOBr OR Br + NO → NOBr\n2\nM1 two equations adding up to the overall equation\nM2 step 1 has one NO and one Br only and is identified as slow step\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 18", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p09.png" ] }, { "id": "9701-2024-mj-43-q06", "question_id": "9701-2024-mj-43-q06", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 15, "status": "available", "reason": null, "text": "6(a)(i) ratio of the concentrations of solute in two solvents 1\n\n6(a)(ii) M1 10.5 = (y ÷ 25) ÷ ((0.74–y) ÷ 40) 2\nM2 y = 0.642 g\n\n6(b) 2\nsp sp2 sp3\nKekulé benzene 0 6 0\nDewar benzene 0 4 2\nLadenburg benzene 0 0 6\nM1 row 1 and 2 correct\nM2 row 3 correct\n\n6(c) M1 120° AND hexagonal/trigonal planar 2\nM2 C-C has -bonds and -bonds AND C-H have -bonds only\n\n6(d) bond strain OR ring strain 1\n\n6(e) 1\npeaks\nDewar benzene 2\nLadenburg benzene 1\ndelocalised benzene 1\n\n6(f)(i) BrCH CH CH CH Br + FeBr → BrCH CH CH CH + + FeBr – 1\n2 2 2 2 3 2 2 2 2 4\n© Cambridge University Press & Assessment 2024 Page 10 of 18\n\n6(f)(ii) 3\nM1 curly arrow from inside the hexagon towards C+\nM2 structure of the intermediate\nM3 curly arrow from C–H bond into the ring AND H+\n© Cambridge University Press & Assessment 2024 Page 11 of 18\n\n6(f)(iii) M1 Y 2\nM2\nZ\nO\nOR\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p10.png", "../answer-assets/9701_s24_ms_43-p11.png", "../answer-assets/9701_s24_ms_43-p12.png" ] }, { "id": "9701-2024-mj-43-q07", "question_id": "9701-2024-mj-43-q07", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 12, "status": "available", "reason": null, "text": "7(a) methyl pentanoate 1\n\n7(b)(i) time between injection and detection 1\n\n7(b)(ii) 29.5% 1\n© Cambridge University Press & Assessment 2024 Page 12 of 18\n\n7(c)(i) 2\nester number of peaks seen number of peaks seen\nin proton (1H) NMR in carbon-13 NMR\nB 5 6\nC 4 5\nM1 row 1 correct\nM2 row 2 correct\n\n7(c)(ii) A and B 1\n© Cambridge University Press & Assessment 2024 Page 13 of 18\n\n7(d) 6\none mark for each structure: F, G, H, J, K and L\n© Cambridge University Press & Assessment 2024 Page 14 of 18\n\n7(d) six alternatives for F\n© Cambridge University Press & Assessment 2024 Page 15 of 18", "source_pages": [ 12, 13, 14, 15 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p12.png", "../answer-assets/9701_s24_ms_43-p13.png", "../answer-assets/9701_s24_ms_43-p14.png", "../answer-assets/9701_s24_ms_43-p15.png" ] }, { "id": "9701-2024-mj-43-q08", "question_id": "9701-2024-mj-43-q08", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 8, "status": "available", "reason": null, "text": "8(a)(i) two 1\n\n8(a)(ii) amide 2\namine\nester\ncarboxylic acid / carboxyl\nany two [1] all four [2]\n\n8(b)(i) M1 hydrolysis 2\nM2 acid–base\n\n8(b)(ii) 3\none mark for each structure\n© Cambridge University Press & Assessment 2024 Page 16 of 18", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p16.png" ] }, { "id": "9701-2024-mj-43-q09", "question_id": "9701-2024-mj-43-q09", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 7, "status": "available", "reason": null, "text": "9(a)(i) C H OH + Na → C H ONa + ½H 1\n6 5 6 5 2\n\n9(a)(ii) 1\nboth structures for one mark\n\n9(b) 3\nM1 correct dipole AND curly arrow from inside the hexagon to the carbon atom\nM2 structure of the intermediate\nM3 curly arrow from C-H bond into the ring AND H+\n© Cambridge University Press & Assessment 2024 Page 17 of 18\n\n9(c)(i) 1\nAll three curly arrows for one mark\n\n9(c)(ii) 1\n© Cambridge University Press & Assessment 2024 Page 18 of 18", "source_pages": [ 17, 18 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_43.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_43-p17.png", "../answer-assets/9701_s24_ms_43-p18.png" ] }, { "id": "9701-2024-mj-51-q01", "question_id": "9701-2024-mj-51-q01", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 0.248(2) (g) 1\n\n1(a)(ii) (500 cm3) volumetric flask 1\n\n1(b)(i) 20.5 1\n100 = 20(.0%)\n5.0\ncorrect working must be shown\n\n1(b)(ii) 1\ngreater no effect smaller\nuncertainty of the\n\nmeasurement\npercentage error of the\n\nmeasurement\n\n1(c)(i) M1 rinse / wash (50 cm3 burette) with (0.0200 mol dm–3 aqueous) sodium thiosulfate (before filling with aqueous sodium 2\nthiosulfate)\nM2 place (aqueous) sodium thiosulfate in the burette (to fill) AND then run some solution out (through the jet) (by opening\nthe tap)\n\n1(b)(ii) M1 2\n add / run Na S O (aq) (into the conical flask)\n2 2 3\n until (permanent) colour change (seen) / end-point reached\nM2 add dropwise (towards the end) (to ensure the end-point is accurate)\n\n1(d) to ensure the reaction(s) is / are complete (in the syringe) 1\n© Cambridge University Press & Assessment 2024 Page 5 of 7\n\n1(e) 2\n(titration) A (titration) B (titration) C (titration) D (titration) E\nfinal (burette)\nreading / cm3\ninitial (burette)\nreading / cm3\ntitre / cm3\nM1 headings and units correct for table\n initial / start and (burette) reading / volume\n final / end and (burette) reading / volume\n titre or volume / (0.002(00) mol dm–3) sodium thiosulfate (solution) and used / added\n units: (cm3) or / cm3 or in cm3 for all headings\nM2 five columns for syringes A–E\n\n1(f) M1 n(Na S O ) = 12.65 / 1000  0.00200 = 2.53  10–5 2\n2 2 3\nM2 n(O ) = M1 / 4 = (2.53  10–5 / 4 = 6.325  10–6 mol)\n2\nconcentration of O = (M1 / 4)  1000 / 30 (= 2.10833  10–4 mol dm–3) (= 2.11  10–4 mol dm–3)\n2\n\n1(g)(i) there was (a small amount of dissolved) oxygen in the other solutions / reagents / reactants (used in the experiment) 1\n\n1(g)(ii) subtract the result (2.26  10–5 mol dm–3) of the distilled water experiment from the final result for dissolved oxygen in 1(f) 1\n\n1(h) chlorine is an oxidising agent 1\nOR chlorine reacts in the same way as oxygen\nOR chlorine reacts with Mn2+ / S O 2– / I– / reactant(s)\n2 3\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_51-p05.png", "../answer-assets/9701_s24_ms_51-p06.png" ] }, { "id": "9701-2024-mj-51-q02", "question_id": "9701-2024-mj-51-q02", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) To ensure the solution(s) (in flasks A and B) are at the same temperature (of the water bath) (before mixing) 1\n© Cambridge University Press & Assessment 2024 Page 6 of 7\n\n2(b)(i) (take the temperature of reaction mixture in flask A) 1\n1 at the start of reaction / step 7 OR (immediately) before step 7 OR step 6 / (immediately) before the reaction OR\n(immediately) after step 5\nAND\n2 at the end of reaction / step 8 / when cross is no longer visible / (immediately) after step 8\n\n2(b)(ii) calculate the mean (temperature during the reaction) 1\n\n2(c) 2\ntemperature, T / C time, t / s temperature, T / K 1 / t / s–1\n15 176 288 0.0057\n24 92 297 0.0109\n39 62 305 0.0161\nM1 column 3 correct M2 column 4 correct\n\n2(d)(i) M1 all points plotted correctly 2\nM2 straight line of best fit line drawn passing close to all points (except marked anomaly)\n\n2(d)(ii) M1 two acceptable coordinates from line if best fit expressed in the form (x,y) 2\nM2 gradient correctly calculated from points listed for M1\nanswer correctly rounded to three significant figures\n\n2(d)(iii) M1 any correct version of the equation: gradient = – 0.434 x E / R 3\nA\nM2 evaluation of expression to give value for E\nA\nM3 units = kJ mol–1\n\n2(d)(iv) yes, there are no / few anomalous points / most points lie on or near the line (of best fit) 1\nOR no, there is an anomalous point(s) / (some) points don’t lie on or near the line (of best fit)\n\n2(e) increase the concentration of one or both of the reactants 1\n© Cambridge University Press & Assessment 2024 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_51-p06.png", "../answer-assets/9701_s24_ms_51-p07.png" ] }, { "id": "9701-2024-mj-52-q01", "question_id": "9701-2024-mj-52-q01", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 15, "status": "available", "reason": null, "text": "1(a) you cannot (easily) measure temperature (change) whilst applying heat 1\n\n1(b)(i) 5.01 g 1\n\n1(b)(ii) M1 order of weighing: 2\n1 weighing boat + CaCO is weighed\n3\n(CaCO added to beaker containing HCl)\n3\n2 weighing boat is weighed after CaCO is added to HCl\n3\nM2 table + units:\n/ g\n1 mass of boat + CaCO (before transfer)\n3\n2 mass of boat (+ residue of CaCO ) (after addition)\n3\nmass of CaCO added\n3\n\n1(c) burette AND burette measures to nearest 0.05 cm3 1\n\n1(d) stir the solution / mixture 1\n\n1(e)(i) M1 10 points plotted correctly from t = 3.5 mins to t = 8 mins 2\nM2 line of best fit drawn involving points from 4 to 8 minutes AND extrapolated back to 3 minutes\n\n1(e)(ii) 38.5 – 19.0 = 19.5 C 1\n\n1(f) the reaction is not complete (at 3.5 minutes) 1\n\n1(g) M1 q = 50 g x 4.18 J g–1 K–1 x 6 K = 12958 J 2\nM2 H = M1 / 0.0500 = 12 958 J / 0.0500 mol = 259 160 J mol–1 = –259 kJ mol–1\n2\n\n1(h) ∆H = ∆H – ∆H = –84 – (–259) = (+)175 (kJ mol–1) 1\nr 1 2\n© Cambridge University Press & Assessment 2024 Page 5 of 6\n\n1(i) M1 weakness: heat loss to the surroundings 2\nM2 improvement: use a lid\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_52-p05.png", "../answer-assets/9701_s24_ms_52-p06.png" ] }, { "id": "9701-2024-mj-52-q02", "question_id": "9701-2024-mj-52-q02", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) M1 volume of solution A = 21.75 cm3 (to nearest 0.05 cm3) 3\nM2 use burette to transfer 21.75 cm3 of solution A and add to a volumetric flask\nM3 top the (100 cm3 volumetric) flask up to 100 cm3 mark with distilled water\n\n2(b) so that concentration of [Fe(SCN)2+] is known (as all solution B / SCN– reacts) 1\n\n2(c)(i) (relative) absorbance 1\n\n2(c)(ii) add ice (to the water in the beaker) 1\n\n2(d)(i) 8.9 and 6.2 in column 3 1\n\n2(d)(ii) 107 and 70 in column 4 1\n\n2(e)(i) the graph is a straight line 1\n\n2(e)(ii) M1 anomalous point ringed 2\nM2 not all solution B was transferred / added (in step 5) (making concentration of FeSCN2+ too low)\n\n2(e)(iii) M1 co-ordinates read and recorded correctly 2\nM2 gradient calculated to three significant figures\n\n2(e)(iv) M1 gradient / 2(e)(iii) = –∆H / 2.303R OR any correct rearrangement of this e.g. ∆H = gradient / 2(e)(iii)  –2.303  8.31 2\nM2 correct calculations of ∆H including conversion to kJ\n© Cambridge University Press & Assessment 2024 Page 6 of 6", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_52-p06.png" ] }, { "id": "9701-2024-mj-53-q01", "question_id": "9701-2024-mj-53-q01", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 0.248(2) (g) 1\n\n1(a)(ii) (500 cm3) volumetric flask 1\n\n1(b)(i) 20.5 1\n100 = 20(.0%)\n5.0\ncorrect working must be shown\n\n1(b)(ii) 1\ngreater no effect smaller\nuncertainty of the\n\nmeasurement\npercentage error of the\n\nmeasurement\n\n1(c)(i) M1 rinse / wash (50 cm3 burette) with (0.0200 mol dm–3 aqueous) sodium thiosulfate (before filling with aqueous sodium 2\nthiosulfate)\nM2 place (aqueous) sodium thiosulfate in the burette (to fill) AND then run some solution out (through the jet) (by opening\nthe tap)\n\n1(b)(ii) M1 2\n add / run Na S O (aq) (into the conical flask)\n2 2 3\n until (permanent) colour change (seen) / end-point reached\nM2 add dropwise (towards the end) (to ensure the end-point is accurate)\n\n1(d) to ensure the reaction(s) is / are complete (in the syringe) 1\n© Cambridge University Press & Assessment 2024 Page 5 of 7\n\n1(e) 2\n(titration) A (titration) B (titration) C (titration) D (titration) E\nfinal (burette)\nreading / cm3\ninitial (burette)\nreading / cm3\ntitre / cm3\nM1 headings and units correct for table\n initial / start and (burette) reading / volume\n final / end and (burette) reading / volume\n titre or volume / (0.002(00) mol dm–3) sodium thiosulfate (solution) and used / added\n units: (cm3) or / cm3 or in cm3 for all headings\nM2 five columns for syringes A–E\n\n1(f) M1 n(Na S O ) = 12.65 / 1000  0.00200 = 2.53  10–5 2\n2 2 3\nM2 n(O ) = M1 / 4 = (2.53  10–5 / 4 = 6.325  10–6 mol)\n2\nconcentration of O = (M1 / 4)  1000 / 30 (= 2.10833  10–4 mol dm–3) (= 2.11  10–4 mol dm–3)\n2\n\n1(g)(i) there was (a small amount of dissolved) oxygen in the other solutions / reagents / reactants (used in the experiment) 1\n\n1(g)(ii) subtract the result (2.26  10–5 mol dm–3) of the distilled water experiment from the final result for dissolved oxygen in 1(f) 1\n\n1(h) chlorine is an oxidising agent 1\nOR chlorine reacts in the same way as oxygen\nOR chlorine reacts with Mn2+ / S O 2– / I– / reactant(s)\n2 3\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_53-p05.png", "../answer-assets/9701_s24_ms_53-p06.png" ] }, { "id": "9701-2024-mj-53-q02", "question_id": "9701-2024-mj-53-q02", "subject": "9701", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) To ensure the solution(s) (in flasks A and B) are at the same temperature (of the water bath) (before mixing) 1\n© Cambridge University Press & Assessment 2024 Page 6 of 7\n\n2(b)(i) (take the temperature of reaction mixture in flask A) 1\n1 at the start of reaction / step 7 OR (immediately) before step 7 OR step 6 / (immediately) before the reaction OR\n(immediately) after step 5\nAND\n2 at the end of reaction / step 8 / when cross is no longer visible / (immediately) after step 8\n\n2(b)(ii) calculate the mean (temperature during the reaction) 1\n\n2(c) 2\ntemperature, T / C time, t / s temperature, T / K 1 / t / s–1\n15 176 288 0.0057\n24 92 297 0.0109\n39 62 305 0.0161\nM1 column 3 correct M2 column 4 correct\n\n2(d)(i) M1 all points plotted correctly 2\nM2 straight line of best fit line drawn passing close to all points (except marked anomaly)\n\n2(d)(ii) M1 two acceptable coordinates from line if best fit expressed in the form (x,y) 2\nM2 gradient correctly calculated from points listed for M1\nanswer correctly rounded to three significant figures\n\n2(d)(iii) M1 any correct version of the equation: gradient = – 0.434 x E / R 3\nA\nM2 evaluation of expression to give value for E\nA\nM3 units = kJ mol–1\n\n2(d)(iv) yes, there are no / few anomalous points / most points lie on or near the line (of best fit) 1\nOR no, there is an anomalous point(s) / (some) points don’t lie on or near the line (of best fit)\n\n2(e) increase the concentration of one or both of the reactants 1\n© Cambridge University Press & Assessment 2024 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9701_s24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-may-june/9701_s24_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s24_ms_53-p06.png", "../answer-assets/9701_s24_ms_53-p07.png" ] }, { "id": "9701-2024-on-41-q01", "question_id": "9701-2024-on-41-q01", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) conjugate acid conjugate base 1\nH PO – HPO 2– [1] BOTH\n2 4 4\n\n1(a)(ii) allows only small changes in pH / resists changes in pH [1] 2\nwhen small amounts of acid or alkali/base / H+ or OH– are added [1]\n\n1(a)(iii) Na+ HPO 2– + H+ → Na+H PO – + Na+ 2\n2 4 2 4\nor HPO 2– + H+ → H PO – [1]\n4 2 4\nNa+H PO – + NaOH → Na+ HPO 2– + H O\n2 4 2 4 2\nor H PO – + OH– → HPO 2– + H O [1]\n2 4 4 2\n\n1(a)(iv) HCO – / hydrogen carbonate [1] 1\n3\n\n1(b)(i) either 10–13.09 or [H+]= 8.1  10–14 seen [1] 2\n8.1x10–14  [OH–] = 1  10–14 seen [1]\n110−14\nOR [OH–] = seen\n8.110−14\nALLOW alternative method\npOH = 0.91 [1]\n[OH–] = 10–0.91 seen [1]\n\n1(b)(ii) 2 moles of OH– in one mole of E / X(OH) 1\n2\nOR\nthere are two hydroxide ions in each formula unit of E / X(OH) [1]\n2\n\n1(b)(iii) 0.0615  0.25 = 0.0154 moles of E 1\nRFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137\nAND barium hydroxide / Ba(OH) [1]\n2\n© Cambridge University Press & Assessment 2024 Page 5 of 14\n\n1(c) K = [Mg2+][OH–]2 OR K = 1.4  10-4  (2  1.4  10-4)2 [1] 3\nsp sp\nK = 1.10  10–11 [1] min 2sf\nsp\ncorrect answer [2]\nmol3 dm–9 [1] ecf from M1\n\n1(d) M1: H and H less exothermic (down the group) 3\nlatt hyd\nOR H and H more exothermic/more negative for Mg [1]\nlatt/LE hyd\nM2: H changes more than H (down the group)\nlatt hyd\nOR H changes more / faster for Mg [1]\nlatt\nM3: H becomes more exothermic (down the group)\nsol\nOR H becomes less exothermic / less negative for Mg [1]\nsol\nmark independently\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p05.png", "../answer-assets/9701_w24_ms_41-p06.png" ] }, { "id": "9701-2024-on-41-q02", "question_id": "9701-2024-on-41-q02", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 14, "status": "available", "reason": null, "text": "2(a) becomes less negative/less exothermic (down the group / from F to I) 2\nAND due to increase in (ionic) radius / size [1]\ndecreased attraction to water\nOR weaker ion-dipole force to water [1]\n\n2(b)(i) CaF (aq) OR Ca2+ (aq) + 2F– (aq) [1] 1\n2\n\n2(b)(ii) • atomisation energy of Ca / H 2\nat(o)m))\n• atomisation energy of F / H\n(2) at(o)m))\nOR F-F bond energy / BE of F-F\n• first ionisation energy / IE of Ca / H\n1 i1\n• second ionisation energy / IE of Ca / H\n2 i2\n• (first) electron affinity /EA of F / H\nea\nany three for [1] ALL five [2]\n© Cambridge University Press & Assessment 2024 Page 6 of 14\n\n2(b)(iii) energy change / energy released 2\nwhen one mole of an ionic compound is formed [1]\nfrom its gaseous ion(s) [1]\n\n2(b)(iv) (H = ) (change) 3 – (change) 1 [1] 1\nlatt\nOR\n(H = ) H(CaF (s)) – H(Ca2+(g)) – 2H(F–(g))\nlatt f 2 f f\n\n2(c) expression involves four correct numbers (13, 1395, 1214, 1650) 2\nAND 2 times [1]\n1395 – 1650 + 2x = –1214+13\nH(F–(g))  = –473 [1] ecf\nf\ncorrect answer [2]\n\n2(d) number of possible arrangements of particles / energy in a system [1] 1\n\n2(e) states or clearly uses G = H - TS [1] 2\nOR 6000 = 30 000 – (298  S)\nS = (+)80.5(4) [1] min 3sf ecf\ncorrect answer [2]\n\n2(f) becomes more soluble 1\nAND S is positive / TS is positive / –TS is negative (as T inc) [1]\n(so G becomes more negative)\n© Cambridge University Press & Assessment 2024 Page 7 of 14", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p06.png", "../answer-assets/9701_w24_ms_41-p07.png" ] }, { "id": "9701-2024-on-41-q03", "question_id": "9701-2024-on-41-q03", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 16, "status": "available", "reason": null, "text": "3(a)(i) rate = k[A][B] [1] 2\nrate = k[A]2 / k[A]2[B]0 AND rate = k[B]2 / k[A]0[B]2 [1] BOTH\n\n3(a)(ii) k = 7.62 / 7.625 / 7.63 [1] ecf from 3a(i) min 3sf 2\nmol–1 dm3 s–1 [1] ecf from 3a(i)\n\n3(a)(iii) reaction is 2nd order 1\nOR reaction is not 1st order\nOR reaction does not have constant half-life\n[1]\n\n3(b)(i) Pt, Pd, Rh any two [1] 1\n\n3(b)(ii) • adsorption (of reactants) 2\n/ bond forms between the surface/catalyst and reactants\n• bond weakening (in reactants)\n• desorption (of reactants / products)\n/ bond breaks between the surface/catalyst and products\nany two [1] all three [2]\n\n3(b)(iii) S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+ [1] any multiple 2\n2 8 4\n2Fe3+ + 2I– → 2Fe2+ + I [1] any multiple\n2\n\n3(b)(iv) (heterogeneous) catalyst and reactants in different states / phases 1\nAND\n(homogeneous) catalyst and reactants in same states / phases [1] BOTH\n\n3(c)(i) similar energy of the (3)d and (4)s subshells / orbitals [1] 1\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n3(c)(ii) M1M2: 4\n• Cl O– chosen\n• O chosen\n2\n• E⦵ values 1.45 (with Cl O–)\ncell\n• E⦵ values 0.96 (with O )\ncell 2\nany two [1] all four [2]\nM3: 2Fe(OH) + Cl O– + H O → Cl– + 2Fe(OH) [1]\n2 2 3\nM4: 4Fe(OH) + O + 2H O → 4Fe(OH) [1]\n2 2 2 3\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p08.png", "../answer-assets/9701_w24_ms_41-p09.png" ] }, { "id": "9701-2024-on-41-q04", "question_id": "9701-2024-on-41-q04", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "4(a) empty (d) orbitals are energetically accessible 1\nOR empty (d) orbitals can form dative bonds with ligands\nOR empty (d) orbitals can accept a lone pair from ligands [1]\n\n4(b)(i) [Co O NH (NH ) ]3+ scores [2] must have Co 2\n2 2 2 3 8 2\n• Co O NH\n2 2 2\n• 8 ammonia so both Co are octahedral (ecf bullet 1)\n• correct charge based on ligands present (ecf Co and NH )\n2\nany two [1] all three [2]\n\n4(b)(ii) 2 1\n3 BOTH [1]\n\n4(b)(iii) E different OR (d–d) energy gap different [1] 2\ndifferent frequency/wavelength/energy from visible light absorbed [1]\n© Cambridge University Press & Assessment 2024 Page 9 of 14\n\n4(c)(i) [Cd(NH ) ]2+ + 4CN– → [Cd(CN) ]2– + 4NH 2\n3 4 4 3\nX Y\ncomplex formed is [Cd(CN) ]2– / Na [Cd(CN) ] [1] equation fully correct\n4 2 4\n[1] ECF\n\n4(c)(ii) [CdCl ]2– [Cd(NH ) ]2+ [Cd(CN) ]2– 1\n4 3 4 4\nOR Z X Y [1]\n\n4(d)(i) Cr O 2– + 8H+ + 3HOOCCOOH → 2Cr3+ + 7H O + 6CO [1] 1\n2 7 2 2\n\n4(d)(ii) • mol Cr O 2– = 0.0500  16.20 / 1000 = 8.10  10–4 2\n2 7\n• mol HOOCCOOH / C O 2– = 8.10  10–4  3 = 2.43  10–3 ecf\n2 4\n• [C O 2–] = 2.43  10–3  40 = 0.0972 mol dm–3 min 2sf ecf\n2 4\n ✓ ✓ [2]\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p09.png", "../answer-assets/9701_w24_ms_41-p10.png" ] }, { "id": "9701-2024-on-41-q05", "question_id": "9701-2024-on-41-q05", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 6, "status": "available", "reason": null, "text": "5(a) bond angle must go from bond to bond 2\nP 109 OR 109.5°\nQ 90° (or 180° if different angle labelled)\nR 90° (or 180° if different angle labelled)\nS 180°\nany two [1] all four [2]\n\n5(b) P tetrahedral 2\nQ square planar\nR octahedral\nS linear\nany two [1] all four [2]\n\n5(c) Q AND R [1] 1\n\n5(d) P AND R [1] 1\n© Cambridge University Press & Assessment 2024 Page 10 of 14", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p10.png" ] }, { "id": "9701-2024-on-41-q06", "question_id": "9701-2024-on-41-q06", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 14, "status": "available", "reason": null, "text": "6(a) C H Cl + AlCl → C H + + AlCl – [1] 1\n2 5 3 2 5 4\nOR C H Cl + FeCl → C H + + FeCl –\n2 5 3 2 5 4\n\n6(b)(i) 1\nAND BOTH [1]\n\n6(b)(ii) alkyl / ethyl group is 2,4 directing 1\nOR ethyl group is electron donating / positive inductive effect\n[1]\n\n6(c)(i) hot alkaline KMnO /MnO – (followed by acid) [1] 1\n4 4\n\n6(c)(ii) C H / C H (C H ) + 12 [O] → C H (COOH) + 2CO + 4H O 2\n10 14 6 4 2 5 2 6 4 2 2 2\nall species correct formulae C H + CO + H O [1]\n10 14 2 2\ncorrect balancing [1]\n\n6(c)(iii) 4 [1] 1\n\n6(d)(i) CDCl does not cause a peak 1\n3\nOR does not interfere with spectrum / peaks [1]\n\n6(d)(ii) • ethylbenzene / C H C H 3\n6 5 2 5\n• triplet\n• CH\n3\n• 2H on neighbouring C / next to CH\n2\n– quartet / quadruplet\n– CH\n2\n– 3H on neighbouring C / next to CH / coupling by CH\n3 3\nany three [1] any five [2] all seven [3]\n© Cambridge University Press & Assessment 2024 Page 11 of 14\n\n6(d)(iii) • benzene-1,2-dioic acid / C H (COOH) 1\n6 4 2\n• H on the benzene / C H\n6 4\n– COOH / carboxylic acid\nALL three [1]\n\n6(d)(iv) • COOH peak disappears OR remove peak at 13.1 (ecf (d)(iii) if  value used) 1\nAND\n• as proton / hydrogen exchanges with deuterium / D\nBOTH [1]\n\n6(e) dehydration / elimination / (auto)condensation [1] 1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p11.png", "../answer-assets/9701_w24_ms_41-p12.png" ] }, { "id": "9701-2024-on-41-q07", "question_id": "9701-2024-on-41-q07", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 15, "status": "available", "reason": null, "text": "7(a) CH CN / ethanenitrile [1] 1\n3\n\n7(b) dilute acid / HCl (aq) 1\nAND hot / heat [1]\nthree components required (acid / aq / heat)\n\n7(c) CH COCl / ethanoyl chloride [1] 1\n3\n\n7(d) CH COOH + SOCl → CH COCl + HCl + SO [1] 1\n3 2 3 2\n\n7(e) C H Br / C H Cl / bromoethane / chloroethane [1] 1\n2 5 2 5\n\n7(f) heat in ethanol AND 1\nunder pressure / in sealed tube [1]\n\n7(g) LiAlH / lithium aluminium hydride [1] 1\n4\n© Cambridge University Press & Assessment 2024 Page 12 of 14\n\n7(h) Examiner checks first that each of 1–6 is only used once and 4\nIGNOREs any repeated unless a CON\nhydrolysis 1\naddition 2\nreduction 2, 6\nsubstitution 3, 4, 5\ntwo correct [1] four correct [2] five correct [3] six correct [4]\n\n7(i) M1 C H NHC H ammonia C H NHCOCH [1] 4\n2 5 2 5 2 5 3\nM2 (basicity linked to)\nlone pair/ p orbital on N\nAND being able accept / donate to a proton/H+\n[1]\nM3 (amine stronger because of)\n+ve inductive effect / electron donating\nalkyl / R / ethyl group [1]\nM4 (amide weaker because)\nlone pair / p orbital on N\nis delocalised into C=O group [1]\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p12.png", "../answer-assets/9701_w24_ms_41-p13.png" ] }, { "id": "9701-2024-on-41-q08", "question_id": "9701-2024-on-41-q08", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 7, "status": "available", "reason": null, "text": "8(a) M1: p orbital / lone pair on the oxygen / O 2\nAND is delocalised into the ring / overlaps with delocalised ring [1]\nM2: O-H weakened (in phenol)\nOR anion / phenoxide ion / conjugate base\nis stabilised/more stable [1]\n\n8(b)(i) 2,4,6-tribromophenol (name) AND hydrogen bromide / HBr [1] 1\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n8(b)(ii) 1\nAND\n\n8(b)(iii) 2C H OH + 2Na → 2C H ONa + H [1] 1\n6 5 6 5 2\n\n8(c) step one: HNO (+ HCl) OR NaNO + HCl 2\n2 2\nAND\nT⩽10 °C\nstep two: H O\n2\nAND\nwarm / T> 10 °C\nany two [1] all four [2]\n© Cambridge University Press & Assessment 2024 Page 14 of 14", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_41.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_41-p13.png", "../answer-assets/9701_w24_ms_41-p14.png" ] }, { "id": "9701-2024-on-42-q01", "question_id": "9701-2024-on-42-q01", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 14, "status": "available", "reason": null, "text": "1(a)(i) Change in concentration = 0.135 [1] 2\n0.135 / 1800 = 7.5  10–5 [1]\n\n1(a)(ii) Rate = k[X] [1] 1\n\n1(a)(iii) k = 0.693 / t / k = 0.693 / 900 / k = ln2 / t / k = ln2 / 900 / t = ln2 / k [1] 1\n½ ½ ½\n\n1(a)(iv) 1.16  10–4 [1] 2\nmol dm–3 s–1 [1]\n\n1(b)(i) Any two from: 1\n• variable oxidation state\n• vacant / empty / unfilled d orbitals [1]\n• can form dative bonds / can accept electrons\n\n1(b)(ii) Iron AND 1\niron is solid, reactants are gases OR\ncatalyst and reactants are in different phases / states [1]\n\n1(b)(iii) Two for one mark, three for two marks: 2\n• adsorption of reactants by Pt\n• bonds of reactants weaken [1]\n• desorption of products [1]\n\n1(b)(iv) NO + SO → NO + SO AND 1\n2 2 3\n2NO + O → 2NO [1]\n2 2\n\n1(c) BiO+ [1] 3\n3H SO + 2BiO+ + H O → 3SO 2- + 8H+ + 2Bi [1]\n2 3 2 4\n0.11 V [1]\n© Cambridge University Press & Assessment 2024 Page 5 of 12", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p05.png" ] }, { "id": "9701-2024-on-42-q02", "question_id": "9701-2024-on-42-q02", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 14, "status": "available", "reason": null, "text": "2(a) H increases from left to right due to increase in charge [1] 3\nhyd\nionic radius decreases from left to right [1]\ncausing increased attractive force to water molecules [1]\n\n2(b)(i) Mg2+ (aq) + 2Cl– (aq) [1] 1\n\n2(b)(ii) Enthalpy change of hydration of magnesium ions and chloride ions [1] 2\nH Mg2+ + 2H Cl– [1]\nhyd hyd\n\n2(b)(iii) Selects 155, 1920 and 364 only [1] 3\n2  364 [1]\nanswer –2493 [1]\n\n2(c) The number of arrangements of the particles and of the energy in the system [1] 1\n\n2(d) G = H – TS [1] 2\nanswer +10.5 [1]\n\n2(e)(i) No, G is positive [1] 1\n\n2(e)(ii) Becomes more / soluble because G becomes more negative / less positive / smaller / closer to zero [1] 1\nQuestion Answer Marks", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p06.png" ] }, { "id": "9701-2024-on-42-q03", "question_id": "9701-2024-on-42-q03", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 19, "status": "available", "reason": null, "text": "3(a)(i) [H+] = 10–12.35 OR [H+] = 4.47  10–13 [1] 2\nK / 4.47  10–13 [1]\nw\nOR\npOH = 1.65 [1]\n[OH–] = 10–1.65 [1]\n\n3(a)(ii) K = (0.0112)(0.0224)2 OR K = [Ca2+][OH–]2 [1] 3\nsp sp\nanswer 5.62  10–6 [1]\nmol3 dm–9 [1]\n© Cambridge University Press & Assessment 2024 Page 6 of 12\n\n3(a)(iii) white solid / white ppt AND pH between 7.01 and 12.34 [1] 3\ncommon ion effect [1]\nhydroxide removed by precipitation OR ppt is Ca(OH) [1]\n2\n\n3(a)(iv) lattice energy and hydration energy greater for CaSO OR 3\n4\nlattice energy and hydration energy decrease down group [1]\nhydration energy decreases more / is dominant factor [1]\nenthalpy of solution is more endothermic for BaSO OR is more endothermic down grp [1]\n4\n\n3(b)(i) Ca + 2CH COOH → Ca(CH COO) + H [1] 1\n3 3 2 2\n\n3(b)(ii) Pair 1: CH COOH and CH COO– [1] 2\n3 3\nPair 2: H O+ and H O OR H O and OH– [1]\n3 2 2\n\n3(b)(iii) [H+][CH COO−] 1\n3 [1]\n[CH COOH]\n3\n\n3(b)(iv) [CH COO–] = 0.788, [CH COOH] = 0.270 2\n3 3\n[H+] = 5.96  10–6 [1]\npH = 5.22 [1]\n\n3(b)(v) CH COO– + H+ → CH COOH 2\n3 3\nOR Ca(CH COO) + 2H+ → 2CH COOH + Ca2+ [1]\n3 2 3\nCH COOH + OH– → CH COO– + H O [1]\n3 3 2\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p06.png", "../answer-assets/9701_w24_ms_42-p07.png" ] }, { "id": "9701-2024-on-42-q04", "question_id": "9701-2024-on-42-q04", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "4(a)(i) red/pink blue [1] 1\n\n4(a)(ii) [Co(H O) ]2+ + 2OH– → Co(H O) (OH) + 2H O [1] 1\n2 6 2 4 2 2\n\n4(a)(iii) acid-base [1] 1\n\n4(b)(i) [CoL ]2+ [1] 1\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 12\n\n4(b)(ii) +2 [1] 1\n\n4(b)(iii) 2 1\n3 [1]\n\n4(b)(iv) Not of the same energy / have different energy [1] 1\n\n4(c)(i) 5Fe2+ + MnO – + 8H+ → 5Fe3+ + Mn2+ + 4H O [1] 1\n4 2\n\n4(c)(ii) 0.02  18.7/1000 = 3.74  10–4 [1] 2\n3.74  10–4  5 = 1.87  10–3 [1]\n\n4(c)(iii) 1.87  10–3  10 = 1.87  10–2 2\n4.18/1.87  10–2 = 224 / 223.5(294) [1]\nn = 2 [1]\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p07.png", "../answer-assets/9701_w24_ms_42-p08.png" ] }, { "id": "9701-2024-on-42-q05", "question_id": "9701-2024-on-42-q05", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 6, "status": "available", "reason": null, "text": "5(a) [[Ni(en) ]2+] [[Ni(en) ]2+] 1\nK = 3 OR 3 [1]\nstab [Ni2+][en]3 [Ni(H O) 2+][en]3\n2 6\n\n5(b)(i) [Ni(en) ]2+ AND 1\n3\nlarger K OR more stable [1]\nstab\n\n5(b)(ii) [Ni(H O) ]2+ + 3en → [Nien ]2+ + 6H O [1] 1\n2 6 3 2\n© Cambridge University Press & Assessment 2024 Page 8 of 12\n\n5(c) 3\noctahedral with correct 3D for one [Ni(en) ]2+ [1]\n3\nsecond optical isomer [1]\noptical isomerism [1]\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p08.png", "../answer-assets/9701_w24_ms_42-p09.png" ] }, { "id": "9701-2024-on-42-q06", "question_id": "9701-2024-on-42-q06", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 12, "status": "available", "reason": null, "text": "6(a)(i) ClOCCOCl [1] 1\n\n6(a)(ii) PCl / PCl [1] 1\n3 5\n\n6(b) oxygen [1] 2\nacidified KMnO [1]\n4\n\n6(c) C=O are electron withdrawing / electronegative [1] 2\nweakens O–H bond OR stabilises anion [1]\n\n6(d)(i) 1\n[1]\n\n6(d)(ii) step 1: CH Cl + AlCl [1] 2\n3 3\nstep 2: hot alkaline KMnO [1]\n4\n© Cambridge University Press & Assessment 2024 Page 9 of 12\n\n6(d)(iii) 2\ncorrectly displayed ester linkage [1]\nrest of structure [1]\n\n6(d)(iv) condensation AND ester [1] 1\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p09.png", "../answer-assets/9701_w24_ms_42-p10.png" ] }, { "id": "9701-2024-on-42-q07", "question_id": "9701-2024-on-42-q07", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 11, "status": "available", "reason": null, "text": "7(a) aluminium chloride OR AlCl [1] 1\n3\n\n7(b)(i) delocalised system / delocalised ring / pi system / pi ring [1] 2\nC–Cl bond [1]\n\n7(b)(ii) C–H bond [1] 2\ndelocalised system / delocalised ring / pi system / pi ring [1]\n\n7(c) 0, 5, 1 [1] 1\n\n7(d) C H + Cl → C H Cl + HCl [1] 1\n6 6 2 6 5\n\n7(e) Cl + 1\nhydrogen / H\nchlorine / Cl [1]\n\n7(f)(i) nucleophilic substitution AND ethanol [1] 1\n\n7(f)(ii) delocalisation of LP of Cl with  system [1] 2\nC–Cl bond is stronger in chlorobenzene / partly double [1]\n© Cambridge University Press & Assessment 2024 Page 10 of 12", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p10.png" ] }, { "id": "9701-2024-on-42-q08", "question_id": "9701-2024-on-42-q08", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "An introduction to A Level organic chemistry", "topic_slug": "9701-topic-29-an-introduction-to-a-level-organic-chemistry", "marks": 12, "status": "available", "reason": null, "text": "8(a) biological activity [1] 1\n\n8(b) rotates plane polarised light 5.0° in anticlockwise direction [1] 1\n\n8(c) enantiomers [1] 1\n\n8(d) racemic [1] 1\n\n8(e) use of chiral catalyst [1] 1\n\n8(f) CH and CH only in column one [1] 3\n2\nCH gives a doublet, CH gives a triplet [1]\n2\ndoublet due to 1 proton on neighbouring carbons\ntriplet due to 2 protons on neighbouring carbons [1]\n\n8(g) 5 [1] 1\n© Cambridge University Press & Assessment 2024 Page 11 of 12\n\n8(h) 3\nat pH 1 [1]\nat pH 3 AND\nor\nat pH 9 [1]\nat pH 14 [1]\n© Cambridge University Press & Assessment 2024 Page 12 of 12", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_42.pdf?download=true", "html": "9701-topic-29-an-introduction-to-a-level-organic-chemistry/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_42-p11.png", "../answer-assets/9701_w24_ms_42-p12.png" ] }, { "id": "9701-2024-on-43-q01", "question_id": "9701-2024-on-43-q01", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) conjugate acid conjugate base 1\nH PO – HPO 2– [1] BOTH\n2 4 4\n\n1(a)(ii) allows only small changes in pH / resists changes in pH [1] 2\nwhen small amounts of acid or alkali/base / H+ or OH– are added [1]\n\n1(a)(iii) Na+ HPO 2– + H+ → Na+H PO – + Na+ 2\n2 4 2 4\nor HPO 2– + H+ → H PO – [1]\n4 2 4\nNa+H PO – + NaOH → Na+ HPO 2– + H O\n2 4 2 4 2\nor H PO – + OH– → HPO 2– + H O [1]\n2 4 4 2\n\n1(a)(iv) HCO – / hydrogen carbonate [1] 1\n3\n\n1(b)(i) either 10–13.09 or [H+]= 8.1  10–14 seen [1] 2\n8.1x10–14  [OH–] = 1  10–14 seen [1]\n110−14\nOR [OH–] = seen\n8.110−14\nALLOW alternative method\npOH = 0.91 [1]\n[OH–] = 10–0.91 seen [1]\n\n1(b)(ii) 2 moles of OH– in one mole of E / X(OH) 1\n2\nOR\nthere are two hydroxide ions in each formula unit of E / X(OH) [1]\n2\n\n1(b)(iii) 0.0615  0.25 = 0.0154 moles of E 1\nRFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137\nAND barium hydroxide / Ba(OH) [1]\n2\n© Cambridge University Press & Assessment 2024 Page 5 of 14\n\n1(c) K = [Mg2+][OH–]2 OR K = 1.4  10-4  (2  1.4  10-4)2 [1] 3\nsp sp\nK = 1.10  10–11 [1] min 2sf\nsp\ncorrect answer [2]\nmol3 dm–9 [1] ecf from M1\n\n1(d) M1: H and H less exothermic (down the group) 3\nlatt hyd\nOR H and H more exothermic/more negative for Mg [1]\nlatt/LE hyd\nM2: H changes more than H (down the group)\nlatt hyd\nOR H changes more / faster for Mg [1]\nlatt\nM3: H becomes more exothermic (down the group)\nsol\nOR H becomes less exothermic / less negative for Mg [1]\nsol\nmark independently\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p05.png", "../answer-assets/9701_w24_ms_43-p06.png" ] }, { "id": "9701-2024-on-43-q02", "question_id": "9701-2024-on-43-q02", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 14, "status": "available", "reason": null, "text": "2(a) becomes less negative/less exothermic (down the group / from F to I) 2\nAND due to increase in (ionic) radius / size [1]\ndecreased attraction to water\nOR weaker ion-dipole force to water [1]\n\n2(b)(i) CaF (aq) OR Ca2+ (aq) + 2F– (aq) [1] 1\n2\n\n2(b)(ii) • atomisation energy of Ca / H 2\nat(o)m))\n• atomisation energy of F / H\n(2) at(o)m))\nOR F-F bond energy / BE of F-F\n• first ionisation energy / IE of Ca / H\n1 i1\n• second ionisation energy / IE of Ca / H\n2 i2\n• (first) electron affinity /EA of F / H\nea\nany three for [1] ALL five [2]\n© Cambridge University Press & Assessment 2024 Page 6 of 14\n\n2(b)(iii) energy change / energy released 2\nwhen one mole of an ionic compound is formed [1]\nfrom its gaseous ion(s) [1]\n\n2(b)(iv) (H = ) (change) 3 – (change) 1 [1] 1\nlatt\nOR\n(H = ) H(CaF (s)) – H(Ca2+(g)) – 2H(F–(g))\nlatt f 2 f f\n\n2(c) expression involves four correct numbers (13, 1395, 1214, 1650) 2\nAND 2 times [1]\n1395 – 1650 + 2x = –1214+13\nH(F–(g))  = –473 [1] ecf\nf\ncorrect answer [2]\n\n2(d) number of possible arrangements of particles / energy in a system [1] 1\n\n2(e) states or clearly uses G = H - TS [1] 2\nOR 6000 = 30 000 – (298  S)\nS = (+)80.5(4) [1] min 3sf ecf\ncorrect answer [2]\n\n2(f) becomes more soluble 1\nAND S is positive / TS is positive / –TS is negative (as T inc) [1]\n(so G becomes more negative)\n© Cambridge University Press & Assessment 2024 Page 7 of 14", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p06.png", "../answer-assets/9701_w24_ms_43-p07.png" ] }, { "id": "9701-2024-on-43-q03", "question_id": "9701-2024-on-43-q03", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 16, "status": "available", "reason": null, "text": "3(a)(i) rate = k[A][B] [1] 2\nrate = k[A]2 / k[A]2[B]0 AND rate = k[B]2 / k[A]0[B]2 [1] BOTH\n\n3(a)(ii) k = 7.62 / 7.625 / 7.63 [1] ecf from 3a(i) min 3sf 2\nmol–1 dm3 s–1 [1] ecf from 3a(i)\n\n3(a)(iii) reaction is 2nd order 1\nOR reaction is not 1st order\nOR reaction does not have constant half-life\n[1]\n\n3(b)(i) Pt, Pd, Rh any two [1] 1\n\n3(b)(ii) • adsorption (of reactants) 2\n/ bond forms between the surface/catalyst and reactants\n• bond weakening (in reactants)\n• desorption (of reactants / products)\n/ bond breaks between the surface/catalyst and products\nany two [1] all three [2]\n\n3(b)(iii) S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+ [1] any multiple 2\n2 8 4\n2Fe3+ + 2I– → 2Fe2+ + I [1] any multiple\n2\n\n3(b)(iv) (heterogeneous) catalyst and reactants in different states / phases 1\nAND\n(homogeneous) catalyst and reactants in same states / phases [1] BOTH\n\n3(c)(i) similar energy of the (3)d and (4)s subshells / orbitals [1] 1\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n3(c)(ii) M1M2: 4\n• Cl O– chosen\n• O chosen\n2\n• E⦵ values 1.45 (with Cl O–)\ncell\n• E⦵ values 0.96 (with O )\ncell 2\nany two [1] all four [2]\nM3: 2Fe(OH) + Cl O– + H O → Cl– + 2Fe(OH) [1]\n2 2 3\nM4: 4Fe(OH) + O + 2H O → 4Fe(OH) [1]\n2 2 2 3\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p08.png", "../answer-assets/9701_w24_ms_43-p09.png" ] }, { "id": "9701-2024-on-43-q04", "question_id": "9701-2024-on-43-q04", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 12, "status": "available", "reason": null, "text": "4(a) empty (d) orbitals are energetically accessible 1\nOR empty (d) orbitals can form dative bonds with ligands\nOR empty (d) orbitals can accept a lone pair from ligands [1]\n\n4(b)(i) [Co O NH (NH ) ]3+ scores [2] must have Co 2\n2 2 2 3 8 2\n• Co O NH\n2 2 2\n• 8 ammonia so both Co are octahedral (ecf bullet 1)\n• correct charge based on ligands present (ecf Co and NH )\n2\nany two [1] all three [2]\n\n4(b)(ii) 2 1\n3 BOTH [1]\n\n4(b)(iii) E different OR (d–d) energy gap different [1] 2\ndifferent frequency/wavelength/energy from visible light absorbed [1]\n© Cambridge University Press & Assessment 2024 Page 9 of 14\n\n4(c)(i) [Cd(NH ) ]2+ + 4CN– → [Cd(CN) ]2– + 4NH 2\n3 4 4 3\nX Y\ncomplex formed is [Cd(CN) ]2– / Na [Cd(CN) ] [1] equation fully correct\n4 2 4\n[1] ECF\n\n4(c)(ii) [CdCl ]2– [Cd(NH ) ]2+ [Cd(CN) ]2– 1\n4 3 4 4\nOR Z X Y [1]\n\n4(d)(i) Cr O 2– + 8H+ + 3HOOCCOOH → 2Cr3+ + 7H O + 6CO [1] 1\n2 7 2 2\n\n4(d)(ii) • mol Cr O 2– = 0.0500  16.20 / 1000 = 8.10  10–4 2\n2 7\n• mol HOOCCOOH / C O 2– = 8.10  10–4  3 = 2.43  10–3 ecf\n2 4\n• [C O 2–] = 2.43  10–3  40 = 0.0972 mol dm–3 min 2sf ecf\n2 4\n ✓ ✓ [2]\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p09.png", "../answer-assets/9701_w24_ms_43-p10.png" ] }, { "id": "9701-2024-on-43-q05", "question_id": "9701-2024-on-43-q05", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 6, "status": "available", "reason": null, "text": "5(a) bond angle must go from bond to bond 2\nP 109 OR 109.5°\nQ 90° (or 180° if different angle labelled)\nR 90° (or 180° if different angle labelled)\nS 180°\nany two [1] all four [2]\n\n5(b) P tetrahedral 2\nQ square planar\nR octahedral\nS linear\nany two [1] all four [2]\n\n5(c) Q AND R [1] 1\n\n5(d) P AND R [1] 1\n© Cambridge University Press & Assessment 2024 Page 10 of 14", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p10.png" ] }, { "id": "9701-2024-on-43-q06", "question_id": "9701-2024-on-43-q06", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 14, "status": "available", "reason": null, "text": "6(a) C H Cl + AlCl → C H + + AlCl – [1] 1\n2 5 3 2 5 4\nOR C H Cl + FeCl → C H + + FeCl –\n2 5 3 2 5 4\n\n6(b)(i) 1\nAND BOTH [1]\n\n6(b)(ii) alkyl / ethyl group is 2,4 directing 1\nOR ethyl group is electron donating / positive inductive effect\n[1]\n\n6(c)(i) hot alkaline KMnO /MnO – (followed by acid) [1] 1\n4 4\n\n6(c)(ii) C H / C H (C H ) + 12 [O] → C H (COOH) + 2CO + 4H O 2\n10 14 6 4 2 5 2 6 4 2 2 2\nall species correct formulae C H + CO + H O [1]\n10 14 2 2\ncorrect balancing [1]\n\n6(c)(iii) 4 [1] 1\n\n6(d)(i) CDCl does not cause a peak 1\n3\nOR does not interfere with spectrum / peaks [1]\n\n6(d)(ii) • ethylbenzene / C H C H 3\n6 5 2 5\n• triplet\n• CH\n3\n• 2H on neighbouring C / next to CH\n2\n– quartet / quadruplet\n– CH\n2\n– 3H on neighbouring C / next to CH / coupling by CH\n3 3\nany three [1] any five [2] all seven [3]\n© Cambridge University Press & Assessment 2024 Page 11 of 14\n\n6(d)(iii) • benzene-1,2-dioic acid / C H (COOH) 1\n6 4 2\n• H on the benzene / C H\n6 4\n– COOH / carboxylic acid\nALL three [1]\n\n6(d)(iv) • COOH peak disappears OR remove peak at 13.1 (ecf (d)(iii) if  value used) 1\nAND\n• as proton / hydrogen exchanges with deuterium / D\nBOTH [1]\n\n6(e) dehydration / elimination / (auto)condensation [1] 1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p11.png", "../answer-assets/9701_w24_ms_43-p12.png" ] }, { "id": "9701-2024-on-43-q07", "question_id": "9701-2024-on-43-q07", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 15, "status": "available", "reason": null, "text": "7(a) CH CN / ethanenitrile [1] 1\n3\n\n7(b) dilute acid / HCl (aq) 1\nAND hot / heat [1]\nthree components required (acid / aq / heat)\n\n7(c) CH COCl / ethanoyl chloride [1] 1\n3\n\n7(d) CH COOH + SOCl → CH COCl + HCl + SO [1] 1\n3 2 3 2\n\n7(e) C H Br / C H Cl / bromoethane / chloroethane [1] 1\n2 5 2 5\n\n7(f) heat in ethanol AND 1\nunder pressure / in sealed tube [1]\n\n7(g) LiAlH / lithium aluminium hydride [1] 1\n4\n© Cambridge University Press & Assessment 2024 Page 12 of 14\n\n7(h) Examiner checks first that each of 1–6 is only used once and 4\nIGNOREs any repeated unless a CON\nhydrolysis 1\naddition 2\nreduction 2, 6\nsubstitution 3, 4, 5\ntwo correct [1] four correct [2] five correct [3] six correct [4]\n\n7(i) M1 C H NHC H ammonia C H NHCOCH [1] 4\n2 5 2 5 2 5 3\nM2 (basicity linked to)\nlone pair/ p orbital on N\nAND being able accept / donate to a proton/H+\n[1]\nM3 (amine stronger because of)\n+ve inductive effect / electron donating\nalkyl / R / ethyl group [1]\nM4 (amide weaker because)\nlone pair / p orbital on N\nis delocalised into C=O group [1]\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p12.png", "../answer-assets/9701_w24_ms_43-p13.png" ] }, { "id": "9701-2024-on-43-q08", "question_id": "9701-2024-on-43-q08", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 7, "status": "available", "reason": null, "text": "8(a) M1: p orbital / lone pair on the oxygen / O 2\nAND is delocalised into the ring / overlaps with delocalised ring [1]\nM2: O-H weakened (in phenol)\nOR anion / phenoxide ion / conjugate base\nis stabilised/more stable [1]\n\n8(b)(i) 2,4,6-tribromophenol (name) AND hydrogen bromide / HBr [1] 1\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n8(b)(ii) 1\nAND\n\n8(b)(iii) 2C H OH + 2Na → 2C H ONa + H [1] 1\n6 5 6 5 2\n\n8(c) step one: HNO (+ HCl) OR NaNO + HCl 2\n2 2\nAND\nT⩽10 °C\nstep two: H O\n2\nAND\nwarm / T> 10 °C\nany two [1] all four [2]\n© Cambridge University Press & Assessment 2024 Page 14 of 14", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_43.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_43-p13.png", "../answer-assets/9701_w24_ms_43-p14.png" ] }, { "id": "9701-2024-on-51-q01", "question_id": "9701-2024-on-51-q01", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "1(a) wear chemically resistant gloves. 1\n\n1(b) M1 Includes the following steps. Measure the mass of.. 2\n1 (Weighing) boat + solid\n2 (Weighing) boat + residue / residual solid (after transfer)\nM2 Table and units. Headings must be unambiguous, and correct units must be included for each heading.\n/ g\nMass of boat + solid (before\ntransfer)\nMass of boat (after transfer)\n(Mass of) solid (transferred)\n\n1(c) M1 Correctly placed vertical condenser including jacket. 2\nM2 (Cold) water in at bottom and water out at the top of the condenser.\n\n1(d)(i) (alkaline aqueous) iodine is in excess. 1\n\n1(d)(ii) to ensure the reaction is complete. 1\n© Cambridge University Press & Assessment 2024 Page 4 of 7\n\n1(d)(iii) to remove soluble substances (from the residue / solid C). 1\nAcceptable named soluble substances in order of diminishing likelihood to be mentioned\n• (aqueous alkaline) iodine\n• Sodium hydroxide\n• Solution B\n• Sodium hydrogencarbonate\n• Sodium iodide\n• C H O - Na+\n7 5 3\n• C H O - Na+\n2 3 2\n• Water washable impurities\n\n1(d)(iv) The residue is less soluble in cold water. 1\n\n1(e)(i) amount of (C H I O) = 0.764 / 687.6 = 1.111  10-3 mol 1\n6 2 2 2\n\n1(e)(ii) amount of C H O = 2 x 1(e)(i) (=2.222  10-3) 1\n9 8 4\nmass of aspirin = 2  1(e)(i)  180 (= 0.400 g)\n\n1(e)(iii) percentage aspirin by mass = (1(e)(ii) / 0.409)  100 (= 97.8%) 1\n\n1(f) amount / mass of solid C would be greater (than the true value) 1\nAND\ncalculated percentage by mass would be greater than (e)(iii)\nQuestion Answer Marks", "source_pages": [ 4, 5 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_51-p04.png", "../answer-assets/9701_w24_ms_51-p05.png" ] }, { "id": "9701-2024-on-51-q02", "question_id": "9701-2024-on-51-q02", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 17, "status": "available", "reason": null, "text": "2(a) 588 nm (589.7) 1\n\n2(b)(i) n (C N H Cl(s)) = 2.5  10-2 x 0.5 dm3 = 1.25  10-2 1\n25 3 30\nmass of C N H Cl(s) = 1.25  10-2  407.5 = 5.09 g\n25 3 30\n© Cambridge University Press & Assessment 2024 Page 5 of 7\n\n2(b)(ii) M1 add a (small) volume of (distilled) water (to the small beaker) 3\nAND\ndissolve the C N H Cl (s)\n25 3 30\nM2 Transfer the solution to a 500 cm3 volumetric flask with washings.\nM3: make up to the (calibration) mark (with distilled water.)\nAND\nthen mix the solution (by inverting the flask)\n\n2(c)(i) 1\nVolume of Volume of\n2.5  10-4 moldm-3 distilled\ncrystal violet water / cm3\n(solution E)\n/ cm3\n0.00 20.00\n4.00 16.00\n8.00 12.00\n12.00 8.00\n16.00 4.00\n20.00 0.00\n\n2(c)(ii) absorbance (of aqueous C N H Cl) 1\n25 3 30\n\n2(d)(i) M1 points plotted correctly. 2\nM2 Straight line of best fit drawn.\n© Cambridge University Press & Assessment 2024 Page 6 of 7\n\n2(d)(ii) M1 Most anomalous point circled. 2\n(Expected to be concentration 1.0  10-4 moldm-3 (solution 3)\nM2 Volume of water added (to mixture) too large\nOR\nVolume of crystal violet solution / solution E added (to mixture) too small.\n\n2(d)(iii) (absorbance is) (directly) proportional (to concentration of crystal violet) (within experimental error). 1\n\n2(d)(iv) Repeat the procedure for the solution giving an anomalous result. (Solution 3) 1\n\n2(e)(i) Solutions being mixed (at t=0) so reading not possible. 1\n\n2(e)(ii) M1 (100, 0.360) and (315, 0.180) 2\nM2 half-life correctly calculated from points listed for M1 (x1–x2) when y = 2y\n2 1\n\n2(e)(iii) first order 1\nAND\nhalf-lives are constant (within experimental error)\n© Cambridge University Press & Assessment 2024 Page 7 of 7", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_51-p05.png", "../answer-assets/9701_w24_ms_51-p06.png", "../answer-assets/9701_w24_ms_51-p07.png" ] }, { "id": "9701-2024-on-52-q01", "question_id": "9701-2024-on-52-q01", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 13.25 g 1\n\n1(a)(ii) M1 dissolve the solid / sodium carbonate (in the beaker) (using a small volume of) (distilled) water 3\nM2 transfer / add to a (250 cm3 volumetric) flask\nand\nrinse (with distilled water)\nM3 top the 250 cm3 volumetric flask up to mark (with distilled water)\n\n1(b)(i) solution Y 1\n\n1(b)(ii) indicator / methyl orange changes colour 1\n\n1(b)(iii) to avoid spitting (of solid when it forms) 1\n\n1(b)(iv) heat to constant mass 1\n\n1(c)(i) 2\nburette reading final 14.85\nburette reading initial 1.35\nvolume of solution X 13.50\nM1 burette readings 14.85 (cm3) and 1.35 (cm3)\nM2 volume of solution X added = 13.50 (cm3)\n\n1(c)(ii) 2  0.05 1\nworking must be shown  100 = 0.741\n13.50\n© Cambridge University Press & Assessment 2024 Page 5 of 7\n\n1(d) M1 M(NaCl) = 58.5 3\nr\nn(NaCl) = 1.17 g / 58.5 = 0.0200\nM2 n(HCl) = 0.02\nc(HCl) (X) = 0.02/0.01335 = 1.50 (mol dm–3)\nM3 n(Na CO ) = 0.02 / 2 = 0.01\n2 3\nc(Na CO ) (Y) = 0.01 / 0.025 = 0.400 (mol dm–3)\n2 3\n\n1(e) M1 calculated concentration of Y will be higher than the true value. 2\nM2 mass / amount of NaCl (in step 7) will be too high.\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_52-p05.png", "../answer-assets/9701_w24_ms_52-p06.png" ] }, { "id": "9701-2024-on-52-q02", "question_id": "9701-2024-on-52-q02", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a)(i) pipeclay triangle 1\n\n2(a)(ii) toxic gas is produced. 1\n\n2(a)(iii) to ensure all of the metal M reacts. 1\n\n2(b) 1\nmass of M 1.4 1.1 0.8 0.5 0.2\n3 4 6 7 9\nmass of S 0.4 0.2 0.2 0.1 0.0\n6 8 0 5 6\n\n2(c) mass of M or mass of metal 1\n\n2(d)(i) 5 points plotted correctly from 2(b) 1\nstraight line of best fit drawn which includes origin 1\n© Cambridge University Press & Assessment 2024 Page 6 of 7\n\n2(d)(ii) most anomalous point selected 1\nnot all of the sulfur had been removed 1\n(in step 4)\n\n2(e) co-ordinates correctly recorded from LOBF 2\ngradient calculated to three significant figures\n\n2(f) not reliable 1\nand\nthere is an anomalous point\n\n2(g)(i) (32.1 / 87.6) = 0.366 1\n\n2(g)(ii) (assuming values in 2(e) and 2(g)(i) are different) 1\nnot correct\nand\ngradient is different\n© Cambridge University Press & Assessment 2024 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_52-p06.png", "../answer-assets/9701_w24_ms_52-p07.png" ] }, { "id": "9701-2024-on-53-q01", "question_id": "9701-2024-on-53-q01", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "1(a) wear chemically resistant gloves. 1\n\n1(b) M1 Includes the following steps. Measure the mass of.. 2\n1 (Weighing) boat + solid\n2 (Weighing) boat + residue / residual solid (after transfer)\nM2 Table and units. Headings must be unambiguous, and correct units must be included for each heading.\n/ g\nMass of boat + solid (before\ntransfer)\nMass of boat (after transfer)\n(Mass of) solid (transferred)\n\n1(c) M1 Correctly placed vertical condenser including jacket. 2\nM2 (Cold) water in at bottom and water out at the top of the condenser.\n\n1(d)(i) (alkaline aqueous) iodine is in excess. 1\n\n1(d)(ii) to ensure the reaction is complete. 1\n© Cambridge University Press & Assessment 2024 Page 4 of 7\n\n1(d)(iii) to remove soluble substances (from the residue / solid C). 1\nAcceptable named soluble substances in order of diminishing likelihood to be mentioned\n• (aqueous alkaline) iodine\n• Sodium hydroxide\n• Solution B\n• Sodium hydrogencarbonate\n• Sodium iodide\n• C H O - Na+\n7 5 3\n• C H O - Na+\n2 3 2\n• Water washable impurities\n\n1(d)(iv) The residue is less soluble in cold water. 1\n\n1(e)(i) amount of (C H I O) = 0.764 / 687.6 = 1.111  10-3 mol 1\n6 2 2 2\n\n1(e)(ii) amount of C H O = 2 x 1(e)(i) (=2.222  10-3) 1\n9 8 4\nmass of aspirin = 2  1(e)(i)  180 (= 0.400 g)\n\n1(e)(iii) percentage aspirin by mass = (1(e)(ii) / 0.409)  100 (= 97.8%) 1\n\n1(f) amount / mass of solid C would be greater (than the true value) 1\nAND\ncalculated percentage by mass would be greater than (e)(iii)\nQuestion Answer Marks", "source_pages": [ 4, 5 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_53-p04.png", "../answer-assets/9701_w24_ms_53-p05.png" ] }, { "id": "9701-2024-on-53-q02", "question_id": "9701-2024-on-53-q02", "subject": "9701", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 17, "status": "available", "reason": null, "text": "2(a) 588 nm (589.7) 1\n\n2(b)(i) n (C N H Cl(s)) = 2.5  10-2 x 0.5 dm3 = 1.25  10-2 1\n25 3 30\nmass of C N H Cl(s) = 1.25  10-2  407.5 = 5.09 g\n25 3 30\n© Cambridge University Press & Assessment 2024 Page 5 of 7\n\n2(b)(ii) M1 add a (small) volume of (distilled) water (to the small beaker) 3\nAND\ndissolve the C N H Cl (s)\n25 3 30\nM2 Transfer the solution to a 500 cm3 volumetric flask with washings.\nM3: make up to the (calibration) mark (with distilled water.)\nAND\nthen mix the solution (by inverting the flask)\n\n2(c)(i) 1\nVolume of Volume of\n2.5  10-4 moldm-3 distilled\ncrystal violet water / cm3\n(solution E)\n/ cm3\n0.00 20.00\n4.00 16.00\n8.00 12.00\n12.00 8.00\n16.00 4.00\n20.00 0.00\n\n2(c)(ii) absorbance (of aqueous C N H Cl) 1\n25 3 30\n\n2(d)(i) M1 points plotted correctly. 2\nM2 Straight line of best fit drawn.\n© Cambridge University Press & Assessment 2024 Page 6 of 7\n\n2(d)(ii) M1 Most anomalous point circled. 2\n(Expected to be concentration 1.0  10-4 moldm-3 (solution 3)\nM2 Volume of water added (to mixture) too large\nOR\nVolume of crystal violet solution / solution E added (to mixture) too small.\n\n2(d)(iii) (absorbance is) (directly) proportional (to concentration of crystal violet) (within experimental error). 1\n\n2(d)(iv) Repeat the procedure for the solution giving an anomalous result. (Solution 3) 1\n\n2(e)(i) Solutions being mixed (at t=0) so reading not possible. 1\n\n2(e)(ii) M1 (100, 0.360) and (315, 0.180) 2\nM2 half-life correctly calculated from points listed for M1 (x1–x2) when y = 2y\n2 1\n\n2(e)(iii) first order 1\nAND\nhalf-lives are constant (within experimental error)\n© Cambridge University Press & Assessment 2024 Page 7 of 7", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9701_w24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2024-oct-nov/9701_w24_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w24_ms_53-p05.png", "../answer-assets/9701_w24_ms_53-p06.png", "../answer-assets/9701_w24_ms_53-p07.png" ] }, { "id": "9701-2025-m-42-q01", "question_id": "9701-2025-m-42-q01", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 17, "status": "available", "reason": null, "text": "1(a) (Ag(2+) is a) forms stable ion(s) 1\nwith incomplete d orbitals / an incomplete d subshell\n\n1(b)(i) Ag(s) → Ag(g) 2\nAg(s) + ½F (g) → AgF(s)\n2\n\n1(b)(ii) H/energy change 2\nwhen 1 mole of an ionic solid / lattice / crystal / compound is formed\nfrom gas phase ions / gaseous ion(s) (under standard conditions)\n\n1(b)(iii) M1: 5 numbers and no multipliers 2\n–203 = +289 + 732 + 79 + EA + (–955)\n1\nM2 correct sign and evaluation ∴ EA = –348 (kJ mol–1) ecf min 3sf\n1\n\n1(c)(i) enthalpy change of solution of AgF = –(–955) + (–464) + (–506) 1\n= –15 (kJ mol–1)\n\n1(c)(ii) AgF is (slightly) soluble / yes 1\nAND (the enthalpy change of solution is slightly) exothermic / negative\necf from Q1(c)(i)\n\n1(d)(i) greater the distance between the nucleus and (the shells of the) electrons 2\nOR atomic radii increases / atomic size increases / more shells\nOR more shielding by inner shells\nthe less attraction between nucleus / protons\nAND incoming electron / added electron\n\n1(d)(ii) halide/X–/ions get larger (down the group) 1\nAND decreasing attraction between ions / weaker ionic bond\n\n1(e)(i) E⦵cell = +0.342 – (+0.222) = (+)0.120 (V) min 2sf 1\nCu2+ + 2Ag + 2Cl– → Cu + 2AgCl 1\n© Cambridge University Press & Assessment 2025 Page 5 of 14\n\n1(e)(ii) M1: [Cl–(aq)] = 10  36.0 / 74.6 = 4.83 (4.8257) mol dm–3 3\n0.059 1\nM2: E = 0.222 + log ecf M1\n1 M1\nM3: E = (+)0.182 (V) min 2sf ecf M1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_42-p05.png", "../answer-assets/9701_m25_ms_42-p06.png" ] }, { "id": "9701-2025-m-42-q02", "question_id": "9701-2025-m-42-q02", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 15, "status": "available", "reason": null, "text": "2(a)(i) [CH COCH ] concentration stays (almost) constant / does not change 1\n3 3\nOR [CH COCH ] concentration does not affect rate\n3 3\n\n2(a)(ii) M1: first order w.r.t. H+ AND rate  4 / 3 and [H+]  4 / 3 (in expts 2 and 3) 2\nM2: first order w.r.t. CH OH\n3\nAND [H+]  1.5, [CH OH]  1.5 and rate  2.25 / 1.52 (expts 1 and 2)\n3\n/ [H+]  2, [CH OH]  1.5 and rate  3 / (2  1.5) (expts 1 and 3)\n3\n\n2(a)(iii) two quoted half-lives within range / 155–170 s AND (roughly) constant 1\n\n2(b) M1: proposed mechanism = 4 3\nM2: slow step contains only species from the rate equation/law\nOR slow step contains CH COCH and H+ from the rate equation/law\n3 3\nM3: stoichiometric / (mole) ratio / amounts\nAND corresponds to order of reaction / to that in the rate equation\n\n2(c)(i) 1\n© Cambridge University Press & Assessment 2025 Page 6 of 14\n\n2(c)(ii) K = 10–3.95 = 1.122  10–4 2\na\n[H+]2 = (1.122  10–4  0.5)\n[H+] = √(1.122  10–4  0.5) = 7.49  10–3\npH = –log (7.49  10–3)\npH = 2.13 min 2sf\n\n2(d)(i) a solution that resists / minimises / opposes changes in pH 2\nwhen small amounts of (strong) acid or base are added\n\n2(d)(ii) M1: moles of conjugate base = 20 / 1000  1.00 = 0.0200 mol 3\nAND moles of acid C left = 50 / 1000  0.500 – 0.0200 = 0.0300 mol\nM2: K = [H+]  0.0200 / 0.0300\na\n[H+] = 3 / 2  10–3.95 = 1.68  10–4 ecf from M1\nM3: pH = –log (1.68  10–4)\npH = 3.77 min 2sf ecf from a calculated [H+] M2\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_42-p06.png", "../answer-assets/9701_m25_ms_42-p07.png" ] }, { "id": "9701-2025-m-42-q03", "question_id": "9701-2025-m-42-q03", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 17, "status": "available", "reason": null, "text": "3(a)(i) (central) metal / transition element atom / ion 1\nAND surrounded by / bonded to ligands\n© Cambridge University Press & Assessment 2025 Page 7 of 14\n\n3(a)(ii) 3\ncoordination formula and charge\ncomplex ion ligand\nnumber of complex\nE Fe2+ NH 6 [Fe(NH ) ]2+\n3 3 6\nF Fe2+ Cl– 4 [FeCl4 ]2–\nG Fe3+ en 6 [Fe(en) ]3+\n3\nany two [1] any four [2] all six [3]\n\n3(a)(iii) three(3) higher AND two(2) lower boxes / energy levels 1\n\n3(b)(i) d–d energy gap / E is different 1\n)\ndifferent frequency / wavelength (of light) absorbed / transmitted / reflected 1\n\n3(b)(ii) [[Fe(H O) SCN]2+] 1\nK = 2 5\nstab [[Fe(H O) ]3+] [SCN]−]\n2 6\n\n3(b)(iii) K = 2.40  105 / 1.40  102 = 1714 min 2sf 1\nc\n\n3(b)(iv) M1: solution turns (violet / purple to) colourless 2\nAND solution then remains / stays colourless / no change\nM2: K of [Fe(H O) F]2+ > [Fe(H O) SCN]2+\nstab 2 5 2 5\nOR [Fe(H O) F]2+ is most stable (more stable than [Fe(H O) SCN]2+)\n2 5 2 5\n© Cambridge University Press & Assessment 2025 Page 8 of 14\n\n3(c)(i) C O 2– → 2CO + 2e– 2\n2 4 2\nMnO – + 8H+ + 5e– → Mn2+ + 4H O\n4 2\n\n3(c)(ii) M1 moles of manganate = 12.20 / 1000  0.0200 = 2.44  10–4 mol 4\nAND moles of ethanedioate = 5 / 2  2.44  10–4 = 6.10  10–4 mol\nM2 moles of K Fe(C O ) = 6.10  10–4 ÷ 3 = 2.03  10–4 mol\n3 2 4 3\nAND mass of K Fe(C O ) = 437.1  2.03  10–4 = 0.0889 g\n3 2 4 3\nM3 mass of water = 0.100 – 0.0889 = 0.0111 g\nAND moles of water = 0.0111 / 18.0 = 6.18  10–4 mol\nM4 x = molar ratio = 6.18  10–4 / 2.03  10–4 = 3\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_42-p07.png", "../answer-assets/9701_m25_ms_42-p08.png", "../answer-assets/9701_m25_ms_42-p09.png" ] }, { "id": "9701-2025-m-42-q04", "question_id": "9701-2025-m-42-q04", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 18, "status": "available", "reason": null, "text": "4(a) propanamide < ammonia < propylamine 4\n(basicity linked to) p orbital of N / lone pair on N\nAND being accept / donate to / coordinate to a proton/H+\n(propanamide) lone pair / p-orbital on N is delocalised\nAND into C=O group / over N-C-O group / across the two electronegative O & N\n(propylamine) R / alkyl / propyl group\nAND is electron donating / positive inductive effect\n\n4(b) K = CH\n3\nCH\n2\nCOCl / propanoyl chloride 3\nL = CH CH CN / propanenitrile\n3 2\nM = LiAlH\n4\nOR H\n2\n/ Ni / correct names\n© Cambridge University Press & Assessment 2025 Page 9 of 14\n\n4(c) 4\nM1: (un)protonated benzylamine\nM2: (un)protonated amine as above\nM3: ethanoic acid\nM4: all amine groups protonated (—NH +) AND all acid groups as —COOH\n3\n\n4(d)(i) C H NO + 6[H] → C H NH + 2H O 1\n7 7 2 7 7 2 2\n\n4(d)(ii) 6\nstep reagents and conditions type of reaction\n1 • Sn & conc. HCl (+ heat) reduction\n2\n• CH\n3\nBr / CH\n3\nCl / CH\n3\nI • (nucleophilic)\n(in ethanol) substitution\n3 • hot MnO –/KMnO • oxidation\n4 4\n4 condensation\n• (+ conc. H SO )\n2 4\n[1] x 6\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_42-p09.png", "../answer-assets/9701_m25_ms_42-p10.png" ] }, { "id": "9701-2025-m-42-q05", "question_id": "9701-2025-m-42-q05", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 19, "status": "available", "reason": null, "text": "5(a) sp2 = 6 / six sp3 = 3 / three 1\n© Cambridge University Press & Assessment 2025 Page 10 of 14\n\n5(b)(i) electrophilic substitution 1\n\n5(b)(ii) AlBr\n3\n1\n\n5(b)(iii) 3\n+ H+\nM1: curly arrow from inside the hexagon ring\nAND towards positively charged carbon atom (in the electrophile (CH ) CH +)\n3 2\nM2: structure of the intermediate\nM3: curly arrow from C–H bond into the ring AND formation / loss of H+\n\n5(b)(iv) 2° carbocation / (CH ) CH+ 1\n3 2\nAND is more stable due to greater positive inductive effect (of alkyl groups)\n\n5(c)(i) S = 146 + 200 – 205 – 278 = –137 (J K–1 mol–1) 1\n\n5(c)(ii) G = H – TS AND T = 298 / 298.15 used 2\nG = –371 – (25+273)(–0.137) = –330 (kJ mol–1) [so feasible] ecf M1\n\n5(d)(i) M1: conditions for phenol = 4\n• aqueous / aq\n• bromine water\n• no AlBr needed\n3\nM2: lone pair (of e–)/p-orbital on oxygen\nAND overlap / delocalised with ring /  system\nM3: electron density in ring increases / ring becomes more electron rich\n(and attracts electrophiles)\nM4: (phenol is more reactive so) polarises Br / bromine / electrophiles more\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 14\n\n5(d)(ii) 1\nQ =\n\n5(d)(iii) 1\nR = / / benzene diazonium chloride\n\n5(d)(iv) azo 1\n\n5(d)(v) structure / name of 2-nitrophenol and 4-nitrophenol 2\n—OH / hydroxyl group is electron donating group\nAND 2,4-directing / ortho / para directing\n© Cambridge University Press & Assessment 2025 Page 12 of 14", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_42.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_42-p10.png", "../answer-assets/9701_m25_ms_42-p11.png", "../answer-assets/9701_m25_ms_42-p12.png" ] }, { "id": "9701-2025-m-42-q06", "question_id": "9701-2025-m-42-q06", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 14, "status": "available", "reason": null, "text": "6(a)(i) 2\nM1: ester group in middle displayed between two monomers\nM2: rest of molecule correct (only one repeat unit)\n\n6(a)(ii) cold acidified dilute KMnO / MnO – 1\n4 4\n\n6(b)(i) condensation OR dehydration OR elimination 1\n\n6(b)(ii) species with two lone pairs / LP of electrons) 2\nthat form dative (covalent) / co-ordinate bond(s)\nto a (central) transition element / metal atom / ion\n\n6(b)(iii) 2\nM1 one correct diagram\nM2 correct diagram of the other optical isomer\n\n6(c)(i) 8 / eight 1\n\n6(c)(ii) to remove peaks / signals / absorptions 1\nfrom acidic protons / OH / COOH / labile protons\n© Cambridge University Press & Assessment 2025 Page 13 of 14\n\n6(c)(iii) Molecule must not be symmetrical (as eight different C environments) 4\nViable structures of X\n• methyl ketone / CH CO group\n3\nbecause (yellow) ppt / CHI formed (with alkaline I (aq))\n3 2\nany two linked statements:\n• two —OH groups\nbecause 1H NMR signals lost in D O\n2\n• ketone / carbonyl / CH CO\n3\nas (13C NMR) signal with  = 205 / 190–220 ppm\n• methyl / carbon next to C=O because\n(13C NMR) signal with  = 30–65 ppm\n© Cambridge University Press & Assessment 2025 Page 14 of 14", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_42-p13.png", "../answer-assets/9701_m25_ms_42-p14.png" ] }, { "id": "9701-2025-m-52-q01", "question_id": "9701-2025-m-52-q01", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a) NO is a toxic gas 1\n2\n\n1(b) To be able to determine the accurate mass of brass that has been transferred (by difference) 1\n\n1(c) (10 cm3) volumetric pipette 1\n\n1(d) 0.0600 mol dm–3 aqueous sodium thiosulfate 1\n\n1(e) M1 Volume of 0.200 mol dm–3 Na S O (aq) = 30.00 cm3 3\n2 2 3\nM2 Using burette transfer 30.00 cm3 of Na S O ((aq)) and add to a volumetric flask.\n2 2 3\nM3 Top the (100 cm3 volumetric) flask up to 100 cm3 mark with distilled water\n\n1(f) 1.00 g 1\n\n1(g)(i) 1\nrough 1 2 3\n20.80 20.30 20.55 20.40\n\n1(g)(ii) (20.30 + 20.40) ÷ 2 = 20.35 (cm3) 1\n\n1(g)(iii) 2  0.05 1\nWorking must be shown  100 = 0.490\n20.40\n\n1(h)(i) = (0.0600  17.35/1000) = 1.04  10–3 / 0.00104 1\n\n1(h)(ii) = 5.21  10–4 / 0.000521 1\n\n1(h)(iii) = 1.04  10–3 / 0.00104 1\n© Cambridge University Press & Assessment 2025 Page 5 of 7\n\n1(h)(iv) 1.041  10–3  10  63.5 = 0.661 1\n\n1(h)(v) 0.661/1.05  100 = 63.0 1\nQuestion Answer Marks", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_52-p05.png", "../answer-assets/9701_m25_ms_52-p06.png" ] }, { "id": "9701-2025-m-52-q02", "question_id": "9701-2025-m-52-q02", "subject": "9701", "year": 2025, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) 2\nM1 Correctly placed vertical condenser including jacket.\nM2 (Cold) water in at bottom and water out at the top of the condenser\n\n2(a)(ii) anti-bumping granules 1\n\n2(b) M1 thymolphthalein 2\nM2 As KOH is used up the pH range of colour change is within the range of change in the pH of mixture\n\n2(c) There is a spot in the reaction mixture chromatogram with same R value as the spot on the ester chromatogram. 1\nf\n\n2(d) fractional distillation 1\n\n2(e) The spectrum does not contain an absorption at 1710–1750 cm–1 1\n© Cambridge University Press & Assessment 2025 Page 6 of 7\n\n2(f) 3\nshift splitting peak structure\narea\n1.2 doublet 6 –CH\n3\n2.2 singlet 1 –OH\n4.0 multiplet 1 >CH–O\nM1 Row 2 correct\nM2 Row 3 correct\nM3 propan-2-ol\n\n2(g) H 2\nH O H C H\nH C C O C H\nH H C H\nH\nM1 correct displayed formula of an ester link\nM2 correct displayed formula of ester X\n\n2(h) The pH remains below 3.0 1\n© Cambridge University Press & Assessment 2025 Page 7 of 7", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2025-March/ms/9701_m25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-march/9701_m25_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m25_ms_52-p06.png", "../answer-assets/9701_m25_ms_52-p07.png" ] }, { "id": "9701-2025-mj-41-q01", "question_id": "9701-2025-mj-41-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "1(a) CaCO → CaO + CO [1] 1\n3 2\n\n1(b) M1: calcium carbonate AND calcium ion / Ca2+ smaller / ionic radius bigger down group 2\nM2: carbonate ion / anion / CO 2– is more distorted/polarised by Ca2+\n3\n\n1(c)(i) calcium hydroxide / Ca(OH) 1\n2\n\n1(c)(ii) M1: A / calcium hydroxide has more exothermic H AND H than B / barium hydroxide 3\nlatt hyd\nM2: difference in H is greater for A / calcium hydroxide than for B / barium hydroxide\nlatt\nM3: H is less exothermic for A / calcium hydroxide than for B / barium hydroxide\nsol\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p07.png" ] }, { "id": "9701-2025-mj-41-q02", "question_id": "9701-2025-mj-41-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i) mol–2 dm6 s–1 1\n\n2(a)(ii) M1: 2.57  10–6 = 26.4  [concentration]3 2\nM2: [concentration] = 4.60  10–3\n\n2(a)(iii) 2.57  10–3 1\n\n2(b)(i) overall 1st order / pseudo 1st order 1\nOR 1st order with respect to Cl AND concentration of NO doesn’t change\n2\n\n2(b)(ii) ln2 / 105.6 or 0.693 / 105.6 1\n\n2(b)(iii) 0.0262 / 2.62  10–2 1\n© Cambridge University Press & Assessment 2025 Page 7 of 16\n\n2(c) M1: homogeneous catalyst OR catalyst because it is regenerated 3\nM2: 2NO + O → 2NO\n2 2\nM3: NO + SO → NO + SO\n2 2 3\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p07.png", "../answer-assets/9701_s25_ms_41-p08.png" ] }, { "id": "9701-2025-mj-41-q03", "question_id": "9701-2025-mj-41-q03", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 13, "status": "available", "reason": null, "text": "3(a)(i) M1: K = [Cr3+] [OH–]3 2\nsp\nM2: mol4 dm–12\n\n3(a)(ii) M1: [OH–] = 3[Cr3+] 3\nM2: 2.47  10–9 mol dm–3\nM3: 2.54  10–7 g dm–3\n\n3(a)(iii) common ion effect 1\n\n3(b)(i) M1: [H+] = 1.23  10–3 2\nM2: pH = 2.91\n\n3(b)(ii) 13.0 1\n\n3(b)(iii) M1: [CH CH CH COOH] = [CH CH CH COO–] 2\n3 2 2 3 2 2\nM2: pH = 4.82\n\n3(c) M1: 0.148 g is left in water 2\nM2: K = 3.01 OR 0.333\npc\n© Cambridge University Press & Assessment 2025 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p08.png" ] }, { "id": "9701-2025-mj-41-q04", "question_id": "9701-2025-mj-41-q04", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) energy change when one mole of gaseous ions dissolves in water 1\n\n4(a)(ii) M1: K+ has larger radius AND K+ has smaller charge 2\nM2: Ca2+ has larger attraction for H O\n2\n\n4(a)(iii) energy change when one mole of ionic solid forms from gaseous ions 1\n\n4(a)(iv) M1: –1650–(2  506) + 2602 2\nM2: –60\n\n4(b) M1: G = H –TS AND T = 298 2\nM2: –174.5 [1]\nQuestion Answer Marks", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p09.png" ] }, { "id": "9701-2025-mj-41-q05", "question_id": "9701-2025-mj-41-q05", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 15, "status": "available", "reason": null, "text": "5(a)(i) (Ar) 3d10 AND (Ar) 3d9 1\n\n5(a)(ii) similar energy of 3d and 4s subshells 1\n\n5(b)(i) pale blue precipitate 1\n\n5(b)(ii) [Cu(H O) ]2+ + 2OH– → Cu(OH) (H O) + 2H O 1\n2 6 2 2 4 2\n\n5(c)(i) deep blue solution 1\n\n5(c)(ii) [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O 1\n2 6 3 3 4 2 2 2\n\n5(c)(iii) ligand exchange 1\n\n5(d)(i) M1: E = Eo +(0.059 / z) log(ox/red) 2\nM2: E = 1.52 + (0.059 / 5)  log (0.002)\n© Cambridge University Press & Assessment 2025 Page 9 of 16\n\n5(d)(ii) 1.15 1\n\n5(d)(iii) 5Cu + 2MnO – + 16H+ → 5Cu2+ + 2Mn2+ + 8H O 1\n4 2\n\n5(d)(iv) Cu2+ / Cu, Cu2+ / Cu, MnO – / Mn2+ / platinum 1\n4\n\n5(e) M1: moles of copper = 0.0120 mol 3\nM2: coulombs required = 2322 C\nM3: current required = 0.129 A\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p09.png", "../answer-assets/9701_s25_ms_41-p10.png" ] }, { "id": "9701-2025-mj-41-q06", "question_id": "9701-2025-mj-41-q06", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "6(a)(i) [Fe(CN) ]3– AND [Fe(CN) ]4– 1\n6 6\n\n6(a)(ii) M1: different E 2\nM2: absorption of different frequency of visible light\nOR wavelength of visible light\n\n6(a)(iii) dative covalent / coordinate 1\n© Cambridge University Press & Assessment 2025 Page 10 of 16\n\n6(b) 4\nEach correct structure = [1]\n© Cambridge University Press & Assessment 2025 Page 11 of 16", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p10.png", "../answer-assets/9701_s25_ms_41-p11.png" ] }, { "id": "9701-2025-mj-41-q07", "question_id": "9701-2025-mj-41-q07", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 11, "status": "available", "reason": null, "text": "7(a) 1,3-diethylbenzene 1\n\n7(b)(i) CDCl 1\n3\n\n7(b)(ii) no difference AND no H that can be exchanged with D 1\n\n7(b)(iii) 3\n / ppm number of group name of\nprotons responsible splitting pattern\n1.3 6 CH triplet\n3\n2.7 4 CH quartet / quadruplet\n2\n7.1 4\nEach correct column = [1]\n\n7(b)(iv) 5 AND 6 1\n\n7(c)(i) C H + 12[O] → C H O + 2CO + 4H O 1\n10 14 8 6 4 2 2\n© Cambridge University Press & Assessment 2025 Page 12 of 16\n\n7(c)(ii) 3\ncompound\nG\ncompound\nJ\ncompound\nK\nEach correct structure = [1]\n© Cambridge University Press & Assessment 2025 Page 13 of 16", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p12.png", "../answer-assets/9701_s25_ms_41-p13.png" ] }, { "id": "9701-2025-mj-41-q08", "question_id": "9701-2025-mj-41-q08", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 11, "status": "available", "reason": null, "text": "8(a)(i) Br + AlBr → Br+ + AlBr – 1\n2 3 4\n\n8(a)(ii) 2\n• reference to methyl group having an effect\n• methyl group is 2,4,6 directing\nCorrect structure and one correct point [1] correct structure and two correct points [2]\n\n8(a)(iii) 3\n→ →\nM1: first curly arrow from inside hexagon towards Br+\nM2: correct intermediate\nM3: second curly arrow from bond into hexagon AND organic product, H+\n\n8(b)(i) chlorobutane AND white precipitate 1\n\n8(b)(ii) M1: C\n4\nH\n9\nCl + H\n2\nO → C\n4\nH\n9\nOH + H+ + Cl – 2\nM2: Cl – + Ag+ → AgCl [1]\n© Cambridge University Press & Assessment 2025 Page 14 of 16\n\n8(b)(iii) M1: delocalisation of lone pair of Cl into benzene ring /  system 2\nM2: strengthens C–Cl bond / causes C–Cl bond to have partial double bond nature", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p14.png", "../answer-assets/9701_s25_ms_41-p15.png" ] }, { "id": "9701-2025-mj-41-q09", "question_id": "9701-2025-mj-41-q09", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "9(a)(i) • nitrobenzene / C H NO 2\n6 5 2\n• Sn and HCl\n• concentrated HCl and heat / boil / reflux\nTwo correct point [1] all three correct points [2]\n\n9(a)(ii) M1: propylamide / propenamide / C H CONH OR propanenitrile / C H CN 2\n2 5 2 2 5\nM2: LiAlH\n4\n\n9(a)(iii) M1: bromine (aq) AND phenylamine / C H NH 2\n6 5 2\nM2: correct structure\n\n9(b) M1: phenylamine ammonia propylamine 4\nM2: a base has a lone pair that can accept a proton / H+\nM3: the lone pair on the nitrogen atom of phenylamine is delocalised into the benzene ring / -system\nM4: propylamine has an electron donating alkyl group\n© Cambridge University Press & Assessment 2025 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p15.png" ] }, { "id": "9701-2025-mj-41-q10", "question_id": "9701-2025-mj-41-q10", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 10, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 7, "status": "available", "reason": null, "text": "10(a) 1\n\n10(b) M1: reagent 1:Fehling’s OR Tollens’ reagent 4\nM2: acidified MnO –\n4\nobservation when treated observation when treated\nwith reagent 1 with reagent 2\npropanoic acid no change no change\nmethanoic acid red ppt OR silver mirror colourless\nethanedioic acid no change colourless\nM3 / M4: One completely correct column [1] both completely correct columns [2]\n\n10(c) M1: chlorine atoms are electron withdrawing / electronegative 2\nM2: weakening O–H bond OR stabilising conjugate base\n© Cambridge University Press & Assessment 2025 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_41.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_41-p16.png" ] }, { "id": "9701-2025-mj-42-q01", "question_id": "9701-2025-mj-42-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 8, "status": "available", "reason": null, "text": "1(a)(i) Ca(NO ) → CaO + 2NO + ½O [1] 1\n3 2 2 2\n\n1(a)(ii) M1: temperature increases down the group 3\nM2: ionic radius OR size of cation / M(2)+ AND increases\nM3: less polarisation / less distortion AND of anion / nitrate ion OR less weakening of NO bond / harder to break NO bond\n\n1(b) M1: moles of SrO = 0.333  103.6 = 3.214  10-3 AND moles of Sr(OH) = 3.214  10-3 OR 9 / 2800 4\n2\nM2: moles of OH– in 250 cm3 = 3.214  10-3  2 = 6.429  10-3 AND moles of OH– in 1 dm3 = 6.429  10-3  4 = 0.0257\nM3: [H+] = 1.00  10–14  0.0257 = 3.89  10–13\nOR pOH = –log[OH–] = 1.59\nM4: pH = –log(3.89  10–13) = 12.41 must be to 2dp\nOR pH = 14 –1.59 = 12.41 must be 2 dp", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p07.png" ] }, { "id": "9701-2025-mj-42-q02", "question_id": "9701-2025-mj-42-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 17, "status": "available", "reason": null, "text": "2(a)(i) A = [Co(H O) ]2+ pink 4\n2 6\nB = [Co(H O) (OH) ] / Co(OH) blue\n2 4 2 2\nC = [Co(NH ) ]2+ straw / yellow-brown\n3 6\nB = precipitation / deprotonation / acid-base\nC = ligand exchange\nAny two [1] any four [2] any six [3] all eight [4]\n\n2(a)(ii) concentrated HCl 1\n\n2(b)(i) • five (3d) orbitals (lines, boxes) in the isolated Co2+ ion of same energy 2\n• splitting three higher and two lower d orbitals\n• energy of all d orbitals in the complex clearly higher than all d orbitals in isolated ion\nany two [1] all three [2]\n© Cambridge University Press & Assessment 2025 Page 7 of 17\n\n2(b)(ii) 1\n\n2(c) M1: H O acting as oxidising agent with Mn(OH) / Mn2+ 4\n2 2 2\nM2: Mn2+ + H O → MnO + 2H+ AND E o = +0.56\n2 2 2 cell\nOR Mn(OH) + H O → MnO + 2H O AND E o = +0.92\n2 2 2 2 2 cell\nM3: H O acting as reducing agent with MnO\n2 2 2\nM4: MnO + 2H+ + H O → Mn2+ + 2H O + O AND Eo = +0.54\n2 2 2 2 2 cell\n\n2(d)(i) colourless to pale pink 1\n\n2(d)(ii) • moles MnO – = 0.0500  18.60 / 1000 = 9.30  10-4 in 25.0 cm3 4\n4\n• moles Fe2+ = 5  9.30  10-4 = 4.65  10-3 in 25.0 cm3\n• moles Fe2+ = 6 x 4.65  10-3 = 2.79  10-2in 150.0 cm3\nM1 / M2: any two of the above bullets [1] all three [2]\nOR\n• moles MnO – = 0.0500  18.60 / 1000 = 9.30  10-4 in 25.0 cm3\n4\n• moles MnO – = 6  9.30  10-4 = 5.58  10-3 in 150.0 cm3\n4\n• moles Fe2+ = 5  5.58  10-3 = 2.79  10-2in 150.0 cm3\nM1 / M2: any two of the above bullets [1] all three [2]\nM3: mass Fe = 55.8  2.79  10-2 = 1.55682 (g) in two tablets\nM4: mass Fe = 0.5  1.55682 = 0.77841 (g) in one tablet AND mass Fe = 7.7841  10–2  1000 = 778.4\n© Cambridge University Press & Assessment 2025 Page 8 of 17", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p07.png", "../answer-assets/9701_s25_ms_42-p08.png" ] }, { "id": "9701-2025-mj-42-q03", "question_id": "9701-2025-mj-42-q03", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 8, "status": "available", "reason": null, "text": "3(a)(i) number of possible arrangements of particles/molecules AND energy in a system 1\n\n3(b)(i) T melting OR solid to liquid 1\n1\nT boiling OR liquid to gas\n2\n\n3(b)(ii) change in disorder from a liquid to a gas is much bigger than solid to liquid 1\nOR\nchange in intermolecular distance between liquid & gas is much bigger than solid to liquid\n\n3(c) M1: ΔHo = (3  –393.5) – ((–824.2) + (3  –110.5)) = –24.8 (kJ mol–1) 3\nr\nM2: ΔGo = ΔHo – TΔSo AND T= 723 used\nM3: (24800 – 36200) ÷ 723 = – [3  213.8 + 2  27.3) – (87.4) – (3x)]\n–15.8 = –608.6 + 3x, so 3x = 592.8, so x = 197.6 / 198 (J K–1 mol–1) min 3sf\n\n3(d) • (as temperature increases) the reaction is less feasible 2\n• as ΔGo becomes less negative / (more) positive\n• due to TS becoming more negative / –TS (becoming more) positive\nM1 / M2: any two of the above bullets [1] all three [2]\nQuestion Answer Marks", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p09.png" ] }, { "id": "9701-2025-mj-42-q04", "question_id": "9701-2025-mj-42-q04", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "4(a)(i) the power/exponent to which a concentration AND of a reactant is raised in the rate equation 1\n\n4(a)(ii) 2\nAny two [1] all three [2]\n© Cambridge University Press & Assessment 2025 Page 9 of 17\n\n4(a)(iii) 1\nnegative gradient curve\n\n4(b)(i) rate = 4  4 = 16 times 1\n\n4(b)(ii) slowest step / slow step in the mechanism / multi-step reaction 1\n\n4(b)(iii) M1: step 1 O + NO → NO + O 2\n3 2 3 2\nM2: step 2 NO + NO → N O AND sum of step 1 and 2 is consistent with equation for reaction 2\n3 2 2 5\n\n4(c)(i) Use of graph k = 12  10–4 / 0.04 = 0.03 1\n\n4(c)(ii) k = 0.693 / t 1\n1/2\nt = 0.693 / 0.030 = 23.1 (s) min 2sf\n1/2\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p09.png", "../answer-assets/9701_s25_ms_42-p10.png" ] }, { "id": "9701-2025-mj-42-q05", "question_id": "9701-2025-mj-42-q05", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 12, "status": "available", "reason": null, "text": "5(a)(i) acid I HClO conjugate base of acid I ClO– 1\nacid II H O conjugate base of acid II OH–\n2\n(a)(ii) M1: [H+] = 10–4.51 = 3.09  10–5 2\nM2: [HClO] = (3.09  10–5)2  (3.70  10–8) = 0.0258 min 2sf\n(a)(iii) 3 HClO → HClO + 2 HCl 1\n3\n© Cambridge University Press & Assessment 2025 Page 10 of 17\n\n5(b)(ii) hydroxide ions / NaOH / OH– AND it reacts with the acid to form the conjugate base / ethanoate ions 1\nOR ethanoate ions / sodium ethanoate AND as it is the conjugate base (of ethanoic acid)\n\n5(d)(i) M1: Use of [Ca2+] = [PO 3–] = 1.14  10–7 in the expression 3\n4\nOR K = (3.42  10–7)3  (2.28  10–7)2\nsp\nM2: K = (3  1.14  10–7)3  (2  1.14  10–7)2\nsp\nK = 2.079  10–33\nsp\nOR 2.08  10–33\nOR 2.1  10–33 min 2sf\nM3: units = mol5 dm–15\n\n5(d)(ii) solubility decreases AND due to the common ion effect 1\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p10.png", "../answer-assets/9701_s25_ms_42-p11.png" ] }, { "id": "9701-2025-mj-42-q06", "question_id": "9701-2025-mj-42-q06", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 19, "status": "available", "reason": null, "text": "6(a)(i) HNO + H SO → NO + + HSO – + H O 1\n3 2 4 2 4 2\nOR HNO + 2H SO → NO + + 2HSO – + H O+\n3 2 4 2 4 3\n© Cambridge University Press & Assessment 2025 Page 11 of 17\n\n6(a)(ii) 3\nM1: curly arrow 1 from inside the hexagon to nitrogen of NO +\n2\nM2: intermediate, usual rules for horseshoe and + charge\nM3: curly arrow 2 from C-H bond into hexagon AND H+\n\n6(b) M1: p-orbital/lone pair from O / oxygen AND is delocalised / overlaps AND into the ring /  system 2\nM2: so electron density of the ring is increased\nOR polarises electrophiles / NO + better\n2\n\n6(c) 1\n\n6(d)(i) (hot) tin / Sn AND concentrated AND HCl 1\n\n6(d)(ii) reduction 1\n© Cambridge University Press & Assessment 2025 Page 12 of 17\n\n6(d)(iii) 1\n\n6(d)(iv) (nucleophilic) addition – elimination 1\n\n6(e) M1: 2-nitrophenol  phenol  water  ethanol 4\nM2: weakens O—H bond / anion stabilised\nAND proton/H+ (more easily) lost / proton / H+ (more easily) donated\nM3: (for 2-nitrophenol / phenol), a p-orbital / lone pair from O / oxygen AND is delocalised / overlaps AND into the ring / \nsystem (and increases electron density in the ring)\nM4: (for ethanol), positive inductive effect / electron donating of alkyl group\n\n6(f)(i) 2\nname of functional group classification of functional group\nalcohol / hydroxyl primary\nalcohol / hydroxyl secondary\namine secondary\nAny three correct [1] all six correct [2]\n© Cambridge University Press & Assessment 2025 Page 13 of 17\n\n6(f)(ii) 1\n\n6(f)(iii) 1\nQuestion Answer Marks", "source_pages": [ 11, 12, 13, 14 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p11.png", "../answer-assets/9701_s25_ms_42-p12.png", "../answer-assets/9701_s25_ms_42-p13.png", "../answer-assets/9701_s25_ms_42-p14.png" ] }, { "id": "9701-2025-mj-42-q07", "question_id": "9701-2025-mj-42-q07", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 11, "status": "available", "reason": null, "text": "7(a) rotate the plane of polarised light equally in the opposite direction 1\n\n7(b)(i) M1: nitrogen / N can accept / bond to a proton / H+ ion 2\nOR donation of lone pair/electrons on the nitrogen / N to a proton / H+ ion\nM2: CH CH(NH )CH CH + H O → CH CH(NH +)CH CH + OH–\n3 2 2 3 2 3 3 2 3\n\n7(b)(ii) M1: more electron donating alkyl / ethyl group on N / NH (in diethylamine) 2\nOR two electron donating alkyl / R / alkane / ethyl group on N / NH (in diethylamine)\nM2: increase electron density on N\nOR make lone pair on N more available (to accept H+)\n\n7(b)(iii) (CH CH ) NH + CH COOH → (CH CH ) NH + CH COO– 1\n3 2 2 3 3 2 2 2 3\n© Cambridge University Press & Assessment 2025 Page 14 of 17\n\n7(b)(iii) (CH CH ) NH + CH COOH → (CH CH ) NH + CH COO– 1\n3 2 2 3 3 2 2 2 3\n\n7(b)(iv) (CH CH ) NH + CH COCl → CH CON(CH CH ) + HCl 1\n3 2 2 3 3 2 3 2\n\n7(c)(i) 1\nmonomer type of polymerisation\naddition\ncondensation\nCH CH(NH )COOH condensation\n3 2\n\n7(c)(ii) 2\nM1: correct displayed amide linkage with an adjacent C=O (to CO) and C(H ) to NH\n2\nM2: rest of the structure correct (only one repeat unit) with continuation bonds\n\n7(c)(iii) they are chemically inert / difficult to hydrolyse / C-C bonds are non-polar 1\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p14.png", "../answer-assets/9701_s25_ms_42-p15.png" ] }, { "id": "9701-2025-mj-42-q08", "question_id": "9701-2025-mj-42-q08", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 6, "status": "available", "reason": null, "text": "8(a) reference: TMS / tetramethylsilane / (CH ) Si 1\n3 4\nAND\nsolvent: D O / CDCl\n2 3\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n8(b) six / 6 peaks 1\n\n8(c) 4\nchemical splitting number of protons on number of 1H atoms\nshift pattern adjacent carbon responsible for the\n / ppm atoms peak\n1.10 triplet 2 3\n1.50 doublet 1 6\n2.45 quartet / quad 3 2\nruplet\n3.75 multiplet 6 1\nAny three [1] any six [2] any nine [3] all eleven [4]\nQuestion Answer Marks", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p15.png", "../answer-assets/9701_s25_ms_42-p16.png" ] }, { "id": "9701-2025-mj-42-q09", "question_id": "9701-2025-mj-42-q09", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 9, "status": "available", "reason": null, "text": "9(a) 2\nM1: correct displayed peptide bond with an adjacent C(H) (to CO) and C(H) to NH\nM2: rest of the dipeptide correct\n\n9(b)(i) pH at which a molecule has no overall charge / no net charge 1\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n9(b)(ii) 3\nM1: correct relative positions of the spots drawn\nM2: ser not moved AND as it is a zwitterion / neutral\nOR ser-lys / lys AND move towards negative (pole) as they are positively charged\nM3: lys moves the furthest / fastest AND as it has the lower M / smaller size (than ser-lys)\nr\n\n9(c)(i) 1\n\n9(c)(ii) M1: step 1 aqueous AND HCl / H SO AND heat / reflux[1] 2\n2 4\nM2: step 2 conc. H SO catalyst AND high temp / heat\n2 4\n© Cambridge University Press & Assessment 2025 Page 17 of 17", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_42-p16.png", "../answer-assets/9701_s25_ms_42-p17.png" ] }, { "id": "9701-2025-mj-43-q01", "question_id": "9701-2025-mj-43-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "1(a) CaCO → CaO + CO [1] 1\n3 2\n\n1(b) M1: calcium carbonate AND calcium ion / Ca2+ smaller / ionic radius bigger down group 2\nM2: carbonate ion / anion / CO 2– is more distorted/polarised by Ca2+\n3\n\n1(c)(i) calcium hydroxide / Ca(OH) 1\n2\n\n1(c)(ii) M1: A / calcium hydroxide has more exothermic H AND H than B / barium hydroxide 3\nlatt hyd\nM2: difference in H is greater for A / calcium hydroxide than for B / barium hydroxide\nlatt\nM3: H is less exothermic for A / calcium hydroxide than for B / barium hydroxide\nsol\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p07.png" ] }, { "id": "9701-2025-mj-43-q02", "question_id": "9701-2025-mj-43-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "2(a)(i) mol–2 dm6 s–1 1\n\n2(a)(ii) M1: 2.57  10–6 = 26.4  [concentration]3 2\nM2: [concentration] = 4.60  10–3\n\n2(a)(iii) 2.57  10–3 1\n\n2(b)(i) overall 1st order / pseudo 1st order 1\nOR 1st order with respect to Cl AND concentration of NO doesn’t change\n2\n\n2(b)(ii) ln2 / 105.6 or 0.693 / 105.6 1\n\n2(b)(iii) 0.0262 / 2.62  10–2 1\n© Cambridge University Press & Assessment 2025 Page 7 of 16\n\n2(c) M1: homogeneous catalyst OR catalyst because it is regenerated 3\nM2: 2NO + O → 2NO\n2 2\nM3: NO + SO → NO + SO\n2 2 3\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p07.png", "../answer-assets/9701_s25_ms_43-p08.png" ] }, { "id": "9701-2025-mj-43-q03", "question_id": "9701-2025-mj-43-q03", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 13, "status": "available", "reason": null, "text": "3(a)(i) M1: K = [Cr3+] [OH–]3 2\nsp\nM2: mol4 dm–12\n\n3(a)(ii) M1: [OH–] = 3[Cr3+] 3\nM2: 2.47  10–9 mol dm–3\nM3: 2.54  10–7 g dm–3\n\n3(a)(iii) common ion effect 1\n\n3(b)(i) M1: [H+] = 1.23  10–3 2\nM2: pH = 2.91\n\n3(b)(ii) 13.0 1\n\n3(b)(iii) M1: [CH CH CH COOH] = [CH CH CH COO–] 2\n3 2 2 3 2 2\nM2: pH = 4.82\n\n3(c) M1: 0.148 g is left in water 2\nM2: K = 3.01 OR 0.333\npc\n© Cambridge University Press & Assessment 2025 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p08.png" ] }, { "id": "9701-2025-mj-43-q04", "question_id": "9701-2025-mj-43-q04", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) energy change when one mole of gaseous ions dissolves in water 1\n\n4(a)(ii) M1: K+ has larger radius AND K+ has smaller charge 2\nM2: Ca2+ has larger attraction for H O\n2\n\n4(a)(iii) energy change when one mole of ionic solid forms from gaseous ions 1\n\n4(a)(iv) M1: –1650–(2  506) + 2602 2\nM2: –60\n\n4(b) M1: G = H –TS AND T = 298 2\nM2: –174.5 [1]\nQuestion Answer Marks", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p09.png" ] }, { "id": "9701-2025-mj-43-q05", "question_id": "9701-2025-mj-43-q05", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 15, "status": "available", "reason": null, "text": "5(a)(i) (Ar) 3d10 AND (Ar) 3d9 1\n\n5(a)(ii) similar energy of 3d and 4s subshells 1\n\n5(b)(i) pale blue precipitate 1\n\n5(b)(ii) [Cu(H O) ]2+ + 2OH– → Cu(OH) (H O) + 2H O 1\n2 6 2 2 4 2\n\n5(c)(i) deep blue solution 1\n\n5(c)(ii) [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O 1\n2 6 3 3 4 2 2 2\n\n5(c)(iii) ligand exchange 1\n\n5(d)(i) M1: E = Eo +(0.059 / z) log(ox/red) 2\nM2: E = 1.52 + (0.059 / 5)  log (0.002)\n© Cambridge University Press & Assessment 2025 Page 9 of 16\n\n5(d)(ii) 1.15 1\n\n5(d)(iii) 5Cu + 2MnO – + 16H+ → 5Cu2+ + 2Mn2+ + 8H O 1\n4 2\n\n5(d)(iv) Cu2+ / Cu, Cu2+ / Cu, MnO – / Mn2+ / platinum 1\n4\n\n5(e) M1: moles of copper = 0.0120 mol 3\nM2: coulombs required = 2322 C\nM3: current required = 0.129 A\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p09.png", "../answer-assets/9701_s25_ms_43-p10.png" ] }, { "id": "9701-2025-mj-43-q06", "question_id": "9701-2025-mj-43-q06", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "6(a)(i) [Fe(CN) ]3– AND [Fe(CN) ]4– 1\n6 6\n\n6(a)(ii) M1: different E 2\nM2: absorption of different frequency of visible light\nOR wavelength of visible light\n\n6(a)(iii) dative covalent / coordinate 1\n© Cambridge University Press & Assessment 2025 Page 10 of 16\n\n6(b) 4\nEach correct structure = [1]\n© Cambridge University Press & Assessment 2025 Page 11 of 16", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p10.png", "../answer-assets/9701_s25_ms_43-p11.png" ] }, { "id": "9701-2025-mj-43-q07", "question_id": "9701-2025-mj-43-q07", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 11, "status": "available", "reason": null, "text": "7(a) 1,3-diethylbenzene 1\n\n7(b)(i) CDCl 1\n3\n\n7(b)(ii) no difference AND no H that can be exchanged with D 1\n\n7(b)(iii) 3\n / ppm number of group name of\nprotons responsible splitting pattern\n1.3 6 CH triplet\n3\n2.7 4 CH quartet / quadruplet\n2\n7.1 4\nEach correct column = [1]\n\n7(b)(iv) 5 AND 6 1\n\n7(c)(i) C H + 12[O] → C H O + 2CO + 4H O 1\n10 14 8 6 4 2 2\n© Cambridge University Press & Assessment 2025 Page 12 of 16\n\n7(c)(ii) 3\ncompound\nG\ncompound\nJ\ncompound\nK\nEach correct structure = [1]\n© Cambridge University Press & Assessment 2025 Page 13 of 16", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p12.png", "../answer-assets/9701_s25_ms_43-p13.png" ] }, { "id": "9701-2025-mj-43-q08", "question_id": "9701-2025-mj-43-q08", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 11, "status": "available", "reason": null, "text": "8(a)(i) Br + AlBr → Br+ + AlBr – 1\n2 3 4\n\n8(a)(ii) 2\n• reference to methyl group having an effect\n• methyl group is 2,4,6 directing\nCorrect structure and one correct point [1] correct structure and two correct points [2]\n\n8(a)(iii) 3\n→ →\nM1: first curly arrow from inside hexagon towards Br+\nM2: correct intermediate\nM3: second curly arrow from bond into hexagon AND organic product, H+\n\n8(b)(i) chlorobutane AND white precipitate 1\n\n8(b)(ii) M1: C\n4\nH\n9\nCl + H\n2\nO → C\n4\nH\n9\nOH + H+ + Cl – 2\nM2: Cl – + Ag+ → AgCl [1]\n© Cambridge University Press & Assessment 2025 Page 14 of 16\n\n8(b)(iii) M1: delocalisation of lone pair of Cl into benzene ring /  system 2\nM2: strengthens C–Cl bond / causes C–Cl bond to have partial double bond nature", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p14.png", "../answer-assets/9701_s25_ms_43-p15.png" ] }, { "id": "9701-2025-mj-43-q09", "question_id": "9701-2025-mj-43-q09", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "9(a)(i) • nitrobenzene / C H NO 2\n6 5 2\n• Sn and HCl\n• concentrated HCl and heat / boil / reflux\nTwo correct point [1] all three correct points [2]\n\n9(a)(ii) M1: propylamide / propenamide / C H CONH OR propanenitrile / C H CN 2\n2 5 2 2 5\nM2: LiAlH\n4\n\n9(a)(iii) M1: bromine (aq) AND phenylamine / C H NH 2\n6 5 2\nM2: correct structure\n\n9(b) M1: phenylamine ammonia propylamine 4\nM2: a base has a lone pair that can accept a proton / H+\nM3: the lone pair on the nitrogen atom of phenylamine is delocalised into the benzene ring / -system\nM4: propylamine has an electron donating alkyl group\n© Cambridge University Press & Assessment 2025 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p15.png" ] }, { "id": "9701-2025-mj-43-q10", "question_id": "9701-2025-mj-43-q10", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 10, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 7, "status": "available", "reason": null, "text": "10(a) 1\n\n10(b) M1: reagent 1:Fehling’s OR Tollens’ reagent 4\nM2: acidified MnO –\n4\nobservation when treated observation when treated\nwith reagent 1 with reagent 2\npropanoic acid no change no change\nmethanoic acid red ppt OR silver mirror colourless\nethanedioic acid no change colourless\nM3 / M4: One completely correct column [1] both completely correct columns [2]\n\n10(c) M1: chlorine atoms are electron withdrawing / electronegative 2\nM2: weakening O–H bond OR stabilising conjugate base\n© Cambridge University Press & Assessment 2025 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_43.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_43-p16.png" ] }, { "id": "9701-2025-mj-44-q01", "question_id": "9701-2025-mj-44-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 19, "status": "available", "reason": null, "text": "1(a)(i) M1: (thermal stability) increases (down the group or from Mg to Ba) 3\nOR carbonates become more stable (down the group or from Mg to Ba)\nM2: ion / cation size / radius increases OR charge density on ion / cation decreases\nM3: less polarised CO (2)– / anion / carbonate ion\n3\nOR weakens the C-O / C=O / covalent bonds less\nOR weakens bonds in the anion less\nOR lattice enthalpy of MO falls (less exothermic) faster than MCO (due to differing sizes of anions)\n3\n\n1(a)(ii) CuCO ⎯⎯→ CuO + CO 1\n3 2\n\n1(b)(i) 1\n[Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿\n\n1(b)(ii) 1\n\n1(b)(iii) the (3)d and (4)s are close / similar in energy 1\n\n1(c)(i) M1: species with / has two lone pairs of electrons 2\nM2: that form dative (covalent) / co-ordinate bond(s) to a (central) transition-element / metal AND atom / ion\nOR donates electron pair(s) to a (central) transition-element / metal AND atom / ion\n© Cambridge University Press & Assessment 2025 Page 7 of 17\n\n1(c)(ii) 3\neach correct structure = [1]\n\n1(c)(iii) optical AND geometrical / cis-trans 1\n\n1(c)(iv) identification of either cis isomer AND dipoles / polar bonds / partial charges do not cancel 1\nOR identification of either cis isomer AND it is asymmetric and has polar bonds\n\n1(d)(i) 1\ncomplex ion in Na Al F Al F 3–\n3 6 6\nligand in Na Al F F–\n3 6\n\n1(d)(ii) M1: charge passed = 1.5  30  60 OR 2700 (C) 4\nM2: n(e–) = 2700 / 96500 OR 2.80  10–2 (mol)\nM3: n(Al) = 2.80  10-2 ÷ 3 OR 9.33  10–3 (mol)\nM4: mass (Al) = (27  9.33  10-3) = 0.25 (g)\n© Cambridge University Press & Assessment 2025 Page 8 of 17", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p07.png", "../answer-assets/9701_s25_ms_44-p08.png" ] }, { "id": "9701-2025-mj-44-q02", "question_id": "9701-2025-mj-44-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 2, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 13, "status": "available", "reason": null, "text": "2(a)(i) white / steamy fumes 1\n\n2(a)(ii) number of (possible) arrangements of particles AND energy in a system 1\n\n2(a)(iii) due to the large number increase in gas molecules formed (in the reaction) 1\nOR there are much more gaseous molecules in the products\nOR six gas moles are produced\n\n2(a)(iv) M1: H = (–859) + (2 x –297) + (4 x –92) – (–1460) – (2 x –246) OR +131 kJ mol–1 3\nr\nM2: use of G = H – TS AND use of 298 (or 273+25) for T\nM3: G = 131 – (298 x 0.768) = –97.9 (kJ mol–1)\n\n2(b)(i) 2Ba2+(aq) + Cr O 2-(aq) + H O(l) ⎯⎯→ 2BaCrO (s) + 2H+(aq) 1\n2 7 2 4\n\n2(b)(ii) M1: d orbital(s) of two different energies / d-d splitting occurs 3\nOR d orbital(s) / d (sub)-shell splits / d-d gap\nOR (inferred from movement of an electron) from lower d to higher d orbital\nM2: electron(s) promoted / excited\nOR electron(s) moves to higher (d–) orbital\nOR electron(s) jumps up (to d– orbital) / jumps to higher (d–orbital)\nM3: wavelength / frequency / light / photon / h / hf absorbed\nOR radiation / energy from visible (region) absorbed AND colour (seen) is complementary\nOR wavelength / frequency / colour / light not absorbed is transmitted / reflected / seen\n\n2(c) M1: H and H decrease / both become less exothermic / less negative 3\nlatt hyd\nM2: H changes more / dominant factor / changes faster\nhyd\nOR H changes less / becomes less exothermic by a smaller extent\nlatt\nM3: H becomes less exothermic / less negative\nsol\nOR H becomes (more) endothermic / (more) positive\nsol\n© Cambridge University Press & Assessment 2025 Page 9 of 17", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p09.png" ] }, { "id": "9701-2025-mj-44-q03", "question_id": "9701-2025-mj-44-q03", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 3, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 15, "status": "available", "reason": null, "text": "3(a)(i) M1: K = [Ni2+][IO –]2 2\nsp 3\nM2: units: mol3 dm–9\n\n3(a)(ii) [Ni2+] = 2.3  10–2; [IO –] = 4.6  10–2 1\n3\nK = (2.3  10–2)  (4.6  10-2)2 = 4.87  10–5\nsp\n\n3(b)(i) Eo = 1.19 – (0.34) = (+)0.85 (V) AND (positive electrode is) right hand side / nickel electrode / platinum / iodate 1\ncell\n\n3(b)(ii) (E would be) less positive / more negative AND as [IO –] is less than 1.0 mol dm–3 / has a lower / smaller concentration 1\ncell 3\n\n3(b)(iii) M1: E is less positive ticked 2\ncell\nM2: [IO –(aq)] is lowered due to the common ion effect\n3\nOR Ni(IO ) precipitating or shown by equation\n3 2\nOR solubility of Ni(IO ) decreasing\n3 2\n\n3(c)(i) [H+] = 10–pH = 10–0.47 = 0.34 / 0.339 (mol dm-3) min 2sf 1\n\n3(c)(ii) [HIO ] = 1.0 – 0.34 = 0.66 (mol dm–3) min 2sf 1\n3eqm\nAND [H+] = [IO –] = 0.34 (mol dm–3) min 2sf\neqm 3 eqm\n\n3(c)(iii) K = (0.34)2 / 0.66 = 0.17– 0.18 (mol dm–3) min 2sf 1\na\n\n3(d)(i) M1: opposes / resists change in pH / controls pH / pH kept within a small range 2\nM2: when small amount of acid / H+ or alkali / base / OH– is added\n\n3(d)(ii) M1: k = {2.1  10–2}/{0.5  (1  10–3)2  0.012} = 4.20  108 min 2sf 2\nM2: mol–4 dm12 s–1\n\n3(d)(iii) rate will be (0.03 / 0.01)2 = 9 times as fast, so rate = 9  2.1  10–2 1\n= 0.19 (0.189) (mol dm–3 s–1) min 2sf\n© Cambridge University Press & Assessment 2025 Page 10 of 17", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p10.png" ] }, { "id": "9701-2025-mj-44-q04", "question_id": "9701-2025-mj-44-q04", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 4, "topic": "Polymerisation", "topic_slug": "9701-topic-35-polymerisation", "marks": 8, "status": "available", "reason": null, "text": "4(a) 4\npolymer type of polymerisation structure of monomer\nX condensation H (CH ) CO H\n2 2 5 2\nOR NH (CH ) COCl\n2 2 5\nY addition CH =CHCO CH\n2 2 3\nZ condensation CH CH(OH)CO H\n3 2\nOR CH CH(OH)COCl\n3\nType of polymerisation (all three) = [1]\nStructure of monomers = [1] for each structure\n\n4(b) pH (where) the species is a zwitterion (is the dominant form) 1\nOR pH (where) the species is (electrically) neutral\nOR pH (where) the species has a (net overall) charge of zero\n\n4(c)(i) 2\nGlu Lys\nGly\n+ -\nStart point\nM1: Gly correct\nM2: Glu and Lys correct\n\n4(c)(ii) it would move towards the positive terminal / anode / side / end or to the left 1\n© Cambridge University Press & Assessment 2025 Page 11 of 17", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-35-polymerisation/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p11.png" ] }, { "id": "9701-2025-mj-44-q05", "question_id": "9701-2025-mj-44-q05", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 5, "topic": "An introduction to A Level organic chemistry", "topic_slug": "9701-topic-29-an-introduction-to-a-level-organic-chemistry", "marks": 10, "status": "available", "reason": null, "text": "5(a)(i) 1\n\n5(a)(ii) the OH / NH groups / O atoms AND can hydrogen bond with water 1\n\n5(a)(iii) C H N O F 1\n33 35 2 5\n\n5(a)(iv) 1\n\n5(a)(v) different / better biological activity 1\n\n5(b)(i) 16 / sixteen 1\n\n5(b)(ii) M1: 4 / four 2\nM2: proton exchange between NH / OH with D\n© Cambridge University Press & Assessment 2025 Page 12 of 17\n\n5(c) groups reacting 2\n• carboxylic acid\n• amide\ngroups forming\n• (primary) alcohol\n• amine\nAny two [1] all four [2]\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-29-an-introduction-to-a-level-organic-chemistry/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p12.png", "../answer-assets/9701_s25_ms_44-p13.png" ] }, { "id": "9701-2025-mj-44-q06", "question_id": "9701-2025-mj-44-q06", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 6, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 15, "status": "available", "reason": null, "text": "6(a) 4\ntest\nI2(aq) warm with\nsodium\nmetal Na 2 CO 3 (aq) 2,4-DNPH + Fehling’s Br 2 (aq)\nOH–(aq) solution\nalcohol / carboxylic carbonyl / (methyl) aldehyde alkene\ncarboxylic acid ketone / ketone / (\nacid aldehyde methyl)\nalcohol\nAny two [1] any three [2] any five [3] all six [4]\n\n6(b)(i) (primary) alcohol 1\n\n6(b)(ii) M1: A → B oxidation/dehydrogenation 3\nM2: C → E reduction/(nucleophilic) addition\nM3: E → F elimination/dehydration\n© Cambridge University Press & Assessment 2025 Page 13 of 17\n\n6(b)(iii) 2\neach correct structure [1]\n\n6(b)(iv) 5\nD F E C\nA any one of the following\neach correct structure [1]\n© Cambridge University Press & Assessment 2025 Page 14 of 17", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p13.png", "../answer-assets/9701_s25_ms_44-p14.png" ] }, { "id": "9701-2025-mj-44-q07", "question_id": "9701-2025-mj-44-q07", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 7, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 11, "status": "available", "reason": null, "text": "7(a) M1: benzoic acid > phenol > ethanol 3\nM2 / M3: any two [1] any three [2]\n• correct link of acidity once AND weakens O—H / OH bond / hydroxyl bond / -O-H\nOR H+ more easily lost / (carboxylate) anion stabilised u / c\n• (benzoic acid) due to negative inductive effect / electron withdrawing effect AND of C=O / COOH / carboxyl\n• (phenol) as lone pair / p-orbital (electrons) on oxygen (on phenol) / AND overlap / delocalised into the ring / π-system\n• (ethanol) alkyl / ethyl / R group AND is electron donating / positive inductive effect\n\n7(b) 2\nEach correct structure = [1]\n\n7(c)(i) ethyl propanoate 1\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(c)(ii) 4\nM1 / M2: any two [1] all four [2]\n• lone pair on O\n• correct arrow from (lone pair) O to C (of C=O)\n• dipole on C=O\n• correct arrow on C=O\nM3: correct intermediate\nM4: arrow from lone pair or charge on O– to C-O bond AND arrow from C-Cl to Cl\n\n7(c)(iii) (nucleophilic) addition – elimination 1\nQuestion Answer Marks", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p15.png", "../answer-assets/9701_s25_ms_44-p16.png" ] }, { "id": "9701-2025-mj-44-q08", "question_id": "9701-2025-mj-44-q08", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 8, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 9, "status": "available", "reason": null, "text": "8(a)(i) M1: energy change / released when 1 mole of a (ionic) solid / lattice / crystal / compound is formed 2\nM2: from gas (phase) ions / gaseous ions (under standard conditions)\n\n8(a)(ii) (enthalpy change when) 1 mole of a substance / solid / solute / molecule AND dissolves in water (to give a solution of infinite 1\ndilution)\n\n8(b) H (NaCl) = H (Na+) + H (Cl–) – H (NaCl) 1\nsol hyd hyd latt\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n8(c) 3\nM1 / M2: any two [1] all four [2]\nM3: all state symbols for the formula are present and correct\n\n8(d) M1: (more exothermic because) Cl– OR NO – because its (ionic) radius / size is smaller 2\n3\nM2: (more exothermic because) (ion-dipole) attraction / bond between it and water is stronger\nOR\nM1: NO – because it has lone pairs on O / more lone pairs\n3\nM2: which can form hydrogen bonds with water\n© Cambridge University Press & Assessment 2025 Page 17 of 17", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_44.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_44-p16.png", "../answer-assets/9701_s25_ms_44-p17.png" ] }, { "id": "9701-2025-mj-51-q01", "question_id": "9701-2025-mj-51-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a) M1 bromobenzene is a skin irritant 2\nM2 toxic gas(es) (are produced)\n\n1(b)(i) warm in an oven 1\n\n1(b)(ii) fractional distillation 1\n\n1(c)(i) arrow pointing inwards at lower entry to condenser with label ‘water in’ 1\n\n1(c)(ii) to prevent water getting into the apparatus 1\n\n1(c)(iii) to avoid pressure build up in the apparatus 1\n\n1(d) M1 order of weighing 1\n1 weighing boat + Mg is weighed\n2 weighing boat is weighed after Mg is transferred\nM2 table + units 1\n/ g\n1 Mass of boat + Mg (before transfer)\n2 Mass of boat (+ residue of Mg) (after transfer)\nMass of Mg powder (transferred)\n\n1(e) 5.23 cm3 1\n\n1(f) to prevent the reaction becoming too violent 1\n\n1(g) hydrochloric acid is in excess 1\n\n1(h) beaker B 1\nAND\nethoxyethane is less dense than water\n© Cambridge University Press & Assessment 2025 Page 7 of 10\n\n1(i)(i) M1 (n(bromobenzene) = 0.0500) 2\nn(ethanal) = 3.00 cm3 × 0.788 g cm–3 / 44.0 = 0.0537 mol\n(Ratio is 1:1)\n(bromobenzene is the limiting reagent)\nM2 0.05 mol 1-phenylethanol has a mass of 0.05 mol × 122.0 g mol–1 = 6.10 g\nPercentage yield = 2.17 g / 6.10 g × 100% = 35.6 %\n\n1(i)(ii) the product contains unreacted ethanal 1\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_51-p07.png", "../answer-assets/9701_s25_ms_51-p08.png" ] }, { "id": "9701-2025-mj-51-q02", "question_id": "9701-2025-mj-51-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) M1 the solution becomes more dilute 2\nM2 the temperature is reduced\n\n2(b) appearance of a (permanent) pink colour 1\n© Cambridge University Press & Assessment 2025 Page 8 of 10\n\n2(c)(i) 2\nTime / mins Titre, V / cm3 (V – V) / cm3 log[(V – V) / cm3]\nt f t f t\n0 12.00 33.70 1.528\n5 16.40 29.30 1.467\n10 22.15 23.55 1.372\n15 23.45 22.25 1.347\n20 26.30 19.40 1.288\n25 28.70 17.00 1.230\n180 45.70\nM1 column 3 correct\nM2 column 4 correct to four significant figures as log(column 3)\n\n2(c)(ii) M1 six points correctly plotted 2\nM2 straight line of best fit drawn\n\n2(c)(iii) anomalous point circled 1\nAND\nthe actual time of the sample removed in step 4 was higher than the recorded time (10 mins)\n\n2(c)(iv) M1 co-ordinates read and recorded correctly 2\nM2 gradient to three significant figures\n\n2(d) M1 gradient (in s–1) = –k / 2.303 OR k = –gradient × 2.303 2\nM2 k = –(c)(iv) × 2.303 / 60\n© Cambridge University Press & Assessment 2025 Page 9 of 10\n\n2(e) not reliable 1\nAND\nthere is an anomalous point\n\n2(f) straight line sketched with more negative gradient with the same intercept 1\n© Cambridge University Press & Assessment 2025 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_51-p08.png", "../answer-assets/9701_s25_ms_51-p09.png", "../answer-assets/9701_s25_ms_51-p10.png" ] }, { "id": "9701-2025-mj-52-q01", "question_id": "9701-2025-mj-52-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 20, "status": "available", "reason": null, "text": "1(a)(i) 8.76 (g) 1\n\n1(a)(ii) M1 dissolve the solid / ethanedioic acid (in the beaker) using (a small volume of) (distilled) water 3\nM2 transfer / add to a (500 cm3 volumetric) flask\nAND\nrinse (with distilled water)\nM3 top the 500 cm3 volumetric flask up to mark (with distilled water)\nAND\n(stopper and) invert the flask\n\n1(b) wear chemically resistant gloves 1\n\n1(c) to remove water from (the surface of) the activated carbon 1\n\n1(d) 4.0 (cm3) 1\n\n1(e) burette 1\n\n1(f) to ensure a uniformly mixed sample 1\n© Cambridge University Press & Assessment 2025 Page 7 of 10\n\n1(g)(i) 1\ntime / mass of non-adsorbed mass of (COOH)\n2\nmins (COOH) in 5.0 cm3 adsorbed by activated\n2\nsample / mg carbon in 5.0 cm3\nsample / mg\n0 2.50 0.00\n5 1.90 0.60\n15 1.52 0.98\n30 1.43 1.07\n45 0.99 1.51\n60 0.88 1.62\n90 0.81 1.69\n120 0.80 1.70\n150 0.80 1.70\n\n1(g)(ii) M1 all points plotted correctly 2\nM2 smooth curved line of best fit passing close to all points except marked anomaly\n\n1(g)(iii) time (of sample removal) 1\n\n1(g)(iv) temperature 1\n\n1(g)(v) anomalous point circled (at 30 mins) 1\nAND\nthe actual time (of removing sample) was less than recorded time (30 minutes)\n© Cambridge University Press & Assessment 2025 Page 8 of 10\n\n1(h)(i) M1 co-ordinates correctly recorded from line given in Fig 1.2 3\nM2 gradient calculated to at least two significant figures\nM3 units = g mg–1\n(h)(ii) q = 1 / gradient in (h)(i) 1\n0\n\n1(h)(i) percentage error of experimental result compared to data book value 1\n(86.0−78.1)\n=  100% = 9.2%\n86.0\nAND\n9.2% > 6.5%\nAND\nother factors contributed towards error\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_52-p07.png", "../answer-assets/9701_s25_ms_52-p08.png", "../answer-assets/9701_s25_ms_52-p09.png" ] }, { "id": "9701-2025-mj-52-q02", "question_id": "9701-2025-mj-52-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 10, "status": "available", "reason": null, "text": "2(a) M1 change in temperature of water / C = 57.3 2\nM2 mass of butane burned / g = 2.10\n\n2(b) 320 g  4.18 J g–1 K–1  57.3 K = 76 600 (J) 1\n\n2(c) M1 number of moles of butane burned = 0.0362 mol 2\nM2 ∆H = –2120 (kJ mol–1)\nc\n\n2(d) stir the water before taking reading 1\n\n2(e) 2.5 / 320  100% = 0.78% 1\n\n2(f) use a lid 1\n\n2(g)(i) percentage error (of temperature change or mass change of butane) increases 1\n© Cambridge University Press & Assessment 2025 Page 9 of 10\n\n2(g)(ii) less heat loss 1\n© Cambridge University Press & Assessment 2025 Page 10 of 10", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_52-p09.png", "../answer-assets/9701_s25_ms_52-p10.png" ] }, { "id": "9701-2025-mj-53-q01", "question_id": "9701-2025-mj-53-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a) M1 bromobenzene is a skin irritant 2\nM2 toxic gas(es) (are produced)\n\n1(b)(i) warm in an oven 1\n\n1(b)(ii) fractional distillation 1\n\n1(c)(i) arrow pointing inwards at lower entry to condenser with label ‘water in’ 1\n\n1(c)(ii) to prevent water getting into the apparatus 1\n\n1(c)(iii) to avoid pressure build up in the apparatus 1\n\n1(d) M1 order of weighing 1\n1 weighing boat + Mg is weighed\n2 weighing boat is weighed after Mg is transferred\nM2 table + units 1\n/ g\n1 Mass of boat + Mg (before transfer)\n2 Mass of boat (+ residue of Mg) (after transfer)\nMass of Mg powder (transferred)\n\n1(e) 5.23 cm3 1\n\n1(f) to prevent the reaction becoming too violent 1\n\n1(g) hydrochloric acid is in excess 1\n\n1(h) beaker B 1\nAND\nethoxyethane is less dense than water\n© Cambridge University Press & Assessment 2025 Page 7 of 10\n\n1(i)(i) M1 (n(bromobenzene) = 0.0500) 2\nn(ethanal) = 3.00 cm3 × 0.788 g cm–3 / 44.0 = 0.0537 mol\n(Ratio is 1:1)\n(bromobenzene is the limiting reagent)\nM2 0.05 mol 1-phenylethanol has a mass of 0.05 mol × 122.0 g mol–1 = 6.10 g\nPercentage yield = 2.17 g / 6.10 g × 100% = 35.6 %\n\n1(i)(ii) the product contains unreacted ethanal 1\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_53-p07.png", "../answer-assets/9701_s25_ms_53-p08.png" ] }, { "id": "9701-2025-mj-53-q02", "question_id": "9701-2025-mj-53-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) M1 the solution becomes more dilute 2\nM2 the temperature is reduced\n\n2(b) appearance of a (permanent) pink colour 1\n© Cambridge University Press & Assessment 2025 Page 8 of 10\n\n2(c)(i) 2\nTime / mins Titre, V / cm3 (V – V) / cm3 log[(V – V) / cm3]\nt f t f t\n0 12.00 33.70 1.528\n5 16.40 29.30 1.467\n10 22.15 23.55 1.372\n15 23.45 22.25 1.347\n20 26.30 19.40 1.288\n25 28.70 17.00 1.230\n180 45.70\nM1 column 3 correct\nM2 column 4 correct to four significant figures as log(column 3)\n\n2(c)(ii) M1 six points correctly plotted 2\nM2 straight line of best fit drawn\n\n2(c)(iii) anomalous point circled 1\nAND\nthe actual time of the sample removed in step 4 was higher than the recorded time (10 mins)\n\n2(c)(iv) M1 co-ordinates read and recorded correctly 2\nM2 gradient to three significant figures\n\n2(d) M1 gradient (in s–1) = –k / 2.303 OR k = –gradient × 2.303 2\nM2 k = –(c)(iv) × 2.303 / 60\n© Cambridge University Press & Assessment 2025 Page 9 of 10\n\n2(e) not reliable 1\nAND\nthere is an anomalous point\n\n2(f) straight line sketched with more negative gradient with the same intercept 1\n© Cambridge University Press & Assessment 2025 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_53-p08.png", "../answer-assets/9701_s25_ms_53-p09.png", "../answer-assets/9701_s25_ms_53-p10.png" ] }, { "id": "9701-2025-mj-54-q01", "question_id": "9701-2025-mj-54-q01", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "54", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 12, "status": "available", "reason": null, "text": "1(a)(i) 0.9976 (g cm–3) 1\n\n1(a)(ii) M1 mass of water = 161.175 – 31.022 g = 130.153 g 2\nM2 volume = mass of water / density in (a)(i) = 130.153 / 0.9976 = 130.466 (cm3)\n\n1(b)(i) to ensure the vapour (Y) (in the conical flask) has reached the temperature (of the boiling water / 100 °C) 1\n\n1(b)(ii) mass of vapour (occupying flask)(at step 7) is used in calculation 1\n\n1(b)(iii) (perform the experiment in a) fume hood 1\n\n1(c)(i) (0.00052) 100 1\n= 3.21  10–3 (%)\n31.123\n\n1(c)(ii) M1 correct value and units used for each quantity 2\n• 99000 Pa\n• 1.2956  10–4 m3 (129.56  10–6)\n• 373 K (273 +100)\nM2 rearranges equation to give n=pV / RT AND correct evaluation\n99000 129.56 10−6\nn=\n8.31 373\n= 4.138055  10–3 (mol)\n\n1(c)(iii) M = 0.306 / (c)(ii) = 73.94778 1\nr\n\n1(d) (calculated Mr would be) greater (than true value) 1\nAND\nmass of vapour measured is greater (all other measurements are the same)\n\n1(e) temperature (of water bath) not high enough (to boil methylbenzene) 1\n© Cambridge University Press & Assessment 2025 Page 7 of 9", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_54.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_54.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_54-p07.png" ] }, { "id": "9701-2025-mj-54-q02", "question_id": "9701-2025-mj-54-q02", "subject": "9701", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "54", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 18, "status": "available", "reason": null, "text": "2(a) 24.15 (g) 1\n\n2(b) M1 add a (small) volume of (distilled) water (to the small beaker) 3\nAND\ndissolve the solid / Cu(NO ) ⦁3H O(s)\n3 2 2\nM2 transfer the solution to a 100 cm3 volumetric flask with washings\nM3 make up to the (calibration) mark / line (with distilled water)\nAND\ninvert the (stoppered) flask (to mix)\n\n2(c) to allow ion(s) to move (between half-cells / two beakers) 1\n\n2(d) 1\nlog [Mn+] electrode potential\nof Mn+(aq) / M(s)\n/ V\n–2.00 –2.958\n–2.30 –2.967\n–3.00 –2.988\n–3.30 –3.006\n–4.00 –3.017\n–5.00 –3.047\n\n2(e) concentration of Mn+ ((aq)) 1\n\n2(f)(i) M1 points plotted correctly 2\nM2 straight line of best fit drawn\n© Cambridge University Press & Assessment 2025 Page 8 of 9\n\n2(f)(ii) point most anomalous to the plotted line of best fit circled 1\nAND\nthe actual concentration of the solution was lower than recorded value (5.00  10–4)\n\n2(f)(iii) repeat the procedure for the solution giving an anomalous result 1\n\n2(g)(i) M1 two acceptable coordinates from line of best fit expressed in the form (x,y) 2\nM2 gradient correctly calculated from points listed for M1 to three significant figures\n\n2(g)(ii) y = E 3\nm = 2.303RT / nF\nc = E⦵\n\n2(g)(iii) M1 correct rearrangement AND values 2\n2.303 8.31  298\nn =\ngradient 9.65 1 04\nM2 correctly calculated value rounded to whole number\n© Cambridge University Press & Assessment 2025 Page 9 of 9", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9701_s25_ms_54.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-may-june/9701_s25_ms_54.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s25_ms_54-p08.png", "../answer-assets/9701_s25_ms_54-p09.png" ] }, { "id": "9701-2025-on-41-q01", "question_id": "9701-2025-on-41-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 15, "status": "available", "reason": null, "text": "1(a)(i) Sr(HCO ) → SrCO + CO + H O 1\n3 2 3 2 2\n\n1(a)(ii) M1: radius of cation / M2+ increases OR charge density of cation / M2+ decreases 2\nM2: less polarisation / less distortion of the anion / carbonate ion\n\n1(b) M1: least soluble CaF < SrF < BaF most soluble 4\n2 2 2\nM2: H and H both become less exothermic / less negative\nlatt hyd\nM3: H changes less / becomes less exothermic by a smaller extent\nhyd\nOR H changes more / becomes less exothermic by a larger extent\nlatt\nM4: H becomes more exothermic / more negative\nsol\n\n1(c)(i) enthalpy change when one mole of gaseous ions dissolve in water to form a solution 1\n\n1(c)(ii) M1: ionic radii AND ionic charge 2\nM2: (ionic) radii increase / charge density decreases AND ∆H decreases/less exothermic AND less attraction between\nhyd\nwater molecules and (gaseous) ions\nOR\nas ionic charge increases / charge density increases AND ∆H increases/more exothermic AND more attraction between\nhyd\nwater molecules and (gaseous) ions\n\n1(d) M1: use of –2957 / –1926 / –505 AND 2  (–505) 2\nM2: correct signs and evaluation\n∆H of MgF (s) = –1926 + (2  –505) – (–2957) = (+)21 kJ mol–1\nsol 2\n\n1(e)(i) M1: K = [Hg 2+] [F–]2 2\nsp 2\nM2: units = mol3 dm–9\n\n1(e)(ii) K = 22  (9.20  10–3)3 = 3.11  10–6 min 2sf 1\nsp\n© Cambridge University Press & Assessment 2025 Page 7 of 19", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p07.png" ] }, { "id": "9701-2025-on-41-q02", "question_id": "9701-2025-on-41-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "2(a) the (3)d and (4)s sub-shells / orbitals / electrons are close / similar in energy 1\n\n2(b) species or ion formed by a central metal atom/ion AND surrounded by/bonded to (one or more) ligands 1\n\n2(c)(i) NaOH / OH–(aq) AND precipitation / ligand exchange / deprotonation / acid–base 1\n\n2(c)(ii) M1: Fe(H O) (OH) is oxidised (from + 2 to + 3 on standing) 3\n2 4 2\nM2: Eo of Fe(H O) (OH) is more negative than Eo of O\n2 3 3 2\nOR Eo = 0.40 – (–0.56) = (+)0.96 V\ncell\nM3: 4Fe(H O) (OH) + O → 4Fe(H O) (OH) + 2H O\n2 4 2 2 2 3 3 2\n\n2(d) ∆Go = –2  1.67  96 500 = –322 310 J mol–1 2\nM1: ∆Go = –nEo F AND n = 2\ncell\nOR –2  1.67  96 500\nM2: ∆Go = –322.3 kJ mol–1\n© Cambridge University Press & Assessment 2025 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p08.png" ] }, { "id": "9701-2025-on-41-q03", "question_id": "9701-2025-on-41-q03", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 11, "status": "available", "reason": null, "text": "3(a) heterogeneous AND MnO is in a different state / phase to the reactants / H O 1\n2 2 2\n\n3(b)(i) M1: (expt 2 and 3) [I–]  2 initial rate  2 so 1st order wrt [I–] 4\nM2: (expt 1 and 2) [I–]  2, [H O ]  0.5 initial rate  1 so 1st order wrt [H O ]\n2 2 2 2\nOR (expt 1 and 4) [I–]  4 initial rate  4 AND [H+]  4 so zero order wrt [H+]\nOR (expt 1 and 3) [I–]  4, [H O ]  0.5 initial rate  2 so 1st order wrt [H O ]\n2 2 2 2\nM3: (expt 1 and 4) [I–] x 4 , [H+]  4 initial rate  4 so zero order wrt [H+]\nOR (expt 3 and 4) [H O ]  2 initial rate  2AND [H+] x 4 so zero order wrt [H+]\n2 2\nOR (expt 3 and 4) [H O ]  2, [H+]  4 initial rate  2 so 1st order wrt [H O ]\n2 2 2 2\nM4: rate = k [H O ] [I–] OR rate = k [H O ] [I–] [H+]0\n2 2 2 2\n\n3(b)(ii) M1: k = 2.42  10–3 ÷ (0.045  0.030) = 1.79 min 2sf 2\nM2: units = dm3 mol-1 s-1\n\n3(c)(i) M1: calculation of one t = 60 s ± 5 s 2\n1/2\nM2: two t calculated that are constant\n1/2\n\n3(c)(ii) k = In2 ÷ 60 = 0.0116 OR k = 0.693 ÷ 60 = 0.0116 min 2sf 1\n\n3(d) increases k and increases the rate of reaction 1", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p09.png" ] }, { "id": "9701-2025-on-41-q04", "question_id": "9701-2025-on-41-q04", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 11, "status": "available", "reason": null, "text": "4(a) M1: potential difference / voltage / EMF AND between two half-cells / two electrodes (in a cell) 2\nM2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa AND (temperature of) 298 K / 25 °C\n© Cambridge University Press & Assessment 2025 Page 9 of 19\n\n4(b)(i) 3\nM1: salt bridge labelled AND voltmeter / V AND wire to electrode and liquid level shown [complete circuit]\nM2: Cu(s) AND Cu2+(aq)\nM3: Zn(s) AND Zn2+(aq)\n\n4(b)(ii) ions AND electrons 1\n\n4(b)(iii) M1: Nernst equation E = Eo + (0.059/z) log(Zn2+/Zn) 2\nM2: E = –0.76 + (0.059 / 2) log(0.25) = –0.778 V\nOR E = –0.76 + (8.31  298) / (96 500  2) ln(0.25) = –0.778 V min 2sf\n\n4(c)(i) (from) +4 (to) +3 1\n\n4(c)(ii) Zn + 2MnO\n2\n→ ZnO + Mn\n2\nO\n3\n1\n\n4(c)(iii) Eo (MnO\n2\n/ Mn\n2\nO\n3\n) = 1.47 – 1.28 = (+)0.19 V 1\n© Cambridge University Press & Assessment 2025 Page 10 of 19", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p09.png", "../answer-assets/9701_w25_ms_41-p10.png" ] }, { "id": "9701-2025-on-41-q05", "question_id": "9701-2025-on-41-q05", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "5(a) 1\n\n5(b) • five 3d orbitals (lines, boxes) in the isolated Cu2+ ion of the same energy 2\n• splitting: three higher and two lower d orbitals\n• energy of all five d orbitals in complex higher than all d orbitals in isolated ion\nany two [1], all three [2]\n\n5(c) M1: more than one (stable) oxidation state / exist in variable oxidation states 2\nM2: vacant / empty (d) orbitals are energetically accessible\nOR vacant / empty (d) orbitals can form dative bonds with ligands\n\n5(d) 2\neach row [1]\n© Cambridge University Press & Assessment 2025 Page 11 of 19\n\n5(e) M1: 2\nM2: (shape) square planar AND (bond angle) 180o\n\n5(f)(i) M1: moles MnO – = 0.075  0.0225 = 1.6875  10–3 3\n4\nM2: moles V2+ = 5 / 3  1.6875  10–3 = 2.8125  10–3\nM3: mass V = 50.9  2.8125  10–3 = 0.143 g\n% of V = 0.143 / 0.250  100 = 57.3 min 2sf\n\n5(f)(ii) 2VO – + 3Zn + 12H+ → 2V2+ + 3Zn2+ + 6H O 2\n3 2\nM1: 2 : 3 ratio on both sides\nM2: rest of the equation correct\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p11.png", "../answer-assets/9701_w25_ms_41-p12.png" ] }, { "id": "9701-2025-on-41-q06", "question_id": "9701-2025-on-41-q06", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 7, "status": "available", "reason": null, "text": "6(a)(i) M1: (R\nf\nvalue) is distance moved by a component / spot / solute AND divided by distance moved by solvent 2\nOR (R value) is the ratio of distance moved by a component / spot / solute AND by distance moved by solvent\nf\nM2: (retention time) is (the time) between injection and detection (of a component)\n© Cambridge University Press & Assessment 2025 Page 12 of 19\n\n6(a)(ii) thin-layer chromatography: polar solvent OR non-polar solvent OR named solvent 1\ngas-liquid chromatography: non-volatile liquid OR high boiling point liquid\n\n6(b) A is more soluble in the mobile phase OR A has less adsorption to the stationary phase 1\n\n6(c)(i) 1\n\n6(c)(ii) M1: Y (two) singlet(s) 2\nM2: Z triplet AND quartet\n© Cambridge University Press & Assessment 2025 Page 13 of 19", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p12.png", "../answer-assets/9701_w25_ms_41-p13.png" ] }, { "id": "9701-2025-on-41-q07", "question_id": "9701-2025-on-41-q07", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 12, "status": "available", "reason": null, "text": "7(a) M1: ClCH\n2\nCOOH BrCH\n2\nCOOH CH\n3\nCOOH CH\n3\nCH\n2\nOH 4\nmost acidic least acidic\nM2: electron withdrawing groups / electronegative groups / negative inductive effect\nAND weakens O–H bond / more stable anion\nOR\nelectron donating groups / positive inductive effect AND strengthens O-H bond / less stable anion\nM3 M4:\n• chlorine is more electron-withdrawing / electronegative than bromine (related to XCH COOH)\n2\n• electron withdrawing of C=O / negative inductive effect of C=O (related to COOH)\n• positive inductive effect / electron donating of alkyl / ethyl / R group (related to ROH)\nany two [1], all three [2]\n\n7(b) M1: 2\nM2: CO\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 19\n\n7(c)(i) 2\nM1: correct displayed amide bond with C=O, (C H )−N and (C–)C=O\n6 5\nM2: rest of the structure correct with continuation bonds\n\n7(c)(ii) 1\n\n7(d)(i) 2\nM1:\nM2:\n© Cambridge University Press & Assessment 2025 Page 15 of 19\n\n7(d)(ii) HNO (and HCl) AND ⩽10 °C 1\n2\nOR\nNaNO AND HCl AND ⩽10 °C\n2\nQuestion Answer Marks", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p14.png", "../answer-assets/9701_w25_ms_41-p15.png", "../answer-assets/9701_w25_ms_41-p16.png" ] }, { "id": "9701-2025-on-41-q08", "question_id": "9701-2025-on-41-q08", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 13, "status": "available", "reason": null, "text": "8(a)(i) NO group directs to 3 (and 5) / meta position AND due to being an electron-withdrawing / electronegative group 1\n2\n\n8(a)(ii) HNO + H SO → NO + + HSO – + H O 1\n3 2 4 2 4 2\nOR HNO + 2H SO → NO + + 2HSO – + H O+\n3 2 4 2 4 3\n\n8(b)(i) 1\n\n8(b)(ii) M1: (reaction 1) ethanoyl chloride / CH\n3\nCOCl AND AlCl3 2\nM2: (reaction 2) 2-bromopropane / (CH ) CHBr AND FeBr\n3 2 3\n\n8(c) M1: 3\nwith C H Br no change / no precipitate AND\n6 5\nwith C H CH Br cream precipitate\n6 5 2\nin C H Br\n6 5\nM2: lone pair / p-orbital on Br delocalised / overlaps with ring /  system\nM3: C–Br bond stronger / has partially double bond character\n© Cambridge University Press & Assessment 2025 Page 16 of 19\n\n8(d)(i) (nucleophilic) addition–elimination 1\n\n8(d)(ii) 4\nM1 M2:\n• lone pair on O\n• curly arrow from (lone pair) O (in H O) to C (of C=O)\n2\n• correct dipole on C=O\n• curly arrow from the C=O bond to O atom\nany two [1], all four [2]\nM3: correct intermediate\nM4: curly arrow from (lone pair on) O(–) to C–O bond AND curly arrow from C–Br to Br\nQuestion Answer Marks", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p16.png", "../answer-assets/9701_w25_ms_41-p17.png" ] }, { "id": "9701-2025-on-41-q09", "question_id": "9701-2025-on-41-q09", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 9, "status": "available", "reason": null, "text": "9(a) M1: (basicity linked to) ability of lone pair / p-orbital to 2\nAND accept / coordinate with a proton / H+\nM2: (lone pair of) electrons on N is delocalised into C=O group (and make amides neutral)\n© Cambridge University Press & Assessment 2025 Page 17 of 19\n\n9(b)(i) M1: M 2\nM2: N\n\n9(b)(ii) LiAlH\n4\n1\n\n9(c) ketone / carbonyl AND secondary / 2o 1\n\n9(d)(i) 2\ncurly arrow 1: from O lone pair to C atom (in C=C)\ncurly arrow 2: from C=C bond to the right-hand O atom in O=O\ncurly arrow 3: from O=O bond to O+\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n9(d)(ii) 1\n© Cambridge University Press & Assessment 2025 Page 19 of 19", "source_pages": [ 17, 18, 19 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_41-p17.png", "../answer-assets/9701_w25_ms_41-p18.png", "../answer-assets/9701_w25_ms_41-p19.png" ] }, { "id": "9701-2025-on-42-q01", "question_id": "9701-2025-on-42-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "1(a) Mg(NO ) → MgO + 2NO + ½O OR 2Mg(NO ) → 2MgO + 4NO + O 1\n3 2 2 2 3 2 2 2\n\n1(b) M1: Mg(NO ) / magnesium nitrate 2\n3 2\nAND magnesium ion / Mg2+\nAND is smaller / has higher charge density\nM2: nitrate ion / anion is more distorted / polarised\n\n1(c)(i) magnesium sulfate and water / MgSO and H O 1\n4 2\n\n1(c)(ii) M1: magnesium sulfate / A has a more exothermic H and H 3\nlatt hyd\nM2: difference in H is greater\nhyd\nM3: magnesium sulfate / A has a more exothermic H\nsol\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p07.png" ] }, { "id": "9701-2025-on-42-q02", "question_id": "9701-2025-on-42-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "2(a)(i) homolytic fission 1\n\n2(a)(ii) 1\n\n2(a)(iii) 115–120o 1\n\n2(b) M1: 3.825  10–3 / 0.003825 / 0.00383 / 0.0038 / 3.8  10–3 2\nM2: units = mol–1 dm3 s–1\n\n2(c) not constant half-life because overall second order 1\n© Cambridge University Press & Assessment 2025 Page 7 of 15\n\n2(d) 8.17 / 8.2 1\n\n2(e) • reactants are adsorbed onto the catalyst surface 2\n• bonds within reactant molecules are weakened\n• products are desorbed from catalyst surface\nAny two [1], all three [2]\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p07.png", "../answer-assets/9701_w25_ms_42-p08.png" ] }, { "id": "9701-2025-on-42-q03", "question_id": "9701-2025-on-42-q03", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 15, "status": "available", "reason": null, "text": "3(a)(i) two species that differ by one proton / H+ 1\n\n3(a)(ii) H PO – AND PO 3– 1\n2 4 4\n\n3(b)(i) M1: [H+] = 2.51  10–4 2\nM2: [CH CH COOH] = 4.67  10–3\n3 2\n\n3(b)(ii) 3.98  10–11 1\n\n3(b)(iii) [HCl] = 2.51  10–4 1\n\n3(b)(iv) M1: top box ticked AND weaker acid 3\nM2: alkyl group is electron donating\nM3: O–H bond strengthened / anion is destabilised\n\n3(c)(i) NaOH AND CH CH COOH + OH– → CH CH COO– + H O 1\n3 2 3 2 2\n\n3(c)(ii) H SO AND conjugate base of CH CH COOH is not present 1\n2 4 3 2\n\n3(d)(i) 3.69  10–5 1\n© Cambridge University Press & Assessment 2025 Page 8 of 15\n\n3(d)(ii) M1: K = [Mn2+][OH–]2 2\nsp\nM2: units = mol3 dm–9\n\n3(d)(iii) 2.0  10–13 1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p08.png", "../answer-assets/9701_w25_ms_42-p09.png" ] }, { "id": "9701-2025-on-42-q04", "question_id": "9701-2025-on-42-q04", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 6, "status": "available", "reason": null, "text": "4(a) energy required when one mole of gaseous atoms is formed from the element 1\n\n4(b) energy released when one mole of gaseous atoms gains one mole of electrons and becomes one mole of gaseous 1– ions 1\n\n4(c) M1: iodine atom has greater radius 2\nM2: less attraction between nucleus and incoming electron in iodine\n\n4(d) M1: ½ (–486 + (2  364) ) 2\nM2: +121\nQuestion Answer Marks", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p09.png" ] }, { "id": "9701-2025-on-42-q05", "question_id": "9701-2025-on-42-q05", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 15, "status": "available", "reason": null, "text": "5(a)(i) Co2+ = [Ar] 3d7 AND Co3+ = [Ar] 3d6 1\n\n5(a)(ii) has vacant d orbitals that are energetically accessible 1\n\n5(a)(iii) • pink 2\n• blue\n• aqueous\nAny two [1], all three [2]\n\n5(a)(iv) [Co(H\n2\nO)\n6\n]2+ + 4HCl → [CoCl4 ]2– + 6H\n2\nO + 4H+ OR [Co(H\n2\nO)\n6\n]2+ + 4Cl– → [CoCl4 ]2– + 6H\n2\nO 1\n© Cambridge University Press & Assessment 2025 Page 9 of 15\n\n5(a)(v) ligand exchange 1\n\n5(a)(vi) [Co(H O) ]2+ + 2OH– → Co(OH) (H O) + 2H O OR 1\n2 6 2 2 4 2\n[Co(H O) ]2+ + 2OH– → Co(OH) + 6H O\n2 6 2 2\n\n5(b)(i) [oxidised] 2\nM1: Nernst equation E = Eo + (0.059 / z) × log( )\n[reduced]\nM2: E = –0.28 + (0.059 / 2) × log(0.02)\n\n5(b)(ii) 1.66 V 1\n\n5(b)(iii) 3Co + Cr O 2– + 14H+ → 3Co2+ + 2Cr3+ + 7H O 1\n2 7 2\n\n5(b)(iv) Co2+ / Co, Co2+ / Co, Cr O 2– / Cr3+ 1\n2 7\n\n5(c) M1: moles of cobalt = 0.547 / 58.9 = 0.00929 3\nM2: coulombs required = 0.00929  96500  2 = 1792\nM3: time required = 1792 / 0.50 = 3584.8 s = 59.7 minutes 3sf\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p09.png", "../answer-assets/9701_w25_ms_42-p10.png" ] }, { "id": "9701-2025-on-42-q06", "question_id": "9701-2025-on-42-q06", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 7, "status": "available", "reason": null, "text": "6(a)(i) Ni(CO) 1\n4\n\n6(a)(ii) • ethanedioate given as C O 2\n2 4\n• formula shows one Ni and three ligands\n• charge = 4–\nany two [1], all three [2]\nNi(C O ) 4– scores [2]\n2 4 3\n© Cambridge University Press & Assessment 2025 Page 10 of 15\n\n6(a)(iii) F AND optical 1\n\n6(b)(i) N atom can donate one lone pair of electrons 1\n\n6(b)(ii) [Cd(en) ]2+ AND has a larger K / is more stable 1\n2 stab\n\n6(b)(iii) [[Cd(CH3NH2)4]2+] 1\n[Cd2+][CH3NH2]4\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p10.png", "../answer-assets/9701_w25_ms_42-p11.png" ] }, { "id": "9701-2025-on-42-q07", "question_id": "9701-2025-on-42-q07", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 11, "status": "available", "reason": null, "text": "7(a) M1: Q CH CH CH(CH )CHO 2\n3 2 3\nM2:\nAND\n\n7(b)(i) T CH CH CH COCH AND V (CH ) CHCOCH 1\n3 2 2 3 3 2 3\n\n7(b)(ii) • yellow solid 2\n• CH CH CH COOH OR (CH ) CHCOOH\n3 2 2 3 2\n• CHI\n3\nAny two [1], all three [2]\n\n7(c)(i) S (CH ) CCHO 1\n3 3\n© Cambridge University Press & Assessment 2025 Page 11 of 15\n\n7(c)(ii) U CH CH COCH CH 1\n3 2 2 3\n\n7(c)(iii) triplet AND two hydrogen atoms on neighbouring carbon atoms 1\n\n7(c)(iv) TMS / tetramethylsilane / Si(CH ) 1\n3 4\n\n7(d) 4 2\n3\n5\n3\nAny two [1], all four [2]\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p11.png", "../answer-assets/9701_w25_ms_42-p12.png" ] }, { "id": "9701-2025-on-42-q08", "question_id": "9701-2025-on-42-q08", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 16, "status": "available", "reason": null, "text": "8(a) the pH at which an amino acid exists as a zwitterion 1\n\n8(b) M1: [HOOCCH(NH )CH CONH ]+ 2\n3 2 2\nM2: [HOOCCH(NH )CH COOH]+\n3 2\n© Cambridge University Press & Assessment 2025 Page 12 of 15\n\n8(c) M1: HOCH CH(NH )CH CH NH 2\n2 2 2 2 2\nM2: HOCH CH(NH )CH CH OH\n2 2 2 2\n\n8(d)(i) HOOCCH\n2\nCOOH + 2SOCl2 → ClOCCH\n2\nCOCl + 2SO\n2\n+ 2HCl [1] 1\n\n8(d)(ii) HOOCCH(CH CONH )NHCOCH CONHCH(CH CONH )COOH 2\n2 2 2 2 2\nM1: three molecules joined, with four correct amide groups only\nM2: whole structure correct as above, displayed formula accepted, skeletal formula as below accepted\n\n8(e) M1: hydrolysis of amide to carboxylate ion 2\nM2: hydrolysis of –COOH to carboxylate ion and rest of molecule\n–OOCCH(NH )CH COO–\n2 2\n© Cambridge University Press & Assessment 2025 Page 13 of 15\n\n8(f) • three monomer residues 3\n• all linkages correct\n• correct trailing bonds\n• repeat unit marked contains correct atoms\nAny two [1], any three [2], all four [3]\nCH CONH CH CONH\n2 2 2 2\n| |\n–COCHNH–[–COCHNH–]–COCHNH–\n|\nCH CONH\n2 2\n\n8(g)(i) • rotate the plane of the plane polarised light 2\n• by the same angle\n• in opposite directions\nAny two [1], all three [2]\n\n8(g)(ii) racemic mixture 1\nQuestion Answer Marks", "source_pages": [ 12, 13, 14 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p12.png", "../answer-assets/9701_w25_ms_42-p13.png", "../answer-assets/9701_w25_ms_42-p14.png" ] }, { "id": "9701-2025-on-42-q09", "question_id": "9701-2025-on-42-q09", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 14, "status": "available", "reason": null, "text": "9(a) • all C atoms are sp2 hybridised 3\n• bonds between C atoms are  and \n• C–C  bonds are formed by overlap of sp2 hybrid orbitals\n• C–C  bonds are formed by overlap of p orbitals\n• bonds between C and H atoms are \n• C–H bonds are formed by overlap of sp2 hybrid orbitals and s orbitals\nAny two [1], any four [2], all six [3]\n© Cambridge University Press & Assessment 2025 Page 14 of 15\n\n9(b) CH\n3\nCl and AlCl3 1\n\n9(c) 3\nM1: curly arrow from within hexagon towards +C\nM2: intermediate, usual rules for horseshoe and + charge\nM3: curly arrow from C–H bond into ring AND H+ product\n\n9(d) M1: HNO AND H SO 2\n3 2 4\nM2: concentrated AND 25 °C ⩽ T ⩽ 60 °C\n\n9(e) alkaline KMnO 1\n4\n\n9(f) 4-nitrobenzoic acid 1\n\n9(g) (hot) concentrated HCl and Sn 1\n\n9(h) M1: perform step 3 before step 2 OR perform step 2 before step 1 2\nM2: COOH is 3-directing OR NO is 3-directing\n2\n© Cambridge University Press & Assessment 2025 Page 15 of 15", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_42-p14.png", "../answer-assets/9701_w25_ms_42-p15.png" ] }, { "id": "9701-2025-on-43-q01", "question_id": "9701-2025-on-43-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 15, "status": "available", "reason": null, "text": "1(a)(i) Sr(HCO ) → SrCO + CO + H O 1\n3 2 3 2 2\n\n1(a)(ii) M1: radius of cation / M2+ increases OR charge density of cation / M2+ decreases 2\nM2: less polarisation / less distortion of the anion / carbonate ion\n\n1(b) M1: least soluble CaF < SrF < BaF most soluble 4\n2 2 2\nM2: H and H both become less exothermic / less negative\nlatt hyd\nM3: H changes less / becomes less exothermic by a smaller extent\nhyd\nOR H changes more / becomes less exothermic by a larger extent\nlatt\nM4: H becomes more exothermic / more negative\nsol\n\n1(c)(i) enthalpy change when one mole of gaseous ions dissolve in water to form a solution 1\n\n1(c)(ii) M1: ionic radii AND ionic charge 2\nM2: (ionic) radii increase / charge density decreases AND ∆H decreases/less exothermic AND less attraction between\nhyd\nwater molecules and (gaseous) ions\nOR\nas ionic charge increases / charge density increases AND ∆H increases/more exothermic AND more attraction between\nhyd\nwater molecules and (gaseous) ions\n\n1(d) M1: use of –2957 / –1926 / –505 AND 2  (–505) 2\nM2: correct signs and evaluation\n∆H of MgF (s) = –1926 + (2  –505) – (–2957) = (+)21 kJ mol–1\nsol 2\n\n1(e)(i) M1: K = [Hg 2+] [F–]2 2\nsp 2\nM2: units = mol3 dm–9\n\n1(e)(ii) K = 22  (9.20  10–3)3 = 3.11  10–6 min 2sf 1\nsp\n© Cambridge University Press & Assessment 2025 Page 7 of 19", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p07.png" ] }, { "id": "9701-2025-on-43-q02", "question_id": "9701-2025-on-43-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "2(a) the (3)d and (4)s sub-shells / orbitals / electrons are close / similar in energy 1\n\n2(b) species or ion formed by a central metal atom/ion AND surrounded by/bonded to (one or more) ligands 1\n\n2(c)(i) NaOH / OH–(aq) AND precipitation / ligand exchange / deprotonation / acid–base 1\n\n2(c)(ii) M1: Fe(H O) (OH) is oxidised (from + 2 to + 3 on standing) 3\n2 4 2\nM2: Eo of Fe(H O) (OH) is more negative than Eo of O\n2 3 3 2\nOR Eo = 0.40 – (–0.56) = (+)0.96 V\ncell\nM3: 4Fe(H O) (OH) + O → 4Fe(H O) (OH) + 2H O\n2 4 2 2 2 3 3 2\n\n2(d) ∆Go = –2  1.67  96 500 = –322 310 J mol–1 2\nM1: ∆Go = –nEo F AND n = 2\ncell\nOR –2  1.67  96 500\nM2: ∆Go = –322.3 kJ mol–1\n© Cambridge University Press & Assessment 2025 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p08.png" ] }, { "id": "9701-2025-on-43-q03", "question_id": "9701-2025-on-43-q03", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 11, "status": "available", "reason": null, "text": "3(a) heterogeneous AND MnO is in a different state / phase to the reactants / H O 1\n2 2 2\n\n3(b)(i) M1: (expt 2 and 3) [I–]  2 initial rate  2 so 1st order wrt [I–] 4\nM2: (expt 1 and 2) [I–]  2, [H O ]  0.5 initial rate  1 so 1st order wrt [H O ]\n2 2 2 2\nOR (expt 1 and 4) [I–]  4 initial rate  4 AND [H+]  4 so zero order wrt [H+]\nOR (expt 1 and 3) [I–]  4, [H O ]  0.5 initial rate  2 so 1st order wrt [H O ]\n2 2 2 2\nM3: (expt 1 and 4) [I–] x 4 , [H+]  4 initial rate  4 so zero order wrt [H+]\nOR (expt 3 and 4) [H O ]  2 initial rate  2AND [H+] x 4 so zero order wrt [H+]\n2 2\nOR (expt 3 and 4) [H O ]  2, [H+]  4 initial rate  2 so 1st order wrt [H O ]\n2 2 2 2\nM4: rate = k [H O ] [I–] OR rate = k [H O ] [I–] [H+]0\n2 2 2 2\n\n3(b)(ii) M1: k = 2.42  10–3 ÷ (0.045  0.030) = 1.79 min 2sf 2\nM2: units = dm3 mol-1 s-1\n\n3(c)(i) M1: calculation of one t = 60 s ± 5 s 2\n1/2\nM2: two t calculated that are constant\n1/2\n\n3(c)(ii) k = In2 ÷ 60 = 0.0116 OR k = 0.693 ÷ 60 = 0.0116 min 2sf 1\n\n3(d) increases k and increases the rate of reaction 1", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p09.png" ] }, { "id": "9701-2025-on-43-q04", "question_id": "9701-2025-on-43-q04", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 11, "status": "available", "reason": null, "text": "4(a) M1: potential difference / voltage / EMF AND between two half-cells / two electrodes (in a cell) 2\nM2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa AND (temperature of) 298 K / 25 °C\n© Cambridge University Press & Assessment 2025 Page 9 of 19\n\n4(b)(i) 3\nM1: salt bridge labelled AND voltmeter / V AND wire to electrode and liquid level shown [complete circuit]\nM2: Cu(s) AND Cu2+(aq)\nM3: Zn(s) AND Zn2+(aq)\n\n4(b)(ii) ions AND electrons 1\n\n4(b)(iii) M1: Nernst equation E = Eo + (0.059/z) log(Zn2+/Zn) 2\nM2: E = –0.76 + (0.059 / 2) log(0.25) = –0.778 V\nOR E = –0.76 + (8.31  298) / (96 500  2) ln(0.25) = –0.778 V min 2sf\n\n4(c)(i) (from) +4 (to) +3 1\n\n4(c)(ii) Zn + 2MnO\n2\n→ ZnO + Mn\n2\nO\n3\n1\n\n4(c)(iii) Eo (MnO\n2\n/ Mn\n2\nO\n3\n) = 1.47 – 1.28 = (+)0.19 V 1\n© Cambridge University Press & Assessment 2025 Page 10 of 19", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p09.png", "../answer-assets/9701_w25_ms_43-p10.png" ] }, { "id": "9701-2025-on-43-q05", "question_id": "9701-2025-on-43-q05", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 14, "status": "available", "reason": null, "text": "5(a) 1\n\n5(b) • five 3d orbitals (lines, boxes) in the isolated Cu2+ ion of the same energy 2\n• splitting: three higher and two lower d orbitals\n• energy of all five d orbitals in complex higher than all d orbitals in isolated ion\nany two [1], all three [2]\n\n5(c) M1: more than one (stable) oxidation state / exist in variable oxidation states 2\nM2: vacant / empty (d) orbitals are energetically accessible\nOR vacant / empty (d) orbitals can form dative bonds with ligands\n\n5(d) 2\neach row [1]\n© Cambridge University Press & Assessment 2025 Page 11 of 19\n\n5(e) M1: 2\nM2: (shape) square planar AND (bond angle) 180o\n\n5(f)(i) M1: moles MnO – = 0.075  0.0225 = 1.6875  10–3 3\n4\nM2: moles V2+ = 5 / 3  1.6875  10–3 = 2.8125  10–3\nM3: mass V = 50.9  2.8125  10–3 = 0.143 g\n% of V = 0.143 / 0.250  100 = 57.3 min 2sf\n\n5(f)(ii) 2VO – + 3Zn + 12H+ → 2V2+ + 3Zn2+ + 6H O 2\n3 2\nM1: 2 : 3 ratio on both sides\nM2: rest of the equation correct\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p11.png", "../answer-assets/9701_w25_ms_43-p12.png" ] }, { "id": "9701-2025-on-43-q06", "question_id": "9701-2025-on-43-q06", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 7, "status": "available", "reason": null, "text": "6(a)(i) M1: (R\nf\nvalue) is distance moved by a component / spot / solute AND divided by distance moved by solvent 2\nOR (R value) is the ratio of distance moved by a component / spot / solute AND by distance moved by solvent\nf\nM2: (retention time) is (the time) between injection and detection (of a component)\n© Cambridge University Press & Assessment 2025 Page 12 of 19\n\n6(a)(ii) thin-layer chromatography: polar solvent OR non-polar solvent OR named solvent 1\ngas-liquid chromatography: non-volatile liquid OR high boiling point liquid\n\n6(b) A is more soluble in the mobile phase OR A has less adsorption to the stationary phase 1\n\n6(c)(i) 1\n\n6(c)(ii) M1: Y (two) singlet(s) 2\nM2: Z triplet AND quartet\n© Cambridge University Press & Assessment 2025 Page 13 of 19", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p12.png", "../answer-assets/9701_w25_ms_43-p13.png" ] }, { "id": "9701-2025-on-43-q07", "question_id": "9701-2025-on-43-q07", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 12, "status": "available", "reason": null, "text": "7(a) M1: ClCH\n2\nCOOH BrCH\n2\nCOOH CH\n3\nCOOH CH\n3\nCH\n2\nOH 4\nmost acidic least acidic\nM2: electron withdrawing groups / electronegative groups / negative inductive effect\nAND weakens O–H bond / more stable anion\nOR\nelectron donating groups / positive inductive effect AND strengthens O-H bond / less stable anion\nM3 M4:\n• chlorine is more electron-withdrawing / electronegative than bromine (related to XCH COOH)\n2\n• electron withdrawing of C=O / negative inductive effect of C=O (related to COOH)\n• positive inductive effect / electron donating of alkyl / ethyl / R group (related to ROH)\nany two [1], all three [2]\n\n7(b) M1: 2\nM2: CO\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 19\n\n7(c)(i) 2\nM1: correct displayed amide bond with C=O, (C H )−N and (C–)C=O\n6 5\nM2: rest of the structure correct with continuation bonds\n\n7(c)(ii) 1\n\n7(d)(i) 2\nM1:\nM2:\n© Cambridge University Press & Assessment 2025 Page 15 of 19\n\n7(d)(ii) HNO (and HCl) AND ⩽10 °C 1\n2\nOR\nNaNO AND HCl AND ⩽10 °C\n2\nQuestion Answer Marks", "source_pages": [ 14, 15, 16 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p14.png", "../answer-assets/9701_w25_ms_43-p15.png", "../answer-assets/9701_w25_ms_43-p16.png" ] }, { "id": "9701-2025-on-43-q08", "question_id": "9701-2025-on-43-q08", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 13, "status": "available", "reason": null, "text": "8(a)(i) NO group directs to 3 (and 5) / meta position AND due to being an electron-withdrawing / electronegative group 1\n2\n\n8(a)(ii) HNO + H SO → NO + + HSO – + H O 1\n3 2 4 2 4 2\nOR HNO + 2H SO → NO + + 2HSO – + H O+\n3 2 4 2 4 3\n\n8(b)(i) 1\n\n8(b)(ii) M1: (reaction 1) ethanoyl chloride / CH\n3\nCOCl AND AlCl3 2\nM2: (reaction 2) 2-bromopropane / (CH ) CHBr AND FeBr\n3 2 3\n\n8(c) M1: 3\nwith C H Br no change / no precipitate AND\n6 5\nwith C H CH Br cream precipitate\n6 5 2\nin C H Br\n6 5\nM2: lone pair / p-orbital on Br delocalised / overlaps with ring /  system\nM3: C–Br bond stronger / has partially double bond character\n© Cambridge University Press & Assessment 2025 Page 16 of 19\n\n8(d)(i) (nucleophilic) addition–elimination 1\n\n8(d)(ii) 4\nM1 M2:\n• lone pair on O\n• curly arrow from (lone pair) O (in H O) to C (of C=O)\n2\n• correct dipole on C=O\n• curly arrow from the C=O bond to O atom\nany two [1], all four [2]\nM3: correct intermediate\nM4: curly arrow from (lone pair on) O(–) to C–O bond AND curly arrow from C–Br to Br\nQuestion Answer Marks", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p16.png", "../answer-assets/9701_w25_ms_43-p17.png" ] }, { "id": "9701-2025-on-43-q09", "question_id": "9701-2025-on-43-q09", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 9, "status": "available", "reason": null, "text": "9(a) M1: (basicity linked to) ability of lone pair / p-orbital to 2\nAND accept / coordinate with a proton / H+\nM2: (lone pair of) electrons on N is delocalised into C=O group (and make amides neutral)\n© Cambridge University Press & Assessment 2025 Page 17 of 19\n\n9(b)(i) M1: M 2\nM2: N\n\n9(b)(ii) LiAlH\n4\n1\n\n9(c) ketone / carbonyl AND secondary / 2o 1\n\n9(d)(i) 2\ncurly arrow 1: from O lone pair to C atom (in C=C)\ncurly arrow 2: from C=C bond to the right-hand O atom in O=O\ncurly arrow 3: from O=O bond to O+\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n9(d)(ii) 1\n© Cambridge University Press & Assessment 2025 Page 19 of 19", "source_pages": [ 17, 18, 19 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_43-p17.png", "../answer-assets/9701_w25_ms_43-p18.png", "../answer-assets/9701_w25_ms_43-p19.png" ] }, { "id": "9701-2025-on-44-q01", "question_id": "9701-2025-on-44-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 19, "status": "available", "reason": null, "text": "1(a) forms (one or more) stable ions / compounds / oxidation states 1\nwith incomplete / partially filled (3)d orbital(s) / d shell / d sub-shell\n\n1(b)(i) (orbitals) are at the same energy 1\n\n1(b)(ii) (1s2) 2s2 2p6 3s2 3p6 3d9 1\n\n1(c)(i) [Cu(NH ) (H O) ]2+(aq) deep / dark / royal blue 1\n3 4 2 2\nAND\n[CuCl ]2–(aq) yellow\n4\n\n1(c)(ii) M1: d orbital(s) of different energy / d-d splitting occurs / d sub-shell splits 3\nM2: electron(s) promoted / excited\nM3: light / wavelength / frequency / photon absorbed AND complementary colour seen\n\n1(d)(i) M1: ligand exchange / replacement / substitution / displacement 2\nM2: [CuCl4 ]2– + 4NH\n3\n+ 2H\n2\nO → [Cu(NH\n3\n)\n4\n(H\n2\nO)\n2\n]2+ + 4Cl–\n\n1(d)(ii) M1: one correct 3D structure of [Cu(H O) (NH ) ]2+ 2\n2 2 3 4\nM2: correct 3D stereoisomer structure of isomer 1\n\n1(d)(iii) cis isomer AND dipoles do not cancel 1\n\n1(e)(i) (two) oxygen and (one) nitrogen (can donate) three lone pair(s) of electrons (to the metal ion) 1\n© Cambridge University Press & Assessment 2025 Page 7 of 16\n\n1(e)(ii) [Co(C H NO ) ]– 1\n4 5 4 2\n\n1(f)(i) equilibrium constant for the formation of the complex (ion) 1\nin a solvent / water / solution OR from its constituent ions or molecules\n\n1(f)(ii) [Ag(CN) ]–(aq) AND largest (value of) K 1\n2 stab\n\n1(g) M1 M2: any two [1], all four [2] 3\n• moles MnO – = 0.015 ×18.70/1000 = 2.805 × 10–4\n4\n• moles SO 2– = 2.805 × 10–4 × 5/2 = 7.0125 × 10–4 (in 10 cm3)\n3\n• moles SO 2– = 1.753 × 10–2 (in 250 cm3)\n3\n• mass Na SO = 1.753 × 10–2 × 126.1 = 2.21 g\n2 3\nM3: % purity = 100 × 2.21 / 3.75 = 58.9 – 59.0 % min 2sf\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p07.png", "../answer-assets/9701_w25_ms_44-p08.png" ] }, { "id": "9701-2025-on-44-q02", "question_id": "9701-2025-on-44-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 2, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 18, "status": "available", "reason": null, "text": "2(a) M1: (solubility) decreases down the group 4\nM2: H and H decrease / become less exothermic / less negative\nlatt hyd\nM3: H decreases / changes more / dominant factor / becomes less exothermic by a larger extent\nhyd\nOR H decreases / changes less / changes slower\nlatt\nM4: H becomes less exothermic / less negative OR H becomes (more) endothermic / (more) positive\nsol sol\n\n2(b)(i) Ag CrO (s) ⇌ 2Ag+(aq) + CrO 2–(aq) 1\n2 4 4\n\n2(b)(ii) M1: K = [Ag+]2 [CrO 2-] 3\nsp 4\nM2: 3√(1.12×10−12÷4) = 6.54  10–5\nM3: [Ag+] = 2  6.54  10–5 = 1.31  10–4 (mol dm–3) min 2sf\n© Cambridge University Press & Assessment 2025 Page 8 of 16\n\n2(c)(i) K = 10–6.49 = 3.236  10–7 2\na\n[H+]2 = 8.08984  10–9\nM1: [H+] = 8.99  10–5\nM2: pH = –log(8.99  10–5) = 4.05 min 2sf\n\n2(c)(ii) conjugate acid H CrO 1\n2 4\nAND\nconjugate base CrO 2–\n4\n\n2(d)(i) (the energy change / released when) one mole of gaseous atoms become 1– ions / gain one mole of electrons 1\n\n2(d)(ii) due to repulsion between negative ion and incoming / gained electron 1\n\n2(d)(iii) 2Ag+(g) S2–(g) → Ag S(s) 1\n2\n\n2(d)(iv) M1: selection of correct six numbers only: –33, 731, 279, –200, 532, –2677 3\nM2: use of  2 as only multiplier with Ag\nM3: correct signs and evaluation of data\n–33 = 2H + (2  731) + 279 + (–200) + 532 + (–2677)\nat\n2H = 571 (kJ mol –1)\nat\nH = +285.5 (kJ mol –1)\nat\n\n2(e) Ag S smaller / less negative / less exothermic lattice neergy AND larger ionic radius (of cation) 1\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 16", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p08.png", "../answer-assets/9701_w25_ms_44-p09.png" ] }, { "id": "9701-2025-on-44-q03", "question_id": "9701-2025-on-44-q03", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 3, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 9, "status": "available", "reason": null, "text": "3(a) number of possible arrangements of particles and energy in a system 1\n\n3(b)(i) 1\nprocess S is negative S is positive\nsteam condensing into water ✓\nsolid KCl dissolving into water ✓\n\n3(b)(ii) M1: three numbers and correct multipliers used 2\nM2: correct signs and evaluation\nSo = (3  203.0) + (223.1) – (2  281.6) = (+)268.9 (J K–1 mol–1)\n\n3(c)(i) positive 1\nAND\ngas produced / on the right OR less / no gas on left\n\n3(c)(ii) M1 and M2: calculation of gradient from the graph for S 4\ngradient = (–)140 / 900 = (–)0.1556\nSo = (–)0.1556  1000\n= (+)155.6 ± 5 (J K–1 mol–1)\nM3: T = 1120 ± 5 (K)\nM4: Ho = (+)172 ± 10 (kJ mol–1)\n© Cambridge University Press & Assessment 2025 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p10.png" ] }, { "id": "9701-2025-on-44-q04", "question_id": "9701-2025-on-44-q04", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 4, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 12, "status": "available", "reason": null, "text": "4(a)(i) 1\nthe order of reaction with respect to [H ] 1\n2\nthe order of reaction with respect to [NO] 2\noverall order of the reaction 3\n\n4(a)(ii) 1 1\nrate \n4\n\n4(a)(iii) rate  27 1\n\n4(a)(iv) a four-particle collision is unlikely 1\n\n4(a)(v) step 1 2NO → N O 2\n2 2\nstep 2 N O + H → N O + H O\n2 2 2 2 2\nstep 3 N O + H → N + H O\n2 2 2 2\n\n4(a)(vi) intermediate formed (step 2) then used up later (step 3) 1\n\n4(b) k = 0.693 / 720 = 9.625  10–4 min 2sf 1\nAND units: s–1\n\n4(c)(i) homogeneous 1\nAND Co3+ / catalyst / it is in the same phase / same state as the reactants\n\n4(c)(ii) M1: (equation 1) 2Co3+ + 2I– → 2Co2+ + I 2\n2\nM2: (equation 2) 2Co2+ + S O 2– → 2Co3+ + 2SO 2–\n2 8 4\n\n4(c)(iii) repulsion of two negative / same charge ions slows the reaction / raises E 1\nA\n© Cambridge University Press & Assessment 2025 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p11.png" ] }, { "id": "9701-2025-on-44-q05", "question_id": "9701-2025-on-44-q05", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 5, "topic": "Hydrocarbons", "topic_slug": "9701-topic-30-hydrocarbons", "marks": 13, "status": "available", "reason": null, "text": "5(a) M1 and M2: any two [1], all three [2] 4\n• (hexagonal ring) planar / (trigonal) planar\n• 120o\n• sp2 hybridised\nM3: p orbitals overlap sideways / laterally (with each other above and below the ring) forming  bonds\nM4:  bonds form when sp2 / hybridised orbitals overlap end-on-end / head on\n\n5(b)(i) M1: (reaction 5) Pt / Ni + H (+ heat) 2\n2\nM2: (reaction 6) (CH\n3\n)\n2\nCHCOCl + AlCl3 (+ heat)\n\n5(b)(ii) reduction / hydrogenation 1\n\n5(c)(i) (CH CH group) directs to 2-, 4- (and 6-) positions AND due to being an electron donating group 1\n2 3\n\n5(c)(ii) CH CH Br + FeBr → CH CH + + FeBr – 1\n3 2 3 3 2 4\n\n5(c)(iii) 3\norganic intermediate\nM1: curly arrow from inside / touching the hexagon to (CH ) group / C+ charge\n2\nM2: structure of the intermediate\nM3: curly arrow from C-H bond into / onto the ring AND formation / loss of H+\n\n5(c)(iv) H+ + FeBr – → HBr + FeBr 1\n4 3\n© Cambridge University Press & Assessment 2025 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-30-hydrocarbons/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p12.png" ] }, { "id": "9701-2025-on-44-q06", "question_id": "9701-2025-on-44-q06", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 6, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 9, "status": "available", "reason": null, "text": "6(a) sp = 1 AND sp2 = 1 AND sp3 = 3 1\n\n6(b)(i) acid-base / neutralisation AND hydrolysis 1\n\n6(b)(ii) * 4\n*\n–\n–\n*\nreaction 7 * –\n*\n–\n–\n*\n–\nreaction 8\nreaction 9 * *\n– –\nany two * [1], any three * [2], any five * [3], all six * [4]\n\n6(c)(i) five / 5 1\n\n6(c)(ii) 2\nproton environment a b c\nname of splitting pattern doublet multiplet doublet\nchemical shift range, 2.2–3.0 3.2–4.0 2.0–3.0\n / ppm\nany three [1], all six [2]\n© Cambridge University Press & Assessment 2025 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p13.png" ] }, { "id": "9701-2025-on-44-q07", "question_id": "9701-2025-on-44-q07", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 7, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 9, "status": "available", "reason": null, "text": "7(a) M1: (most acidic) benzoic acid 4\n4-methylphenol\nwater\n(least acidic) phenylmethanol\nM2: correct link of acidity to\nweakens O—H / H+ more easily lost / anion stabilised / conjugate base stabilised\nM3 and M4: any two [2]\n• for benzoic acid: negative inductive effect of C=O / electronegative C=O / two electronegative O / –ve charge delocalised\n(across two O) / resonance in –COO–\n• for 4-methylphenol: as lone pair on oxygen delocalised into the ring /  system / delocalised system\nOR p orbital on oxygen overlaps with ring /  system / delocalised system\n• for phenylmethanol: positive inductive effect of alkyl group / R / CH\n2\n\n7(b) 1\n\n7(c)(i) 1\n\n7(c)(ii) (reaction 10) addition–elimination / condensation / esterification 1\nAND\n(reaction 11) addition–elimination / condensation (polymerisation) / esterification\n© Cambridge University Press & Assessment 2025 Page 14 of 16\n\n7(c)(iii) 2\nM1: ester linkage shown\nM2: rest of the structure correct with continuation bonds\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p14.png", "../answer-assets/9701_w25_ms_44-p15.png" ] }, { "id": "9701-2025-on-44-q08", "question_id": "9701-2025-on-44-q08", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "44", "question_number": 8, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 11, "status": "available", "reason": null, "text": "8(a) • chlorobenzene is less reactive than ethanoyl chloride 2\n• p orbital / lone pair on Cl / –Cl will overlap / delocalise into the ring\nOR C of COCl has most electron deficient OR has an electronegative oxygen atom / two electronegative atoms / electron\nwithdrawing C=O group\n• due to partial double C–Cl bond\nOR C–Cl bond strengthened (more) / stronger bond between Cl and benzene / ring\nOR C–Cl weakened in ROCl\nany two [1], all three [2]\n\n8(b)(i) • nitrile 2\n• amide\n• ketone / carbonyl\n• ester\nany two [1], all four [2]\n\n8(b)(ii) 8 / eight 1\n\n8(b)(iii) no need to separate optical isomers 1\n© Cambridge University Press & Assessment 2025 Page 15 of 16\n\n8(c) 3\nM1: amide bond hydrolysed and COO– and NH formed\n2\nM2: nitrile bond hydrolysed and COO– formed\nM3: ester bond hydrolysed and COO– and OH formed\n\n8(d)(i) percentage = 100  28 / (28 + 58 + 13 + 42) = 19.9(%) min 2sf 1\n\n8(d)(ii) D is attracted more strongly to the stationary phase / forms stronger intermolecular forces with the stationary phase 1\n© Cambridge University Press & Assessment 2025 Page 16 of 16", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_44.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_44-p15.png", "../answer-assets/9701_w25_ms_44-p16.png" ] }, { "id": "9701-2025-on-51-q01", "question_id": "9701-2025-on-51-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 2.12 g 1\n\n1(a)(ii) M1 add a (small) volume of (distilled) water (to the small beaker) 3\nAND\ndissolve the solid / AgNO (s)\n3\nM2\nTransfer the solution and washings (using a funnel)\nAND\ninto a 250 cm3 volumetric flask.\nM3 Make up to the (calibration) mark / line with distilled water.\nAND\nthen invert the (stoppered) flask (several times)\n\n1(b)(i) (20(.0) cm3) pipette. 1\n\n1(b)(ii) barium (ions) could react with chromate (ions). 1\n\n1(b)(iii) Potassium chromate(VI) / K CrO (aq) is an irritant / corrosive 1\n2 4\n\n1(c)(i) They are (the only ones that are) concordant. 1\n\n1(c)(ii) 20.05 1\n100 = 0.48(54%)\n20.60\nCorrect working must be shown.\n\n1(d)(i) 20.65  0.05 / 1000 = 0.00103(25) 1\n\n1(d)(ii) = ((d)(i) / 2) *(250 / 20) 1\n= (d)(i)  6.25\n© Cambridge University Press & Assessment 2025 Page 7 of 10\n\n1(d)(iii) M1 M = 1.58 / (d)(ii) 2\nr\nM2 x= (M1 – 208.3) / 18\nAND\nx must be an integer.\nSome working must be shown.\n\n1(e) Heat to drive off water of crystallisation. 3\nM1 Apparatus\n1 Bunsen burner.\n2 Crucible (supported by pipeclay triangle)\nM2 Measurements\n1 Measure mass of crucible (+ lid) (empty).\n2 Measure mass crucible (+ lid) + solid before heating\n3 Measure mass of crucible (+ lid) + residue (after heating.)\nM3 Key step in method\n• Description of heat to constant mass.\nOR\nProduce a suitable precipitate using a viable reaction.\nM1 Apparatus and method\n• Chooses suitable reagent.\n• Filter the precipitate.\n• Wash and (allow to) dry\nM2 Measurements\n• Measure mass of hydrated barium chloride used.\n• Measure mass of precipitate.\nM3 Key step in method\n• Description of dry to constant mass.\n© Cambridge University Press & Assessment 2025 Page 8 of 10", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_51-p07.png", "../answer-assets/9701_w25_ms_51-p08.png" ] }, { "id": "9701-2025-on-51-q02", "question_id": "9701-2025-on-51-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) To ensure no air / other gas remains in the syringe 1\n\n2(b)(i) 2\n1\n0.707 0.500 0.222 0.158 0.109\nMr\nRate of effusion /cm3s-1 4.63 3.27 1.45 1.26 0.71\n\n2(b)(ii) Time (taken). 1\n\n2(b)(iii) size of hole 1\n\n2(c) M1 all 5 points plotted correctly. 2\nM2 Straight line of best fit drawn\n\n2(d) Point most anomalous to the plotted line of best fit circled. 1\nAND\nIf point above LOBF\n(Measured rate of effusion is higher than the true rate predicted by LOBF.)\nfor this point …\nmeasured / recorded (effusion) time < true / actual time\nOR\nIf point below LOBF\n(Measured rate of effusion is lower than the true rate predicted by LOBF.)\nfor this point …\nmeasured / recorded (effusion) time > true / actual time\n\n2(e) Yes AND The graph / LOBF… 1\nis a straight-line AND it goes through the origin.\n\n2(f) The results are not reliable as there is an anomalous result. 1\n© Cambridge University Press & Assessment 2025 Page 9 of 10\n\n2(g)(i) M1 Effusion rate = 50/31.6 = (1.58) 1\n1\nCorrect value for read from x-axis\nMr\nM2 Value correctly converted to M 1\n.\n\n2(g)(ii) If calculated M > 16 1\n(The average) M(of the other components) is greater than 16 / calculated value for g(i).\n\n2(h) (Rate of effusion) increases 1\nAND\n(At higher temperature) gas particles move more quickly.\n© Cambridge University Press & Assessment 2025 Page 10 of 10", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_51-p09.png", "../answer-assets/9701_w25_ms_51-p10.png" ] }, { "id": "9701-2025-on-52-q01", "question_id": "9701-2025-on-52-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 17, "status": "available", "reason": null, "text": "1(a) 0.486 g 1\n\n1(b) M1 Dissolve the solid / KSCN(s) in (the beaker) (using a small volume of) (distilled) water 3\nM2 Transfer/add to a (250 cm3 volumetric) flask\nand\nRinse (with distilled water)\nM3 Top the 250 cm3 volumetric flask up to mark (with distilled water)\nand\nInvert the (stoppered) flask\nAllow ‘250 volumetric flask’ (i.e. units need not be seen)\n\n1(c)(i) to avoid oxidation (of contents of flask) 1\n\n1(c)(ii) to ensure that equilibrium has been reached 1\n\n1(c)(iii) Silver / Ag 1\n\n1(d)(i) (yes) 1\nand\ntwo titres (2 and 3) are within 0.10 cm³ (of each other)\n\n1(d)(ii) Working must be shown 1\n2  0.05\n=  100 = 0.457%\n21.90\n\n1(d)(iii) percentage error (in larger titre) is lower 1\n\n1(e)(i) 0.0437 mol dm–3 1\n\n1(e)(ii) 0.00630 mol dm–3 1\n© Cambridge University Press & Assessment 2025 Page 8 of 11\n\n1(e)(iii) M1 K = 0.00630 / (0.04370)2 = 3.30 1\nc\nM2 units = dm3 mol–1 1\n\n1(f)(i) No 1\nand\nAs temperature increases, K\nc\ndecreases\n\n1(f)(ii) (data points that obey a line of best fit are reliable because) there are no anomalous points. 1\n\n1(f)(iii) exothermic 1\nand\nAs temperature increases:\nK decreases\nc\nor\nequilibrium shifts to left\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_52-p08.png", "../answer-assets/9701_w25_ms_52-p09.png" ] }, { "id": "9701-2025-on-52-q02", "question_id": "9701-2025-on-52-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "2(a)(i) (25.0 cm3) volumetric pipette 1\n\n2(a)(ii) (wear) chemically resistant gloves 1\n© Cambridge University Press & Assessment 2025 Page 9 of 11\n\n2(b)(i) 2\nTotal volume Corrected\nin beaker / conductivity\ncm3 / S cm–1\n25.00 37000\n30.00 27600\n35.00 16800\n40.00 3680\n45.00 9000\n50.00 24000\n55.00 39600\n60.00 50400\n65.00 63700\nM1 Total volume in beaker values\nM2 Corrected conductivity values\n\n2(b)(ii) volume of 0.100 mol dm–3 Ba(OH) (aq) added to beaker 1\n2\n\n2(c) M1: all points 1–4 and 5–9 plotted correctly 2\nM2: two straight lines of best fit line drawn going through most points.\nlines must intersect.\nPlace M2 tick in ‘V’ of graph\nPlace M1 tick or cross on left hand side of grid\nPlace cross (or highlight) against mis-plot\n© Cambridge University Press & Assessment 2025 Page 10 of 11\n\n2(d)(i) intersection point volume (read from within grid) 1\n\n2(d)(ii) nSO 2– =0.100  ans (d)(i) / 1000 = mol 1\n4\n[SO 2–(aq)] = [0.100  (d)(i) / 1000] / 0.025\n4\n\n2(e)(i) diagram showing labelled filter paper in a labelled (filter) funnel 1\n\n2(e)(ii) wash the residue with distilled water 1\n\n2(e)(iii) heat to constant mass 1\n\n2(e)(iv) higher value 1\nand\n(water would cause) (measured) mass / amount (of residue) would be greater\n© Cambridge University Press & Assessment 2025 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_52-p09.png", "../answer-assets/9701_w25_ms_52-p10.png", "../answer-assets/9701_w25_ms_52-p11.png" ] }, { "id": "9701-2025-on-53-q01", "question_id": "9701-2025-on-53-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 2.12 g 1\n\n1(a)(ii) M1 add a (small) volume of (distilled) water (to the small beaker) 3\nAND\ndissolve the solid / AgNO (s)\n3\nM2\nTransfer the solution and washings (using a funnel)\nAND\ninto a 250 cm3 volumetric flask.\nM3 Make up to the (calibration) mark / line with distilled water.\nAND\nthen invert the (stoppered) flask (several times)\n\n1(b)(i) (20(.0) cm3) pipette. 1\n\n1(b)(ii) barium (ions) could react with chromate (ions). 1\n\n1(b)(iii) Potassium chromate(VI) / K CrO (aq) is an irritant / corrosive 1\n2 4\n\n1(c)(i) They are (the only ones that are) concordant. 1\n\n1(c)(ii) 20.05 1\n100 = 0.48(54%)\n20.60\nCorrect working must be shown.\n\n1(d)(i) 20.65  0.05 / 1000 = 0.00103(25) 1\n\n1(d)(ii) = ((d)(i) / 2) *(250 / 20) 1\n= (d)(i)  6.25\n© Cambridge University Press & Assessment 2025 Page 7 of 10\n\n1(d)(iii) M1 M = 1.58 / (d)(ii) 2\nr\nM2 x= (M1 – 208.3) / 18\nAND\nx must be an integer.\nSome working must be shown.\n\n1(e) Heat to drive off water of crystallisation. 3\nM1 Apparatus\n1 Bunsen burner.\n2 Crucible (supported by pipeclay triangle)\nM2 Measurements\n1 Measure mass of crucible (+ lid) (empty).\n2 Measure mass crucible (+ lid) + solid before heating\n3 Measure mass of crucible (+ lid) + residue (after heating.)\nM3 Key step in method\n• Description of heat to constant mass.\nOR\nProduce a suitable precipitate using a viable reaction.\nM1 Apparatus and method\n• Chooses suitable reagent.\n• Filter the precipitate.\n• Wash and (allow to) dry\nM2 Measurements\n• Measure mass of hydrated barium chloride used.\n• Measure mass of precipitate.\nM3 Key step in method\n• Description of dry to constant mass.\n© Cambridge University Press & Assessment 2025 Page 8 of 10", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_53-p07.png", "../answer-assets/9701_w25_ms_53-p08.png" ] }, { "id": "9701-2025-on-53-q02", "question_id": "9701-2025-on-53-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) To ensure no air / other gas remains in the syringe 1\n\n2(b)(i) 2\n1\n0.707 0.500 0.222 0.158 0.109\nMr\nRate of effusion /cm3s-1 4.63 3.27 1.45 1.26 0.71\n\n2(b)(ii) Time (taken). 1\n\n2(b)(iii) size of hole 1\n\n2(c) M1 all 5 points plotted correctly. 2\nM2 Straight line of best fit drawn\n\n2(d) Point most anomalous to the plotted line of best fit circled. 1\nAND\nIf point above LOBF\n(Measured rate of effusion is higher than the true rate predicted by LOBF.)\nfor this point …\nmeasured / recorded (effusion) time < true / actual time\nOR\nIf point below LOBF\n(Measured rate of effusion is lower than the true rate predicted by LOBF.)\nfor this point …\nmeasured / recorded (effusion) time > true / actual time\n\n2(e) Yes AND The graph / LOBF… 1\nis a straight-line AND it goes through the origin.\n\n2(f) The results are not reliable as there is an anomalous result. 1\n© Cambridge University Press & Assessment 2025 Page 9 of 10\n\n2(g)(i) M1 Effusion rate = 50/31.6 = (1.58) 1\n1\nCorrect value for read from x-axis\nMr\nM2 Value correctly converted to M 1\n.\n\n2(g)(ii) If calculated M > 16 1\n(The average) M(of the other components) is greater than 16 / calculated value for g(i).\n\n2(h) (Rate of effusion) increases 1\nAND\n(At higher temperature) gas particles move more quickly.\n© Cambridge University Press & Assessment 2025 Page 10 of 10", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_53-p09.png", "../answer-assets/9701_w25_ms_53-p10.png" ] }, { "id": "9701-2025-on-54-q01", "question_id": "9701-2025-on-54-q01", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "54", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a) 38.87 g 1\n\n1(b) M1 Dissolve the solid / MnSO .H O / manganese sulfate (in the beaker) using (a small volume of) distilled water 3\n4 2\nM2 transfer / add to a (100 cm3 volumetric) flask\nAND\nrinse (with distilled water)\nM3 top up the 100 cm3 volumetric flask to mark (with distilled water)\nAND\nInvert the (stoppered) flask)\n\n1(c)(i) (wear) chemically resistant gloves. 1\n\n1(c)(ii) dissolving NaOH(s) is (very) exothermic 1\n\n1(d)(i) Oxygen in trapped air could dissolve into the water sample 1\n\n1(d)(ii) (25.0 cm3) volumetric pipette 1\n\n1(d)(iii) To observe the end-point more clearly 1\n\n1(e)(i) M1 1\nrough 1 2 3\n12.70 13.05 13.15\nM2 13.10 cm3 1\n\n1(e)(ii) titrations 2 and 3 are within 0.10 cm3 of each other 1\n© Cambridge University Press & Assessment 2025 Page 7 of 11\n\n1(e)(iii) working must be shown 1\n2  0.05\n 100 = 0.760\n13.15\n\n1(f)(i) (e)(i) / (2  106) = 6.55  10–6 (mol) 1\n\n1(f)(ii) (f)(i)  10 = 6.55  10–5 (mol) 1\n\n1(f)(iii) (f)(ii)  1.28  105 = 8.38 (mg dm–3) 1\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_54.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_54.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_54-p07.png", "../answer-assets/9701_w25_ms_54-p08.png" ] }, { "id": "9701-2025-on-54-q02", "question_id": "9701-2025-on-54-q02", "subject": "9701", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "54", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a)(i) to increase the rate of drying / removal of (distilled) water 1\n\n2(a)(ii) iodine is not (very) soluble (in water) 1\n\n2(a)(iii) mass change is (too / very) small 1\n\n2(a)(iv) repeat without any iodine 1\n\n2(b) 1\ntime / s mass of tin reacted\n/ g\n0 0.000\n100 0.069\n200 0.120\n300 0.218\n400 0.291\n500 0.359\n© Cambridge University Press & Assessment 2025 Page 8 of 11\n\n2(c)(i) M1 points plotted correctly 1\ncentre of all crosses should be on line for  value.\ny value should be on the line for point 1, 3\npoint 2, 6 – just below line / on line\n4 – midway between two lines\n5 – just above line / on line\nM2 straight line of best fit drawn 1\n\n2(c)(ii) most anomalous point circled 1\n(expected to be at time 200 s, point 3, below line of best fit)\nAND\nthe balance/mass reading was taken before 200 s / too early\n\n2(d) M1 two acceptable coordinates from line of best fit expressed in the form (x,y). 2\nM2 gradient correctly calculated from points listed for M1.\nanswer correctly rounded to 3 significant figures.\n© Cambridge University Press & Assessment 2025 Page 9 of 11\n\n2(e)(i) 1\nvolume of volume of\n0.400 mol dm–3 methylbenzene\nI solution used / cm3\n2\nused / cm3\n100.0 0.0\n75.0 25.0\n50.0 50.0\n25.0 75.0\n\n2(e)(ii) [I ] 1\n2\n\n2(e)(iii) M1 no 1\nAND\nthe rate of reaction is proportional to the concentration of iodine / a conclusion based on calculations done for M2\nM2 explained using actual values from columns 3 and 4 in Table 2.2 1\n© Cambridge University Press & Assessment 2025 Page 10 of 11", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9701_w25_ms_54.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2025-oct-nov/9701_w25_ms_54.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_w25_ms_54-p08.png", "../answer-assets/9701_w25_ms_54-p09.png", "../answer-assets/9701_w25_ms_54-p10.png" ] }, { "id": "9701-2026-m-42-q01", "question_id": "9701-2026-m-42-q01", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 23, "status": "available", "reason": null, "text": "1(a)(i) (d-block element which forms one or more) stable ions/compounds with an incomplete d orbital(s)/d sub-shell 1\n\n1(a)(ii) 1\n3d 4s 4p\n[Ar] ↿ ↿ ↿ ↿ ↿ ↿\n\n1(a)(iii) 1\n\n1(a)(iv) more than one (stable) oxidation state / exist in variable oxidation states 2\nvacant/empty (d) orbitals are energetically accessible\nOR\nvacant/empty (d) orbitals can form dative bonds with ligands\nOR\nvacant/empty (d) orbitals can accept lone pair to bond with ligands\n\n1(b)(i) M1: emf / potential / voltage of an electrode / half-cell compared / connected to SHE / hydrogen half-cell 2\nM2: 1 mol dm–3, 1 atm, 298 K\n© Cambridge University Press & Assessment 2026 Page 7 of 18\n\n1(b)(ii) M1: (Cr2+ ions) can be oxidised to Cr3+ 2\nM2: by O in acidic conditions / O with H+ because E⦵ =(+)1.64 (V)\n2 2 cell\nOR\nby O in acidic conditions / O with H+ as E⦵ O / H+ is more positive than Cr3+ / Cr2+\n2 2 2\nOR\nby O in alkaline conditions / O with OH– because E⦵ = (+) 1.66 (V) OR +0.81 V\n2 2 cell\nOR\nby O in alkaline conditions / O with OH– as E⦵ O / OH– is more positive than Cr3+ / Cr2+\n2 2 2\n\n1(b)(iii) for green solid 1\nCr3+(aq) + 3OH–(aq) → Cr(OH) (s)\n3\n\n1(b)(iv) for yellow solution 2\n4Cr(OH) (s) + 8OH–(aq) + 3O (g) → 4CrO 2–(aq) + 10H O(l)\n3 2 4 2\ncorrect species [1]\ncorrect balancing [1]\n\n1(c)(i) gaseous products are formed 1\nOR\nincrease in moles / molecules of gas\n\n1(c)(ii) 915 = 192 + 81 + 4S – 114 2\n∴ S = ¼  756 = (+) 189 (J K–1 mol–1)\n\n1(c)(iii) G always / needs / is to be negative / <0 because H < 0 / negative / exothermic S > 0 / positive / –TS is negative 2\nhigh activation energy / E prevents reaction\na\n\n1(d)(i) (enthalpy change/energy released when) 2\n• 1 mole of\n• gaseous atoms accept\n• one mole of e–\nany two [1], all three [2]\n© Cambridge University Press & Assessment 2026 Page 8 of 18\n\n1(d)(ii) M1: (2  398 + 3248 + 2  5233 + 3  702) 2\nM2: –1128 – 14112 = –15240 (kJ mol–1) min 4 sf\n\n1(d)(iii) ● Cr O less exothermic / less negative 2\n2 3\n● Cr3+ (has same charge but) larger radius / is less charge dense\n● less attraction between Cr3+ and O2– / the different ions OR weaker bond between Cr3+ and O2–\nany two [1], all three [2]\n© Cambridge University Press & Assessment 2026 Page 9 of 18", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_42-p07.png", "../answer-assets/9701_m26_ms_42-p08.png", "../answer-assets/9701_m26_ms_42-p09.png" ] }, { "id": "9701-2026-m-42-q02", "question_id": "9701-2026-m-42-q02", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 18, "status": "available", "reason": null, "text": "2(a)(i) M1: moles of manganate = 23.80 / 1000  0.0500 OR 1.19  10–3 mol 3\nOR initial moles of Fe2+ = 17.70 ÷ 277.9 OR 6.37  10–2 mol\nM2: moles of Fe2+ in aliquot = 5  1.19  10–3 OR 5.95  10–3 mol\nAND\ntotal moles of Fe2+ = 5.95  10–3  10 OR 5.95  10–2 mol\nOR\nmass of FeSO ·7H O left = 5.95  10–3  277.9 OR 1.6535 g (in 25 cm3)\n4 2\nM3: mass of FeSO ·7H O left = 1.6535 g  10 = 16.535 g (in 250 cm3)\n4 2\nOR\nmass of FeSO ·7H O left = 277.9  5.95  10–2 = 16.535 g\n4 2\npercentage oxidised = 100%  (17.70 – 16.535) / 17.70\n= 100%  1.165 / 17.70 = 6.58 % min 2sf\n\n2(a)(ii) MnO – / manganate is a (self-)indicator 1\n4\nOR\nMnO – / manganate changes colour / purple to colourless / decolourises\n4\nOR\nsolution changes (from colourless) to pink / purple at the end-point\n\n2(b)(i) 1\n\n2(c) manganese is both oxidised and reduced (to a single oxidation state) 2\nfrom +VII / +7 to +III / +3 AND +II / +2 to +III / +3\n\n2(d)(i) K = [Mn2+] [OH–]2 1\nsp\n© Cambridge University Press & Assessment 2026 Page 10 of 18\n\n2(d)(ii) [Mn2+(aq)] = 3.20  10–3 / 88.9 OR 3.60  10–5 mol dm–3 3\nK = 3.60  10–5  (2  3.60  10–5)2 = 4  (3.60  10–5)3\nsp\n= 1.87  10–13 min 2sf\nmol3 dm–9\n\n2(e)(i) M1: d orbital(s) of two different energies / d-d splitting occurs 3\nOR\nd orbital(s)/d (sub)-shell splits / d-d gap\nM2: electron(s) promoted / excited\nOR\nelectron(s) moves to higher (d–) orbital\nM3: wavelength / frequency / light / photon / h / hf absorbed\nOR\nradiation / energy from visible (region) absorbed AND colour (seen) is complementary\nOR\nwavelength / frequency / colour / light not absorbed is transmitted / reflected / seen\n© Cambridge University Press & Assessment 2026 Page 11 of 18\n\n2(e)(ii) 4\nformula [Mn(C O ) ]3–\n2 4 3\nformula of\nC O 2–\nligand 2 4\nshape octahedral\ncoordination\n6\nnumber\ntype of\noptical\nstereoisomerism\nstructures\nand\nQuestion Answer Marks", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_42-p10.png", "../answer-assets/9701_m26_ms_42-p11.png", "../answer-assets/9701_m26_ms_42-p12.png" ] }, { "id": "9701-2026-m-42-q03", "question_id": "9701-2026-m-42-q03", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 16, "status": "available", "reason": null, "text": "3(a) 2\n\n3(b) HO(CH ) COOH + 2PCl → Cl (CH ) COCl + 2POCl + 2HCl 2\n2 3 5 2 3 3\norganic product Cl (CH ) COCl [1]\n2 3\nrest of equation [1]\n© Cambridge University Press & Assessment 2026 Page 12 of 18\n\n3(c)(i) [H+] = (K  [HA])½ = (1.91  10–5  0.150)½ OR 1.69  10–3 (mol dm–3) 2\na\npH = –log [H+] = 2.77 min 2sf\n\n3(c)(ii) (a solution that) resists / opposes / minimises changes in pH 2\nwhen small (amounts / volumes) of acid / H+ and base / alkali / OH– are added to it\n\n3(c)(iii) HO(CH ) COO– 1\n2 3\n\n3(c)(iv) M1: moles of HA initially = (0.150  25.0 / 1000) OR 3.75  10–3 mol 3\nAND\nmoles of NaOH / moles of A– = (0.200  10.0 / 1000) OR 2.00  10–3 mol\nM2: remaining moles of HA = 3.75  10–3 – 2.00  10–3 OR 1.75  10–3 mol\nM3: [H+] = K [HA]/[A–] = 1.91  10–5 \na\n© Cambridge University Press & Assessment 2026 Page 13 of 18\n(\n(\n1\n2\n. 7\n. 0\n5\n0\n\n\n1\n1\n0\n0\n−\n−\n3\n3\n\n\n0\n0\n. 0\n. 0\n3\n3\n5\n5\n)\n)\n= 1.67  10–5\nAND pH = –log 1.67  10–5 = 4.78 min 2sf\n10\n\n3(d) (an excess of) acidified K Cr O / KMnO AND heat / boil / hot / reflux 1\n2 2 7 4\n\n3(e)(i) (the) ratio of the concentrations of a solute between two solvents at equilibrium 1\n\n3(e)(ii) M1: [acid(ether)] = 0.0375 mol dm–3 AND [acid(aq)] = 0.2775 mol dm–3 2\nM2: K = [acid(ether)] / [acid(aq)] = 0.135\npc", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_42-p12.png", "../answer-assets/9701_m26_ms_42-p13.png" ] }, { "id": "9701-2026-m-42-q04", "question_id": "9701-2026-m-42-q04", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Carboxylic acids and derivatives", "topic_slug": "9701-topic-33-carboxylic-acids-and-derivatives", "marks": 14, "status": "available", "reason": null, "text": "4(a)(i) two separate half-life readings quoted t ≅ 6000 (s) 250 2\n½\nsame / constant half-life (∴ first order with respect to [ester])\n\n4(a)(ii) tangent seen at t = 0 AND initial rate (from gradient of tangent) ≅ 0.24 / 8750 = 2.74  10–5 (mol dm–3 s–1) min 2sf 1\n\n4(a)(iii) rate = k [CH COOCH ] [H+] 2\n3 3\nmol–1 dm3 s–1\n\n4(b)(i) • reaction 6 / CH COCl goes to completion / non-reversible 2\n3\n• reaction 7 / CH COOH is an equilibrium (reaction) / reversible\n3\n• reaction 6 has a gas formed AND has a greater S / entropy is increased (so more feasible)\nany two [2]\n\n4(b)(ii) 4\nM1: correct arrow from lone pair on O (in CH OH) to C (of C=O)\n3\nM2: correct dipole on C=O AND correct arrow from the C=O bond to O atom\nM3: correct intermediate (both charges present)\nM4: arrow from (lone pair on) O(–) to C–O bond AND arrow from C-Cl bond to Cl\n© Cambridge University Press & Assessment 2026 Page 14 of 18\n\n4(c) M1: chlorobenzene < chloroethane < ethanoyl chloride 3\nM2 & M3: Any two from:\n(acyl chlorides easier than alkyl chlorides)\n• has two electronegative atom groups/an extra electronegative atom / an electronegative oxygen / electron withdrawing\nC=O AND so C (of C=O) highly / more electron deficient / C-Cl bond weak(est)\n(aryl chlorides hardest)\n• LP / lone pair / p-orbital on Cl/chlorine AND overlaps / delocalised with ring /  system AND (C-Cl) bond is\nstrengthened / has partial double bond character\n(alkyl chlorides easier than aryl chlorides)\n• carbon is bonded to an electronegative atom / Cl / not bonded to electronegative O AND so has less electron deficient\ncarbon\nOR\n• C-Cl bond strengthened AND by positive inductive effect of alkyl / R group\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-33-carboxylic-acids-and-derivatives/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_42-p14.png", "../answer-assets/9701_m26_ms_42-p15.png" ] }, { "id": "9701-2026-m-42-q05", "question_id": "9701-2026-m-42-q05", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 17, "status": "available", "reason": null, "text": "5(a)(i) HNO + H SO → +NO + H O + HSO – 1\n3 2 4 2 2 4\nOR\nHNO + 2H SO → +NO + H O+ + 2HSO –\n3 2 4 2 3 4\n© Cambridge University Press & Assessment 2026 Page 15 of 18\n\n5(a)(ii) 3\n+ H+\nM1: curly arrow 1 from touching / inside the hexagon to nitrogen of NO +\n2\nM2: correct intermediate\nM3: curly arrow 2 from C-H bond into hexagon (touching / inside) AND H+\n\n5(a)(iii) CH C H NO + 6[H] → CH C H NH + 2H O 1\n3 6 4 2 3 6 4 2 2\n\n5(a)(iv) Stage 1 M1: HNO OR NaNO with (dilute strong) acid AND ⩽ 10 °C 2\n2 2\nStage 2 M2: warming with H O / aqueous\n2\n\n5(a)(v) CH Cl AND Al Cl 2\n3 3\nNO directs to the meta / 3(,5)-position (in the ring so 4-methylation unlikely)\n2\n\n5(b)(i) it is inert / unreactive 1\n\n5(b)(ii) a high-boiling point / non-volatile AND (non-polar) liquid / solvent 1\n\n5(b)(iii) it is the least polar 1\nOR\nit is more attracted to the stationary phase\nOR\nforms a strong IMF to the stationary phase\nOR\nattracted to the stationary phase for a longer time\nOR\nit is the least volatile / highest bp\n© Cambridge University Press & Assessment 2026 Page 16 of 18\n\n5(b)(iv) % = 100%  22 / (29 + 34 + 22) = 25.88 1\n% = 25.9 OR 26 min 2sf\n\n5(c) M1: weakest = ethanol … water … phenol = strongest 4\nM2: why acids are stronger or weaker\nstrongest acid has weakest O—H\nOR\nH+ more easily lost\nOR\nanion stabilised / conjugate base stabilised\nM3: for phenol:\nas p-orbital / lone pair on oxygen overlaps / delocalised into the ring /  system / delocalised system\nM4: for ethanol:\npositive inductive effect/electron donating of alkyl / R group\nQuestion Answer Marks", "source_pages": [ 15, 16, 17 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_42-p15.png", "../answer-assets/9701_m26_ms_42-p16.png", "../answer-assets/9701_m26_ms_42-p17.png" ] }, { "id": "9701-2026-m-42-q06", "question_id": "9701-2026-m-42-q06", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 12, "status": "available", "reason": null, "text": "6(a)(i) 2\n\n6(a)(ii) pH at which a molecule is neutral / has no overall charge / charge of zero / no net charge 1\n\n6(a)(iii) 1\n© Cambridge University Press & Assessment 2026 Page 17 of 18\n\n6(b) M1: 3\nM2: direction in terms of charge of species\nZ moves towards positive (pole) AND is negatively charged\nOR\nX / Y move towards negative (pole) AND is positively charged\nM3: distance in terms of size\nX moves farther / faster (than Y) because of lower M / smaller molecular size\nr\n\n6(c)(i) M1: species with / has two lone pairs / LP (of electrons) 2\nM2: that form dative (covalent) / co-ordinate bond(s) to a (central) transition-element / metal AND atom / ion\nOR\ndonates electron pair(s) to a (central) transition-element / metal AND atom / ion\n\n6(c)(ii) 1\n\n6(c)iii) better / improved biological activity 1\nOR\nreduction of dosage\nOR\nthe other isomer may have other / different / unwanted biological activity\nOR\nthe other isomer may have different / unwanted side effects / be toxic / be harmful\n\n6(c)(iv) [Cu(H O) ]2+ + 2Z– → [Cu(Z) (H O) ] + 4H O 1\n2 6 2 2 2 2\n© Cambridge University Press & Assessment 2026 Page 18 of 18", "source_pages": [ 17, 18 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_42-p17.png", "../answer-assets/9701_m26_ms_42-p18.png" ] }, { "id": "9701-2026-m-52-q01", "question_id": "9701-2026-m-52-q01", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 18, "status": "available", "reason": null, "text": "1(a) 1.200 1\n\n1(b)(i) (2  0.0005 / 7.235)  100 = 0.0138 1\nFull working must be shown\n\n1(b)(ii) Increase the mass of X used 1\n\n1(c) M1 Dissolve the solid / X (in the beaker) (using a small volume of) (distilled) water 3\nM2 Transfer / add to a (250 cm3 volumetric) flask\nAND\nRinse (with distilled water)\nM3 Top the 250 cm3 volumetric flask up to mark (with distilled water)\nAND\nInvert the (stoppered) flask (several times)\n\n1(d)(i) (wear) chemically resistant gloves 1\n\n1(d)(ii) 25 cm3 volumetric pipette 1\n\n1(d)(iii) NaB(C H ) is added in excess. 1\n6 5 4\n\n1(d)(iv) to increase the mass of precipitate / KB(C H ) 1\n6 5 4\n\n1(d)(v) to dry the crucible 1\n\n1(d)(vi) 0.01(00) 1\n\n1(d)(vii) 0.746 1\n\n1(e)(i) 0.015(0) 1\n\n1(e)(ii) 0.293 OR 0.2925 1\n\n1(e)(iii) 0.1615 OR 0.162 1\n© Cambridge University Press & Assessment 2026 Page 7 of 9\n\n1(f) 1\neffect on calculated mass of KCl in 1(d)(vii) Higher\neffect on calculated mass of NaCl in 1(e)(ii) Lower\n\n1(g) thermal decomposition of the residue / KB(C H ) 1\n6 5 4\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_52-p07.png", "../answer-assets/9701_m26_ms_52-p08.png" ] }, { "id": "9701-2026-m-52-q02", "question_id": "9701-2026-m-52-q02", "subject": "9701", "year": 2026, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 12, "status": "available", "reason": null, "text": "2(a)(i) 1\nVol of KCl Vol of distilled water added\n0.00 50.00\n0.50 49.50\n1.00 49.00\n1.50 48.50\n2.00 48.00\n2.50 47.50\n3.00 47.00\n3.50 46.50\n4.00 46.00\n4.50 45.50\n5.00 45.00\n\n2(a)(ii) (50 cm3) burette 1\n© Cambridge University Press & Assessment 2026 Page 8 of 9\n\n2(a)(iii) relative light intensity 1\n\n2(a)(iv) M1 all 10 points plotted correctly. 2\nM2 both lines of best fit drawn correctly.\n\n2(a)(v) The line of best fit is straight 1\nAND\npasses through (0,0)\n\n2(a)(vi) Point at 6.00  10–3 circled 1\nAND\ntoo much distilled water was added (to the prepared solution).\nOR\ntoo little KCl (aq) was added (to the prepared solution).\n\n2(b)(i) Correct value of [K+] in diluted sample from graph 1\n\n2(b)(ii) Within the healthy range 1\nAND\nstated undiluted blood plasma has a concentration of 2(b)(i)  100 g dm–3\n\n2(b)(iii) Other (cat)ions (in blood plasma) will emit (other) colours 1\n\n2(c) All concentrations above 8.00  10–3 g dm–3 show the same relative light intensity. 1\nOR\nThe calibration graph is flat above [K+] = 8.00 g dm–3\n\n2(d) Not reliable 1\nAND\nthere is an anomalous point\n© Cambridge University Press & Assessment 2026 Page 9 of 9", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2026-March/ms/9701_m26_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-march/9701-chemistry-march-2026-mark-scheme-52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_m26_ms_52-p08.png", "../answer-assets/9701_m26_ms_52-p09.png" ] }, { "id": "9701-2026-mj-41-q01", "question_id": "9701-2026-mj-41-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 14, "status": "available", "reason": null, "text": "1(a) Ca2+(g) + S2–(g) → CaS(s) 2 If reverse lattice energy equation is given\nM1 all species CaS(s) → Ca2+(g) + S2–(g) award [1] if all\nM2 all state symbols species and state symbols are correct.\n\n1(b) (ionic) bonds are formed / the ions bond 1 A attractive force  distance.\nA electrostatic attraction  distance.\nI electrostatic attraction alone.\nI exothermic / energy released.\nI ions come together.\nI formation of the lattice / solid / CaS.\nI reference to energy required to break\nbonds.\nCON energy required to form bonds.\nCON bond formed is clearly covalent, e.g.\nCa–S.\n\n1(c) M1 MgO CaS KCl 3\nA for all four charges\nM2 CaS (and MgO) the ions / particles are more highly charged than KCl I nuclear charge\nor KCl has ions with the smallest charge ora I charge density for M2 and M3\nM3 MgO the ions/particles are smaller than CaS\nor MgO has the smallest ions ora\n\n1(d)(i) (enthalpy of) atomisation (of Cl) or bond energy (Cl–Cl ) and 1 If this is given in symbols we will accept\nH , H or H . If symbols and\nat atom atomisation\nwords are given, I the symbols.\n(first) electron affinity (of chlorine) If this is given in symbols we will accept\nH , H , H or\nEA EA1 electron affinity\nH . If symbols and words are\nfirst electron affinity\ngiven, I the symbols.\nCON electron affinity of chloride / Cl – / Cl\n2\n© Cambridge University Press & Assessment 2026 Page 7 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(d)(ii) M1 chlorine value doubled / –486 seen / 2x – 243 seen 2 ecf use of 1x – 243 for ans = –400\nM2 answer –643 (working needed).\n2336 +(2x – 243) –2493 = –643 –643 scores [2]\n\n1(e) M1 Correct G = H – TS and use of T = 873 3 CON G for G. A  .\nGibbs\nM2 G = –36 / –35.8 / –35.78 / –35.775 ecf T only If T = 600 then [1] ecf for +12.\nCON G for G but do not penalise twice\nM3 it is (feasible) and If in words we need ‘change’.\nG is negative / G is less than zero UNC\n(Note: G now in bold) I G is exothermic.\nM3 is UNConditional on the basis it is a correct answer on the answer line. If A G in words ‘Gibbs Free Energy\nthe calculation has not been worked through to an answer, or has given a change’\nnegative value for G, there is no M3 for ‘not feasible because G is positive’.\n\n1(f) • higher 2 ⚫✓✓ any two for one mark.\nAll three for two marks.\n• Ca2+ larger (than Mg2+) ora\nor ionic radius / size I atomic\n• (In CaCO ) less distortion / polarisation of anion / CO y– ora or group 2 trend but up / down the group\n3 x\nmust be clear.\nall standalone I references to ionic bond strength.\nora for MgCO 3 but must be stated or Ca2+ less polarising on anion / CO x y–\nI ‘depolarisation’.\nor C-O / C=O bond (must be identified,\ne.g. C–O) is less weakened by Ca2+.\n© Cambridge University Press & Assessment 2026 Page 8 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p07.png", "../answer-assets/9701_s26_ms_41-p08.png" ] }, { "id": "9701-2026-mj-41-q02", "question_id": "9701-2026-mj-41-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 11, "status": "available", "reason": null, "text": "2(a)(i) –log[H+] 1 [ ] essential.\nCON –log(H+), –log [H+] , –ln[H+].\ne\n\n2(a)(ii) [H+][OH–] / K xK 1 [ ] essential but do not penalise in (i) and\na b\n(ii).\nI 1  10–14.\n\n2(b)(i) (10–2.9) 1\n0.0013/0.00126/1.3x10–3/1.26x10–3\n\n2(b)(ii) 2bi2 / 0.105 0.001262 / 0.105 ecf bi 1\n1.5  10–5 / 1.51  10–5 / 1.53  10–5 ecf\n\n2(c)(i) M1 [CH CH CH COOH] = 0.15 3 If [CH CH CH COOH] = 0.45 [0] but then\n3 2 2 3 2 2\nM2 [CH CH CH COO–Na+] in solution D = 0.036 / 0.0359 [salt] in mixture = 0.108 [1] ecf\n3 2 2\n[salt] in solution C = 0.162 [1] ecf VERIFY\nM3 [CH CH CH COO–Na+] in solution C = 0.054 / 0.0538 / 0.0539 / 0.0540 with working.\n3 2 2\nAnother possible approach goes via moles of acid to get moles of salt,\nM1 moles acid = 0.045 mol\nM2 moles salt = 0.0108 / 0.0107 mol\nM3 original [salt] = 0.0538 / 0.0539 / 0.0540\n\n2(c)(ii) CH CH CH COO–Na+ + H+ → CH CH CH COOH + Na+ 1\n3 2 2 3 2 2\nor CH CH CH COO– + H+ → CH CH CH COOH\n3 2 2 3 2 2\nor CH CH CH COO– + H SO → CH CH CH COOH + HSO –\n3 2 2 2 4 3 2 2 4\nor 2CH CH CH COO–Na+ + H SO → 2CH CH CH COOH + Na SO\n3 2 2 2 4 3 2 2 2 4\n\n2(c)(iii) CH CH CH COOH + OH– → CH CH CH COO– + H O 1\n3 2 2 3 2 2 2\nor CH CH CH COOH + KOH → CH CH CH COOK + H O\n3 2 2 3 2 2 2\nor CH CH CH COOH + K+OH– → CH CH CH COO–K+ + H O\n3 2 2 3 2 2 2\n© Cambridge University Press & Assessment 2026 Page 9 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(d) M1 2,2-dichlorobutanoic and 2\nchlorine is electron withdrawing / negative inductive effect / electronegative\nI references to C=O, O, alkyl groups.\nM2 weakens O–H bond / stabilises anion / stabilises conjugate base / stabilises A –O–H bond and OH bond but CON –\ncarboxylate ion OH bond, C–O–H bond.\nQuestion Answer Marks Guidance", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p09.png", "../answer-assets/9701_s26_ms_41-p10.png" ] }, { "id": "9701-2026-mj-41-q03", "question_id": "9701-2026-mj-41-q03", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "3(a) 2 1\n1\n3\n\n3(b) M1 1.4x10–6 2\nM2 mol–2dm6s–1 I mol–2dm6s– .\n\n3(c)(i) (overall) first order and 1 I ‘[H ] first order’.\n2\n[NO] is (almost) constant I [NO] zero order.\nI [NO] not a limiting factor.\nI [H ] is a limiting factor.\n2\nA [NO] pseudo zero order.\n\n3(c)(ii) M1 33 000 / 32 500 / 32 542 / 32 540 / 32 535 / 3.25  104, etc.\nM2 130 000 / 130 141 / 1.3  105 / 1.32  105, etc. ecf\n© Cambridge University Press & Assessment 2026 Page 10 of 21\nt\n12\n2 32 542 if ln2 used.\nanswer = t\n12\n 4\n\n3(d)(i) M1 one N O and one H on L or one N O and one H O on R 2 (M1 is for either correct LHS or RHS).\n2 2 2 2 2\nM2 N O + H → N O + H O\n2 2 2 2 2\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(d)(ii) no H involved in step one (and H in rate equation) or 1 I ‘does not agree with rate equation’.\n2 2\nonly one species from rate equation or I reference to NO without reference to\ndoes not contain species in the same molar proportion as rate equation or order of reaction.\nif step 1 is RDS then order wrt [H ] = 0 or I N O not in the rate equation.\n2 2 2\nstep 1 does not include all species in rate equation or\nrate would be second order wrt NO only or\nreaction would be overall second order\nQuestion Answer Marks Guidance", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p10.png", "../answer-assets/9701_s26_ms_41-p11.png" ] }, { "id": "9701-2026-mj-41-q04", "question_id": "9701-2026-mj-41-q04", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 10, "status": "available", "reason": null, "text": "4(a) Go = –nF Eo 1 A Go = – nF E .\ncell cell\n\n4(b) Cu2+ + 2V2+ → Cu + 2V3+ 2 CON cancellable species / electrons\nIf ⇌ read forwards\n(+) 0.60 I 0.6\n+ not needed but – is CON.\n\n4(c) M1 Cu 2 or in words if unambiguous.\nM2 V2+\n\n4(d) M1 Cl / chlorine 2 CON if both oxidised and reduced form\n2\ngiven.\nM2 V / vanadium CON if both oxidised and reduced form\ngiven.\n\n4(e) All candidates get three marks 3\n© Cambridge University Press & Assessment 2026 Page 11 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p11.png" ] }, { "id": "9701-2026-mj-41-q05", "question_id": "9701-2026-mj-41-q05", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "5(a) similarity in energy of (3)d and (4)s (subshells / orbitals) 1 this is a syllabus statement – do not allow\nA same much variation of meaning, I statements\nthat can be understood to refer to 3d only.\n\n5(b) 3d7 4s2 / 4s2 3d7 1 A full configurations with no errors.\n3d7 / 3d7 4s0\n\n5(c) M1 the d orbitals 3\nare split / become non-degenerate / reference to E within d-subshell\nor E between d orbitals\nM2 electrons are promoted / excited / jump to higher level owtte\nM3 absorbing (energy of) light / photon / frequency / wavelength / h and ‘complementary’ may be described but\ncomplementary colour is seen / reflected / transmitted ‘different colour’ is not enough.\nM3 CON emits / energy released.\n\n5(d) M1 blue to red / blue to pink 2 A named shades of blue and pink.\nM2 CoCl + 6H O → [Co(H O) ]2+ + 2Cl – or\n2 2 2 6\nCoCl 2– + 6H O → [Co(H O) ]2+ + 4Cl – I CoCl + 6H O → CoCl 6H O .\n4 2 2 6 2 2 2 2\nI Co2+ + 6H O → [Co(H O) ]2+.\n2 2 6\n\n5(e)(i) (white / brown / off white) precipitate / ppt / solid and 1 CON clearly wrong colours, e.g.\nbrown (solution) black / blue / green / purple\nor wrong observations, e.g. fizzing.\n\n5(e)(ii) starch and 1\nblue / black to colourless / white / off-white colourless / white to blue / black is CON.\n\n5(e)(iii) redox 1 A oxidation-reduction (both needed).\n© Cambridge University Press & Assessment 2026 Page 12 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(e)(iv) • moles of S O 2– = 0.2  37.6 / 1000 = 7.52  10–3 3 ⚫✓✓✓ two points for one mark\n2 3\n• moles of I = 3.76  10–3 ecf\n2\n• moles of Cu = 7.52  10–3 ecf\ncorrect answer scores [3].\n• mass of Cu = 0.48 / 0.478 / 0.4775 / 0.47752 g ecf\n0.239 g, 0.955 g or 1.41 g on the answer line, without any correct,\nunderstandable working, scores [0]\nQuestion Answer Marks Guidance", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p12.png", "../answer-assets/9701_s26_ms_41-p13.png" ] }, { "id": "9701-2026-mj-41-q06", "question_id": "9701-2026-mj-41-q06", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "6(a) M1 two lone pairs / LP 2\nM2 donated / form dative bond(s) / coordinate bond(s) CON ‘2 or more dative bonds’.\nto (central) metal / TM / TE\natom / ion\n\n6(b)(i) [Fe(C\n2\nO\n4\n)\n3\n]3– or Fe(C\n2\nO\n4\n)\n3\n3– 1\n\n6(b)(ii) M1 One correct isomer of [Fe(C 2 O 4 ) 3 ]3– drawn with 3D and oct 2 Correct 3D. octahedral, different\nThese two renditions of octahedral are\nM2 Two optical isomers drawn with 3D and octahedral only acceptable.\nI charges.\n\n6(b)(iii) optical only 1\n© Cambridge University Press & Assessment 2026 Page 13 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c)\n© Cambridge University Press & Assessment 2026 Page 14 of 21\n\n\nH\n H\ng\ng\n2 +\nI\n4\n\n2 \n I\n\n−\n−\n\n\n4\n2 A [HgI 2–]. 4\nM1 A [Hg2+] with complexing H 2 O.\nThe charges shown here are essential.\nFinal set of brackets must be square 3.\nPenalise two or more round outer\nbrackets.\nM2 mol–4 dm12 unc ecf only after K with 4th power.\nstab\nQuestion Answer Marks Guidance", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p13.png", "../answer-assets/9701_s26_ms_41-p14.png" ] }, { "id": "9701-2026-mj-41-q07", "question_id": "9701-2026-mj-41-q07", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 14, "status": "available", "reason": null, "text": "7(a)(i) HNO and H SO and conc once and 25° ⩽ T ⩽ 60° 1 Acids may be named.\n3 2 4\nAny dilute or aq is CON.\nTemperature must be numerical and\nwithin range.\nI heat or warm.\nCON boil or reflux.\n\n7(a)(ii) 0° ⩽ T ⩽ 10° and 1 A ‘T below 10o’ etc.\nCON heat or warm.\nHNO or NaNO + HCl I sodium nitrate.\n2 2\nA nitrous acid.\n\n7(a)(iii) benzenediazonium chloride / phenyl diazonium chloride 1 A very minor errors, e.g. word order.\nCON benzyl’benzoyl.\nI diazo.\n\n7(b)(i) 0 1 2 1 A blank for zero.\n0 6 3\n\n7(b)(ii) 6 1\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(iii) 1\nI + on the H+.\nCurly arrow must start on C=O.\nAny incorrect curly arrow is CON.\nA in two steps, e.g.\ndipole\nwith dipole.\nCurly arrow from C=O to H+\nAdditional charges are CON.\nor\n© Cambridge University Press & Assessment 2026 Page 15 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(iv) 3 ‘Usual rules’ for K.\nCentre of + charge must be above a line\njoining C3 and C5.\n‘Horseshoe’ must include C6, C1 and C2,\nmay include one or both of C3 and C5,\nmust not include C4.\nI curly arrow showing delocalisation of\nphenolic LP (LP on oxygen of phenol).\nCON additional curly arrows\nM1 CA on leftmost diagram Curly arrow must start on or within\nhexagon and may cross hexagon\nM2 K, showing horseshoe, + charge and H anywhere.\nM3 curly arrow on C–H bond in K back into ring\nCurly arrow must touch hexagon.\nUse of quinoid intermediate:\nM1 as above – curly arrow on leftmost diagram\nM2 for the intermediate shown at the bottom of guidance. The curly arrow for\nM3 is shown on this diagram but is not necessary for M2.\nM3 as above – curly arrow on C–H bond in K back into ring\n© Cambridge University Press & Assessment 2026 Page 16 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(v) • less readily 2 ⚫✓⚫✓ two points for one mark,\n• lone pair (LP) on O (of phenol) / p-orbital on O four points for two marks.\n• is delocalised into the ring / overlaps with delocalised ring / -system A LP on OH I LP in phenol.\n• activates ring / more attractive to electrophile / electron (charge) density ‘Delocalised’ or ‘joins  system’ needed\nincreased (‘electron’ inferred if ‘LP donated to ring’) / becomes more may be awarded without LP if linked to O.\nelectron rich\nI references to polarisation of electrophile\n(it has a full + charge, stated in the Q)\nI makes ring more reactive / more\nsusceptible to electrophiles / more\nnegative.\nI ‘O is electron donating’.\nI ‘attracts nucleophile more’.\n\n7(c)(i) 2 A OH of tertiary alcohol group substituted\nto bromide as well as ring substitutions\nand equation with HBr and H O on the\n2\nright.\n+ 2Br → + 2HBr\n2\nM1 structure of product (compound H)\nM2 equation balanced, must have HBr\n→ product above + H O + HBr\n2\n\n7(c)(ii) electrophilic substitution 1 CON nucleophilic substitution unless –OH\non sidechain substituted with Br when it\nscroll and SEEN becomes I.\n© Cambridge University Press & Assessment 2026 Page 17 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 14, 15, 16, 17 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p14.png", "../answer-assets/9701_s26_ms_41-p15.png", "../answer-assets/9701_s26_ms_41-p16.png", "../answer-assets/9701_s26_ms_41-p17.png" ] }, { "id": "9701-2026-mj-41-q08", "question_id": "9701-2026-mj-41-q08", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "8(a)(i) CH COOH + SOCl → CH COCl + SO + HCl 1 CON aq.\n3 2 3 2\n\n8(a)(ii) PCl or PCl 1 A names, phosphorus(III) chloride,\n3 5\nphosphorus trichloride, phosphorus(V)\nchloride, phosphorus pentachloride.\nCON aq, but not in both 8(a)(i) and\n8(a)(ii).\n\n8(a)(iii) 1 Or any unambiguous structural or skeletal\nformula.\nor CH CONHCH(CH )\n3 3 2\n\n8(a)(iv) 1 No ecf from 8(a)(iii) – Q specifies L is an\namide and gives the structure of M.\nor CH CHNH CH\n3 2 3\nor skeletal\n\n8(a)(v) (nucleophilic) addition-elimination 1 No ecf from 8(a)(iii) – Q specifies L is an\namide.\nI ‘condensation’ – correct but not a\nmechanism.\n\n8(a)(vi) LiAl H 1 No ecf from 8(a)(iii) – Q specifies L is an\n4\namide.\nCON aq.\n\n8(b)(i) • r – triplet 2 ⚫✓⚫✓ two points for one mark,\n• s – quartet/quadruplet four points for two marks.\n• t – doublet\n• u – multiplet A septet, heptet, septuplet.\n© Cambridge University Press & Assessment 2026 Page 18 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(b)(ii) M1 one more peak and reference to N-H 2 A ‘NH peak is present’.\nA ‘NH peak is not present in D O’.\n2\nSubject of Q is ‘in CDCl ’ so assume this if no solvent specified. A reference to five peaks or five proton\n3\nSee first two ALLOWs in guidance. environments and reference to N-H.\nA ‘it is exchanged with D’ if NH peak\nM2 H (of NH) is exchanged with D (in D O) / NH + D → ND + H / already referred to and you are confident\n2\nH (of NH) is not exchanged with D (in CDCl ) they refer to ‘in D O’.\n3 2\nA ‘it is not exchanged with D’ if NH peak\nthe bits in brackets for M2 are not essential but if clearly wrong it is CON already referred to and you are confident\nthey refer to ‘in CDCl ’.\n3\nI ‘D O replaces the H’.\n2\n© Cambridge University Press & Assessment 2026 Page 19 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 18, 19 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p18.png", "../answer-assets/9701_s26_ms_41-p19.png" ] }, { "id": "9701-2026-mj-41-q09", "question_id": "9701-2026-mj-41-q09", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 11, "status": "available", "reason": null, "text": "9(a) M1 One pair of diagrams of P with correct use of 3D showing mirror images of 3\ntwo optical isomers Q may be given above, and P may be\ngiven below.\nMirror plane not essential.\nM2 One pair of diagrams showing correct E and Z of Q\nM3 conditional on optical linked to an\nattempt on P and geometrical / cistrans\nM3 Optical or enantiomerism in the right place and / EZ linked to an attempt on Q in M1 and\nGeometric(al) or E / Z or A cis/trans in the right place M2.\n\n9(b)(i) the pH at which (the amino acid is) 1 A clear and correct description of\na zwitterion / no net charge / no overall charge / no total charge zwitterion for second component.\nelectrically neutral / charge is neutral\n© Cambridge University Press & Assessment 2026 Page 20 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n9(b)(ii) 3\npH 4.0 pH 5.5 pH 8.0\n⚫✓⚫✓ ⚫✓\nT\ntwo structures for one mark,\nfour structures for two marks,\nsix structures for three mark.\nPhe at 5.5 can be\nU\nTable left is repeated larger after 9(c)(iii).\n\n9(c)(i) 2 Fully correct [2].\nM1 – amide links correct\nM2 – remainder is fully correct, e.g.\nterminal groups not trailing bonds\nneutral, zwitterion, 1+ or 1–\nside/R groups CH – H – CH\n3 3\nall H atoms on -carbons\n\n9(c)(ii) hydrolysis 1 I acid / alkaline / enzymatic / NS.\n\n9(c)(iii) condensation 1 I addition-elimination.\n© Cambridge University Press & Assessment 2026 Page 21 of 21\nTrace ID: PC-F9EA0E463D\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 20, 21 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-41.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_41-p20.png", "../answer-assets/9701_s26_ms_41-p21.png" ] }, { "id": "9701-2026-mj-42-q01", "question_id": "9701-2026-mj-42-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 20, "status": "available", "reason": null, "text": "1(a)(i) • they behave as catalysts 1 I high melting point / high densities.\nI form complex ions.\n• (they form) coloured compounds / complexes / ions I variable / multiple oxidation states.\nBOTH [1]\n\n1(a)(ii) the d and s sub-shells / orbitals are close / similar in energy 1 I use of ‘shells’ for orbitals.\nA (n-1)d and ns if any numbers see.n\nA electrons in d and s sub-shells / orbitals available for bonding\n\n1(b)(i) 1 A use of .\n\n1(b)(ii) dative (covalent) / coordinate 1\n\n1(c) vacant (d) orbital(s) are energetically accessible 1 I subshells / shells / energy levels.\nor vacant (d) orbital(s) can form dative bonds with ligands I partially filled d-orbital(s) are energy\nor vacant (d) orbital(s) can accept a (lone) pair of electrons to form a accessible.\ndative bond A ‘empty’ / ‘unoccupied’ for vacant.\n\n1(d)(i) NaOH / OH– 1 A NH for NaOH.\n3\nand precipitation / ligand exchange / A ligand substitution / ligand.\n/ neutralisation / deprotonation / acid-base BOTH replacement / ligand displacement.\nI precipitate.\n\n1(d)(ii) M1 [Cr(H O) (OH) ] + 6NH → [Cr(NH ) ]3+ + 3H O + 3OH– 2 A use [Cr(NH ) ](OH) as a product.\n2 3 3 3 3 6 2 3 6 3\nM2 ligand exchange / replacement / substitution / displacement\n© Cambridge University Press & Assessment 2026 Page 7 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(e) M1 one correct 3D structure of [Cr(H O) (OH) ] 2 CON use of O H once only then apply ecf.\n2 3 3 2\nM2 correct 3D structure of the other isomer One isomer - two water & two OH ligands at\n180°.\nOne isomer – all water & all OH ligands at 90°.\n\n1(f)(i) M1 [Fe(H O) ]3+ / Fe3+ weakens O–H bonds / polarises O–H bonds 2 A M1 [Fe(H O) ]3+ undergoes\n2 6 2 6\nhydrolysis / deprotonation.\nM2 [Fe(H O) ]3+ → [Fe(H O) OH]2+ + H+ A M1 one of the H O ligands loses a H+.\n2 6 2 5 2\nA M1 H O in the complex can donate a H+.\n2\nA other alternatives for M2 for example\n[Fe(H O) ]3+ + H O → [Fe(H O) OH]2+ +\n2 6 2 2 5\nH O+ .\n3\n\n1(f)(ii) Fe3+ has lower pH / Fe2+ has a higher pH 1 A Fe3+ is more acidic / Fe2+ is less acidic\nand for ‘lower pH / higher pH”.\nFe3+ has a higher charge / smaller radii / higher charge density (so easier\nto hydrolyse)\nBOTH ora\n\n1(g) M1 (Ag+) d-subshell is full / complete or d10 or d-orbitals are full 2 A use of 3d instead of 4d.\nI d orbital is full for M1.\nI M1 no empty d subshells.\nM2 no d-d(*) transition or no d electrons promoted / excited [1] CON M2 no d-d splitting / no E gap.\nor cannot absorb light in the visible spectrum A M2 cannot absorb energy in the visible\nspectrum.\n(must imply promotion) ‘d’ can be qualified from M1 for M2\n© Cambridge University Press & Assessment 2026 Page 8 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(h) M1 M2 5 A route 1 via 500 cm3 in M3 & M4\n• moles MnO – = 0.025  0.02180 = 5.45  10–4 moles C O 2– = 1.3625  10–3  20 = 0.02725\n4 2 4\n• moles C O 2– = 2.5  5.45  10–4 = 1.3625  10–3 moles H C O = 1.02  10–3  20 = 0.0204\n2 4 2 2 4\necf (moles MnO –  2.5) moles Na C O = 0.02725 – 0.0204 = 0.00685.\n4 2 2 4\n• moles OH– = 0.150  0.0136 = 2.04  10–3\n• moles H 2 C 2 O 4 = ½  2.04  10–3 = 1.02  10–3 mass Na 2 C 2 O 4 = 0.00685  134 = 0.9179 g\necf (moles OH– x ½)\nany two [1] all four [2] A route 2 in M4 & M3\nmoles C O 2– = 1.3625  10–3  20 = 0.02725\n2 4\nM3 moles Na C O = 1.3625  10–3 – 1.02  10–3 = 3.425  10–4 ecf moles H C O = 1.02  10–3  20 = 0.0204\n2 2 4 2 2 4\n(in 25 cm3 A) [1] mass C O 2– = 0.02725  134 = 3.6515 g\n2 4\nmoles H C O = 0.0204  134 = 2.7336 g.\n2 2 4\nM4 moles Na C O = 3.425  10–4  20 = 6.85  10–3 ecf\n2 2 4\n(in 500 cm3 A) (M3) mass Na C O = 3.6515 – 2.7336 =\n2 2 4\nmass Na C O = 6.85  10–3  134 = 0.9179 g [1] 0.9179 g.\n2 2 4\nM5 % = 0.9179 / 5.00  100 = 18.4 / 18.358 [1] min 2sf ecf A ecf M4 answer to calculated value  20 \ncorrect answer scores [5] 134.\nA 18.2(24) (from use of 3sf 1.36  10–3) for correct answer A ecf M5 for (a calculated mass ÷ 5.0)  100.\n© Cambridge University Press & Assessment 2026 Page 9 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p07.png", "../answer-assets/9701_s26_ms_42-p08.png", "../answer-assets/9701_s26_ms_42-p09.png" ] }, { "id": "9701-2026-mj-42-q02", "question_id": "9701-2026-mj-42-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 10, "status": "available", "reason": null, "text": "2(a) 1 I LP drawn / any charges on diagram.\nARROW 1 from O– to C–O bond\n(arrow can start anywhere on O).\nor BOTH arrows [1]\nARROW 2 from C–O bond to the other O– .\n\n2(b) • PbCO (most) CaCO CuCO (least) 2 A Pb(2+)(most) Ca(2+) Cu(2+) (least).\n3 3 3\n• (stability increases) as (ionic) radii increases / size of M2+ increases / NO ecf.\nor Pb2+ largest radii ora Bullets are all u / c.\nA bullet 2 charge density of M2+ decreases.\n• less polarisation / distortion of carbonate ion / anion / CO (2)–\n3\nor less weakening of CO bond / C-O / C=O  ✓ ✓ CON bullet 3 stronger ionic bond linked to\nlarger ionic radii.\nI C=O is less polarised.\nI weaker ionic bond.\n\n2(c) M1 H and H (both) become less exothermic / less negative [1] 3 A M1 H and H decrease.\nlatt hyd latt hyd\nA M1 H and H become more\nlatt hyd\nendothermic / positive.\nM2 H changes more / changes faster / dominant factor / becomes less CON missing  or H once only then ecf.\nhyd\nexothermic by a larger extent [1] ora\nM3 H becomes less exothermic / less negative I M3 H decreases.\nsol sol\nor H becomes more endothermic / more positive I M3 H becomes less exothermic.\nsol solubility\n\n2(d)(i) M1 (K for Mg(OH) ) = [Mg2+][OH–]2 [1] 2\nsp 2\nM2 units mol3 dm–9 [1] ecf from M1 I state symbols.\n\n2(d)(ii) [H+] = 10–10.35 = 4.47  10–11 1\n[OH–] = 1  10–14 ÷ 4.47  10–11 = 2.24  10–4 min 2sf\n© Cambridge University Press & Assessment 2026 Page 10 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(d)(iii) K = (0.5  2.24  10–4)  (2.24  10–4)2 1 A ecf\nsp\nK = 5.61(2) / 5.62  10–12 u / c min 2sf ecf for 2(d)(i) and 2(d)(ii) correct calculated answer of\nsp\nK = (0.5  2(d)(ii))  (2(d)(ii))2.\nsp\nI units even if incorrect\nNO ecf from K = [Mg2+][OH–].\nsp\n© Cambridge University Press & Assessment 2026 Page 11 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p10.png", "../answer-assets/9701_s26_ms_42-p11.png" ] }, { "id": "9701-2026-mj-42-q03", "question_id": "9701-2026-mj-42-q03", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 10, "status": "available", "reason": null, "text": "3(a)(i) M1 Exp 1 & 2 [CH OH]  1.5 rate  1.5 or [CH OH]  2 / 3 rate  2 / 3 4 Reasoning must be linked to correct expts –\n3 3\nand so first order to [CH OH] [1] u / c CON once only for M1-M3.\n3\nM2 Exp 3 & 4 [H+]  1.33 rate  1.33 or [H+]  3 / 4 rate  3 / 4 A any viable mathematical approach.\nand so first order to [H+] [1] u / c\nM3 Exp 2 & 3 [CH CHO]  0.5, [CH OH]  2 and [H+]  3 rate  3 A any correct alternates to M3.\n3 3\nand so first order to [CH CHO] [1] ecf –all conditional of orders in M1M2 e.g.\n3\nExp 2 & 4 M3 [CH\n3\nCHO]  2, [CH\n3\nOH]  0.5\nA second part of M1 M2 M3 and [H+]  1 / 4 rate  1 / 4\n‘and first order’ can be read in from their rate equation and so first order to [CH CHO].\n3\nM4 Rate = k[CH OH][H+][CH CHO] [1] u / c ecf Exp 1 & 3 M3 [CH CHO]  0.5, [CH OH]  3\n3 3 3 3\nor rate = 1.25 [CH OH][H+][CH CHO] and [H+]  3 rate  9 / 2\n3 3\nand so first order to [CH CHO].\n3\nI any annotations on the table unless directed by the candidates Exp 1 & 4 M3 [CH CHO]  0.5, [CH OH]  3\n3 3\nand [H+]  4 rate  6\nand so first order to [CH CHO].\n3\n\n3(a)(ii) mol–2dm6s–1 u / c ecf from 3(a)(i) 1\n\n3(a)(iii) initial rate = 0.0195 u / c ecf from 3(a)(i) min 2sf 1 A methods with calculations of k first then rate.\n\n3(b) 2 I any curly arrows even if incorrect.\nA H+ + Br- for HBr.\neach step 2 x [1]\n© Cambridge University Press & Assessment 2026 Page 12 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(c) • (reactants) adsorb (to surface of catalyst) 2\nI absorbed.\n• bonds (in reactant) weaken A bullet 1 ‘binds to the surface of the catalyst’\nfor ‘adsorbed’.\n• (reaction occurs and the) product(s) are desorbed\nor reaction occurs and substances are desorbed I lowers E a .\nA product(s) deadsorbed\n● ✓ ✓ I bonds between reactants.\nI ‘bonds are broken’ (happens in all reactions).\n© Cambridge University Press & Assessment 2026 Page 13 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p12.png", "../answer-assets/9701_s26_ms_42-p13.png" ] }, { "id": "9701-2026-mj-42-q04", "question_id": "9701-2026-mj-42-q04", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 6, "status": "available", "reason": null, "text": "4(a) (the) ratio of the concentrations of a solute / substance / compound 1 CON if in terms of concentration of ‘solutes’.\nbetween\ntwo (immiscible) solvents / two liquids\n(at) equilibrium\n\n4(b) EITHER (3 Bullet 3 subsumes bullet 2 → [1].\nSolution 1 Bullet 4 subsumes bullets 1,2,3 → [2].\nM1M2\n• Initial moles of (COOH) = 2.7 / 90 or 0.03 A ecf for bullet 4 from calculated values of\n2\n• moles of NaOH = 0.0165  0.200 or 0.0033 bullets 1 and 3, for example using the incorrect\n• moles of (COOH) 2 in ether = 0.0033 / 2 or 0.00165 M r / the wrong ratio.\n• moles of (COOH) in water = 0.0300 – 0.00165 or 0.02835 ecf\n2\nNO ecf for M3 from bullets 1 and 2\n● ✓● ✓\nM3 K pc = 0.00165 / 0.02835 = 0.058(2) min 2sf ecf from bullets 3 & 4 only for K pc = [0.033 / 0.03].\nor 3)\nSolution 2\nM1M2\n• moles of NaOH = 0.0165  0.200 or 0.0033\n• moles of (COOH) in ether = 0.0033 / 2 or 0.00165\n2\n• mass of (COOH) in ether = 0.00165  90 or 0.1485 g\n2\n• mass of (COOH) in water = 2.70 – 0.1485 or 2.5515 g ecf\n2\n● ✓● ✓\nA reverse ratio = 17(.2) [3].\nM3 K = (0.1485 ÷ 100) ÷ (2.5515 ÷ 100) = 0.058(2) min 2sf ecf from\npc\nbullets 3 & 4 only\n© Cambridge University Press & Assessment 2026 Page 14 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(c) 2 I in terms of hydrophilic / phobic.\n• K would be higher / larger (if K < 1 in 4(b))\npc pc\n(for butanedioic acid) (ecf from 4(b)) ora A bullet 1 K would be lower (if K > 1\npc pc\nin 4(b)).\n• butanedioic acid is less polar / more non-polar ora\nA ‘organic solvent’ / ‘ether’ for ethoxyethane.\n• so less soluble in water / more soluble in ethoxyethane ora\nor dissolves less in water / dissolves more in ethoxyethane I bullet 2 ‘butanedioic acid is non-polar’ alone.\n● ✓✓ I bullet 3 concentration of butanedioic acid is\nlower in water.\nBullet 2 / 3 – comparison needed.\n© Cambridge University Press & Assessment 2026 Page 15 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p14.png", "../answer-assets/9701_s26_ms_42-p15.png" ] }, { "id": "9701-2026-mj-42-q05", "question_id": "9701-2026-mj-42-q05", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 11, "status": "available", "reason": null, "text": "5(a) M1 emf / potential difference / voltage 2 A half-cell for electrode.\nand between / compared to\nand a standard hydrogen electrode and another electrode CON hydrogen cell.\nM2 under standard conditions of\n• 1 atm / 101 kPa / 100 kPa\n• 298 K / 25 °C,\n• (all solutions being) 1M / 1 mol dm–3 u / c\n\n5(b)(i) M1M2 3 A M3 two calculated E\ncell\n• Zn + 2Fe3+ → Zn2+ + 2Fe2+ • bullet 1 E = 1.53 (V)\ncell\n• NO – + 10H+ + 8Fe2+ → NH + + 3H O + 8Fe3+ • bullet 2 E = +0.10 (V)\n3 4 2 cell\n• NO – + 10H+ + 4Zn → NH + + 3H O + 4Zn2+ • bullet 3 E = +1.63 (V).\n3 4 2 cell\n● ✓✓\nM3 E is positive (to be feasible) I M3 use of ‘greater’ for more positive.\ncell\nor Eo of Fe3+ / Fe2+ and NO – / NH + is more positive\n3 4\nor Eo of redox system 1 and 2 are more positive ora u / c\n\n5(b)(ii) high E for the reaction OWTTE 1 I rate is too slow.\na\nI non-standard conditions are used.\n© Cambridge University Press & Assessment 2026 Page 16 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b)(iii) 3\nI inclusion negative ion with SHE like Cl–, NO –,\n3\nSO 2–.\n4\nCON bullet 2 & 9 battery in the circuit.\nA bullet 6 / 7 graphite / C for Pt.\nState symbols not required.\n• labelled salt bridge\n• voltmeter / V / potentiometer\n• H\n2\n• good delivery system (of H )\n2\n• H+(aq) / HCl(aq) solution labelled [state symbol not required]\n• Pt electrode\n• Pt electrode\n• NO and NH + and H+ solution labelled\n3 4\nor HNO and NH NO\n3 4 3\n• complete circuit (salt bridge touches the solution & wires from V go\nto the electrodes)\n● ● ✓ ● ● ✓● ● ✓\n\n5(c)  Go = -nE F 2 A M1 use of Go = –nEF and use of –645.6.\ncell\n–645.6  1000 = –3  E  96 500\nM2 E = (+)2.23 (V) min 2sf A ecf [1] from\ncell\nGo = nE F → E = –2.23.\ncell cell\n© Cambridge University Press & Assessment 2026 Page 17 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p16.png", "../answer-assets/9701_s26_ms_42-p17.png" ] }, { "id": "9701-2026-mj-42-q06", "question_id": "9701-2026-mj-42-q06", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 14, "status": "available", "reason": null, "text": "6(a)(i) Observations: 1 A initial colour brown / yellow for ‘orange’ and\nwhite ppt. variants of these.\nand (solution) decolourises / turns colourless / orange colour disappears CON fizzing.\nI steamy fumes.\nI discolouration / colour discharged.\nI ‘colour disappears’.\n\n6(a)(ii) 1\nstructure\n\n6(b) M1 lone pair on oxygen / O delocalised / overlaps 2 I donated / incorporated in reference to\nand into the ring /  system / delocalised system ‘delocalised / overlaps’.\nor p-orbital on oxygen / O delocalised / overlaps\nand with ring /  system / delocalised system [1]\nI ‘H+ more easily lost’ for more acidic.\nM2 weakens O—H and phenol is more acidic\nor (phenol forms) anion / conjugate base / phenoxide ion is more I O–H is weak alone – comparison needed.\nstable\nand phenol is more acidic [1] A phenol is a stronger (acid) for ‘more acidic’.\n\n6(c)(i) 1 I connectivity errors to COOH.\nPhenol must be 1,2-substituted and drawn.\nFully skeletal structures must have brackets if\nsolid line continuation bonds .\n© Cambridge University Press & Assessment 2026 Page 18 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c)(ii) 2 A M2 if structure (partly / fully) displayed / all the\ncarbons shown with ‘–’ continuation bond.\nIf structure fully skeletal: continuation bonds –\nmust be dashed / wavy / different or have\nthrough brackets.\nPhenol must be 1,2-substituted and drawn.\nM1 ester linkage shown between phenol & COOH [1]\nM2 rest of the structure correct [1] DEP on M1\n\n6(d) 2 A M1 for LiAlH reduction [conjugated\n4\nunsaturated carboxylic acid].\nM1M2\n© Cambridge University Press & Assessment 2026 Page 19 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(e)(i) D E 2 D and E are unconditional.\nOH O\nOH OH\nCH\n3\nO\nOR O CH 3\nCH\nO 3\nO CH 3\nM1M2\n\n6(e)(ii) Mark based on 6(e)(i) 3\nA CH Br / CH I for alkylation.\n3 3\nstep 1 CH Cl (and AlCl ) [1] CON (aq), I any halogen carrier for alkylation.\n3 3\nCON other reagents\nI heat / any temp for alkylation & acylation.\nstep 2 hot / reflux / heat / boil / T ⩾ 60 °C\nand (alkaline / acidified / neutral) MnO – / KMnO [1] CON use of KMNO for oxidation.\n4 4 4\nCON (aq) once only then apply ecf.\nstep 3 CH COCl [1] CON (aq) CON UV for acylation.\n3\nCON any other reagent e.g. cH SO\n2 4\n© Cambridge University Press & Assessment 2026 Page 20 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 18, 19, 20 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p18.png", "../answer-assets/9701_s26_ms_42-p19.png", "../answer-assets/9701_s26_ms_42-p20.png" ] }, { "id": "9701-2026-mj-42-q07", "question_id": "9701-2026-mj-42-q07", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 11, "status": "available", "reason": null, "text": "7(a)(i) (a substance able to) rotate plane polarised light 1 A (a substance able to) rotate plane of (plane)\npolarised light.\nI rotating an amino acid instead of plane of\npolarised light.\n\n7(a)(ii) (a mixture containing) equal amounts / proportions / concentrations 1\nand\nof optical isomers / enantiomers / stereoisomers\n\n7(b)(i) pH at (which a molecule / amino acid) has 1\nand I ‘pH when the molecule has no charge’ alone.\nno overall charge / is neutral / is a zwitterion [1]\n\n7(b)(ii) CH CH(NH )COO– 1\n3 2\n\n7(c) 2 A ecf for 1 mark here for M2 – repeat error.\nAlso ecf M2 if H omitted on both N – repeat\nM1 two peptide linkage shown\nerror.\n(C=O needs to be displayed but allow NH not displayed)\nCON M1 -CO-HN -.\nM2 rest of the structure correct\nDEP on M1 (except for specific ecfs in guidance)\nA M1 if more than 2 amide links but M2 not\nA carboxylate ion / zwitterion / NH + ion\n3 available.\n© Cambridge University Press & Assessment 2026 Page 21 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(d)(i) 2\nM1M2\n\n7(d)(ii) M1 step 1 CH COCl CON (aq) 3 ecf penalise (aq) once only in step 1,2 and 3.\n3\nCON additional reagent e.g. AlCl , cH SO\n3 2 4\nA M1 CH COOCOCH for ‘CH COCl’.\n3 3 3\nM2 step 2 PCl + heat or SOCl or PCl CON (aq)\n3 2 5\nAll u / c marking points.\nM3 step 3 C H and Al Cl CON (aq)\n6 6 3\nA any viable halogen carrier for acylation\n(e.g. FeBr ).\n3\n© Cambridge University Press & Assessment 2026 Page 22 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 21, 22 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p21.png", "../answer-assets/9701_s26_ms_42-p22.png" ] }, { "id": "9701-2026-mj-42-q08", "question_id": "9701-2026-mj-42-q08", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 8, "status": "available", "reason": null, "text": "8(a) tetramethylsilane 1 Only answer – correct spelling required.\nI TMS, (CH ) Si .\n3 4\n\n8(b)(i) 5 / five 1\n\n8(b)(ii) 2 A douplet.\nproton environment name of splitting pattern\nI duplet, double, multiple, triple.\nw doublet\nA heptet / septet for multiplet.\nx multiplet\ny triplet\nz triplet\n● ✓ ● ✓\n\n8(b)(iii)   5 –12 [1] 2 M1 requires this range instead a single value.\nM2 (N-H peak appears and) A M2 (N-H) proton / hydrogen exchanges with\nno proton / hydrogen exchanges / replaces / substitutes deuterium / D / D+ in D O.\n2\nwith deuterium / D / D+ ora I proton / hydrogen is taken away / removed by\nthe deuterium.\n\n8(c) 2 I carboxylic acid (products isolated at pH 10).\nA carboxylate ion.\nA -ONa.\nCON –O–Na (covalent bond).\nM1M2\n© Cambridge University Press & Assessment 2026 Page 23 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 23 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p23.png" ] }, { "id": "9701-2026-mj-42-q09", "question_id": "9701-2026-mj-42-q09", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "9(a) M1 Q ammonia P 4\nmost basic least basic\nexplanation\nM2 (basicity linked to) ability of lone pair / p-orbital to I ‘attract’ for accept.\naccept / donate to / coordinate\nwith a proton / H+\nA explanations in terms of hyperconjugation of\nExplanation M3 M4 alkyl group.\n“alkyl group can donate electron density into\n• lone pair / p-orbital from N (in P) empty (p) orbital on positively-charged carbon”.\nand delocalised / overlaps with (-)ring / benzene\nI incorporated / donated.\n• electron donating / +I\nand R / alkyl / CH grp (in Q) LP with N can be qualified anywhere in their\n2\nanswer.\n• increases electron density on N or N is more electron rich\nI ‘charge density’ for electron density.\nor N LP is more available (in Q) ora ● ✓✓\n\n9(b)(i) M1 step 1 HNO and H SO and conc. / concentrated 2 CON omission of conc. once.\n3 2 4\nM2 step 2 Sn and conc. HCl I temp.\nbut CON 10 °C or lower –CON once.\nCON (aq) in step 1 only.\nCON Sn catalyst in step 2.\nI NaOH with step 2.\n\n9(b)(ii) 1 CON 6 [H+]\n© Cambridge University Press & Assessment 2026 Page 24 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n9(c)(i) or 1 A –N Cl or –N + form.\n2 2\nA –N≡NCl or –N+≡N form.\nA –N=NCl or –N=N+ form.\nif a Cl present - must be electrically neutral.\nPosition of + charge must be correct if seen.\nCON covalent N–Cl bond seen.\n\n9(c)(ii) M1 step 1 NaNO with HCl or HNO 2 CON heat at 10 oC.\n2 2\nand ⩽10 °C [1] I NaNO .\n3\nA ‘HNO / H SO ’ for HCl\n3 2 4\nM2 step 2 alkaline / NaOH A M2 respective phenoxide ion\nand for both reagents in step 2.\nI ‘base’ alone.\nA sulfonic acid phenol as an alternative with\nNaOH.\nA structures with SO – (no Na+).\n3\nI temp for M2.\n© Cambridge University Press & Assessment 2026 Page 25 of 25\nTrace ID: PC-9D76E91B48\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 24, 25 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-42.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_42-p24.png", "../answer-assets/9701_s26_ms_42-p25.png" ] }, { "id": "9701-2026-mj-43-q01", "question_id": "9701-2026-mj-43-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 14, "status": "available", "reason": null, "text": "1(a) Ca2+(g) + S2–(g) → CaS(s) 2 If reverse lattice energy equation is given\nM1 all species CaS(s) → Ca2+(g) + S2–(g) award [1] if all\nM2 all state symbols species and state symbols are correct.\n\n1(b) (ionic) bonds are formed / the ions bond 1 A attractive force  distance.\nA electrostatic attraction  distance.\nI electrostatic attraction alone.\nI exothermic / energy released.\nI ions come together.\nI formation of the lattice / solid / CaS.\nI reference to energy required to break\nbonds.\nCON energy required to form bonds.\nCON bond formed is clearly covalent, e.g.\nCa–S.\n\n1(c) M1 MgO CaS KCl 3\nA for all four charges\nM2 CaS (and MgO) the ions / particles are more highly charged than KCl I nuclear charge\nor KCl has ions with the smallest charge ora I charge density for M2 and M3\nM3 MgO the ions/particles are smaller than CaS\nor MgO has the smallest ions ora\n\n1(d)(i) (enthalpy of) atomisation (of Cl) or bond energy (Cl–Cl ) and 1 If this is given in symbols we will accept\nH , H or H . If symbols and\nat atom atomisation\nwords are given, I the symbols.\n(first) electron affinity (of chlorine) If this is given in symbols we will accept\nH , H , H or\nEA EA1 electron affinity\nH . If symbols and words are\nfirst electron affinity\ngiven, I the symbols.\nCON electron affinity of chloride / Cl – / Cl\n2\n© Cambridge University Press & Assessment 2026 Page 7 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(d)(ii) M1 chlorine value doubled / –486 seen / 2x – 243 seen 2 ecf use of 1x – 243 for ans = –400\nM2 answer –643 (working needed).\n2336 +(2x – 243) –2493 = –643 –643 scores [2]\n\n1(e) M1 Correct G = H – TS and use of T = 873 3 CON G for G. A  .\nGibbs\nM2 G = –36 / –35.8 / –35.78 / –35.775 ecf T only If T = 600 then [1] ecf for +12.\nCON G for G but do not penalise twice\nM3 it is (feasible) and If in words we need ‘change’.\nG is negative / G is less than zero UNC\n(Note: G now in bold) I G is exothermic.\nM3 is UNConditional on the basis it is a correct answer on the answer line. If A G in words ‘Gibbs Free Energy\nthe calculation has not been worked through to an answer, or has given a change’\nnegative value for G, there is no M3 for ‘not feasible because G is positive’.\n\n1(f) • higher 2 ⚫✓✓ any two for one mark.\nAll three for two marks.\n• Ca2+ larger (than Mg2+) ora\nor ionic radius / size I atomic\n• (In CaCO ) less distortion / polarisation of anion / CO y– ora or group 2 trend but up / down the group\n3 x\nmust be clear.\nall standalone I references to ionic bond strength.\nora for MgCO 3 but must be stated or Ca2+ less polarising on anion / CO x y–\nI ‘depolarisation’.\nor C-O / C=O bond (must be identified,\ne.g. C–O) is less weakened by Ca2+.\n© Cambridge University Press & Assessment 2026 Page 8 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p07.png", "../answer-assets/9701_s26_ms_43-p08.png" ] }, { "id": "9701-2026-mj-43-q02", "question_id": "9701-2026-mj-43-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 11, "status": "available", "reason": null, "text": "2(a)(i) –log[H+] 1 [ ] essential.\nCON –log(H+), –log [H+] , –ln[H+].\ne\n\n2(a)(ii) [H+][OH–] / K xK 1 [ ] essential but do not penalise in (i) and\na b\n(ii).\nI 1  10–14.\n\n2(b)(i) (10–2.9) 1\n0.0013/0.00126/1.3x10–3/1.26x10–3\n\n2(b)(ii) 2bi2 / 0.105 0.001262 / 0.105 ecf bi 1\n1.5  10–5 / 1.51  10–5 / 1.53  10–5 ecf\n\n2(c)(i) M1 [CH CH CH COOH] = 0.15 3 If [CH CH CH COOH] = 0.45 [0] but then\n3 2 2 3 2 2\nM2 [CH CH CH COO–Na+] in solution D = 0.036 / 0.0359 [salt] in mixture = 0.108 [1] ecf\n3 2 2\n[salt] in solution C = 0.162 [1] ecf VERIFY\nM3 [CH CH CH COO–Na+] in solution C = 0.054 / 0.0538 / 0.0539 / 0.0540 with working.\n3 2 2\nAnother possible approach goes via moles of acid to get moles of salt,\nM1 moles acid = 0.045 mol\nM2 moles salt = 0.0108 / 0.0107 mol\nM3 original [salt] = 0.0538 / 0.0539 / 0.0540\n\n2(c)(ii) CH CH CH COO–Na+ + H+ → CH CH CH COOH + Na+ 1\n3 2 2 3 2 2\nor CH CH CH COO– + H+ → CH CH CH COOH\n3 2 2 3 2 2\nor CH CH CH COO– + H SO → CH CH CH COOH + HSO –\n3 2 2 2 4 3 2 2 4\nor 2CH CH CH COO–Na+ + H SO → 2CH CH CH COOH + Na SO\n3 2 2 2 4 3 2 2 2 4\n\n2(c)(iii) CH CH CH COOH + OH– → CH CH CH COO– + H O 1\n3 2 2 3 2 2 2\nor CH CH CH COOH + KOH → CH CH CH COOK + H O\n3 2 2 3 2 2 2\nor CH CH CH COOH + K+OH– → CH CH CH COO–K+ + H O\n3 2 2 3 2 2 2\n© Cambridge University Press & Assessment 2026 Page 9 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(d) M1 2,2-dichlorobutanoic and 2\nchlorine is electron withdrawing / negative inductive effect / electronegative\nI references to C=O, O, alkyl groups.\nM2 weakens O–H bond / stabilises anion / stabilises conjugate base / stabilises A –O–H bond and OH bond but CON –\ncarboxylate ion OH bond, C–O–H bond.\nQuestion Answer Marks Guidance", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p09.png", "../answer-assets/9701_s26_ms_43-p10.png" ] }, { "id": "9701-2026-mj-43-q03", "question_id": "9701-2026-mj-43-q03", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 9, "status": "available", "reason": null, "text": "3(a) 2 1\n1\n3\n\n3(b) M1 1.4x10–6 2\nM2 mol–2dm6s–1 I mol–2dm6s– .\n\n3(c)(i) (overall) first order and 1 I ‘[H ] first order’.\n2\n[NO] is (almost) constant I [NO] zero order.\nI [NO] not a limiting factor.\nI [H ] is a limiting factor.\n2\nA [NO] pseudo zero order.\n\n3(c)(ii) M1 33 000 / 32 500 / 32 542 / 32 540 / 32 535 / 3.25  104, etc.\nM2 130 000 / 130 141 / 1.3  105 / 1.32  105, etc. ecf\n© Cambridge University Press & Assessment 2026 Page 10 of 21\nt\n12\n2 32 542 if ln2 used.\nanswer = t\n12\n 4\n\n3(d)(i) M1 one N O and one H on L or one N O and one H O on R 2 (M1 is for either correct LHS or RHS).\n2 2 2 2 2\nM2 N O + H → N O + H O\n2 2 2 2 2\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(d)(ii) no H involved in step one (and H in rate equation) or 1 I ‘does not agree with rate equation’.\n2 2\nonly one species from rate equation or I reference to NO without reference to\ndoes not contain species in the same molar proportion as rate equation or order of reaction.\nif step 1 is RDS then order wrt [H ] = 0 or I N O not in the rate equation.\n2 2 2\nstep 1 does not include all species in rate equation or\nrate would be second order wrt NO only or\nreaction would be overall second order\nQuestion Answer Marks Guidance", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p10.png", "../answer-assets/9701_s26_ms_43-p11.png" ] }, { "id": "9701-2026-mj-43-q04", "question_id": "9701-2026-mj-43-q04", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 10, "status": "available", "reason": null, "text": "4(a) Go = –nF Eo 1 A Go = – nF E .\ncell cell\n\n4(b) Cu2+ + 2V2+ → Cu + 2V3+ 2 CON cancellable species / electrons\nIf ⇌ read forwards\n(+) 0.60 I 0.6\n+ not needed but – is CON.\n\n4(c) M1 Cu 2 or in words if unambiguous.\nM2 V2+\n\n4(d) M1 Cl / chlorine 2 CON if both oxidised and reduced form\n2\ngiven.\nM2 V / vanadium CON if both oxidised and reduced form\ngiven.\n\n4(e) All candidates get three marks 3\n© Cambridge University Press & Assessment 2026 Page 11 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p11.png" ] }, { "id": "9701-2026-mj-43-q05", "question_id": "9701-2026-mj-43-q05", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "5(a) similarity in energy of (3)d and (4)s (subshells / orbitals) 1 this is a syllabus statement – do not allow\nA same much variation of meaning, I statements\nthat can be understood to refer to 3d only.\n\n5(b) 3d7 4s2 / 4s2 3d7 1 A full configurations with no errors.\n3d7 / 3d7 4s0\n\n5(c) M1 the d orbitals 3\nare split / become non-degenerate / reference to E within d-subshell\nor E between d orbitals\nM2 electrons are promoted / excited / jump to higher level owtte\nM3 absorbing (energy of) light / photon / frequency / wavelength / h and ‘complementary’ may be described but\ncomplementary colour is seen / reflected / transmitted ‘different colour’ is not enough.\nM3 CON emits / energy released.\n\n5(d) M1 blue to red / blue to pink 2 A named shades of blue and pink.\nM2 CoCl + 6H O → [Co(H O) ]2+ + 2Cl – or\n2 2 2 6\nCoCl 2– + 6H O → [Co(H O) ]2+ + 4Cl – I CoCl + 6H O → CoCl 6H O .\n4 2 2 6 2 2 2 2\nI Co2+ + 6H O → [Co(H O) ]2+.\n2 2 6\n\n5(e)(i) (white / brown / off white) precipitate / ppt / solid and 1 CON clearly wrong colours, e.g.\nbrown (solution) black / blue / green / purple\nor wrong observations, e.g. fizzing.\n\n5(e)(ii) starch and 1\nblue / black to colourless / white / off-white colourless / white to blue / black is CON.\n\n5(e)(iii) redox 1 A oxidation-reduction (both needed).\n© Cambridge University Press & Assessment 2026 Page 12 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(e)(iv) • moles of S O 2– = 0.2  37.6 / 1000 = 7.52  10–3 3 ⚫✓✓✓ two points for one mark\n2 3\n• moles of I = 3.76  10–3 ecf\n2\n• moles of Cu = 7.52  10–3 ecf\ncorrect answer scores [3].\n• mass of Cu = 0.48 / 0.478 / 0.4775 / 0.47752 g ecf\n0.239 g, 0.955 g or 1.41 g on the answer line, without any correct,\nunderstandable working, scores [0]\nQuestion Answer Marks Guidance", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p12.png", "../answer-assets/9701_s26_ms_43-p13.png" ] }, { "id": "9701-2026-mj-43-q06", "question_id": "9701-2026-mj-43-q06", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 8, "status": "available", "reason": null, "text": "6(a) M1 two lone pairs / LP 2\nM2 donated / form dative bond(s) / coordinate bond(s) CON ‘2 or more dative bonds’.\nto (central) metal / TM / TE\natom / ion\n\n6(b)(i) [Fe(C\n2\nO\n4\n)\n3\n]3– or Fe(C\n2\nO\n4\n)\n3\n3– 1\n\n6(b)(ii) M1 One correct isomer of [Fe(C 2 O 4 ) 3 ]3– drawn with 3D and oct 2 Correct 3D. octahedral, different\nThese two renditions of octahedral are\nM2 Two optical isomers drawn with 3D and octahedral only acceptable.\nI charges.\n\n6(b)(iii) optical only 1\n© Cambridge University Press & Assessment 2026 Page 13 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c)\n© Cambridge University Press & Assessment 2026 Page 14 of 21\n\n\nH\n H\ng\ng\n2 +\nI\n4\n\n2 \n I\n\n−\n−\n\n\n4\n2 A [HgI 2–]. 4\nM1 A [Hg2+] with complexing H 2 O.\nThe charges shown here are essential.\nFinal set of brackets must be square 3.\nPenalise two or more round outer\nbrackets.\nM2 mol–4 dm12 unc ecf only after K with 4th power.\nstab\nQuestion Answer Marks Guidance", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p13.png", "../answer-assets/9701_s26_ms_43-p14.png" ] }, { "id": "9701-2026-mj-43-q07", "question_id": "9701-2026-mj-43-q07", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Hydroxy compounds", "topic_slug": "9701-topic-32-hydroxy-compounds", "marks": 14, "status": "available", "reason": null, "text": "7(a)(i) HNO and H SO and conc once and 25° ⩽ T ⩽ 60° 1 Acids may be named.\n3 2 4\nAny dilute or aq is CON.\nTemperature must be numerical and\nwithin range.\nI heat or warm.\nCON boil or reflux.\n\n7(a)(ii) 0° ⩽ T ⩽ 10° and 1 A ‘T below 10o’ etc.\nCON heat or warm.\nHNO or NaNO + HCl I sodium nitrate.\n2 2\nA nitrous acid.\n\n7(a)(iii) benzenediazonium chloride / phenyl diazonium chloride 1 A very minor errors, e.g. word order.\nCON benzyl’benzoyl.\nI diazo.\n\n7(b)(i) 0 1 2 1 A blank for zero.\n0 6 3\n\n7(b)(ii) 6 1\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(iii) 1\nI + on the H+.\nCurly arrow must start on C=O.\nAny incorrect curly arrow is CON.\nA in two steps, e.g.\ndipole\nwith dipole.\nCurly arrow from C=O to H+\nAdditional charges are CON.\nor\n© Cambridge University Press & Assessment 2026 Page 15 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(iv) 3 ‘Usual rules’ for K.\nCentre of + charge must be above a line\njoining C3 and C5.\n‘Horseshoe’ must include C6, C1 and C2,\nmay include one or both of C3 and C5,\nmust not include C4.\nI curly arrow showing delocalisation of\nphenolic LP (LP on oxygen of phenol).\nCON additional curly arrows\nM1 CA on leftmost diagram Curly arrow must start on or within\nhexagon and may cross hexagon\nM2 K, showing horseshoe, + charge and H anywhere.\nM3 curly arrow on C–H bond in K back into ring\nCurly arrow must touch hexagon.\nUse of quinoid intermediate:\nM1 as above – curly arrow on leftmost diagram\nM2 for the intermediate shown at the bottom of guidance. The curly arrow for\nM3 is shown on this diagram but is not necessary for M2.\nM3 as above – curly arrow on C–H bond in K back into ring\n© Cambridge University Press & Assessment 2026 Page 16 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(v) • less readily 2 ⚫✓⚫✓ two points for one mark,\n• lone pair (LP) on O (of phenol) / p-orbital on O four points for two marks.\n• is delocalised into the ring / overlaps with delocalised ring / -system A LP on OH I LP in phenol.\n• activates ring / more attractive to electrophile / electron (charge) density ‘Delocalised’ or ‘joins  system’ needed\nincreased (‘electron’ inferred if ‘LP donated to ring’) / becomes more may be awarded without LP if linked to O.\nelectron rich\nI references to polarisation of electrophile\n(it has a full + charge, stated in the Q)\nI makes ring more reactive / more\nsusceptible to electrophiles / more\nnegative.\nI ‘O is electron donating’.\nI ‘attracts nucleophile more’.\n\n7(c)(i) 2 A OH of tertiary alcohol group substituted\nto bromide as well as ring substitutions\nand equation with HBr and H O on the\n2\nright.\n+ 2Br → + 2HBr\n2\nM1 structure of product (compound H)\nM2 equation balanced, must have HBr\n→ product above + H O + HBr\n2\n\n7(c)(ii) electrophilic substitution 1 CON nucleophilic substitution unless –OH\non sidechain substituted with Br when it\nscroll and SEEN becomes I.\n© Cambridge University Press & Assessment 2026 Page 17 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 14, 15, 16, 17 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-32-hydroxy-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p14.png", "../answer-assets/9701_s26_ms_43-p15.png", "../answer-assets/9701_s26_ms_43-p16.png", "../answer-assets/9701_s26_ms_43-p17.png" ] }, { "id": "9701-2026-mj-43-q08", "question_id": "9701-2026-mj-43-q08", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 10, "status": "available", "reason": null, "text": "8(a)(i) CH COOH + SOCl → CH COCl + SO + HCl 1 CON aq.\n3 2 3 2\n\n8(a)(ii) PCl or PCl 1 A names, phosphorus(III) chloride,\n3 5\nphosphorus trichloride, phosphorus(V)\nchloride, phosphorus pentachloride.\nCON aq, but not in both 8(a)(i) and\n8(a)(ii).\n\n8(a)(iii) 1 Or any unambiguous structural or skeletal\nformula.\nor CH CONHCH(CH )\n3 3 2\n\n8(a)(iv) 1 No ecf from 8(a)(iii) – Q specifies L is an\namide and gives the structure of M.\nor CH CHNH CH\n3 2 3\nor skeletal\n\n8(a)(v) (nucleophilic) addition-elimination 1 No ecf from 8(a)(iii) – Q specifies L is an\namide.\nI ‘condensation’ – correct but not a\nmechanism.\n\n8(a)(vi) LiAl H 1 No ecf from 8(a)(iii) – Q specifies L is an\n4\namide.\nCON aq.\n\n8(b)(i) • r – triplet 2 ⚫✓⚫✓ two points for one mark,\n• s – quartet/quadruplet four points for two marks.\n• t – doublet\n• u – multiplet A septet, heptet, septuplet.\n© Cambridge University Press & Assessment 2026 Page 18 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(b)(ii) M1 one more peak and reference to N-H 2 A ‘NH peak is present’.\nA ‘NH peak is not present in D O’.\n2\nSubject of Q is ‘in CDCl ’ so assume this if no solvent specified. A reference to five peaks or five proton\n3\nSee first two ALLOWs in guidance. environments and reference to N-H.\nA ‘it is exchanged with D’ if NH peak\nM2 H (of NH) is exchanged with D (in D O) / NH + D → ND + H / already referred to and you are confident\n2\nH (of NH) is not exchanged with D (in CDCl ) they refer to ‘in D O’.\n3 2\nA ‘it is not exchanged with D’ if NH peak\nthe bits in brackets for M2 are not essential but if clearly wrong it is CON already referred to and you are confident\nthey refer to ‘in CDCl ’.\n3\nI ‘D O replaces the H’.\n2\n© Cambridge University Press & Assessment 2026 Page 19 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 18, 19 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p18.png", "../answer-assets/9701_s26_ms_43-p19.png" ] }, { "id": "9701-2026-mj-43-q09", "question_id": "9701-2026-mj-43-q09", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 11, "status": "available", "reason": null, "text": "9(a) M1 One pair of diagrams of P with correct use of 3D showing mirror images of 3\ntwo optical isomers Q may be given above, and P may be\ngiven below.\nMirror plane not essential.\nM2 One pair of diagrams showing correct E and Z of Q\nM3 conditional on optical linked to an\nattempt on P and geometrical / cistrans\nM3 Optical or enantiomerism in the right place and / EZ linked to an attempt on Q in M1 and\nGeometric(al) or E / Z or A cis/trans in the right place M2.\n\n9(b)(i) the pH at which (the amino acid is) 1 A clear and correct description of\na zwitterion / no net charge / no overall charge / no total charge zwitterion for second component.\nelectrically neutral / charge is neutral\n© Cambridge University Press & Assessment 2026 Page 20 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n9(b)(ii) 3\npH 4.0 pH 5.5 pH 8.0\n⚫✓⚫✓ ⚫✓\nT\ntwo structures for one mark,\nfour structures for two marks,\nsix structures for three mark.\nPhe at 5.5 can be\nU\nTable left is repeated larger after 9(c)(iii).\n\n9(c)(i) 2 Fully correct [2].\nM1 – amide links correct\nM2 – remainder is fully correct, e.g.\nterminal groups not trailing bonds\nneutral, zwitterion, 1+ or 1–\nside/R groups CH – H – CH\n3 3\nall H atoms on -carbons\n\n9(c)(ii) hydrolysis 1 I acid / alkaline / enzymatic / NS.\n\n9(c)(iii) condensation 1 I addition-elimination.\n© Cambridge University Press & Assessment 2026 Page 21 of 21\nTrace ID: PC-F2CDA313A1\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 20, 21 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-43.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_43-p20.png", "../answer-assets/9701_s26_ms_43-p21.png" ] }, { "id": "9701-2026-mj-44-q01", "question_id": "9701-2026-mj-44-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 1, "topic": "Electrochemistry", "topic_slug": "9701-topic-24-electrochemistry", "marks": 7, "status": "available", "reason": null, "text": "1(a) M1 Oxygen / O 2 I ‘O’ alone.\n2\nM2 4OH– → 2H O + O + 4e– I state symbols.\n2 2\nA multiples, use of ‘⇌’.\nA 4OH– – 4e– → 2H O + O .\n2 2\n\n1(b)(i) M1 2.4  12  60 or 1728 (C) 2 A 0.00896 or 0.008955.\nM2 1728 / (2  96 500) = 0.00895 / 0.0090 mol min 2sf 1.4945  10–4 (no  60) [1].\n0.0179 [1] (no ÷ 2).\nor Correct answer [2].\n1728 0.009 [1] (1sf).\n=0.00897 min 2sf ecf\n2\n( 1.610−196.0221023)\n\n1(b)(ii)\n© Cambridge University Press & Assessment 2026 Page 7 of 21\nV =\nn R\np\nT\n→\n1 ( b ) ( i)\n1\n\n0\n8\n1\n.\n0\n3\n0\n1\n0\n 6 5 3\n= 4 . 8 1  1 0 − 4 m 3\n1 ECF from 1(b)(i),\ne.g. if mol in 1(b)(i) = 0.0179, volume =\n0.962.\n4.81×10–4 m3 × 1000 = 0.481 (dm3)\n0.482 if 0.00897 mol.\nanswer must be 3sf\n0.484 if 0.0090 mol.\n0.962 if 0.0179 mol.\nA use of their unrounded previous value,\ne.g. 0.962 if 0.018 mol (i.e. they are using\n0.0179).\ncorrect answer [1].\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(c) Full expression:\nM1\n© Cambridge University Press & Assessment 2026 Page 8 of 21\n1 . 6\n3 6\n\n0\n1\n0\n0\n0\n1 −\n\n9 \n2 3\n8 . 5 1\n= 6 . 0 8  1 0 2 3 ( m o −l 1 )\n2 I 6  1023, 6.1  1023 as not 3 sf\n6.08  1023 scores [2].\nA – ve sign if seen in working, but\n36 000 ÷ 1.6  10–19 or 2.25  1023 electrons / atoms Na –6.08  1023 scores max [1].\nM2 ans ÷ (8.51 / 23) = 6.08  1023 must be 3sf\nECF from use of incorrect A for M2.\nr\nM for NaCl used instead of Na gives\nor (using F = –Le) r\nL = 1.55  1024 [1]\nM1 8.51 ÷ 23.0 = 0.370 moles of electrons / atoms Na and use of atomic number instead of A r gives\n36 000 ÷ 0.370 or 97 297 (value for F) L = 2.91  1023 [1]\nM2 ans ÷ 1.6  10–19 = 6.08  1023 must be 3sf  rather than ÷ A r gives 8.33  1022 [1].\nQuestion Answer Marks Guidance", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-24-electrochemistry/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p07.png", "../answer-assets/9701_s26_ms_44-p08.png" ] }, { "id": "9701-2026-mj-44-q02", "question_id": "9701-2026-mj-44-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 2, "topic": "Equilibria", "topic_slug": "9701-topic-25-equilibria", "marks": 16, "status": "available", "reason": null, "text": "2(a) M1 (strong acid) fully ionised / dissociated and 2 I fully / partially dissolves.\n(weak acid) partially ionised / dissociated I ‘not fully’ for ‘partially’\ndissociated / ionised\nM2 to give / donate H+ / proton or is a proton donor (mentioned once) I ‘strongly dissociates’, ‘dissociates easily’.\n\n2(b) pair 1: HOOCCOOH and HOOCCOO– and 1 CON Na+ salts\npair 2: HOOCCOO– and –OOCCOO– (question asks for organic pairs) .\npairs can be in either order\nA other representations if correct,\ne.g. (COOH) , (COO–) , (OOCCOO)2–,\n2 2\nC O 2– but not OOCCOO2–.\n2 4\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(c)(i) M1 Two reactants that will work (ignore stoichiometry) 2 I Ca2+ as a reactant.\nM2 Correct balanced equation M2 dependent on M1.\ne.g: I state symbols.\nCa + 2CH COOH → Ca(CH COO) + H or\n3 3 2 2\nCaO + 2CH COOH → Ca(CH COO) + H O or pass other correct examples to TL.\n3 3 2 2\nCa(OH) + 2CH COOH → Ca(CH COO) + 2H O or\n2 3 3 2 2\nI where a chemical reaction is not\nCaCO + 2CH COOH → Ca(CH COO) + CO + H O or\n3 3 3 2 2 2\nhappening, e.g.\nCa(HCO ) + 2CH COOH → Ca(CH COO) + 2CO + 2H O\n3 2 3 3 2 2 2\nCaCl + 2CH COOH → Ca(CH COO) +\n2 3 3 2\n2HCl.\n\n2(c)(ii) (a solution that) resists / opposes / minimises changes in pH 1 A keeps pH almost constant.\nwhen small amounts / volumes A pH does not change much.\nof acid / H+ or base / alkali / OH– are added to it A ‘does not change significantly’ for\nresists.\nA ‘controls pH’ as it is in 2(c)(iii)\nI ‘minimises pH’ alone.\nCON ‘kept constant’ / ‘maintains pH’ but\nA maintains pH in a small range.\nI moderate for ‘small’.\n\n2(c)(iii) M1 CH COOH + OH– → CH COO– + H O 2 If full salt formula used, must be correct.\n3 3 2\nA use of ‘⇌’ but read in forward direction.\nM2 CH 3 COO– + H+ → CH 3 COOH or A use of HCl, NaOH, etc.\nCa(CH 3 COO) 2 + 2H+ → 2CH 3 COOH + Ca2+ I CH 3 COOH → CH 3 COO– + H+ and/or H+\n+ OH– → H O.\n2\nIf more than two eqns given, mark\nanswers on answer lines.\n© Cambridge University Press & Assessment 2026 Page 9 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(d) M1 In NaOH : [H+] = 10–11.34 or 4.57  10–12 3 11.6 scores [3].\n11.04 [2] with working and one mistake\nM2 In NaOH: [OH–] = 10–14 ÷ 4.57  10–12 or 2.1910–3 seen\nIn Ba(OH) : [OH–] = 2.19  10–3 × 2 or 4.376  10–3 ecf throughout\n2\nA use of pOH:\nM3 In Ba(OH) : [H+] =10–14 ÷ 4.376  10–3 or 2.29  10–12 pOH = 14 – 11.34 or 2.66. [1]\n2\npH = 11.6(41) min 1dp\n[OH-] = 10-2.66 or 2.188  10-3\nAlternative method: [OH-] in Ba(OH) 2 = 2  2.188  10-3\nM1 [H+] = 10–11.34 or 4.57  10–12 or 4.376  10-3 [1]\nM2 4.57  10–12 ÷ 2 = 2.29  10–12\nM3 pH = 11.6(41) (No mention of [OH–]) new pOH = -log(4.376  10-3) or 2.36\nnew pH = 14 – 2.36 = 11.64 [1]\n\n2(e)(i) (the) ratio of the concentrations of a solute / substance / compound 1 A eqm for equilibrium.\nbetween two (immiscible) solvents / liquids (at) equilibrium I equations.\nA ‘Ratio’ may be given using mathematical\nexpression.\nthe equilibrium constant that relates\n‘Solvents / liquids’ may be clearly implied.\nthe concentration of a solute / substance / compound (partitioned)\nI ‘solubility’, ‘species’.\nbetween two (immiscible) solvents / liquids\nI ‘solutions’ for solvent.\n\n2(e)(ii) (0.68 / 5.32 =) 0.13 / 0.128 / 0.1278 min 2sf 1 Only answer\n0.1278195489.\n\n2(e)(iii) • less than / lower (ecf if K from 2(e)(ii) is the wrong way up) 3 (inverse K = 7.82)\npc pc\n• solute / A must be polar / form H-bonds / ionic (as more soluble in water) Bullets 2, 3 and 4 are unconditional.\n• cyclohexane is non-polar / less polar than butan-1-ol or No ecf, marks are unconditional\ncyclohexane does not form H-bonds throughout.\n• A is less soluble in cyclohexane (than in butan-1-ol) or\nless (A) will dissolve in cyclohexane or\nmore (A) will dissolve in water (than cyclohexane)\n✓✓✓\n⚫\n© Cambridge University Press & Assessment 2026 Page 10 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-25-equilibria/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p08.png", "../answer-assets/9701_s26_ms_44-p09.png", "../answer-assets/9701_s26_ms_44-p10.png" ] }, { "id": "9701-2026-mj-44-q03", "question_id": "9701-2026-mj-44-q03", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 3, "topic": "Reaction kinetics", "topic_slug": "9701-topic-26-reaction-kinetics", "marks": 8, "status": "available", "reason": null, "text": "3(a) 1 / first 1 Only answer.\n1 / first\n2 / second\n\n3(b)(i) M1 … 0.120 … unc 2 No efc from 3(a).\nM2 … … 6.90  10–6 unc\nmin 2sf\n\n3(b)(ii) M1 0.0046 / 4.6  10–3 min 2sf unc 2 No ecf.\nM2 mol–1 dm3 s–1 unc Any order, but no work left for examiner to\ndo so I s–1 / mol1 dm–3 or mol–1 dm3 / s CON\nmol– or s–\n\n3(c) 69 / 68.6 / 68.61 / 68.63 min 2sf 1 68.63 if ln2 used rather than 0.693.\nA any answer between 68.61–68.63.\n\n3(d) M1 e.g. NO + F → NO F or NO + F → NO F + F 2 Equations must be balanced for charge as\n2 2 2 2 2 2 2\nNO + NO F → 2NO F F + NO → NO F well as species for M2.\n2 2 2 2 2 2\nstep 1 step 1 I radical dots.\nI equations with cancellable species.\nfirst step has one NO and one F only as reactants (ignore products) and step\n2 2\n1 / their equation for step 1 is slowest\nM2 two steps add up to equation\n© Cambridge University Press & Assessment 2026 Page 11 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-26-reaction-kinetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p11.png" ] }, { "id": "9701-2026-mj-44-q04", "question_id": "9701-2026-mj-44-q04", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 4, "topic": "Chemical energetics", "topic_slug": "9701-topic-23-chemical-energetics", "marks": 11, "status": "available", "reason": null, "text": "4(a)(i) one mole of gaseous atoms and 1 A one e– is added to each atom in 1 mole\ngains one mole of electrons / forms one mole of gaseous 1– ions of gaseous atoms.\nCON energy required/needed.\nI equations.\nCON incorrect near misses, e.g. ‘1 mol of\ne– is added to each atom…’ or ‘1 e– is\nadded to 1 mol of gaseous atom…’.\n\n4(a)(ii) (first electron affinity is) 2\n• less exothermic / less –ve down group OWTTE A decreases / less exothermic.\nI less energy needed.\n• (atomic) radius / size increasing or more shells or I charge density.\nmore shielding CON ionic radius for second bullet.\n• less attraction of nucleus for\nvalence / incoming / outer shell electron A nuclear charge / protons for nucleus.\n✓✓\n⚫\n\n4(b) M1 any number more negative than –670 for NaCl and 4 e.g. –700 for NaCl and –750 for MgO.\neven more negative than that for MgO\nA NaCl has smaller ionic radius for M2.\nM2 explanation for NaCl: correct reference to I charge density unless incorrect.\nsmaller ions / cation and anion I higher nuclear charge.\nM3 explanation for MgO: I atomic radius throughout.\ncorrect reference to higher charge(s) of Mg(2+) and / or O(2–) or\ncorrect reference to smaller radius / radii of cation and / or anion A comparison with NaCl or KBr for MgO.\nCON similar radii for M3.\nM4 ONE correct reference to no ecf.\nstronger forces / attraction between ions or\nstronger ionic bonds\n© Cambridge University Press & Assessment 2026 Page 12 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(c)(i) number of (possible) arrangements of particles / molecules / atoms 1 I references to disorder / chaos, etc.\nand energy in a system I ‘arrangement of particles’ without\n‘number of’.\nI randomness / extent.\nI of a substance.\nI ‘measure of’ for ‘number of’.\nI total.\n\n4(c)(ii) decrease in number of gaseous particles OWTTE 1 A ‘gas forms solid’.\nA ‘more gas in reactants’ or ‘less gas in\nanswer needs to either mention both sides of the equation or use a products’.\ncomparative term, e.g. larger / smaller A ‘1 mol of gas in reactants and same mol\ne.g. ignore ‘there are no gas molecules in the products’ of solid on both sides’.\nbut allow ‘there are fewer gas molecules in the products’ I decrease in disorder.\nI ‘Cl / a gas is consumed’ alone.\n2\nI ‘number of mols / particles decreases’.\n\n4(c)(iii) • the reaction is less spontaneous 2\nA ‘∆G more likely to be positive’\n• as ∆G or (∆H – T∆S) I ∆G or ∆S becomes less exothermic\nbecomes less negative / (more) positive I increases / decreases, etc.\n• due to ∆S being negative or A ΔS < 0.\nT∆S (becoming more) negative or I ↑ or ↓ for more / less positive.\n–T∆S (becoming more) positive\n✓✓\n⚫\n© Cambridge University Press & Assessment 2026 Page 13 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-23-chemical-energetics/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p12.png", "../answer-assets/9701_s26_ms_44-p13.png" ] }, { "id": "9701-2026-mj-44-q05", "question_id": "9701-2026-mj-44-q05", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 5, "topic": "Group 2", "topic_slug": "9701-topic-27-group-2", "marks": 7, "status": "available", "reason": null, "text": "5(a)(i)\nCa(NO ) → CaO + 2NO +\n3 2 2\n© Cambridge University Press & Assessment 2026 Page 14 of 21\n1\n2\n1 A other multiples.\nO or\n2 I state symbols.\n2Ca(NO ) → 2CaO + 4NO + O\n3 2 2 2\n\n5(a)(ii) M1 increases / more stable (down the group) and 2 I atomic radius.\nionic / M2+ radius increases / charge density of ion / M2+ decreases CON charge density increasing.\nI nitrate less polarised / distorted.\nM2 anion / nitrate ion / NO – less polarised / distorted or I less polarisation ‘between’ ions.\n3\nN=O / N–O less polarised / distorted/weakened A N–O bond stronger / harder to break.\nA distortion ‘with’ or ‘to’ NO –.\n3\nA NO 2–.\n3\nCON M2 for stronger ionic\nbond / attraction between ions.\nI ‘nitrate group’ for NO – .\n3\n\n5(b)(i) MgSO more soluble than BaSO 1 A MgSO is soluble and BaSO is\n4 4 4 4\ninsoluble.\nMust explicitly be a comparison of the sulfates A solubility of Gp2 sulfates decreases\n(down the group).\nI comparison of basicity.\nI comparisons between Mg(OH) and\n2\nBa(OH) .\n2\nI all discussion regarding ΔH , etc.\nlatt\n\n5(b)(ii) M1 H and H less negative / exothermic (down group 2) 3 A ora throughout.\nlatt hyd\nM2 H changes more / is dominant factor / significant A decreases for M1.\nlatt\nM3 H more exothermic down group A ‘of oxides’ as a slip.\nsol\nM3 I ‘solubility more exothermic’.\nor in terms of the compounds in the question: I HE and LE unless defined.\nM1 H and H are less negative for Ba(OH) Penalise missing  or H once only.\nlatt hyd 2\nM2 H latt is less negative by a larger amount OWTTE I H solubility .\nI any discussion of sulfates.\nM3 H is more exothermic for Ba(OH) (than for Mg(OH) )\nsol 2 2\nI discussion of ionic radii / charge density\nthroughout.\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-27-group-2/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p14.png" ] }, { "id": "9701-2026-mj-44-q06", "question_id": "9701-2026-mj-44-q06", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 6, "topic": "Chemistry of transition elements", "topic_slug": "9701-topic-28-chemistry-of-transition-elements", "marks": 13, "status": "available", "reason": null, "text": "6(a) (ion formed from a central) metal atom / ion 1 A transition element for metal.\n(surrounded by, or bonded to, one or more) ligands\n\n6(b)(i) M1 x = 1, y = 5 2 A negative or positive signs.\nM2 x = 4, y = 2\n\n6(b)(ii) 6 / six 1 Only answer.\n\n6(b)(iii) 2\nNo mark for either diagram with wrong 3D.\nM1 one diagram with No mark for diagrams with bidentate\ncorrect 3D and ligands.\nformula I incorrect ligand attachment (i.e. though\nthe H of NH\n3\n).\nM2 Penalise incorrect monodentate ligands\nonce only, then A.\nsecond diagram\nshows correct 3D One diagram must have all NH 3 ligands\nand other isomer cis, one diagram must have two NH 3\nligands trans\ni.e one facial, one meridial.\n\n6(b)(iv) yes no 1 A crosses as ticks if other box on row is\nstereoisomerism ✓ blank.\noptical isomerism ✓\ngeometric isomerism ✓ Three ticks only, all correct.\n\n6(c)(i) equilibrium constant 1 I ‘the ratio of…’. They have been given the\nformation of a complex expression and are just describing it.\nin a solvent / water / from its (constituent) ions / molecules I ‘in a solution’ for solvent.\n\n6(c)(ii) mol–6 dm18 any order 1 Only answer.\n© Cambridge University Press & Assessment 2026 Page 15 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c)(iii) M1 working makes it clear all terms substituted into\nthe expression correctly\n© Cambridge University Press & Assessment 2026 Page 16 of 21\n4 . 8 0  1 0 7 =\n( 0 . 4 5 5 ) 6\n0\n\n.\n\n0\n\n7\nN\n6\ni\n2\n( H\n2\nO )\n6\n 2 + \n,   N i ( H\n2\nO )\n6\n 2 +  =\n( 0 . 4 5 5\n0\n)\n.\n6\n0\n\n7\n4\n6\n.\n2\n8 0  1 0 7\n2 only ecf is M2 for TE errors in the\nconcentrations / K .\nstab\nCorrect answer scores [2]\nM2 1.8 / 1.79  10–7 min 2sf\n\n6(c)(iv) [Ni(H O) ]2+ + 6 NH → (1) [Ni(NH ) ]2+ + 6 H O 1 Only answer.\n2 6 3 3 6 2\n\n6(c)(v) Ligand exchange / displacement / replacement / substitution 1\nQuestion Answer Marks Guidance", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-28-chemistry-of-transition-elements/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p15.png", "../answer-assets/9701_s26_ms_44-p16.png" ] }, { "id": "9701-2026-mj-44-q07", "question_id": "9701-2026-mj-44-q07", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 7, "topic": "Organic synthesis", "topic_slug": "9701-topic-36-organic-synthesis", "marks": 13, "status": "available", "reason": null, "text": "7(a)(i) 120° and 1 Only answers.\nsp2\n\n7(a)(ii) • sp2 (and sp2) 2 CON s orbital unless clearly referring to C-\n• head on / direct / end-on / linear / acceptable description H, where you ignore s orbital.\n• p (and p) A orbital type written in overlap space if\n• sideways / lateral / acceptable description not written in ‘orbitals involved’ space.\nOtherwise consider only orbitals on the\n✓ ✓ ‘orbitals involved lines’ and only overlap\n⚫ ⚫\non the ‘orbital overlap’ lines.\nRead ‘6p orbitals’ as the number of p\norbitals.\n\n7(b)(i) conc HNO + conc H SO , T = 25–60 °C 1 Conc, con or c seen once.\n3 2 4\nCON aq.\nA any T in range.\nI heat / warm / reflux.\n\n7(b)(ii) C H NH 1 A C H NH + or C H NH Cl.\n6 5 2 6 5 3 6 5 3\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)(iii) M1 HNO / HCl + NaNO , T = 0–10 °C 2 A any T in range.\n2 2\nI HNO / NaNO .\n3 3\nM2 H O / (aq) and warm / heat / T > 10 °C A [1] for all correct reagents or both\n2\ncorrect temperatures.\nM2 CON addition of NaOH but ignore\nacids.\n\n7(c)(i) M1 curly arrow from within / touching hexagon towards C+ 3 CON additional curly arrows in M1.\nI additional curly arrows in M3.\nCON curly arrow from H rather than\nmiddle of bond for M3\nintermediate – normal rules apply.\n‘Horseshoe’ must include C3, C4 and C5,\nmay include C2 and / or C6, CON\nincluding C1.\n+ charge must be above a line joining C2\nand C6.\nM2 intermediate\nM3 curly arrow from C–H bond goes within hexagon\nand H+\nI connectivity of side group if drawn as\nOCCH for M2.\n3\nA Kekulé for M2 if correct.\nA HCl for M3 if Cl – introduced earlier.\n\n7(c)(ii) C H COCH + 4[O] → C H COOH + CO + H O 1\n6 5 3 6 5 2 2\n© Cambridge University Press & Assessment 2026 Page 17 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(d) M1 NaOH(aq) / OH–(aq) / alkaline and iodine / I 2 A NaOH and I (aq).\n2 2\nA Cl / OH–.\n2\nM2 any acid / H+ unc\nA HCl, H SO .\n2 4\n(pass alternative (correct) method to TL for advice, e.g. LiAlH then alkaline I CON additional reagents.\n4 2\nfollowed by acid)\nQuestion Answer Marks Guidance", "source_pages": [ 16, 17, 18 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-36-organic-synthesis/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p16.png", "../answer-assets/9701_s26_ms_44-p17.png", "../answer-assets/9701_s26_ms_44-p18.png" ] }, { "id": "9701-2026-mj-44-q08", "question_id": "9701-2026-mj-44-q08", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 8, "topic": "Analytical techniques", "topic_slug": "9701-topic-37-analytical-techniques", "marks": 10, "status": "available", "reason": null, "text": "8(a) TMS / tetramethylsilane / (CH ) Si 1\n3 4\n\n8(b)(i) M1 Q and (peak at) 5.4 / 4.5–7.0 = Ar-OH / phenol 2 I Q and 4 peaks for 4 (proton)\nenvironments\nM2 Q and (peak(s)) at 6.8–7.3 / 6.0–9.0 = aromatic / H–Ar (question asks for protons responsible).\nIf explanation is blank, check spectra.\nA benzene ring for aromatic.\n\n8(b)(ii) • triplet 2\n• two protons/H on neighbouring/adjacent C A neighbouring CH 2 or CH 3 for adjacent.\n• quartet A quadruplet for quartet.\n• three protons/H on neighbouring/adjacent C A near-miss spellings (e.g. one letter\nmissing / incorrect only or ‘tirplet’, etc.).\n✓ ✓\n⚫ ⚫\n\n8(c) peak (at 5.4) disappears 1 A chemical shift disappears.\nand I it would have 3 or 4 peaks as we don’t\nthe OH / 5.4 proton / H is liaible or know if they are counting the TMS peak.\nthe OH / 5.4 proton / H exchanges / replaced with D or I OH exchanges with D O.\n2\nequation, e.g. ROH + D O → ROD + DHO I (OH) reacts with D O.\n2 2\nI OH exchanges with OD.\nCON incorrect peak / spectra.\n© Cambridge University Press & Assessment 2026 Page 18 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(d) ethanoic acid phenol ethanol 4\n• correct reference to O-H bond strength / anion stability A ecf for matching O–H strength / anion\nstability correctly to incorrect order for\nbullet 1.\n• (ethanoic acid) C=O has electron withdrawing / –I effect CON bullet 1 for incorrect reference to\nO–H bond strength / anion stability.\nI COOH / COO– is an electron withdrawing\ngroup.\nA e– / negative charge delocalises over\nCOO– / C=O.\n• (phenol) LP/p-orbital on O overlaps/is delocalised and\nwith/into ring/ system\n• (ethanol) electron donating/+I effect of alkyl/R/CH /CH /C H\n2 3 2 5\n✓ ✓ ✓ for four bullets\n⚫", "source_pages": [ 18, 19 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-37-analytical-techniques/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p18.png", "../answer-assets/9701_s26_ms_44-p19.png" ] }, { "id": "9701-2026-mj-44-q09", "question_id": "9701-2026-mj-44-q09", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 9, "topic": "Nitrogen compounds", "topic_slug": "9701-topic-34-nitrogen-compounds", "marks": 9, "status": "available", "reason": null, "text": "9(a) • asparagine is negative / contains COO– 3 A ecf for bullets 2 and 6 if charges and\n• asparagine moves left / towards the positive/anode directions are reversed.\n• histidine is a zwitterion / neutral / uncharged A bullets 2, 4 and 6 from diagram only if\n• histidine does not move directions not mentioned.\n• lysine is positive / contains NH +\n3\n• lysine moves right / towards the negative/cathode\n✓ ✓ ✓\n⚫ ⚫ ⚫\n© Cambridge University Press & Assessment 2026 Page 19 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n9(b)(i) 1 Correct 3D.\nCorrect formulae.\nDiagrams are different isomers.\nNo connectivity penalty for any group\nmay see RH structure same orientation as\nLH structure with any 2 groups switched.\n\n9(b)(ii) racemic / racemate 1 A near misses, e.g. racemix.\n\n9(b)(iii) M1 effect on/rotation of plane polarised light 2 I angle of rotation, optical activity.\nM2 biological activity / interaction with enzyme A biological reactivity / biochemistry.\n/ biological effects / bioactivity.\nA interaction with chiral reagents.\nI comments on structure.\nI chemical biology, biological active, etc.\n\n9(b)(iv) H NCH(CH OH)COOH + 2PCl → 2POCl + 2HCl + H NCH(CH Cl)COCl 2 A [1] if molecular formula used,\n2 2 5 3 2 2\nC H NOCl as only mistake.\n3 5 2\nM1 Correct organic product A any unambiguous structure.\nA [1] if only one OH substituted and\nM2 all correct otherwise all correct.\nA ammonium salt but beware balancing with HCl, e.g.\nH NCH(CH OH)COOH + 2PCl → 2POCl + HCl + ClH NCH(CH Cl)COCl\n2 2 5 3 3 2\n© Cambridge University Press & Assessment 2026 Page 20 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 19, 20 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-34-nitrogen-compounds/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p19.png", "../answer-assets/9701_s26_ms_44-p20.png" ] }, { "id": "9701-2026-mj-44-q10", "question_id": "9701-2026-mj-44-q10", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "44", "question_number": 10, "topic": "Polymerisation", "topic_slug": "9701-topic-35-polymerisation", "marks": 6, "status": "available", "reason": null, "text": "10(a) ethan(e)dioyl (di)chloride / oxalyl chloride 1 A ethdioyl (di)chloride, oxal(di)oyl\n(di)chloride, oxalic acid dichloride, oxalic\ndichloride\nI numbers.\nI ethanoyl dichloride, ethanediacyl\n(di)chloride, diethanoyl (di)chloride.\n\n10(b)(i) 2 A skeletal, but beware leading/trailing\nbonds.\nA [1] if NH has incorrect number of H’s.\nM1 one correct amide link displayed A additional atoms if outside of bracketed\nM2 exactly one RU, trailing bonds, no errors repeat unit.\nA –NH–.\ncorrect amide link should be attached to a\nC at each end,\ne.g. C–CO–NH–C, and not Cl–CO–NH–.\n\n10(b)(ii) condensation and 1 I CONH as linkage.\n(poly) amide (group/bond/linkage) A polyamide for amide.\nI peptide.\n\n10(c) LiAlH or acceptable name 1 CON aqueous / (aq).\n4\n(A H and Ni / Pt / Pd and Heat and\n2\npressure).\n\n10(d) ammonia 1 CON aqueous / (aq) / alkaline.\nheat\nsealed tube/pressure\n© Cambridge University Press & Assessment 2026 Page 21 of 21\nTrace ID: PC-0F078E40FE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 21 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_44.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-44.pdf?download=true", "html": "9701-topic-35-polymerisation/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_44-p21.png" ] }, { "id": "9701-2026-mj-51-q01", "question_id": "9701-2026-mj-51-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "1(a) pipeclay triangle 1 A clay triangle / clay pipe triangle / triangle clay pipe\nI tripod, gauze, triangular holder\n\n1(b) to prevent spitting 1 A to prevent cracking / damaging / breaking crucible\nA to prevent losing solid / (mixture) X / solid jumping out\n(of crucible)\nA to prevent splitting\nA CaCl or Ca(NO ) for solid\n2 3 2\nR burning\n\n1(c) NO is a hazardous gas 1 A NO gas is toxic / poisonous / acidic / corrosive\n2 2\nA NO is an irritant to eyes / by inhaling\n2\nA NO (g) for NO gas\n2 2\nI flammable, harmful\nR answers about O\n2\n\n1(d)(i) M1 mass lost on heating is 17.81 – 16.53 g = 1.28 g 2 award 2 marks for 0.0237\nrelative molecular mass of (4NO + O ) = 216.0 award 1 marks for 0.0278, 0.0618, 0.135 and 0.0156\n2 2\n1.28 g is 1.28 / 216 mol = 0.005926 mol of (4NO + O )\n2 2\n1.28 g is NO and O in the ratio of 4:1\n2 2\nrelative mass of (4NO + O ) = 4  14.0 + 10  16.0 =\n2 2\n216.0 g mol–1\n1.28 g is 1.28 / 216 mol = 0.005926 mol of (4NO + O )\n2 2\nNO is 4 this.\n2\nA 2 or more SF\nM2 number of moles NO = 4  0.005926 = 0.0237 mol ecf from incorrect M1:\n2\nA a number of moles  4\n[A if assume all gas released to be NO = 0.0278 mol\n2\nA if 5.07 g is assumed to be all Ca(NO ) then\n3 2\n5.07 / 164.1 = 0.0309 mol Ca(NO ) so 2  0.0309 =\n3 2\n0.0618 mol\nA if 3.79 g is assumed to be CaO so 3.79 / 56.1 =\n0.06756 mol CaO then 2  0.06756 = 0.135 mol\nA if 1.28 g is assumed to be Ca(NO ) then 1.28 / 164.1\n3 2\n= 0.00780 mol Ca(NO ) so 2  0.0078 = 0.0156 mol]\n3 2\n© Cambridge University Press & Assessment 2026 Page 7 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(d)(ii) Correct working must be shown. 1 A the following values are amongst acceptable values\nfor the answer following a correct expression 0.06,\npercentage error = (2  0.005 / 17.81)  100% = 0.0561 0.056, 0.05 etc.\nI ignore missing  100 if correct answer is given\n(2  0.005) may be seen as (0.01 / 2  2)\n\n1(d)(iii) increase the mass of mixture X used 1 A any wording that implies increasing the mass of X\ne.g. weigh a larger sample / reactant / contents\nA use a bigger amount of X / Use a larger quantity of X /\nuse more X etc.\nA use a larger mass of ‘crucible + X’\nA ‘use a larger mass’ or more than 5.07 g\nI ‘use a large mass’ alone\nI repeat and take mean\nI changes to heating e.g. heat for longer\nR more accurate balance or any other change to\napparatus e.g. increase mass of crucible alone\nR incorrect use of weight for mass\nR heat to constant mass\n\n1(e) M1 mol of Ca(NO ) in original sample = 0.5  0.0223 mol = 2 award 2 marks for 36.7\n3 2\n0.01115 mol A more SF e.g. 1.829715 g 36.741265%\nA 36.7 rounded to 37 for M2\nI rounding errors after 3SF.\nM2 mass Ca(NO ) in original sample = 0.01115 mol  If M1 is rounded to 0.0112 mol and then used in M2\n3 2\n164.1 g mol–1 = 1.829715 g award M1 only\n% by mass Ca(NO ) = 1.829715 / 4.98  100% = 36.7% ecf for M2 from incorrect M1 and allow final answer to\n3 2\nbe rounded to 2SF\n© Cambridge University Press & Assessment 2026 Page 8 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(f)(i) heat to constant mass 1 A description of the process of heating to constant\nmass e.g. reheat and reweigh until mass remains\nconstant\nA repeat steps 3 and 4 to constant mass\nresponse must include (re)heat and constant mass\n\n1(f)(ii) The percentage by mass will be smaller. 1 A NO or O or both for gas\n2 2\nAND A amount / mol for mass\nmass of gas lost is smaller (than the maximum possible)\nOR I not all NO released\n2\nmass of solid remaining after heating is higher (than actual mass) I calculated mass of Ca(NO ) (that decomposes) is\n3 2\nlower than actual value\n\n1(g) M1 attempted use of ice / cold water for condensing the gas 2 M1 A coolant/cool water for ice\nwithout loss of the NO as gas\n2\nIf a condenser is seen, max 1 mark\n• must be angled downwards towards unsealed\nM2 in an unsealed collection vessel surrounded by iced / cold collection vessel\nwater • must show water in and out of jacket\n• ignore position of water in and out\n• condensed NO must be delivered into a vessel\n2\nsurrounded by ice/cold water.\nIf a condenser is delivering into a u-tube and ice, then 2\nmarks can be awarded.\n© Cambridge University Press & Assessment 2026 Page 9 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_51-p07.png", "../answer-assets/9701_s26_ms_51-p08.png", "../answer-assets/9701_s26_ms_51-p09.png" ] }, { "id": "9701-2026-mj-51-q02", "question_id": "9701-2026-mj-51-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 17, "status": "available", "reason": null, "text": "2(a) only (direct) light from the LED should be allowed to reach the 1 A to prevent outside light entering (the colorimeter)\nphotocell A to prevent outside light reaching the photocell\nfor outside light A external light / background light\n/surrounding light\nfor photocell A detector\nA no (indirect) light via reflections\nA the only source of light is from the LED\nI other light for outside light\nI light absorbed unless it is absorbed by the photocell\n\n2(b)(i) 1 Must be 3 DP\nconcentration of resistance of\nabsorbance, A\nI –(aq) / mol dm–3 photocell, R / k\n3 s\n3.60  10-6 34.28 0.839\n2.94  10-6 23.42 0.684\n2.20  10-6 16.01 0.519\n1.52  10-6 9.68 0.300\n7.90  10-7 7.50 0.189\n\n2(b)(ii) distilled water (with few drops of starch solution) 1 A (acidified) I–(aq)/solution (with few drops of starch\nsolution)\nA acidified I– ions (with few drops of starch solution)\nA I – solution/(aq) with no starch\n3\n\n2(b)(iii) concentration of I – 1\n3\n© Cambridge University Press & Assessment 2026 Page 10 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b)(iv) M1 all 5 points plotted correctly 2 Place plotting tick on grid above line of best fit.\nPlace line tick in left-hand margin\nPlace a cross beside an incorrect point\nM2 straight line of best fit through origin passing through or close ecf from 2(b)(i)\nto all points except anomaly (point 2)\nR if point 2 is on a vertical gridline\nR if point 4 is on a horizontal gridline\nR line of best fit that does not extend from the origin to\ncover all plotted points.\nR if line of best fit is one square or more from origin on\neither axis\nA line of best fit between the first and last (correctly\nplotted) points is acceptable as long as it goes close\nenough to the origin. All points except anomaly should\nbe on the line of best fit.\nPoint marked as anomalous (ringed) should not be\nused in determining the line of best fit.\nA ecf acceptable line of best fit drawn for incorrectly\nplotted points.\nR line of best fit if the points (except anomaly) not\napproximately evenly distributed on both sides of line\nI extra points added which are used in later calculations\nI remnants of lines that the candidate has tried to erase\nR curved line\nR double or “sketchy” lines.\n© Cambridge University Press & Assessment 2026 Page 11 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b)(v) most anomalous point circled 1 Expected answer is point 2 at (1.52  10–6, 0.300 below\nthe line of best fit)\nAND\nA light intensity (in colorimeter) increased\nany one reason from:\n• not enough starch added (in step 2) A ora for ecf point above line of best fit\n• less than 10.0 cm3 solution transferred (from boiling tube to\nmeasuring cylinder in step 4) A light got into the colorimeter/black box not shut (no\n• solution (from step 2) was left for less than 10 minutes (in ora for point above line of best fit)\nstep 3)\nI reading of absorbance was too low\nI comments about temperature/heat\nI starch not mixed properly\nI systematic error\nI evaporation\nI reference to rates\n\n2(c)(i) to allow all the I – ions form a complex (with the starch) 1 A for complete reaction (of the I – ions and starch)\n3 3\nA enough time for the complex to form completely\nI to reach equilibrium\nI reference to rate of reaction\nI all the starch reacts\n\n2(c)(ii) 10 cm3 volumetric pipette 1 A graduated pipette as long as volume is not < 10 cm3\nA 10 cm3 pipette\nA volumetric pipette\nI ‘pipette’ alone\nR burette\n\n2(c)(iii) M1 absorbance = 0.640 2 A 0.64 with working shown\nConcentration should be read from the graph to nearest\n½ square.\nM2 concentration read from the graph to nearest ½ square and I any figures after 3SF\ngiven to 3 SF I trailing zeros (allow 2.7 for 2.70)\nA non-standard form\n2.75  10–6 mol dm–3 is a common answer\n© Cambridge University Press & Assessment 2026 Page 12 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(c)(iv) (15 / 1000)  2(c)(iii) 1 A 2 SF minimum\nA calculator value from 2(c)(iii) (e.g. 2.725 used for 2.73\ngiven in 2(c)(iii))\nI trailing zeros\nA non-standard form\n\n2(c)(v) 2(c)(iv)  (1000 / 7500) 1 If using the given value, then:\n3.23  10–8  1000 / 7500 = 4.31  10–9 mol dm–3\nA calculator value from 2(c)(iv)\nA 2 SF minimum\nA non-standard form\n\n2(c)(vi) increases 1\nAND\n(the value of) the concentration of I – increases (at same\n3\nabsorbance)\n\n2(c)(vii) 9.87 + 0.14  9.87 = 11.25 AND 13.66 – 0.14  13.66 = 11.75 1 A any set of calculations which gives 11.25 and 11.75\nAND\nthe concentration of O is different (because the extremes of the\n3\nranges do not overlap)\n\n2(c)(viii) R increases 1 A light travels through a deeper solution\ns\nAND A light interacts with more I – particles\n3\nthe light has a longer path length (through the solution) A absorbance increases with depth (of solution)\nI light interacts with more I – solution(aq)\n3\nI more light is absorbed alone\nI measuring cylinder is deeper\n\n2(c)(ix) concentration of I – will be higher giving a lower percentage error 1 A amount / mol of I – for concentration of I –\n3 3 3\nI uncertainty for percentage error\nA uncertainty percentage error for percentage error\nalone\n© Cambridge University Press & Assessment 2026 Page 13 of 13\nTrace ID: PC-54A496C6B0\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/51 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 10, 11, 12, 13 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-51.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_51-p10.png", "../answer-assets/9701_s26_ms_51-p11.png", "../answer-assets/9701_s26_ms_51-p12.png", "../answer-assets/9701_s26_ms_51-p13.png" ] }, { "id": "9701-2026-mj-52-q01", "question_id": "9701-2026-mj-52-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a)(i) 7.61 g 1 0.300  100/1000  253.8\nA 7.614\nR 7.62 or 127 or 254 used in working\nA 7.6 but only if there is no evidence of 7.62 / 127 / 254\n© Cambridge University Press & Assessment 2026 Page 7 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(a)(ii) M1 dissolve the solid / I (s) (in the beaker) (using a small 3 M1 is the dissolving mark\n2\nvolume of) methanol M2 is the transfer with rinsing mark into a flask\nM3 is volumetric apparatus and inversion mark (we need\n100 cm3 AND volumetric flask seen somewhere) M1 and M2\ncould still be awarded if another type of flask (e.g. conical) is\nused – but must be a flask not a beaker / measuring cylinder\nM1 I ‘weigh the solid’ I mass / mol I if seen\n2\nI ‘place in a beaker’ (assume they are repeating the\nquestion) but if they transfer the solid to a second container,\nevidence of rinsing is needed to allow M1\nM1 R if it is assumed I is aqueous/in methanol\n2\nM1 R if suggestion is to add more than 50 cm3 of methanol\nor to fill the beaker to dissolve solid in the beaker. Any\nvolume quoted must be 50 cm3 or less.\nA prepare / make a solution for dissolve.\nIf dissolve not seen here, A ‘transfer the solution’ in M2 to be\nequivalent to ‘dissolve’ in M1 ie they must have dissolved in\nM1 if they are transferring a solution in M2.\nM2 transfer/add to a (100 cm3 volumetric) flask M2 A graduated flask for volumetric flask\nAND A extra transfers as long as the exact number of moles of\nrinse (with methanol) solid end up in the vol flask via rinsing.\nAssume ‘rinse’ / ‘wash’ is into flask unless stated I lack of\nrinsing the funnel\nM2 R if transfer of solution from M1 is via burette/pipette\nassume ‘rinse the residue’ refers to residual solution in the\nbeaker\nM3 top the 100 cm3 volumetric flask up to mark (with M3 A make up to 100 cm3 with methanol. I ‘top up’ without\nmethanol) reference to 100\nAND A ‘fill the flask to 100 cm3 / mark’ but R ‘fill the flask’ (alone)\ninvert the (stoppered) flask for ‘invert the flask’ A ‘shake’, ‘swirl’ etc\nI Homogenise (alone) I ‘mix’ (alone) /‘mix the flask’ (we need\nto know how they mix/homogenise)\nR stir in the flask\n© Cambridge University Press & Assessment 2026 Page 8 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(a)(ii) A max 2 marks if solid is added directly into the volumetric\nflask\nM1 is not awarded\nM2 rinsing of beaker is still needed to ensure all the solid is\nput into the vol flask\nM3 OK\nIf they fill volumetric flask to the mark / add 100 cm3 of\nmethanol and then add solid A max 1 mark:\nM1 is not awarded\nM2 is not awarded (because they cannot rinse into a full\nflask)\nM3 still possible for 100 cm3 volumetric flask AND methanol\nAND invert to gain credit\n‘100 cm3 volumetric flask’ needs to be seen in M2 or M3 for\nM3 to be awarded\nMethanol needs to be seen at least once in M1, M2 or M3\nwithout methanol 3 marks = 2 marks, 2 marks = 1 and 1\nmark = 0 marks.\nmethanol must be seen once to award M1 if not directly\nused in M1\nmax 2 marks if (distilled) water is used as the solvent\n© Cambridge University Press & Assessment 2026 Page 9 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b)(i) methanol is flammable 1 A inflammable for flammable R non-flammable\nA other comments implying the flammability of methanol\ne.g. methanol vapours ignite\nA solution / I (meth) / substance is flammable A solvent is\n2\nflammable\nI I (alone) is flammable\n2\nA tetramethylammonium iodide / (CH ) NI is flammable A\n3 4\nreactants are flammable\nA contents are flammable\nA polyiodide salt / (CH ) NI is flammable A product(s) are\n3 4 2n+1\nflammable\nI ‘(it is) flammable’ alone (not specifying whether it is the\ncontents or the flask which is flammable)\nI references to explosions\nI to reduce evaporation of methanol / prevent methanol\nboiling\nI (CH ) NI / reactant will (thermally) decompose\n3 4\nI to avoid spitting\nI flask shattering / flask breaking\n\n1(b)(ii) to remove any soluble impurities 1 A to remove impurites / to remove reactants / I / I (meth)\n2 2\n/ (CH ) NI\n3 4\nA to remove solute(s) / A to remove solution(s)\nA to remove (methanol-)soluble substances (no need to tell\nus what the substances are)\nI to remove substances (alone)\nI to remove filtrate\nR to remove product / residue / solid impurities\nR to remove methanol / solvent / water\n© Cambridge University Press & Assessment 2026 Page 10 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b)(iii) leave (to dry) 1 A description of leaving until constant mass\ne.g. keep on weighing until constant mass\nAND e.g. weigh every 5 minutes until constant mass\ne.g. dry to constant mass\nuntil constant mass e.g. repeat drying and weighing until a constant mass is\nobtained\ne.g. wait until constant mass\nR heating or methods which do not reflect step 6 (where\nheating does not occur)\ne.g. heat to constant mass / place in an oven until constant\nmass / leave in a warm place\nI use of cobalt chloride paper / desiccator\nR leave until constant volume\nR constant weight for constant mass\n© Cambridge University Press & Assessment 2026 Page 11 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b)(iv) M1 suitable results table showing rows which include: 2 mark table first\n• mass of (evaporating) basin the table may be vertical as well as horizontal\n• mass of (evaporating) basin + crystals A ‘weight’ for ‘mass’ in this question only\n• mass of crystals ‘mass’ must be seen in table\n• with column to enter values\nA dish for basin\n• units i.e. ‘g’\nA any table that has:\ne.g.\nmass of dish / weighing boat / any container + crystals\nmass of (evaporating) basin / g mass of dish / weighing boat / any container (ignore ‘+\nresidue’)\nmass of (evaporating) basin + crystals / g\nmass of crystals / product / (CH ) NI / solid\n3 4 2n+1\nunits i.e. ‘g’\nmass of crystals / g\nA rows 1 and 2 in any order\nI other rows\nR tables which do not have a container\nR tables which do not measure mass of crystals (e.g. tables\nwhich only measure mass change due to methanol/water\nloss e.g. ‘wet crystals to dry crystals)\nI mass change due to solvent loss as extra row(s)\nR tables which measure mass of stated substance other\nthan crystal/product e.g. I\n2\nallow units in value box (as shown)\nmass\nMass of evaporating g\nbasin\n© Cambridge University Press & Assessment 2026 Page 12 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b)(iv) M2 description of correct method used in obtaining mass of A as an alternative description of correct method for M2:\ndry crystals\n• weigh an evaporating basin + crystals\n• Weigh an evaporating basin • rinse and dry the evaporating basin and reweigh\n• (Transfer the dry crystals from the filter paper to the I descriptions of maths involved in determining mass of\nevaporating basin and) reweigh the evaporating basin + crystals\ncrystals.\nother methods which involve transfer of crystal from the\noriginal basin into another container (might be a basin) will\nnot score M2\ne.g.\n• weigh a weighing boat (or other container)\n• transfer the crystals from the evaporating basin to the\nweighing boat / container and reweigh weighing\nboat/container\n\n1(c)(i) 0.0200 mol dm–3 Na S O (aq) 1 A Na S O (aq) / aqueous sodium thiosulfate / 0.02 mol dm–3\n2 2 3 2 2 3\nNa S O\n2 2 3\nA sodium thiosulfate / Na S O alone without aqueous\n2 2 3\nA rinse with solution to be used in the burette\nI slips in concentration e.g. 0.2 mol dm–3\nI rinsing with distilled water or organic drying solvent e.g.\npropanone before rinsing with Na S O (aq).\n2 2 3\nR rinsing with anything else after Na S O (aq).\n2 2 3\nR incorrect formulae\nR ‘rinse with water/ Na S O (aq)’ (too vague)\n2 2 3\n© Cambridge University Press & Assessment 2026 Page 13 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(c)(ii) Working must be shown.\n=\n© Cambridge University Press & Assessment 2026 Page 14 of 22\n2\n1\n\n3\n0\n.\n.\n3\n0\n5\n5\n1 (0.05  2) may be seen separate or as (0.1 / 2  2)\nI 0.1 as numerator without evidence of 0.05\nA Any correct answer that rounds to 0.7 i.e. A\n 100 = 0.749%\n0.7 / 0.74 / 0.749\nA 0.75 / 0.750\nI missing ‘x 100’ in calculation if correct answer seen\nI Rounding errors beyond 1st SF\nR 0.0749 this must mean 0.005 was used in the working\nR incorrect answers with correct working\nTitre used must be 13.35\n\n1(c)(iii) 13.30 cm3 1 I 13.3\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(c)(iv) M1 mol of Na S O in mean titre 3 A 3 marks for 0.338 if no evidence of incorrect working. I\n2 2 3\n= 13.30 cm3  0.02 / 1000 minor transcription errors\n= 2.66  10-4 mol A 0.338 up to calculator value (0.337554) I RE after 3rd SF\nA 2SF as final answer if 3SF or more correct answer seen in\nworking\nM2 mol of I in 10 cm3 of solution A Otherwise mark by step.\n2\n= 2.66  10-4 / 2=1.33  10-4 mol M2 subsumes M1\nand\nA ecf for mean titre from 1(c)(iii)\nmol of I in 100 cm3 of solution A\n2\nA M2 for ecf of M1\n=1.33  10-4 mol  100 / 10 =1.33  10-3 mol\nA M3 for ecf of M2\nM3 mass of I in 100 cm3 of solution A = 1.33  10-3  253.8\n2 A Use of 254\n=0.338 g\nA ecf for use of 126.9 / 127 for M I if used in 1(a)(i)\nr 2\nScaling up by 100 / 10 appears in step 2, but can appear in\nany step.\nA mole ratio methods involving concentration  volume of\nsolutions\nC  V C = [I (aq)] in solution A V = volume solution A\n1 1 1 2 1\nC  V C = [Na S O (aq)] V = volume Na S O (aq)\n2 2 2 2 2 3 2 2 2 3\nconcentration of I (in soln A)  volume I (soln A) = 1\n2 2\nconcentration of Na S O  volume Na S O 2\n2 2 3 2 2 3\nconcentration of I  10.00 = 1\n2\n0.0200  volume Na S O 2\n2 2 3\nThis leads to [I (aq)] in solution A = 0.0133 mol dm–3 M1 ✓\n2\nMoles of I in 100 cm3 of solution A = 0.0133 100 / 1000 =\n2\n1.33  10–3 M2 ✓\n(M2 subsumes M1)\n1.33  10-3  253.8 = 0.338 g M3 ✓\nOther methods may be used.\n© Cambridge University Press & Assessment 2026 Page 15 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(c)(v) Working must be shown. 1 The mass (or substitute value) of I from 1(c)(iv) must be\n2\nused or a derived number of moles from this value must be\nmol I = 0.338 / 253.8 = 0.00133 mol I seen (moles may be seen in M2 of 1(c)(iv))\n2 2\nmass (CH ) NI = 0.472 – 0.338 = 0.134 g A ecf from 1(c)(iv) for 0.338 (mass of I )\n3 4 2\nmol (CH 3 ) 4 NI = 0.134/200.9 = 0.000667 mol A ecf for M r I 2 = 126.9 / 127 if 126.9 / 127 was used in 1(a)(i)\nratio = 0.00133 = 1.99… (= 2) Apply the following formulas to help\n0.000667 If 253.8 is used as M I\nr 2\nn = 0.7916  a where a = answer to (c)(iv)\nAND (0.472–a)\nn = 2\nIf 126.9 is used as M I\nr 2\nn = 1.583 x a where a = answer to (c)(iv)\n(0.472–a)\nA as minimum seen for working:\n0.338 / 253.8 and (0.472 – 0.338) / 200.9 or 0.134 / 200.9\nor\n0.00133 mol I and 0.000667 mol of (CH ) NI\n2 3 4\nor\n1.99…\nA mol of I = 0.00133 or mol of I derived from incorrect\n2 2\nmass of I in 1(c)(iv)\n2\nand\nA mol of (CH ) NI = 0.000667 or mol of (CH ) NI derived\n3 4 3 4\nfrom incorrect mass of I in 1(c)(iv)\n2\nUsing given 0.374 g as the mass of I gives n = 3.\n2\nOther methods may be used.\n© Cambridge University Press & Assessment 2026 Page 16 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_52-p07.png", "../answer-assets/9701_s26_ms_52-p08.png", "../answer-assets/9701_s26_ms_52-p09.png", "../answer-assets/9701_s26_ms_52-p10.png", "../answer-assets/9701_s26_ms_52-p11.png", "../answer-assets/9701_s26_ms_52-p12.png", "../answer-assets/9701_s26_ms_52-p13.png", "../answer-assets/9701_s26_ms_52-p14.png", "../answer-assets/9701_s26_ms_52-p15.png", "../answer-assets/9701_s26_ms_52-p16.png" ] }, { "id": "9701-2026-mj-52-q02", "question_id": "9701-2026-mj-52-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a) M1 labelled gas syringe 2 M1 gas syringe must be attached (directly to) side arm\nR M1if delivery to gas syringe goes through extra apparatus\neg condenser etc\nA ‘syringe’ / ‘glass syringe’ as label\nA for displacement of non-aqueous named liquid (instead of\nwater) into an inverted labelled measuring cylinder\nM2\nM2 sealed apparatus using a bung / cork / stopper A for any apparatus that is sealed and is capable of\ncollecting a measured volume of gas\ne.g. displacement of water using a measuring cylinder\ne.g. gas syringe coming from hole in bung and side arm\nsealed\nA ‘lid’ as label for a clearly drawn bung. A ‘bung’ as label for\nwhat looks like a lid R lid as a label if it looks like a lid\nR if ‘sealed’ apparatus includes addition of solution that\nwould push air into the gas syringe\ne.g. burette delivering a solution through the bung.\nI attempts at mixing solutions within sealed apparatus e.g.\ntest-tube dangling on a string\nI lines across delivery tubes\nA for loose diagrams - if it looks like a bung at the top of the\nflask assume it to be sealed\nA non-tight connection of delivery tubes to side arm\nA loose piston in the gas syringe\nR sealed apparatus that would pop apart due to increased\npressure\n© Cambridge University Press & Assessment 2026 Page 17 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a) If M2 (or M1) is given for a displacement of water apparatus\nbe lenient with position of delivery tube at the bottom of the\nmeasuring cylinder.\nBut it should not ‘enter’ the measuring cylinder above the\nwater line or go through the wall of the water trough\nR M1 and M2 if a syringe being used to deliver reactants\n\n2(b) time 1 I units R ‘time intervals’\n\n2(c) the volume (of CO ) was measured / recorded too 1 A (recorded) time is greater than actual time\n2\nearly / before 1000s for actual time A true time / real time\nfor recorded time A time read / the time (seen) in the results\ntable\nA reading was taken / recorded / measured at (880 –\n920) s / (120 – 80) s early\nR timer was started / stopped too early / late\nI comments on volume error\nassume ‘the time was recorded’ means ‘the recorded time’\ni.e. ‘time was recorded too late’ is correct\ni.e. ‘time was recorded too early’ is incorrect\n\n2(d) a smooth curve that: 1 I points drawn on curve\n• starts at the origin A curve to start within first four small squares of y-axis and\nat 0 on the x-axis\n• is to the left of the original curve R curves which cross original graph\nR curves which clearly stray outside y-axis\n• flattens at 10 cm3 and before 1300 s A curves to stray up to a ¼ of a square above 10 cm3 as a\n‘hump AND return to 10 cm3\nR curves (with humps) that go too high above 10 cm3\n• ends at 1600s\nR curves where horizontal part stops before 1400s\n© Cambridge University Press & Assessment 2026 Page 18 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(e)(i) 2 all values must be to 2 DP\n[Na S O ] rate log log\n2 2 8 A ecf for M2 if all –ve signs are missing in both columns\n/ moldm−3 / moldm−3s−1 [Na S O ] (rate)\n2 2 8 A ecf for -1.93 in column 4 if –2.43 is seen in column 3\nA ecf for both –2.4 AND –2.6 seen in column 4 if –2.(0) AND\n1.00 x 10–2 0.02760 –2.00 –1.56\n–2.8 is seen in column 3\n5.74 x 10–3 0.01160 –2.24 –1.94\n3.67 x 10–3 0.00549 –2.44 –2.26\n2.88 x 10–3 0.00398 –2.54 –2.40\n2.08 x 10–3 0.00250 –2.68 –2.60\n1.57 x 10–3 0.00169 –2.80 –2.77\n✓ ✓\nM1 column 3\nM2 column 4\n© Cambridge University Press & Assessment 2026 Page 19 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(e)(ii) M1 all 6 points correctly plotted 2 If a point should be on a line, it must be on a line.\nIf a point should not be on a line, it must not be on a line.\nIf not supposed to be on the line, points must be within half\na small square of the correct position and in the correct\nsmall square.\nBlobs/thick crosses should be less than ½ a square across\nand be correctly centred.\nM2 line of best fit correctly drawn\nA ecf for acceptable line of best fit for incorrectly plotted\npoints\nI remnants of lines that have been rubbed out\nplace tick for M1 on graph\nplace tick for M2 to the left-hand side of graph\nIndicate any incorrect plots with colour tool once\nline of best fit must cover all the points\nline of best fit guidance:\n• line should meet right-hand side of grid above –1.6\n• point at (–2.24, –1.94) should not be significantly above\nthe line.\n\n2(e)(iii) M1 co-ordinates correctly recorded from line of best fit 2 range at least 0.4 units on x-axis\ntable points may be used but must be on line of best fit\n(beware (–2.00, –1.56))\nCo-ordinates should be read to nearest ½ square\nI any (lack of) trailing zeroes on co-ordinates e.g. A -2.00 as\n-2\nIf x, y values are reversed, R ecf of x / y but A M2 for a\ncorrect gradient (y / x)\nM2 gradient calculated and correctly rounded to three I units\nsignificant figures\n© Cambridge University Press & Assessment 2026 Page 20 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(e)(iv) correct values of y, m and x for use in y = mx + c where 1 A ecf of incorrect x and y values from co-ordinates in 2(e)(iii)\ny and x are coordinates of a point of the line m is gradient\ncalculated in 2(e)(iii) E.g.\n–1.94 = [2(e)(iii)  –2.24] + log k\nAND Log k = –1.94 – [2(e)(iii)  –2.24]\ndetermination of log k k = 10–1.94 – [2(e)(iii) x –2.24]\nAND A 1.55 to be used in place of 2(e)(iii)\ndetermination of k A gradient from 2(e)(iii) expressed as a fraction or calculator\nvalue\nA correct answer to 2 or more SF\nI RE after 2nd SF\n\n2(e)(v) Method 1: 1 A other suitable methods\n[2(e)(iii) – 1.5]  100 OR [1.5 – 2(e)(iii)] = Q%\n1.5 1.5 A use of 3/2 instead of 1.5 for true gradient\nAND\n(if Q% less than 4.5) AND (measurement) error A 1.55 for their value of 2(e)(iii)\nOR\n(if Q% greater than 4.5) AND other factors If their gradient in 2(e)(iii) is 1.50\nA 0% to be called ‘an error’\nMethod 2: A statement ‘there is no error’\n4.5 % of 1.5 = 0.0675 AND 1.5 + 0.0675 = 1.5675 AND / OR\n1.5 – 0.0675 = 1.4325 But a qualifying statement about ‘measurement’ error is\nAND required.\n2(e)(iii) between 1.4325 and 1.5 AND / OR between 1.5\nAND 1.5675 AND measurement error\nOR\n2(e)(iii) is outside this range AND other factors\n© Cambridge University Press & Assessment 2026 Page 21 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(e)(vi) it is reliable 1 The answer must match the graph\nAND\nno anomalous results/points A ‘(all) the points are on or (very) near to line (of best fit)’\nA ‘outliers’ for ‘anomalous points’\nA ‘(straight) line passing through / touching all points’ but I\n‘straight line passing all points’\nOR I ‘line fits the data’\nA there is (only) one anomalous point, which must be\nidentified, which can be ignored (and all the other points are\non (or very near) line of best fit)\nit is not reliable I most of the points are on the line of best fit\nAND Not reliable lines of best fit will come from incorrect data\nthere is an anomalous point plotting causing an anomalous point.\nPoints near but not on the line are not anomalous unless the\ncandidate has indicated this by a circle etc or by identifying\nthe point using co-ordinates.\n© Cambridge University Press & Assessment 2026 Page 22 of 22\nTrace ID: PC-3F04F08B6B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/52 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 17, 18, 19, 20, 21, 22 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-52.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_52-p17.png", "../answer-assets/9701_s26_ms_52-p18.png", "../answer-assets/9701_s26_ms_52-p19.png", "../answer-assets/9701_s26_ms_52-p20.png", "../answer-assets/9701_s26_ms_52-p21.png", "../answer-assets/9701_s26_ms_52-p22.png" ] }, { "id": "9701-2026-mj-53-q01", "question_id": "9701-2026-mj-53-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 13, "status": "available", "reason": null, "text": "1(a) pipeclay triangle 1 A clay triangle / clay pipe triangle / triangle clay pipe\nI tripod, gauze, triangular holder\n\n1(b) to prevent spitting 1 A to prevent cracking / damaging / breaking crucible\nA to prevent losing solid / (mixture) X / solid jumping out\n(of crucible)\nA to prevent splitting\nA CaCl or Ca(NO ) for solid\n2 3 2\nR burning\n\n1(c) NO is a hazardous gas 1 A NO gas is toxic / poisonous / acidic / corrosive\n2 2\nA NO is an irritant to eyes / by inhaling\n2\nA NO (g) for NO gas\n2 2\nI flammable, harmful\nR answers about O\n2\n\n1(d)(i) M1 mass lost on heating is 17.81 – 16.53 g = 1.28 g 2 award 2 marks for 0.0237\nrelative molecular mass of (4NO + O ) = 216.0 award 1 marks for 0.0278, 0.0618, 0.135 and 0.0156\n2 2\n1.28 g is 1.28 / 216 mol = 0.005926 mol of (4NO + O )\n2 2\n1.28 g is NO and O in the ratio of 4:1\n2 2\nrelative mass of (4NO + O ) = 4  14.0 + 10  16.0 =\n2 2\n216.0 g mol–1\n1.28 g is 1.28 / 216 mol = 0.005926 mol of (4NO + O )\n2 2\nNO is 4 this.\n2\nA 2 or more SF\nM2 number of moles NO = 4  0.005926 = 0.0237 mol ecf from incorrect M1:\n2\nA a number of moles  4\n[A if assume all gas released to be NO = 0.0278 mol\n2\nA if 5.07 g is assumed to be all Ca(NO ) then\n3 2\n5.07 / 164.1 = 0.0309 mol Ca(NO ) so 2  0.0309 =\n3 2\n0.0618 mol\nA if 3.79 g is assumed to be CaO so 3.79 / 56.1 =\n0.06756 mol CaO then 2  0.06756 = 0.135 mol\nA if 1.28 g is assumed to be Ca(NO ) then 1.28 / 164.1\n3 2\n= 0.00780 mol Ca(NO ) so 2  0.0078 = 0.0156 mol]\n3 2\n© Cambridge University Press & Assessment 2026 Page 7 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(d)(ii) Correct working must be shown. 1 A the following values are amongst acceptable values\nfor the answer following a correct expression 0.06,\npercentage error = (2  0.005 / 17.81)  100% = 0.0561 0.056, 0.05 etc.\nI ignore missing  100 if correct answer is given\n(2  0.005) may be seen as (0.01 / 2  2)\n\n1(d)(iii) increase the mass of mixture X used 1 A any wording that implies increasing the mass of X\ne.g. weigh a larger sample / reactant / contents\nA use a bigger amount of X / Use a larger quantity of X /\nuse more X etc.\nA use a larger mass of ‘crucible + X’\nA ‘use a larger mass’ or more than 5.07 g\nI ‘use a large mass’ alone\nI repeat and take mean\nI changes to heating e.g. heat for longer\nR more accurate balance or any other change to\napparatus e.g. increase mass of crucible alone\nR incorrect use of weight for mass\nR heat to constant mass\n\n1(e) M1 mol of Ca(NO ) in original sample = 0.5  0.0223 mol = 2 award 2 marks for 36.7\n3 2\n0.01115 mol A more SF e.g. 1.829715 g 36.741265%\nA 36.7 rounded to 37 for M2\nI rounding errors after 3SF.\nM2 mass Ca(NO ) in original sample = 0.01115 mol  If M1 is rounded to 0.0112 mol and then used in M2\n3 2\n164.1 g mol–1 = 1.829715 g award M1 only\n% by mass Ca(NO ) = 1.829715 / 4.98  100% = 36.7% ecf for M2 from incorrect M1 and allow final answer to\n3 2\nbe rounded to 2SF\n© Cambridge University Press & Assessment 2026 Page 8 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(f)(i) heat to constant mass 1 A description of the process of heating to constant\nmass e.g. reheat and reweigh until mass remains\nconstant\nA repeat steps 3 and 4 to constant mass\nresponse must include (re)heat and constant mass\n\n1(f)(ii) The percentage by mass will be smaller. 1 A NO or O or both for gas\n2 2\nAND A amount / mol for mass\nmass of gas lost is smaller (than the maximum possible)\nOR I not all NO released\n2\nmass of solid remaining after heating is higher (than actual mass) I calculated mass of Ca(NO ) (that decomposes) is\n3 2\nlower than actual value\n\n1(g) M1 attempted use of ice / cold water for condensing the gas 2 M1 A coolant/cool water for ice\nwithout loss of the NO as gas\n2\nIf a condenser is seen, max 1 mark\n• must be angled downwards towards unsealed\nM2 in an unsealed collection vessel surrounded by iced / cold collection vessel\nwater • must show water in and out of jacket\n• ignore position of water in and out\n• condensed NO must be delivered into a vessel\n2\nsurrounded by ice/cold water.\nIf a condenser is delivering into a u-tube and ice, then 2\nmarks can be awarded.\n© Cambridge University Press & Assessment 2026 Page 9 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_53-p07.png", "../answer-assets/9701_s26_ms_53-p08.png", "../answer-assets/9701_s26_ms_53-p09.png" ] }, { "id": "9701-2026-mj-53-q02", "question_id": "9701-2026-mj-53-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 17, "status": "available", "reason": null, "text": "2(a) only (direct) light from the LED should be allowed to reach the 1 A to prevent outside light entering (the colorimeter)\nphotocell A to prevent outside light reaching the photocell\nfor outside light A external light / background light\n/surrounding light\nfor photocell A detector\nA no (indirect) light via reflections\nA the only source of light is from the LED\nI other light for outside light\nI light absorbed unless it is absorbed by the photocell\n\n2(b)(i) 1 Must be 3 DP\nconcentration of resistance of\nabsorbance, A\nI –(aq) / mol dm–3 photocell, R / k\n3 s\n3.60  10-6 34.28 0.839\n2.94  10-6 23.42 0.684\n2.20  10-6 16.01 0.519\n1.52  10-6 9.68 0.300\n7.90  10-7 7.50 0.189\n\n2(b)(ii) distilled water (with few drops of starch solution) 1 A (acidified) I–(aq)/solution (with few drops of starch\nsolution)\nA acidified I– ions (with few drops of starch solution)\nA I – solution/(aq) with no starch\n3\n\n2(b)(iii) concentration of I – 1\n3\n© Cambridge University Press & Assessment 2026 Page 10 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b)(iv) M1 all 5 points plotted correctly 2 Place plotting tick on grid above line of best fit.\nPlace line tick in left-hand margin\nPlace a cross beside an incorrect point\nM2 straight line of best fit through origin passing through or close ecf from 2(b)(i)\nto all points except anomaly (point 2)\nR if point 2 is on a vertical gridline\nR if point 4 is on a horizontal gridline\nR line of best fit that does not extend from the origin to\ncover all plotted points.\nR if line of best fit is one square or more from origin on\neither axis\nA line of best fit between the first and last (correctly\nplotted) points is acceptable as long as it goes close\nenough to the origin. All points except anomaly should\nbe on the line of best fit.\nPoint marked as anomalous (ringed) should not be\nused in determining the line of best fit.\nA ecf acceptable line of best fit drawn for incorrectly\nplotted points.\nR line of best fit if the points (except anomaly) not\napproximately evenly distributed on both sides of line\nI extra points added which are used in later calculations\nI remnants of lines that the candidate has tried to erase\nR curved line\nR double or “sketchy” lines.\n© Cambridge University Press & Assessment 2026 Page 11 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b)(v) most anomalous point circled 1 Expected answer is point 2 at (1.52  10–6, 0.300 below\nthe line of best fit)\nAND\nA light intensity (in colorimeter) increased\nany one reason from:\n• not enough starch added (in step 2) A ora for ecf point above line of best fit\n• less than 10.0 cm3 solution transferred (from boiling tube to\nmeasuring cylinder in step 4) A light got into the colorimeter/black box not shut (no\n• solution (from step 2) was left for less than 10 minutes (in ora for point above line of best fit)\nstep 3)\nI reading of absorbance was too low\nI comments about temperature/heat\nI starch not mixed properly\nI systematic error\nI evaporation\nI reference to rates\n\n2(c)(i) to allow all the I – ions form a complex (with the starch) 1 A for complete reaction (of the I – ions and starch)\n3 3\nA enough time for the complex to form completely\nI to reach equilibrium\nI reference to rate of reaction\nI all the starch reacts\n\n2(c)(ii) 10 cm3 volumetric pipette 1 A graduated pipette as long as volume is not < 10 cm3\nA 10 cm3 pipette\nA volumetric pipette\nI ‘pipette’ alone\nR burette\n\n2(c)(iii) M1 absorbance = 0.640 2 A 0.64 with working shown\nConcentration should be read from the graph to nearest\n½ square.\nM2 concentration read from the graph to nearest ½ square and I any figures after 3SF\ngiven to 3 SF I trailing zeros (allow 2.7 for 2.70)\nA non-standard form\n2.75  10–6 mol dm–3 is a common answer\n© Cambridge University Press & Assessment 2026 Page 12 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(c)(iv) (15 / 1000)  2(c)(iii) 1 A 2 SF minimum\nA calculator value from 2(c)(iii) (e.g. 2.725 used for\n2.73 given in 2(c)(iii))\nI trailing zeros\nA non-standard form\n\n2(c)(v) 2(c)(iv)  (1000 / 7500) 1 If using the given value, then:\n3.23  10–8  1000 / 7500 = 4.31  10–9 mol dm–3\nA calculator value from 2(c)(iv)\nA 2 SF minimum\nA non-standard form\n\n2(c)(vi) increases 1\nAND\n(the value of) the concentration of I – increases (at same\n3\nabsorbance)\n\n2(c)(vii) 9.87 + 0.14  9.87 = 11.25 AND 13.66 – 0.14  13.66 = 11.75 1 A any set of calculations which gives 11.25 and 11.75\nAND\nthe concentration of O is different (because the extremes of the\n3\nranges do not overlap)\n\n2(c)(viii) R increases 1 A light travels through a deeper solution\ns\nAND A light interacts with more I – particles\n3\nthe light has a longer path length (through the solution) A absorbance increases with depth (of solution)\nI light interacts with more I – solution(aq)\n3\nI more light is absorbed alone\nI measuring cylinder is deeper\n\n2(c)(ix) concentration of I – will be higher giving a lower percentage error 1 A amount / mol of I – for concentration of I –\n3 3 3\nI uncertainty for percentage error\nA uncertainty percentage error for percentage error\nalone\n© Cambridge University Press & Assessment 2026 Page 13 of 13\nTrace ID: PC-B8B10956DE\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/53 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 10, 11, 12, 13 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-53.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_53-p10.png", "../answer-assets/9701_s26_ms_53-p11.png", "../answer-assets/9701_s26_ms_53-p12.png", "../answer-assets/9701_s26_ms_53-p13.png" ] }, { "id": "9701-2026-mj-54-q01", "question_id": "9701-2026-mj-54-q01", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "54", "question_number": 1, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 16, "status": "available", "reason": null, "text": "1(a) (re)heat and (re)weigh to constant mass 1 A “heat to constant mass”\nA description of the process of heating to constant mass\nA “crucible and residue” or “sample” for “crucible and\ncontents”\nR “heat to constant weight”\n\n1(b)(i) (mass of Na CO •xH O(s) added / g =) 4.89 1 R any other values.\n2 3 2\n(mass of residue / g =) 2.06\nAND\n(mass of water lost =) 2.83\n\n1(b)(ii) increase the mass of Na CO •xH O(s) used 1 A use more (than 4.89 g / mass calculated in 1(b)(i) of)\n2 3 2\nNa CO •xH O(s)\n2 3 2\nA use of “amount” for “mass”\nA use of “solid” or “crystals” or “contents” or “reactant” or\n“sample” or “hydrated sodium carbonate” or “sodium\ncarbonate crystals” for “Na CO •xH O(s)”\n2 3 2\nA increase the “mass of crucible and contents”\nI changes to heating\nI repeat and take mean\nR changing the crucible. e.g. increase the mass of crucible\nalone\nR “anhydrous sodium carbonate” / “sodium carbonate”\nalone / “powder” / Na CO for “Na CO •xH O (s)”\n2 3 2 3 2\nR incorrect use of “weight” for “mass” but A ecf from 1(a).\nR heat to constant mass\n© Cambridge University Press & Assessment 2026 Page 7 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(c) M1 n(H O) = 2.83 / 18 = (0.157(222))\n2\nAND\nn(Na CO ) = 2.06 / 106 = (0.0194(34))\n2 3\nM2 x =\n© Cambridge University Press & Assessment 2026 Page 8 of 19\nn\nn\n( N\n( H\na\n2\n2\nO\nC O\n)\n3 )\n2 M1\nA ecf from 1b(i)\nA rounding of amounts e.g. n(H O) = 0.16 and n(Na CO ) =\n2 2 3\n0.02\nA valid alternative calculation routes\nM2\n= (0.157 / 0.0194 = 8.09 → 8) A ecf from M1 if values used in M2 are calculated n(H O) and\n2\nn(Na CO ).\n2 3\nR non-integer value for x. e.g. 8.0\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(d)(i) M1 add a (small) volume of (distilled) water (to the small 3 A ‘solution' produced in M1 or M2 for “dissolve the solid”.\nbeaker) A “solid disappears” for dissolves.\nAND A extra transfers as long as the exact number of moles of\ndissolve the solid / Na CO •xH O(s) solid end up in the volumetric flask via rinsing. (Allow greater\n2 3 2\nvolume added when accommodated by vessel used.)\nA prepare / make a solution for dissolve.\nI use of measuring cylinder / stirring / glass rod\nI description of measuring mass / incorrect mass used\nR water volume > 50 cm3 if specified OR fill (50 cm3) beaker\nM2 A “rinse the beaker” or “wash the beaker” in place of\nTransfer the solution and washings (using a funnel) “washings”\nAND BOD assume rinsing and washing is of the beaker into the\ninto a 250 cm3 volumetric flask. flask if not clearly indicated otherwise e.g. to wash the\napparatus\nA 250 volumetric flask / 250 cm3 graduated flask / 250 cm3\nstandard flask\nA M2 If solid added directly to 250 cm3 volumetric flask AND\nbeaker (+ funnel) are rinsed, and washings added to flask\nAND the solid dissolved in distilled / deionised water in flask\nbefore M3\nI use of funnel / glass rod\nI rinse the volumetric flask (with (distilled) water)(before\nadding solid / solution)\nR transfer solution with burette or pipette\nR weighing solid into flask\n© Cambridge University Press & Assessment 2026 Page 9 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(d)(i) M3 Make up to the (calibration) mark / line (with distilled A ‘250 cm3 calibration’, ‘line’ or other similar for ‘mark’.\nctd. water). A deionised water, DW, dH O, diH O for ‘distilled water’.\n2 2\nAND A make up to 250 cm3 (with distilled water).\nthen invert (the stoppered flask) (several times) A ‘shake’, ‘swirl’ and ‘invert’ for ‘invert flask’. Must describe\naction, not ‘mix’ alone.\nA M2 and M3 if all correct except volume used to dissolve in\nM1 > 50 cm3 < 250 cm3 if specified.\nA use of pipette / burette to fill to mark.\nI top up / fill alone.\nI homogenise / mix\nR stir\nR M3 if solid not dissolved before making up to the mark. If\npurpose of inverting / shaking is to dissolve solid.\nA ecf M3 for volume > 250 cm3 (i.e. I volume from M1 and\nM2 when marking M3)\nM3\nDistilled / deionised water only needs to be mentioned once\nto infer its use throughout. If not mentioned, then M3 not\nawarded.\n\n1(d)(ii) (0.01(00) mol dm–3) hydrochloric acid / HCl((aq)). 1 A (rinse with the) solution to be used (for the titration in the\nburette)\nA dilute hydrochloric acid\nI wash with distilled water / propanone before wash with\n(0.0100 mol dm-3) HCl\nR wash with distilled water / propanone after wash with\n(0.0100 mol dm-3) HCl\n© Cambridge University Press & Assessment 2026 Page 10 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(e)(i) (it is calculated from) titrations 1 and 3 1 A titrations 1 and 3 (are used), which are within 0.1(0) (cm³)\nof each other / are ±0.1(0) (cm3)\nAND A the difference between titres 1 and 3 is 0.1(0) (cm³)\nA the difference between titres 1 and 3 is equal or less than\nwhich are concordant. 0.1(0) (cm³)\nA the difference between titres 1 and 3 is within 0.1(0) (cm³)\n(of each other)\nA values for titrations 1 and 3. e.g. 25.25 and 25.15\nA values clearly indicated and annotated in Table 1.1\nI mention of not using titration 2\nR if titration 2 is used\n\n1(e)(ii) M1 n(HCl) in mean titre= 0.0252  0.0100 4 A ecf throughout\n= 2.52  10–4 (mol) A valid alternative calculation routes\nM2 n(Na CO ) in 25.0 cm3 of solution\n2 3\n= (2.52  10-4 ÷ 2) = 1.26  10–4 (mol)\nM3 n(Na 2 CO 3 ) in 250.0 cm3 of solution M3 A m(Na 2 CO 3 ) in 0.365 g = 1.26  10–4  10  106\n= 1.26  10–4  10 = 1.26  10–3 (mol) = 0.134 g\nM (Na CO •xH O) = 0.365 / 1.26  10–3 = 289.68 m(H 2 O) = 0.365 – 0.134 = 0.231 g\nr 2 3 2\nn(H O) = 0.231 / 18 = 0.012858\n2\nM4 x = (289.68 – 106) / 18 = 10.20 → 10 M4 x= 0.012858 / 1.26  10–3 = 10.20 → 10\nA ECF for M4 if values used are calculated n(H\n2\nO) and\nn(Na\n2\nCO\n3\n).\nMissing factor of 10 in M3 leads to x = 155.\nM4 still available. M4 R non-integer value for x, but A ecf from e.g. 8.0 in 1(c)\nfor e.g.10.0.\nMissing ÷ 2 in M2 leads to x = 2. M3 and M4 still available.\n© Cambridge University Press & Assessment 2026 Page 11 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(e)(iii) correct working must be shown 1 2 × ( ½ × 0.10)\nA  100 = 0.3976\n25.15\n2 × 0.05\nmaximum percentage error =  100 = 0.3976(14) 0.05 + 0.05\n25.15 A  100 = 0.3976\n25.15\nA the following values for the answer following a correct\nexpression – 0.4, 0.40, 0.39, 0.397 etc\nI missing  100 if correct answer given\nR incorrectly calculated answer with correct working\nR (0.005  2 / 25.15)  100 = 0.0398\n\n1(f) (effect on value of =) increases 1 A ‘higher’ or ‘greater’ or ‘overestimated’ for ‘increases’ etc\nA “decreases” or “smaller” or “underestimated” for “lower”\nAND etc\nA ‘sodium carbonate’ or ‘Na CO •xH O(s)’ for ‘Na CO ’\n2 3 2 2 3\nn(Na CO ) is lower OR n(H O) is greater (than true value). A ‘number of moles’ or ‘mole number’ or ‘amount’ for n\n2 3 2\nI powder / crystals\nI measured volume of solution A / Na CO (aq) is less\n2 3\nOR I recorded volume of solution A / Na CO (aq) is larger\n2 3\n(effect on value of x =) no / negligible effect A ‘slight increase’ for ‘negligible’\nAND\nvalue of x is to the nearest integer so (small) change in\nvolume does not affect value", "source_pages": [ 7, 8, 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_54.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-54.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_54-p07.png", "../answer-assets/9701_s26_ms_54-p08.png", "../answer-assets/9701_s26_ms_54-p09.png", "../answer-assets/9701_s26_ms_54-p10.png", "../answer-assets/9701_s26_ms_54-p11.png", "../answer-assets/9701_s26_ms_54-p12.png" ] }, { "id": "9701-2026-mj-54-q02", "question_id": "9701-2026-mj-54-q02", "subject": "9701", "year": 2026, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "54", "question_number": 2, "topic": "Planning, analysis and evaluation", "topic_slug": "9701-experimental-skills", "marks": 14, "status": "available", "reason": null, "text": "2(a)(i) n(NH Cl) = 5.00 / 53.5 = 0.0935 1 R any answer other than 3.46 g\n4\nmass of Ca(OH) = 0.0935 / 2  74.1 = 3.46 g\n2\n© Cambridge University Press & Assessment 2026 Page 12 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a)(ii) hazardous gas (ammonia) produced 1 A toxic / corrosive / irritant for ‘hazardous’\nI flammable\nI bad smell\nI harmful / dangerous / ’harmful to health’\nI alkaline\n\n2(a)(iii) to remove water (vapour) (from (ammonia) gas) 1 A to absorb water (from gas)\nA to avoid water passing into gas jar\nA to dry (the gas)\nA to ensure (ammonia) gas does not contain water\nI to flush any air out of the tube.\nI dehydrate / dehydration\nI to purify\nI to prevent ammonia dissolving in water.\nI ‘CaO reacts with water’ alone (no reference to removing\nwater)\n(no mark – to make sure only ammonia reaches the gas jar)\nR to remove chlorine\nR to react with ammonia\nR “to dry water” alone\n\n2(a)(iv) ammonia is less dense than (most gases in) air 1 A ‘NH ’for ‘ammonia’\n3\nA “ammoniaisless dense than oxygen and nitrogen”\nA “ammonia has a smaller density than (most gases in) air”\nA ammonia has a smaller M than mean M for air\nr r\nI ‘lighter’\nI it will displace the air\nI atmosphere\nI ‘less dense than oxygen alone or nitrogen alone’\nR less dense than any other named gas. e.g. chlorine\n© Cambridge University Press & Assessment 2026 Page 13 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b) ammonia is soluble (in water) 1 A ‘NH ’for ‘ammonia’\n3\nA ammonia would react with water\nA correct species in balanced equation for reaction of\nammonia with water, ignore state symbols (allow =, → or ⇌)\ne.g. NH + H O ⇌ NH + + OH−\n3 2 4\nA suck back could occur\nA ammonia would / can / will dissolve (in the water)\nI incorrect reaction equations, products etc\n© Cambridge University Press & Assessment 2026 Page 14 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(c)(i) M1 all six points plotted correctly 2 M1\nIf point should be on a line, it must be on the line.\nIf it should not be on the line, it must not be so.\nBlobs/thick crosses should be less than ½ a small square\nacross and be correctly centred.\nM1 not awarded if centre of at least one cross is clearly off a\nline it should be on, on a line it should not be or in the wrong\nsmall grid square.\nNote: Other points are correctly plotted.\nIf not clearly plotted at a different y value, assume point at\n0s is present and obscured by the point at same time for\ntungsten. (In this case there is no need to check for this\npoint).\nM2\nM2 line of best fit drawn\nA ecf acceptable best fit line drawn for incorrectly plotted\npoints\nI remnants of lines that have been rubbed out\nR double or “sketchy” lines.\nR line that does not extend across full range of x-values for\nplotted points. (A ½ small square tolerance at each end)\n\n2(c)(ii) [NH ] (in mol dm–3) 1 A ‘concentration of ammonia’ or ‘concentration of NH ’ for\n3 3\n‘[NH ]’\n3\nA [NH ] / mol dm–3\n3\nI type of catalyst\n© Cambridge University Press & Assessment 2026 Page 15 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(d)(ii) there are no anomalous results 1 A there are no points off the line (of best fit)\nA (all) the points are (very) close to / on the line (of best fit)\nA (straight) line passes through / touching (all) points\nR (straight) line passing all points\nA “outliers” for “anomalous results”\nA “anomalous” to mean “anomaly” or “anomalous result”\nA “plots” for “points”\nI the line fits the data\nI “values” for “points”\nI most points are on / close to the line of best fit\n\n2(d)(iii) yes (it does support the student’s conclusion) 1 A inferred agreement with student e.g. saying the reaction is\nzero order\nAND R no, (it does not support the student’s conclusion)\n(the line has) constant gradient A the line is straight\nA the graph / line is linear\nA t decreases (with time)\n½\nA rate (of decrease of [NH ]) is constant (over the time\n3\nperiod shown by the graph)(at a particular temperature)\nA [NH ] has a linear relationship with time\n3\n© Cambridge University Press & Assessment 2026 Page 16 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(e) For straight line of best fit:\nctd. M1 any two different points from line of best fit correctly\nrecorded (x ,y ) and (x ,y ).\n1 1 2 2\nCan be awarded it is clear an attempt has been made to\ncalculate a gradient.\nM2\nCalculates gradient of line correctly. (y -y ) / (x -x )\n2 1 2 1\nAND\nCalculates t correctly using gradient and y = mx + c to work\n½\nout value of x (t ) at half value of y (y at t=0).\n½ 0\nt = ((y / 2) – c) /m\n½ 0\nFor a curved line of best fit:\nM1 any two different points from line of best fit correctly\nrecorded (x ,y ) and (x ,y ).\n1 1 2 2\nCan be awarded if a clear attempt has been made to use\ncalculation below.\nM2 states “assuming first order” owtte\nAND\nhalf-life correctly calculated as follows:\ncalculate 𝑘 using coordinates from line of best fit\nln(y1/ y2)\nk =\nx2−x1\nexpected value k 1.4110−3 s−1\nfor a first-order reaction:\n© Cambridge University Press & Assessment 2026 Page 18 of 19\nt1\n/2\n=\n0 . 6\nk\n9 3\nTolerance x value ± half a square (10 s) from stated y-value.\nA coordinates 1 and 2 recorded in either order\nM2 A ecf for correctly evaluated t from points on line of best\n½\nfit if only error in M1 is that one of the x values has been\nincorrectly read from graph.\nM2 A ecf correctly evaluated t from otherwise correct LOBF\n½\npoints both written (y, x)\nM2 A ecf for minor slip in recording correct coordinates. e.g.\nmissing  10–3 (M1 not awarded)\nNo ecf M2 if any two mistakes made in recorded coordinates\nR negative value for M2\nexpected value t = 491 s approx.\n½\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(f) (four half-lives: 4  550 =) 2200 s 1\n© Cambridge University Press & Assessment 2026 Page 19 of 19\nTrace ID: PC-3193057FA4\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP97a01p/5a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/54 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/", "source_pages": [ 12, 13, 14, 15, 16, 18, 19 ], "source_pdf": "_source-pdfs/2026-May-June/ms/9701_s26_ms_54.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/chemistry-9701/2026-may-june/9701-chemistry-june-2026-mark-scheme-54.pdf?download=true", "html": "9701-experimental-skills/answers.html", "image_paths": [ "../answer-assets/9701_s26_ms_54-p12.png", "../answer-assets/9701_s26_ms_54-p13.png", "../answer-assets/9701_s26_ms_54-p14.png", "../answer-assets/9701_s26_ms_54-p15.png", "../answer-assets/9701_s26_ms_54-p16.png", "../answer-assets/9701_s26_ms_54-p18.png", "../answer-assets/9701_s26_ms_54-p19.png" ] } ] }