Cambridge A Level Chemistry 9701 — 2025 Oct/Nov Paper 5 · Variant 4

This document has 12 pages. [Turn over Cambridge International AS & A Level * 6 6 8 8 5 9 2 5 4 4 * DC (DE) 360152/3 © UCLES 2025 CHEMISTRY 9701/54 Paper 5 Planning, Analysis and Evaluation October/November 2025 1 hour 15 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You may use a calculator. ● You should show all your working and use appropriate units. INFORMATION ● The total mark for this paper is 30. ● The number of marks for each question or part question is shown in brackets [ ]. ● The Periodic Table is printed in the question paper. ● Important values, constants and standards are printed in the question paper. ,










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2 9701/54/O/N/25 © UCLES 2025 1 A student uses a technique called the Winkler method to determine the mass of oxygen dissolved in a sample of water from a lake. Two solutions, X and Y, are prepared. Solution X is 2.30 mol dm–3 aqueous manganese(II) sulfate, MnSO4(aq). Solution Y is alkaline aqueous potassium iodide, KI(aq). (a) Calculate the mass of solid hydrated manganese(II) sulfate, MnSO4•H2O(s), needed to make 100.0 cm3 of solution X. Give your answer to two decimal places. mass of MnSO4•H2O(s) = … g [1] (b) The student is given a small beaker containing the mass of MnSO4•H2O(s) calculated in (a). Describe how the student should prepare exactly 100.0 cm3 of solution X. Include the names and capacities of each piece of key apparatus used. Write your answer using a series of numbered steps. … … … … … … … [3] * 0000800000002 * ,










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3 9701/54/O/N/25 © UCLES 2025 [Turn over (c) Solution Y is prepared as follows. step 1 Place 100 cm3 of distilled water in a 250 cm3 beaker. step 2 Add about 8 g of solid sodium hydroxide, NaOH(s), and stir to dissolve. step 3 Cool the solution to room temperature using an ice‑bath. step 4 Repeat steps 2 and 3 until a total of 32 g of NaOH(s) has been dissolved. step 5 Dissolve about 14 g of potassium iodide, KI(s), into the solution formed in step 4. (i) Solution Y is corrosive. Other than wearing safety goggles, state one safety precaution that the student should take when preparing solution Y. … [1] (ii) Suggest why the solution is cooled in step 3. … … … [1] * 0000800000003 * ,










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4 9701/54/O/N/25 © UCLES 2025 (d) The student uses the following procedure to determine the mass of oxygen dissolved in a sample of water from the lake. step 1 Collect a 250 cm3 sample of lake water in a bottle. step 2 Add 1 cm3 of solution X and 1 cm3 of solution Y to the bottle. step 3 Immediately stopper the bottle, ensuring as little air as possible is trapped. step 4 Shake the bottle to mix its contents. A brown precipitate, manganese(III) hydroxide, Mn(OH)3(s), is formed. step 5 Add 1.5 cm3 of concentrated sulfuric acid to the contents of the bottle. The precipitate dissolves, and iodine is formed. step 6 Dilute this solution to exactly 500.0 cm3 using distilled water to form solution Z. step 7 Transfer 25.0 cm3 of solution Z into a conical flask, and titrate with 1.00 × 10–3 mol dm–3 aqueous sodium thiosulfate, Na2S2O3(aq). Add 1 cm3 of starch solution near to the end‑point. step 8 Repeat step 7 as many times as necessary. (i) Suggest why it is important to avoid trapping air inside the bottle in step 3. … … [1] (ii) Identify the piece of apparatus that the student should use to transfer the 25.0 cm3 of solution Z in step 7. … [1] (iii) Suggest why starch solution is added in step 7. … [1] * 0000800000004 * ,










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5 9701/54/O/N/25 © UCLES 2025 [Turn over (e) The student records the results shown in Table 1.1. Table 1.1 rough titration titration 1 titration 2 titration 3 final burette reading / cm3 13.60 12.75 26.20 14.50 initial burette reading / cm3 0.00 0.05 13.15 1.35 titre / cm3 13.60 (i) Complete Table 1.1 and calculate the mean titre. mean titre = … cm3 [2] (ii) Explain why the student does not need to carry out any further titrations. … … [1] (iii) Calculate the percentage error in the measurement of the titre for titration 3. Show your working. percentage error = … % [1] * 0000800000005 * ,










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6 9701/54/O/N/25 © UCLES 2025 (f) The following equations show the reactions that take place during the procedure in (d). steps 2, 3 and 4 4Mn2+(aq) + 8OH–(aq) + O2(aq) + 2H2O(l) 4Mn(OH)3(s) step 5 2Mn(OH)3(s) + 2I–(aq) + 6H+(aq) I2(aq) + 6H2O(l) + 2Mn2+(aq) step 7 I2(aq) + 2S2O3 2–(aq) 2I–(aq) + S4O6 2–(aq) (i) Calculate the amount, in mol, of iodine, I2(aq), in 25.0 cm3 of solution Z. amount of I2(aq) = … mol [1] (ii) Use your answer to (f)(i) and the equations given to calculate the amount, in mol, of dissolved oxygen, O2(aq), in 500.0 cm3 of solution Z. amount of O2(aq) in 500.0 cm3 of solution Z = … mol [1] (iii) Dissolved oxygen content, mg dm–3, is the mass of oxygen dissolved in water. Use your answer to (f)(ii) to calculate the dissolved oxygen content in the lake water collected in step 1. [If you were unable to obtain an answer to (f)(ii), then use amount of O2(aq) in 500.0 cm3 of solution Z = 7.12 × 10–5 mol. This is not the correct answer.] dissolved oxygen content = … mg dm–3 [1] [Total: 16] * 0000800000006 * ,










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7 9701/54/O/N/25 © UCLES 2025 [Turn over 2 A student uses the following method to investigate the kinetics of the reaction between iodine and tin to produce tin(IV) iodide, SnI4. step 1 Rinse a block of tin with distilled water and then rinse it with propanone. step 2 Place 50 cm3 of a 0.400 mol dm–3 solution of iodine dissolved in methylbenzene in a 100 cm3 beaker. step 3 Suspend the block of tin from a three decimal place balance as shown in Fig. 2.1. Start a timer. step 4 Record the balance reading every 100 seconds. block of tin wire iodine dissolved in methylbenzene balance 4.979 Fig. 2.1 (a) (i) Suggest why the student rinses the block of tin with propanone after rinsing it with distilled water in step 1. … … [1] (ii) Suggest why water is not used as the solvent for iodine. … … [1] * 0000800000007 * ,










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8 9701/54/O/N/25 © UCLES 2025 (iii) Suggest why a three decimal place balance is more suitable than a two decimal place balance for this experiment. … [1] (iv) Suggest a control experiment that could be used to verify that the loss in mass of tin is caused by reaction with iodine and not any other factor. … … [1] (b) The student’s results are shown in Table 2.1. Complete Table 2.1. Table 2.1 time / s balance reading / g total mass of tin reacted / g 0 4.979 0.000 100 4.910 200 4.859 300 4.761 400 4.688 500 4.620 [1] (c) (i) Use the results from Table 2.1 to plot a graph on the grid in Fig. 2.2 to show the relationship between total mass of tin reacted and time. Use a cross (×) to plot each data point. Draw a straight line of best fit. * 0000800000008 * ,










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9 9701/54/O/N/25 © UCLES 2025 [Turn over total mass of tin reacted / g time / s 0 100 200 300 400 500 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Fig. 2.2 [2] (ii) Circle the point on the graph in Fig. 2.2 that you consider to be most anomalous. Suggest one reason why this anomaly may have occurred during this experimental procedure. Assume all measurements of mass are accurate. … … [1] * 0000800000009 * ,










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10 9701/54/O/N/25 © UCLES 2025 (d) Use your graph in Fig. 2.2 to determine the gradient of the line of best fit. State the coordinates of both points you used in your calculation. These must be selected from your line of best fit. Give your gradient to three significant figures. coordinates 1 … coordinates 2 … gradient = … [2] (e) Another student makes various concentrations of solutions of iodine dissolved in methylbenzene by dilution of the 0.400 mol dm–3 I2 solution. The student repeats the experiment at a different temperature using these solutions. Table 2.2 volume of 0.400 mol dm–3 I2 solution used / cm3 volume of methylbenzene used / cm3 [I2] / mol dm–3 relative rate of reaction 100.0 0.0 0.400 4.76 0.300 3.57 0.200 2.35 0.100 1.15 (i) Complete Table 2.2 by adding the volumes of solutions that are mixed to make 100.0 cm3 of a solution of iodine dissolved in methylbenzene for each required concentration. [1] (ii) Identify the independent variable in this experiment. … [1] * 0000800000010 * ,










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11 9701/54/O/N/25 © UCLES 2025 (iii) The student concludes that the rate equation for the reaction between iodine and tin is as follows. rate = k [I2]2 State whether the results support the student’s conclusion. Explain your answer using values from Table 2.2. … … … … … … [2] [Total: 14] Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) * 0000800000011 * ,









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12 9701/54/O/N/25 © UCLES 2025 To avoid the issue of disclosure of answer‑related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Group The Periodic Table of Elements 1 H hydrogen 1.0 2 He helium 4.0 1 2 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 3 Li lithium 6.9 4 Be beryllium 9.0 atomic number atomic symbol Key name relative atomic mass 11 Na sodium 23.0 12 Mg magnesium 24.3 19 K potassium 39.1 20 Ca calcium 40.1 37 Rb rubidium 85.5 38 Sr strontium 87.6 55 Cs caesium 132.9 56 Ba barium 137.3 87 Fr francium – 88 Ra radium – 5 B boron 10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 81 Tl thallium 204.4 6 C carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sn tin 118.7 82 Pb lead 207.2 22 Ti titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium 178.5 104 Rf rutherfordium – 23 V vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium – 24 Cr chromium 52.0 42 Mo molybdenum 95.9 74 W tungsten 183.8 106 Sg seaborgium – 25 Mn manganese 54.9 43 Tc technetium – 75 Re rhenium 186.2 107 Bh bohrium – 26 Fe iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium – 27 Co cobalt 58.9 45 Rh rhodium 102.9 77 Ir iridium 192.2 109 Mt meitnerium – 28 Ni nickel 58.7 46 Pd palladium 106.4 78 Pt platinum 195.1 110 Ds darmstadtium – 29 Cu copper 63.5 47 Ag silver 107.9 79 Au gold 197.0 111 Rg roentgenium – 30 Zn zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200.6 112 Cn copernicium – 114 Fl flerovium – 116 Lv livermorium – 7 N nitrogen 14.0 15 P phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 83 Bi bismuth 209.0 8 O oxygen 16.0 16 S sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium – 9 F fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 53 I iodine 126.9 85 At astatine – 10 Ne neon 20.2 18 Ar argon 39.9 36 Kr krypton 83.8 54 Xe xenon 131.3 86 Rn radon – 113 Nh nihonium – 115 Mc moscovium – 117 Ts tennessine – 118 Og oganesson – 21 Sc scandium 45.0 39 Y yttrium 88.9 57–71 lanthanoids 89–103 actinoids 57 La lanthanum 138.9 89 Ac lanthanoids actinoids actinium – 58 Ce cerium 140.1 90 Th thorium 232.0 59 Pr praseodymium 140.9 91 Pa protactinium 231.0 60 Nd neodymium 144.2 92 U uranium 238.0 61 Pm promethium – 93 Np neptunium – 62 Sm samarium 150.4 94 Pu plutonium – 63 Eu europium 152.0 95 Am americium – 64 Gd gadolinium 157.3 96 Cm curium – 65 Tb terbium 158.9 97 Bk berkelium – 66 Dy dysprosium 162.5 98 Cf californium – 67 Ho holmium 164.9 99 Es einsteinium – 68 Er erbium 167.3 100 Fm fermium – 69 Tm thulium 168.9 101 Md mendelevium – 70 Yb ytterbium 173.1 102 No nobelium – 71 Lu lutetium 175.0 103 Lr lawrencium – * 0000800000012 * ,









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This document consists of 11 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/54 Paper 5 Planning, Analysis and Evaluation October/November 2025 MARK SCHEME Maximum Mark: 30 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 11 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alon gside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 11 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thre sholds or grade descriptors in mind. Science-Specific Marking Principles 1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks should not be awarded if the keywords are used incorrectly. 2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored. 3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation fro m other syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection). 4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where necessary and any exceptions to this general principle will be noted. 5 ‘List rule’ guidance For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided. • Any response marked ignore in the mark scheme should not count towards n. • Incorrect responses should not be awarded credit but will still count towards n. • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response. • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 11 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 11 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standard isation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning Unclear Information missing or insufficient for credit Benefit of the doubt given Contradiction in response otherwise markworthy, mark not given Incorrect point or mark not awarded Part of the correct answer has been seen. Full credit has not been awarded. Error carried forward applied Highlighted text Highlighting areas of text Incorrect or insufficient point ignored while marking the rest of the response On-page comment box Allows comments to be entered on the page

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 11 Annotation Meaning Off-page comment box Allows comments to be entered at the bottom of the RM Assessor marking window and then displayed when the associated question item is navigated to Rounding error Repeat error Blank page or part of script seen Error in number of significant figures Transcription error Correct point or mark awarded

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 11 Question Answer Marks 1(a) 38.87 g 1 1(b) M1 Dissolve the solid / MnSO4.H2O / manganese sulfate (in the beaker) using (a small volume of) distilled water M2 transfer / add to a (100 cm3 volumetric) flask AND rinse (with distilled water) M3 top up the 100 cm3 volumetric flask to mark (with distilled water) AND Invert the (stoppered) flask) 3 1(c)(i) (wear) chemically resistant gloves. 1 1(c)(ii) dissolving NaOH(s) is (very) exothermic 1 1(d)(i) Oxygen in trapped air could dissolve into the water sample 1 1(d)(ii) (25.0 cm3) volumetric pipette 1 1(d)(iii) To observe the end-point more clearly 1 1(e)(i) M1 rough 1 2 3 12.70 13.05 13.15 1 M2 13.10 cm3 1 1(e)(ii) titrations 2 and 3 are within 0.10 cm3 of each other 1

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 11 Question Answer Marks 1(e)(iii) working must be shown 2 0.05 13.15   100 = 0.760 1 1(f)(i) (e)(i) / (2  106) = 6.55  10–6 (mol) 1 1(f)(ii) (f)(i)  10 = 6.55  10–5 (mol) 1 1(f)(iii) (f)(ii)  1.28  105 = 8.38 (mg dm–3) 1 Question Answer Marks 2(a)(i) to increase the rate of drying / removal of (distilled) water 1 2(a)(ii) iodine is not (very) soluble (in water) 1 2(a)(iii) mass change is (too / very) small 1 2(a)(iv) repeat without any iodine 1 2(b) time / s mass of tin reacted / g 0 0.000 100 0.069 200 0.120 300 0.218 400 0.291 500 0.359 1

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 11 Question Answer Marks 2(c)(i) M1 points plotted correctly centre of all crosses should be on line for  value. y value should be on the line for point 1, 3 point 2, 6 – just below line / on line 4 – midway between two lines 5 – just above line / on line 1 M2 straight line of best fit drawn 1 2(c)(ii) most anomalous point circled (expected to be at time 200 s, point 3, below line of best fit) AND the balance/mass reading was taken before 200 s / too early 1 2(d) M1 two acceptable coordinates from line of best fit expressed in the form (x,y). M2 gradient correctly calculated from points listed for M1. answer correctly rounded to 3 significant figures. 2

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 11 Question Answer Marks 2(e)(i) volume of 0.400 mol dm–3 I2 solution used / cm3 volume of methylbenzene used / cm3 100.0 0.0 75.0 25.0 50.0 50.0 25.0 75.0 1 2(e)(ii) [I2] 1 2(e)(iii) M1 no AND the rate of reaction is proportional to the concentration of iodine / a conclusion based on calculations done for M2 1 M2 explained using actual values from columns 3 and 4 in Table 2.2 1

9701/54 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 11 Working for answers 1(a) n(MnSO4•H2O) = c  V = 2.3  0.1 = 0.23 mol mass MnSO4•H2O = 0.23  169.0 = 38.87 g 1(f)(i) n(S2O32-) = (13.10  0.001) / 1000 = 1.31  10–5 mol n(I2) in 25.00 cm3 of solution Z = n(S2O32–)/2 = 6.55  10–6 mol 1(f)(ii) n(O2) in 25.00 cm3 of Z = (f)(i) / 2 = 3.275  10–6 mol n (O2) in 500 cm3 of Z = 3.275  10-6  500 / 25 = 6.55  10–5 mol 1(f)(iii) n(O2) in 1dm3 of lake water = (f)(ii)  1000 / 250 = 2.62  10–4 mol dissolved oxygen content = (2.62  10–4  32.0  1000) = 8.38 mg dm–3