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Chemistry 9701/36 October/November 2023: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 3 Advanced Practical Skills 2, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_w23_qp_369701_w23_ms_369701/36/O/N/23

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Paper
Paper 3 Advanced Practical Skills 2
Total marks
40
Time
2 hours
Questions
3
Code on the paper
9701/36/O/N/23
9701_w23_ms_36.pdf

9701/36 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 112 marks

Question

When hydrogen peroxide is decomposed by a suitable catalyst, oxygen and water are produced. You will determine the concentration of a solution of hydrogen peroxide by decomposing it and measuring the mass of oxygen produced. FB 1 is aqueous hydrogen peroxide, H2O2. FB 2 is the catalyst, manganese(IV) oxide, MnO2. (a) Method ● Use the measuring cylinder to transfer 25.0 cm3 of FB 1 into a conical flask. ● Weigh the conical flask containing FB 1. Record the mass. ● Weigh the container with FB 2. Record the mass. ● Tip all the FB 2 into the conical flask. Swirl the flask gently. ● Weigh the container with any residual FB 2. Record the mass. ● Calculate the mass of FB 2 added. Record the mass. ● Leave the conical flask and its contents to stand for 30 minutes. Swirl the flask occasionally. While the reaction is taking place, begin work on Question 2 or Question 3. ● When the reaction is complete, weigh the flask and its contents. Record the mass. You will use FB 1 in Questions 2 and 3. I II III [3] (b) Calculations (i) Calculate the mass of oxygen liberated. mass of oxygen = …… g [1] (ii) Give the equation for the decomposition of hydrogen peroxide in (a). Include state symbols. …… [1] (iii) Calculate the concentration, in mol dm–3, of hydrogen peroxide in FB 1. concentration of H2O2 in FB 1 = …… mol dm–3 [2] (iv) The “volume strength” of hydrogen peroxide is equal to the volume of oxygen, in dm3, produced at room conditions, when 1.00 dm3 of a solution of hydrogen peroxide is completely catalytically decomposed. Use your answer in (b)(iii) to calculate the volume, in dm3, of oxygen produced when 1.00 dm3 of FB 1 decomposes at room conditions. Your answer to this calculation is numerically equal to the “volume strength”, in vol, of solution FB 1. Show your working. “volume strength” of H2O2 in FB 1 = …… vol [1] (c) (i) Describe the observation to determine whether or not the decomposition of hydrogen peroxide in FB 1 is complete. Explain your answer. …… [1] (ii) A student suggests that the experiment in (a) would be more accurate if a loose-fitting plug of cotton wool was pushed gently into the mouth of the conical flask during the reaction. State whether the student is correct. Explain your answer. …… [2] (iii) Deduce the mass of manganese(IV) oxide, FB 2, left in the conical flask at the end of the experiment. Explain your answer. …… [1] [Total: 12]

Mark scheme

1(a) I Unambiguous headings and units for four weighings (g), / g, in g or g next to each entry 3 • (Mass of) conical / flask + H2O2 / solution/FB 1 = W • (Mass of) container + solid / catalyst / MnO2 / FB 2 = X • (Mass of) container (empty / with residual MnO2) = Y • (Mass of) conical / flask and contents (at end / after reaction / after 30 minutes) = Z II Weighings and mass of FB 2 recorded. • All weighings (minimum 3) recorded to same decimal places (either 2 dp or 3 dp) • mass of FB 2, correctly calculated. (X – Y) III: Accuracy (Q) mark in 1(a) Award III if candidate’s mass loss (calculated to 2 dp) is within, or equal to, 25% of supervisor’s mass loss (calculated to 2 dp). Mass loss = W + (X – Y) – Z 1(b)(i) Mass of oxygen correctly calculated 1 Mass loss = weighing 1 + mass of FB 2 added – weighing 4 = W + (X – Y) – Z and answer given to a minimum of 2 significant figures (sf) 1(b)(ii) Equation with state symbols 1 2H2O2(aq) → 2H2O(l) + O2(g) 1(b)(iii) Correctly uses (b)(i) to calculate [H2O2(aq)] in FB 1 2 M1: correct expression/answer for amount of O2 liberated: n(O2) = (b)(i) / 32 mol (does not need to be evaluated) M2: concentration of H2O2 and answer given to 2–4 sf [H2O2(aq)] = 40 × 2 × answer to step 1 (final answer is approx = 2.5 × (b)(i)) mol dm−3) 1(b)(iv) Correct use of (b)(iii) to calculate volume strength of H2O2 1 Volume strength = (b)(iii) × 0.5 × 24.0 (= (b)(iii) × 12) vol and answer given to 2–4 sf 1(c)(i) EITHER 1 (Continuing) effervescence shows that reaction is not complete. OR no (more) fizzing shows that reaction has finished / is complete. or mass of flask and contents is constant shows reaction is complete (owtte) 1(c)(ii) M1: Plug prevents escape of liquid droplets (owtte) 1 or plug stops solution / FB 1 fizzing out (owtte) or plug traps aerosol formed M2: Student is correct (must be stated explicitly) 1 and provided that plug is pre-weighed 1(c)(iii) EITHER 1 Mass is same as listed in (a) (or value stated) and MnO2 is a catalyst OR MnO2 is a catalyst so its mass will not change (during the reaction)

Question 214 marks

Question

The concentration of a solution of hydrogen peroxide can also be determined by titration with acidified potassium manganate(VII). The equation for the reaction is shown below. 2MnO4–(aq) + 5H2O2(aq) + 6H+(aq) 2Mn2+(aq) + 5O2(g) + 8H2O(l) FB 1 is aqueous hydrogen peroxide, H2O2. FB 3 is 0.0180 mol dm–3 potassium manganate(VII), KMnO4. FB 4 is dilute sulfuric acid, H2SO4. (a) Method Dilution of FB 1 ● Pipette 10.0 cm3 of FB 1 into the 250 cm3 volumetric flask. ● Make the solution up to the mark using distilled water. ● Shake the flask thoroughly. ● This diluted solution of hydrogen peroxide is FB 5. Titration ● Fill the burette with FB 3. ● Pipette 25.0 cm3 of FB 5 into a clean conical flask. ● Rinse out the measuring cylinder with distilled water. Use the measuring cylinder to transfer 20 cm3 of FB 4 into the same flask. ● Perform a rough titration and record your burette readings in the space below. The rough titre is …… cm3. ● Carry out as many accurate titrations as you think necessary to obtain consistent results. ● Make sure any recorded results show the precision of your practical work. ● Record, in a suitable form below, all your burette readings and the volume of FB 3 added in each accurate titration. I II III IV V VI VII [7] (b) From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtained the mean value. 25.0 cm3 of FB 5 required …… cm3 of FB 3. [1] (c) Calculations (i) Give your answers to (c)(ii), (c)(iii) and (c)(iv) to the appropriate number of significant figures. [1] (ii) Calculate the amount, in mol, of potassium manganate(VII) present in the volume of FB 3 calculated in (b). amount of KMnO4 = …… mol [1] (iii) Use your answer to (c)(ii) and the equation on page 4 to calculate the amount, in mol, of hydrogen peroxide in FB 5 used in each titration. amount of H2O2 = …… mol [1] (iv) Calculate the concentration of hydrogen peroxide in FB 1, in mol dm–3. concentration of H2O2 in FB 1 = …… mol dm–3 [1] (d) (i) State which procedure, the method you used in 1(a) or the method you used in 2(a), gives a more accurate value for the concentration of H2O2 in FB 1. Explain your answer. …… [1] (ii) The uncertainty in reading a 25 cm3 pipette is ±0.06 cm3. Explain why this pipette is more accurate than a burette for measuring 25.0 cm3 of a solution. …… [1] [Total: 14] Qualitative Analysis For each test you should record all your observations in the spaces provided. Examples of observations include: ● colour changes seen ● the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added ● the formation of any gas and its identification (where appropriate) by a suitable test. You should record clearly at what stage in a test an observation is made. Where no change is observed, you should write ‘no change’. Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given. If any solution is warmed, a boiling tube must be used. Rinse and reuse test-tubes and boiling tubes where possible. No additional tests should be attempted.

Mark scheme

2(a) I All the following data are recorded • two burette readings and the titre for the rough titration • initial and final burette readings for two (or more) accurate titrations II Appropriate headings and units in the accurate titration table and titre values recorded for accurate titrations • initial / start and (burette) reading / volume • final / end and (burette) reading / volume • titre or volume / FB 3 and used / added • unit: / cm3 or (cm3) or in cm3 (for each heading) or cm3 unit given for each volume recorded III All accurate burette readings are recorded to the nearest 0.05 cm3. The requirement to record to 0.05 applies to burette readings, including 0.00 cm3 (if this was the initial reading), but it does not apply to the titre. IV: The final accurate titre recorded is within 0.10 cm3 of any other accurate titre. For assessment of accuracy (Q) marks, the Examiner should round all burette readings to the nearest 0.05 cm3. Check and correct subtractions for supervisor and candidates. Then select the ‘best’ titres using the hierarchy: • two (or more) accurate identical titres (ignoring any that are labelled ‘rough’), then • two (or more) accurate titres within 0.05 cm3, then • two (or more) accurate titres within 0.10 cm3, etc. These best titres should be used to calculate the mean titre, expressed to nearest 0.01 cm3. Calculate the difference (d) between the candidate’s mean titre and the supervisor’s mean titre. Write and ring the supervisor’s value near each candidate’s table of results. Write the value of d on each script. Award the accuracy (Q) marks as shown below. Award V if d ⩽ 0.80 (cm3) 7 Award VI if d ⩽ 0.50 Award VII if d ⩽ 0.30 Tolerance for low titres: If supervisor’s titre is < 10.0 cm3, tolerances are 0.15, 0.25, 0.40 cm3. 2(b) Correctly calculates the mean titre. 1 • Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm3. • Working / explanation must be shown or ticks must be shown next to the two (or more) accurate readings selected. • The mean should be quoted to 2 dp, and be rounded to nearest 0.01 cm3. (e.g. 26.625 cm3 must be rounded to 26.63 cm3) • Allow the mean expressed to 1 dp, if all accurate burette readings were given to 1 dp and the mean is exactly correct. (e.g. 26.0 and 26.2 = 26.1 is allowed) (e.g. 26.0 and 26.1 = 26.1 is wrong – should be 26.05) 2(c)(i) Answers to parts (c)(ii), (c)(iii) and (c)(iv) all quoted to 3 or 4 significant figures. 1 2(c)(ii) Correct calculation of amount of KMnO4 1 Amount of KMnO4 = vol in (b) × 0.018 / 1000 mol 2(c)(iii) Correct use of (c)(ii) to calculate amount of H2O2. 1 Amount of H2O2 = 2.5 × (c)(ii) mol 2(c)(iv) Correct use of (c)(iii) to calculate concn of H2O2 in FB 1. 1 [H2O2] = (c)(iii) × 250 / 10 × 1000 / 25 = 1000 × (c)(iii) mol dm−3 2(d)(i) Comparing accuracy of the two methods 1 Expt 2(a) is more accurate and one reason (from the list below): • Two (or more) consistent titres are obtained in 2(a) • Mass loss in 1(a) is small, giving high(er) percentage error. • The reaction in 2(a) has definite / sudden / sharp end / finish point • The reaction in 1(a) is not complete (provided this agrees with answer 1(c)(i)) 2(d)(ii) Errors compared 1 (Error in volume measured by burette =) 2 × 0.05 > 0.06 FB 6 is dilute HNO3; FB 8 is FeSO4(aq)

Question 314 marks

Question

(a) FB 6 is a solution containing one cation and one anion. The anion is listed in the Qualitative analysis notes. FB 7 is aqueous sodium thiosulfate, Na2S2O3. (i) To a 1 cm depth of FB 7 in a test-tube, add an equal volume of FB 6. Record the first change you observe. Wash out your test-tube with plenty of tap water as soon as you finish this test. …… [1] (ii) Carry out tests to identify the anion in FB 6. The anion does not contain carbon or sulfur. Record details of your tests and all your observations in a suitable form below. [3] (iii) Give the formula of FB 6. FB 6 is …… . [1] (b) FB 8 is a solution containing one cation and one anion, both of which are listed in the Qualitative analysis notes. FB 1 is aqueous hydrogen peroxide. (i) Use a 1 cm depth of FB 8 for the following tests. Record your observations in Table 3.1. Table 3.1 test observation Test 1 Add aqueous ammonia. Test 2 Add aqueous barium nitrate or aqueous barium chloride, then add dilute hydrochloric acid. Test 3 Add an equal volume of FB 1, then add aqueous sodium hydroxide. Test 4 Add a few drops of acidified aqueous potassium manganate(VII). [5] (ii) Give the ionic equation for the reaction of FB 8 in Test 1. Include state symbols. …… [2] (iii) Test 3 involves a redox reaction. Using your observations, justify the statement that a redox reaction has taken place. …… [2] [Total: 14] Qualitative analysis notes 1 Reactions of cations cation reaction with NaOH(aq) NH3(aq) aluminium, Al 3+(aq) white ppt. soluble in excess white ppt. insoluble in excess ammonium, NH4+(aq) no ppt. – ammonia produced on warming barium, Ba2+(aq) faint white ppt. is observed unless no ppt. [Ba2+(aq)] is very low calcium, Ca2+(aq) white ppt. unless [Ca2+(aq)] is very no ppt. low chromium(III), Cr3+(aq) grey-green ppt. soluble in excess grey-green ppt. insoluble in excess giving dark green solution copper(II), Cu2+(aq) pale blue ppt. insoluble in excess pale blue ppt. soluble in excess giving dark blue solution iron(II), Fe2+(aq) green ppt. turning brown on green ppt. turning brown on contact with air contact with air insoluble in excess insoluble in excess iron(III), Fe3+(aq) red-brown ppt. insoluble in excess red-brown ppt. insoluble in excess magnesium, Mg2+(aq) white ppt. insoluble in excess white ppt. insoluble in excess manganese(II), Mn2+(aq) off-white ppt. rapidly turning brown off-white ppt. rapidly turning brown on contact with air on contact with air insoluble in excess insoluble in excess zinc, Zn2+(aq) white ppt. soluble in excess white ppt. soluble in excess 2 Reactions of anions anion reaction carbonate, CO32– CO2 liberated by dilute acids chloride, Cl –(aq) gives white ppt. with Ag+(aq) (soluble in NH3(aq)) bromide, Br –(aq) gives cream / off-white ppt. with Ag+(aq) (partially soluble in NH3(aq)) iodide, I–(aq) gives pale yellow ppt. with Ag+(aq) (insoluble in NH3(aq)) nitrate, NO3–(aq) NH3 liberated on heating with OH–(aq) and Al foil nitrite, NO2–(aq) NH3 liberated on heating with OH–(aq) and Al foil; decolourises acidified aqueous KMnO4 sulfate, SO42–(aq) gives white ppt. with Ba2+(aq) (insoluble in excess dilute strong acids); gives white ppt. with high [Ca2+(aq)] sulfite, SO32–(aq) gives white ppt. with Ba2+(aq) (soluble in excess dilute strong acids); decolourises acidified aqueous KMnO4 thiosulfate, S2O32–(aq) gives off-white / pale yellow ppt. slowly with H+ 3 Tests for gases gas test and test result ammonia, NH3 turns damp red litmus paper blue carbon dioxide, CO2 gives a white ppt. with limewater hydrogen, H2 ‘pops’ with a lighted splint oxygen, O2 relights a glowing splint 4 Tests for elements element test and test result iodine, I2 gives blue-black colour on addition of starch solution 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 = 6.022 × 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) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 12 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.4 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge

Mark scheme

3(a)(i) Off-white / cream / yellow precipitate (formed) 1 3(a)(ii) Test for nitrogen anion 3 M1: Add (excess) NaOH + heat + aluminium M2: Effervescence / gas / NH3 turns (red) litmus blue Test to distinguish between nitrate and nitrite M3: Add (a few drops of acidified) KMnO4 (manganate(VII) if name used) and purple colour remains / (KMnO4) is not decolourised 3(a)(iii) FB 6 is HNO3 (formula required) 1 3(b)(i) 11 Observations shown in the table 5 Test 1 (+ ammonia) • green precipitate • insoluble / no change in excess (ammonia) • (precipitate / mixture) turns brown (on surface) Test 2 (+ aq Ba2+ then HCl) • (with Ba2+) white precipitate (not ‘off-white’: soluble in excess is a CON) • (with acid) (ppt) is insoluble (in acid) or no change / no reaction Test 3 (+ FB 1 then NaOH) • (with H2O2) (solution) turns / turns yellow / light brown / pale brown • Effervescence / bubbling (in either box) • (Gas) relights glowing splint • (with NaOH) red-brown / brown / rust precipitate • Precipitate is insoluble / does not dissolve / no change in excess (NaOH) Test 4 (KMnO4) • (KMnO4) decolourised or changes (from purple) to colourless / pale yellow 3(b)(ii) Ionic equation 2 M1: OH– is used in the equation M2: Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) 3(b)(iii) EITHER 2 M1: Iron(II) oxidised to iron(III) / Fe2+ → Fe3+ + e– or H2O2 oxidises Fe2+ M2: (Brown) ppt is Fe(OH)3 / ppt (produced) is red-brown OR M1: H2O2 disproportionates or H2O2 is oxidised and reduced. M2: Oxygen is produced OR M1: 2Fe2+(aq) + 2H+(aq) + H2O2(aq) → 2Fe3+(aq) + 2H2O(l) M2: it is stated which species is oxidised / reduced

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Where can I find the 9701/36 October/November 2023 mark scheme?

On this page, free: the mark scheme for AS & A Level Chemistry 9701/36, October/November 2023 (9701_w23_ms_36), and the question paper (9701_w23_qp_36), to read or download, with each question's marking points written out underneath.

How many marks is 9701/36 out of?

Chemistry 9701/36 for October/November 2023 is out of 40 marks, with 2 hours to answer it, across 3 questions.

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