Chemistry 9701/53 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 5 Planning, Analysis and Evaluation, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The concentration of dissolved oxygen in a sample of water can be measured using the following method.
Manganese(II) hydroxide, Mn(OH)2, is oxidised by the oxygen dissolved in a sample of water to form manganese(III) hydroxide, Mn(OH)3.
O2(aq) + 4Mn(OH)2(s) + 2H2O(l) 4Mn(OH)3(s)
The manganese(III) hydroxide then reacts with iodide ions to produce aqueous iodine.
2Mn(OH)3(s) + 6HCl (aq) + 2KI(aq) 2MnCl 2(aq) + 2KCl (aq) + I2(aq) + 6H2O(l)
The amount of iodine produced is proportional to the amount of dissolved oxygen.
25.0 cm3 of the solution containing aqueous iodine is transferred into a conical flask and titrated against 1.00 × 10–3 mol dm–3 sodium thiosulfate, Na2S2O3.
I2(aq) + 2Na2S2O3(aq) Na2S4O6(aq) + 2NaI(aq)
(a) (i) Complete Table 1.1 and determine the mean titre to be used in calculating the concentration of dissolved oxygen.
Table 1.1
trial run run 1 run 2 run 3 final burette
27.30 28.10 28.25 26.95 reading / cm3 initial burette
0.00 1.10 1.55 0.15 reading / cm3 titre / cm3
mean titre = …… cm3 [2]
(ii) Calculate the concentration of dissolved oxygen in the 25.0 cm3 of solution.
Show your working.
concentration of dissolved oxygen in 25.0 cm3 of solution = …… mol dm–3 [3]
(b) Suggest a suitable piece of apparatus for the transfer of 25.0 cm3 of the solution containing aqueous iodine.
…… [1]
(c) Water samples are collected in full sealed flasks.
Explain why the sealed flask must be completely full.
…… [1]
(d) The concentration of oxygen in water at different temperatures is shown in Table 1.2. The concentration value is missing for 25 °C.
Table 1.2
concentration of temperature / °C oxygen × 10–4 / mol dm–3
0 4.58
5 3.97
10 3.20
15 3.13
20 2.82
25
30 2.33
35 2.15
40 2.05
(i) Plot a graph of concentration of oxygen (y-axis) against temperature (x-axis) on the grid.
Use a cross (×) to plot each data point. Draw a smooth curve of best fit.
5.00
4.50 concentration of oxygen × 10–4 / mol dm–3
4.00
3.50
3.00
2.50
2.00
0 5 10 15 20 25 30 35 40 temperature / °C
[2]
(ii) Use the graph to deduce the concentration of oxygen at 25 °C.
concentration of oxygen at 25 °C = …… mol dm–3 [1]
(iii) Circle the most anomalous point on the graph.
Suggest an explanation for this anomaly. Assume that there was no error in measuring oxygen concentration.
…… [2]
[Total: 12]
Mark scheme
1(a)(i) M1: 27.30, 27.00, 26.70, 26.80 2
M2: correct averaging of two (or more) titres within 0.10 cm3 of each other.
(Expected answer = (26.70 + 26.80) / 2 = 26.75)
1(a)(ii) M1: n(Na2S2O3(aq)) = (a(i) / 1000) × 0.001 3
(Expected answer = 26.75 / 1000 x 0.001 = 2.675 × 10-5)
M2: n(O2(aq)) = M1 / 4
(Expected answer = 2.675 × 10-5 / 4 = 6.6875 x 10-6)
M3: concentration of oxygen = M2 × 1000 / 25 = M2 x 40
(Expected answer = 6.6875 × 10-6 × 1000 / 25
= 2.675 × 10-4 mol dm-3)
Answer to at least 3SF
1(b) (25.0 cm3) volumetric pipette 1
1(c) to avoid any oxygen from (trapped) air (in the flask) reacting with / oxidising Mn(OH)2 1
1(d)(i) M1: All points plotted correctly. 2
M2: smooth curve LOBF drawn.
1(d)(ii) Correct value read from candidate’s graph. 1
Value given to at least 3SF.
(Expected value – 2.57 × 10-4 moldm-3)
1(d)(iii) M1 Point most anomalous to the plotted line of best fit circled. 2
M2:
For anomalous point below the LOBF
The (actual) temperature > recorded / measured temperature (in the table).
For anomalous point above the LOBF.
The (actual) temperature < recorded / measured temperature (in the table).
Benzenediazonium chloride, C6H5N2Cl , decomposes in water as shown in the following equation.
C6H5N2Cl (aq) + H2O(l) C6H5OH(aq) + N2(g) + HCl (aq)
A solution of 0.0750 mol dm–3 of C6H5N2Cl (aq) decomposes at a constant temperature of 50 °C.
The volume of nitrogen gas, N2(g), collected is recorded every 5 minutes for 45 minutes.
(a) Draw a labelled diagram to show how the apparatus could be set up to carry out this experiment.
[3]
(b) Using this method, a student obtains the graph shown in Fig. 2.1.
60
50
40
30 volume N2
/ cm3
20
10
00 5 10 15 20 25 30 35 40 45 time / min
Fig. 2.1
(i) On Fig. 2.1, draw a tangent to the curve at t = 0 mins. Calculate the initial rate of reaction in cm3 min–1.
initial rate of reaction = …… cm3 min–1 [2]
(ii) Explain why the initial rate of reaction is calculated at t = 0 mins rather than dividing the total volume of gas produced by the time taken to produce it.
…… [1]
(iii) Describe how the curve in Fig. 2.1 would be different, if at all, if the atmospheric pressure increases. All other conditions stay the same.
…… [1]
(iv) On Fig. 2.1, draw a second curve to show the graph produced if the same volume of
0.0375 mol dm–3 C6H5N2Cl (aq) decomposes at a constant temperature of 50 °C. All other conditions stay the same. [1]
(c) Another student investigates the effect of changing the concentration of C6H5N2Cl (aq) at 50 °C. He measures the time taken to collect 0.0150 dm3 of N2(g) and calculates the rate of
N2 production by dividing 0.0150 dm3 by the time taken. The results are shown in Table 2.1.
Table 2.1
concentration of time taken to rate of N2
C6H5N2Cl (aq) collect 0.0150 dm3 production
/ mol dm–3 of N2 / s / dm3 s–1
0.500 21
0.400 33
0.300 48
0.200 64
0.100 122
(i) Complete the table to calculate the values for the rate of N2 production.
Give your answers to three significant figures. [1]
(ii) The reaction is first order and obeys the following rate equation.
rate = k [C6H5N2Cl ]
Explain how the data in Table 2.1 supports this statement.
…… [1]
(iii) State the dependent variable in this investigation.
…… [1]
(iv) The student wants to perform a similar experiment using 0.200 mol dm–3 C6H5N2Cl (aq).
Describe how the student should make a standard solution of 100.0 cm3 of 0.200 mol dm–3
C6H5N2Cl (aq) starting from a solution of 0.500 mol dm–3 C6H5N2Cl (aq).
Give the name and size of any key apparatus which should be used and describe how the student should ensure the volume is exactly 100.0 cm3.
Write your answer using a series of numbered steps.
…… [3]
(v) Explain why (iv) must be carried out at a temperature below 5 °C.
…… [1]
(d) C6H5N2Cl is used in the manufacture of synthetic dyes. A student prepares a sample of the dye using the reaction scheme shown in Fig. 2.2.
NH2
step 1 step 2
NaNO2 N
NH2 + HCl N≡N Cl N
< 5 °C + C6H5NH2
Fig. 2.2
In step 1, phenylamine, C6H5NH2, is converted into C6H5N2Cl .
In step 2, C6H5N2Cl is reacted with more C6H5NH2 to produce the solid dye, which is then filtered.
(i) The student’s final yield was 52%. They had not spilled any reagents or products.
Suggest two reasons why the student’s yield was lower than 100%. Assume no errors were made in the measurement of any substances.
reason 1: ……
reason 2: ……
[2]
(ii) Explain why the solid dye should be dried before assessing the melting point.
…… [1]
[Total: 18]
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
2(a) M1: Leakproof reaction vessel capable of delivering a sample of gas. 3
M2: Suitable method of collection and measurement of the gas produced. (eg. over water or gas syringe).
M3: method for monitoring & maintaining temperature (thermostatically controlled water bath).
2(b)(i) M1: A tangent line which is… 2
• a straight line
• starts at / passes through 0,0.
• keeps to the left of the curve of the original line.
• covers at least 25 cm3 (5 big squares) on y-axis.
• passes between 4.0 and 5.5 minutes (inc.) at 25 cm3
M2: correct calculation of the gradient of tangent line drawn.
2(b)(ii) The total volume of gas produced divided by the time taken is the mean rate (rather than the initial rate). 1
2(b)(iii) It (the line / curve) would be below / lower (than the one on the graph) (starting at 0,0). 1
2(b)(iv) A second curve which: 1
• starts at 0,0 (± 1 small square)
• is less steep than curve for 0.0750 moldm-3. (before any plateau)
• produces 27.5 cm3 gas in total by 45 minutes. (± half a small square).
• Extends from 0 – 45 minutes.
2(c)(i) 1
concentration rate
0.5 0.000714 7.14 ×10-4
0.4 0.000455 4.55 ×10-4
0.3 0.000313 3.13 ×10-4
0.2 0.000234 2.34 ×10-4
0.1 0.000123 1.23 ×10-4
2(c)(ii) explanation using numbers from the table that whatever change is made to the concentration causes the same change in 1
the rate (within experimental error).
2(c)(iii) time (taken to collect 0.0150dm3 of N2) 1
2(c)(iv) M1: Correct volume, 40(.0) cm3, (of 0.5(00) moldm-3 ) C6H5N2Cl(aq) is used / measured. 3
M2: measure the volume from M1 C6H5N2Cl using a (50 cm3) burette into a 100 cm3 volumetric flask.
M3: make up to the (calibration) mark with distilled water.
AND
then mix the solution (by inverting the flask)
2(c)(v) to make sure the concentration does not change (from 0.200 moldm-3) (by making sure the C6H5N2Cl does not decompose) 1
2(d)(i) Two acceptable reasons for low yield. 2
1 Some C6H5N2Cl had decomposed / reacted with water (before reacting with phenylamine) OR temperature rose above
5 oC (in step 1) OR phenylamine used was impure.
2 Some dye / solid is not recovered / is lost through / left in filter OR some of the dye dissolves in the water / is aqueous (in
step 2).
2(d)(ii) (Residual) water causes solid to have a different / inaccurate (lower) melting point (to the dry solid). 1
OR
To ensure all water has been driven off to increase accuracy in measurement of melting point
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Chemistry 9701/53 for October/November 2023 is out of 30 marks, with 1 hour 15 minutes to answer it, across 2 questions.
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