Chemistry 9701/52 February/March 2022: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 5 Planning, Analysis and Evaluation, February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A student has a sample of copper(II) sulfate crystals, CuSO4•xH2O. The student wants to show that the value of x is 5.
The student uses the following method.
step 1 Weigh a clean crucible on a balance reading to two decimal places. Record the mass.
step 2 Place the sample of CuSO4•xH2O into the crucible. Record the mass.
step 3 Heat the crucible gently for about 1 minute then strongly for about 4 minutes.
step 4 Weigh the crucible and contents. Record the mass.
(a) Identify the instruction that is missing between step 3 and step 4.
. …… [1]
(b) Explain why gentle heating takes place in step 3.
. …… [1]
(c) Name the apparatus that should be used to hold the crucible during heating.
. …… [1]
(d) The method is incomplete.
State the step(s) that should be carried out to complete the method.
. ……
. ……
. …… [1]
(e) The student records their results in Table 1.1.
Table 1.1
mass / g mass of crucible 13.60 mass of crucible + contents before heating 21.09 mass of crucible + contents at the end of experiment 17.94
(i) Calculate the experimental value of x from these results.
experimental value of x = …… [3]
(ii) Suggest why the experimental value of x varies from the expected value of 5.
If you were unable to obtain an answer to (e)(i), use the experimental value x = 6.9. This is not the correct answer.
. ……
. …… [1]
(f) The empty crucible weighs 13.60 g.
Calculate the percentage error in this measurement.
Show your working.
percentage error = …… [1]
[Total: 9]
Mark scheme
1(a) allow (crucible / contents) to cool 1
1(b) to prevent spitting of crystals 1
1(c) pipe-clay triangle 1
1(d) heat to constant mass 1
1(e)(i) M1: (mass H2O) = 3.15 (g) AND (mass CuSO4) = 4.34 (g) 3
M2: n(H2O) = 3.15 / 18 = 0.175
AND
n(CuSO4) = 4.34 / 159.6 = 0.02719…
M3: 0.02719… / 0.02719… AND 0.175 / 0.02719… = 6.44
1(e)(ii) If value of x in 1(e)(i) is above 5: 1
CuSO4 must have decomposed / crystals were wet before heating / some crystals were lost when heated (e.g. during
spitting) / impurity that decomposes or evaporates
If value of x in 1(e)(i) is below 5:
did not heat it enough / not all water driven off / unreactive impurity
1(f) [(2 × 0.005) / 13.60] × 100 = 0.0735 (%) 1
The conductivity of an ionic solution can be determined by passing an electric current through the solution and measuring the conductivity using a conductivity meter.
(a) A student carries out an experiment to measure the conductivity of different solutions of ethanoic acid, CH3COOH, which is a weak acid.
The acid dissociation constant, Ka, can be determined from this experiment.
The student makes standard solution A, 250.0 cm3 of 2.00 mol dm–3 CH3COOH(aq).
(i) State what is meant by a standard solution.
. ……
. …… [1]
(ii) Describe how the student should make standard solution A from pure ethanoic acid.
The concentration of standard solution A should be 2.00 mol dm–3 to the nearest three significant figures.
Your answer should state the name and capacity of any apparatus that the student should use. A balance is not available.
Pure ethanoic acid is a liquid with a density of 1.05 g cm–3 at room temperature.
You may wish to write your answer using a series of numbered steps.
[Mr: CH3COOH, 60.0]
. ……
. ……
. ……
. ……
. ……
. ……
. ……
. ……
. …… [4]
(b) The student wears chemically resistant gloves throughout this procedure.
Suggest why.
. …… [1]
(c) The student dilutes standard solution A with distilled water to make solutions of different concentrations. The conductivity of these solutions is measured.
Table 2.1 shows the results.
Table 2.1
1 2 3 4
1
[CH3COOH] conductivity molar conductivity, ΛM
/ mol dm–3 / S dm–1 / dm2 S mol–1
/ dm1.5 mol–0.5
0.0500 3.76 × 10–3
0.0250 2.74 × 10–3
0.0125 1.92 × 10–3
0.00625 1.33 × 10–3
0.003125 1.15 × 10–3
0.0015625 6.68 × 10–4
In order to determine Ka the results must be used to obtain two sets of data.
Column 3 is the reciprocal of the square root of the concentration (column 1).
Column 4 is the molar conductivity, ΛM, which is found by dividing the conductivity (column 2) by the concentration (column 1).
(i) Complete columns 3 and 4 in Table 2.1. Give all values to three significant figures. [2]
(ii) Identify the dependent variable in this experiment.
. …… [1]
(iii) The student decided to measure the conductivity of distilled water at the start of the experiment.
Suggest why.
. ……
. …… [1]
(iv) State one variable that should be controlled.
. …… [1]
(d)
Plot a graph on the grid to show the relationship between molar conductivity, ΛM, and
1
. Use a cross (×) to plot each data point. Draw a line of best fit that includes the
origin.
0.45
0.40
0.35
0.30
0.25 molar conductivity, M
/ dm2 S mol–1
0.20
0.15
0.10
0.05
0.00
0 5 10 15 20 25 30
1
/ dm1.5 mol–0.5
[2]
(e) Circle one point on the graph that you consider to be most anomalous.
The conductivity meter was correctly functioning.
Suggest one reason why this anomaly may have occurred during this experimental procedure.
. …… [2]
(f) The equation for the line of best fit is shown.
Ka
ΛM =
Λ∞ = 3.91 dm2 S mol–1
(i) Use your graph 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 the gradient to three significant figures.
coordinates 1 …… coordinates 2 ……
gradient = …… dm0.5 S mol–0.5
[2]
(ii) Determine the acid dissociation constant, Ka, of ethanoic acid. Include units in your answer.
Ka = ……
units = ……
[2]
(g) The student repeats the experiment with propanoic acid. The numerical value of the Ka of propanoic acid is experimentally determined as 1.28 × 10–5.
The theoretical numerical value is 1.34 × 10–5.
Comment on the validity of the experimental result. Assume the maximum total percentage error from the measurements made is 6.5%.
. ……
. ……
. …… [2]
[Total: 21]
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)(i) solution of known concentration 1
2(a)(ii) M1: measure 28.55 cm3 of CH3COOH 4
M2: using burette / graduated pipette
M3: (transfer/add CH3COOH to) a 250 cm3 volumetric flask
M4: top the (volumetric) flask up to (250 cm3) mark with distilled water
2(b) (pure) CH3COOH is corrosive 1
2(c)(i) 2
3 4
4.47 0.0752
6.32 0.110
8.94 0.154
12.6 0.213
17.9 0.368
25.3 0.428
M1: column 3 all values to 3 SF
M2: column 4 all values to 3 SF
2(c)(ii) conductivity 1
2(c)(iii) to calibrate the conductivity meter 1
2(c)(iv) temperature 1
2(d) M1: 6 points plotted correctly 2
M2: line of best fit drawn
2(e) M1: 5th point circled at 17.9, 0.368 2
M2: ethanoic acid was diluted incorrectly
2(f)(i) M1: co-ordinates read and recorded correctly 2
M2: gradient calculated to 3 SF
2(f)(ii) M1: (gradient calculated in (f)(i) / 3.91)2 2
(0.0169 / 3.91)2 = 1.87 × 10–5
M2: mol dm–3
2(g) M1: percentage error of experimental value = [(1.34 – 1.28) / 1.34] × 100 = 4.5% 2
M2: valid and the calculated actual percentage error is less than the maximum percentage error
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Chemistry 9701/52 for February/March 2022 is out of 30 marks, across 2 questions.
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