Chemistry 9701/52 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 5 Planning, Analysis and Evaluation, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Solution X is dilute hydrochloric acid of unknown concentration. A student uses titration with aqueous sodium carbonate, Na2CO3(aq), to determine the concentration of solution X.
The student plans to prepare 250.0 cm3 of 0.500 mol dm–3 aqueous sodium carbonate,
Na2CO3(aq).
(a) (i) Calculate the mass of Na2CO3(s) needed to make 250.0 cm3 of 0.500 mol dm–3
Na2CO3(aq), using a two decimal place balance.
mass of Na2CO3(s) = …… g [1]
(ii) Describe how the student should make 250.0 cm3 of 0.500 mol dm–3 Na2CO3(aq) starting from the mass of Na2CO3(s) calculated in (i) supplied in a 50 cm3 beaker.
Give the name and size of any key apparatus to be used.
Write your answer using a series of numbered steps.
…… [3]
(b) The student incorrectly makes up the 250.0 cm3 of Na2CO3(aq) and the concentration is not
0.500 mol dm–3. The student calls this solution Y.
The student uses the following method to determine both the concentration of solution X and the concentration of solution Y.
step 1 Transfer 25.0 cm3 of solution Y into a conical flask using a volumetric pipette. Add a few drops of methyl orange.
step 2 Titrate the sample in the conical flask with solution X.
step 3 Transfer a fresh 25.0 cm3 portion of solution Y into a second conical flask, but do not add methyl orange.
step 4 Use the burette to add the volume of solution X used in step 2 to the second conical flask.
step 5 Measure and record the mass of a dry evaporating basin.
step 6 Transfer the contents of the second conical flask into the evaporating basin.
step 7 Use a water bath to heat the solution in the evaporating basin until all the water is evaporated and only solid remains.
step 8 Measure and record the mass of the evaporating basin and solid residue.
(i) In step 1, 25.0 cm3 of solution Y is transferred using a volumetric pipette.
State what the volumetric pipette should be rinsed with before carrying out step 1.
…… [1]
(ii) Describe how the student detects the end-point of the titration in step 2.
…… [1]
(iii) Explain why the evaporating basin is not heated directly with a Bunsen burner in step 7.
…… [1]
(iv) Suggest what the student should do to ensure all of the water has evaporated from the residue before completing step 8.
…… [1]
(c) The student’s burette readings taken in step 2 are shown in Fig. 1.1.
1 13
2 14
3 15
initial final
Fig. 1.1
(i) Use Fig. 1.1 to complete Table 1.1.
Table 1.1
burette reading (final) / cm3
burette reading (initial) / cm3
volume of solution X added / cm3
[2]
(ii) Calculate the percentage error in the volume of solution X calculated in (i).
Show your working.
percentage error = …… [1]
(d) A second student repeats the experiment in (b) using solution X and 25.0 cm3 of solution Y.
The results are shown in Table 1.2.
Table 1.2
volume of solution X added in titration / cm3 13.35 mass of dry evaporating basin / g 44.52 mass of evaporating basin and solid residue / g 45.69
The reaction that takes place in the titration is:
2HCl (aq) + Na2CO3(aq) 2NaCl (aq) + H2O(l) + CO2(g)
Using the results in Table 1.2, calculate the following:
• the amount, in mol, of sodium chloride
amount of sodium chloride = …… mol
• the concentration of solution X and the concentration of solution Y.
concentration of solution X = …… mol dm–3
concentration of solution Y = …… mol dm–3
[3]
(e) State what happens to the value obtained for the concentration of solution Y if not all the water is evaporated in step 7.
Explain your answer.
…… [2]
[Total: 16]
Mark scheme
1(a)(i) 13.25 g 1
1(a)(ii) M1 dissolve the solid / sodium carbonate (in the beaker) (using a small volume of) (distilled) water 3
M2 transfer / add to a (250 cm3 volumetric) flask
and
rinse (with distilled water)
M3 top the 250 cm3 volumetric flask up to mark (with distilled water)
1(b)(i) solution Y 1
1(b)(ii) indicator / methyl orange changes colour 1
1(b)(iii) to avoid spitting (of solid when it forms) 1
1(b)(iv) heat to constant mass 1
1(c)(i) 2
burette reading final 14.85
burette reading initial 1.35
volume of solution X 13.50
M1 burette readings 14.85 (cm3) and 1.35 (cm3)
M2 volume of solution X added = 13.50 (cm3)
1(c)(ii) 2× 0.05 1
working must be shown × 100 = 0.741
13.50
1(d) M1 Mr(NaCl) = 58.5 3
n(NaCl) = 1.17 g / 58.5 = 0.0200
M2 n(HCl) = 0.02
c(HCl) (X) = 0.02/0.01335 = 1.50 (mol dm–3)
M3 n(Na2CO3) = 0.02 / 2 = 0.01
c(Na2CO3) (Y) = 0.01 / 0.025 = 0.400 (mol dm–3)
1(e) M1 calculated concentration of Y will be higher than the true value. 2
M2 mass / amount of NaCl (in step 7) will be too high.
A group of students uses the following method to investigate the change in mass that takes place when metal M combines with sulfur to form a metal sulfide.
step 1 Weigh a clean crucible and lid using a balance. Record the mass.
step 2 Place a coiled length of metal wire into the crucible and weigh the crucible, lid and wire.
step 3 Cover the metal wire in the crucible with a large quantity of powdered sulfur and replace the lid.
step 4 Heat the crucible in a fume hood until the crucible glows red. Continue heating strongly
until no more sulfur can be seen.
step 5 Allow the apparatus to cool and weigh the crucible, lid and residue.
lid coiled length of metal wire covered with sulfur crucible
tripod
heat
Fig. 2.1
(a) (i) Fig. 2.1 shows how the apparatus is set up when heating the crucible in step 4.
Identify the missing piece of apparatus used to support the crucible during heating.
…… [1]
(ii) Suggest why the students heat their crucibles in a fume hood in step 4.
…… [1]
(iii) Suggest why a large quantity of powdered sulfur is used in step 3.
…… [1]
(b) Five students each use a different mass of M.
Their results are shown in Table 2.1.
Complete Table 2.1 by inserting values for the mass of M and the mass of sulfur which reacts for each student.
Table 2.1
student 1 student 2 student 3 student 4 student 5
mass of crucible and lid / g 34.15 38.28 35.68 33.70 36.84
mass of crucible, lid and M / g 35.58 39.42 36.54 34.27 37.13
mass of crucible, lid and
36.04 39.70 36.74 34.42 37.19 residue after heating / g
mass of M / g
mass of sulfur which reacts / g
[1]
(c) Identify the independent variable in this experiment.
…… [1]
(d) (i) Plot a graph on the grid in Fig. 2.2 to show the relationship between mass of sulfur which reacts and mass of M.
Use a cross (×) to plot each data point. Draw a straight line of best fit which includes the origin.
0.50
0.45
0.40
0.35
0.30
mass of sulfur which 0.25 reacts
/g
0.20
0.15
0.10
0.05
0.00
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 mass of M / g
Fig. 2.2
[2]
(ii) Circle the point on the graph you consider to be most anomalous.
Suggest one reason why this anomaly may have occurred during this experimental procedure.
Assume no error was made in the measurement of any mass.
…… [2]
(e) Determine the gradient of your 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]
(f) Use your line of best fit in Fig. 2.2 to determine if the results obtained by the students are reliable.
Explain your answer.
…… [1]
(g) Another student suggests that the metal M used in the experiment is strontium, Sr, which forms strontium sulfide, SrS, when heated with sulfur.
(i) Deduce the gradient of the line of best fit for the graph of mass of sulfur which reacts against mass of strontium for the compound SrS.
gradient …… [1]
(ii) Use your answer to (i) to explain if the results of the experiment support the student’s suggestion.
…… [1]
[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 = 6.022 × 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
3 4 5 6 7 8 9 10 11 12
23.0 24.3 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
138.9 140.1 140.9 144.2 – 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) pipeclay triangle 1
2(a)(ii) toxic gas is produced. 1
2(a)(iii) to ensure all of the metal M reacts. 1
2(b) 1
mass of M 1.4 1.1 0.8 0.5 0.2
3 4 6 7 9
mass of S 0.4 0.2 0.2 0.1 0.0
6 8 0 5 6
2(c) mass of M or mass of metal 1
2(d)(i) 5 points plotted correctly from 2(b) 1
straight line of best fit drawn which includes origin 1
2(d)(ii) most anomalous point selected 1
not all of the sulfur had been removed 1
(in step 4)
2(e) co-ordinates correctly recorded from LOBF 2
gradient calculated to three significant figures
2(f) not reliable 1
and
there is an anomalous point
2(g)(i) (32.1 / 87.6) = 0.366 1
2(g)(ii) (assuming values in 2(e) and 2(g)(i) are different) 1
not correct
and
gradient is different
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Chemistry 9701/52 for October/November 2024 is out of 30 marks, with 1 hour 15 minutes to answer it, across 2 questions.
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