CIE A-Level Chemistry 9701 Paper 3 Practice Paper (Advanced Practical Skills) - 40 marks, 2 h, with answers and mark scheme

CIE A-Level Chemistry 9701 Paper 3 Practice Paper (Advanced Practical Skills) – 40 marks, 2 h, with answers and mark scheme

2026-10-30

TutorZone original practice paper – CIE A-Level Chemistry 9701 Paper 3 (Advanced Practical Skills)

2 questions, 11 parts – 40 marks – 1 hour 15 minutes – calculator allowed

Total 40 marks · 2 h · 11 parts

Marks distribution 15 + 15 = 30

Level A2 (second year of A-Level)

Paper comparison

Paper	Time	Marks	Role
Paper 3 (advanced practical skills)	2 h	40	this paper
Paper 4 (A Level structured)	2 hours	100	other A Level paper
Paper 3 (AS practical skills)	2 hours	40	AS paper
Paper 2 (AS structured)	2 h	60	AS paper

Data and formulae

Quantity	Value
Gas constant R	8.31 J K-1 mol-1
Avogadro constant L	6.02 x 1023 mol-1
Planck constant h	6.63 x 10-34 J s
Speed of light c	3.00 x 108 m s-1
Elementary charge e	1.60 x 10-19 C
Relative atomic masses H, C, O, Na, S	1.0, 12.0, 16.0, 23.0, 32.1

Questions

Total: 40 marks · 2 h · calculator allowed

Q 1 · Titration: purity of the iron(II) salt FA 1 using potassium manganate(VII) [14]

(a) A student titrates 25.0 cm3 of FA 1 (iron(II) sulfate solution acidified with dilute sulfuric acid) with 0.0200 mol dm-3 KMnO4(aq). The burette readings are shown. Calculate the titre of each titration and state which one must be rejected.

Titration	Initial reading / cm3	Final reading / cm3
rough	0.00	24.20
1	0.15	24.85
2	0.00	24.10
3	0.00	24.15

[2]

(b) Calculate the mean titre from the concordant readings (cm3, 2 decimal places). [2]

(c) Calculate the number of moles of KMnO4 in the mean titre. [1]

(d) Write the ionic equation for the reaction between MnO4- and Fe2+, and calculate the moles of Fe2+ in 25.0 cm3 of FA 1. [2]

(e) FA 1 was made by dissolving 9.45 g of the hydrated salt in water and making the volume up to 250 cm3. Calculate the moles of Fe2+ in 250 cm3 and hence the relative molecular mass Mr of the salt. [2]

(f) Calculate the mass of iron in FA 1 and the percentage by mass of iron in the sample. [2]

(g) The salt is (NH4)2Fe(SO4)2.6H2O, Mr = 392.1, containing 14.23% iron. Compare your answers with these values, comment on the purity of the sample and suggest one improvement to increase accuracy. [3]

Q 2 · Qualitative analysis: identifying the ions in the unknown salt FA 4 [12]

(a) FA 4 is a hydrated salt. Complete the table with the observations for each test (colour, precipitate, solubility, gas).

Test	Observation
1. Dissolve FA 4 in distilled water
2. Add a little NaOH(aq), then excess, then warm
3. Add BaCl2(aq), then excess dilute HCl
4. Add a little aqueous ammonia, then excess ammonia

[4]

(b) From the observations, state which three ions FA 4 contains and justify each one. [3]

(c) FA 4 is NH4Al(SO4)2.12H2O. Using Ar: N 14.0, H 1.0, Al 27.0, S 32.1, O 16.0, calculate its Mr and the percentage by mass of water of crystallisation. [3]

(d) Explain how test 4 distinguishes Al3+ from Zn2+. [2]

Q 3 · Calorimetry: enthalpy change of neutralisation of sodium hydroxide with hydrochloric acid [14]

(a) 25.0 cm3 of 1.00 mol dm-3 hydrochloric acid and 25.0 cm3 of 1.00 mol dm-3 sodium hydroxide are mixed in an insulated cup and the temperature is recorded every 30 s, as shown. State the maximum temperature and the temperature rise, and explain why the maximum, not the last reading, is used.

Time / s	0	30	60	90	120	150	180	210	240	270
Temperature / C	20.0	20.0	20.5	24.0	26.0	26.7	26.7	26.6	26.5	26.4

[3]

(b) Assuming the solution density is 1.00 g/cm3 and its specific heat capacity is 4.18 J g-1 K-1, calculate the energy released by the reaction in J. [3]

(c) Calculate the moles of water formed and hence the enthalpy change of neutralisation DH in kJ/mol, with its sign. [3]

(d) State two reasons why the experimental value is not accurate. [2]

(e) Suggest two improvements and explain why your value (-56.0 kJ/mol) is smaller in magnitude than the Data Booklet value of -57.3 kJ/mol. [3]

Answers and mark scheme

Q	Part	Answer	Mark points	marks
1	(a)	The titres are 24.20, 24.70, 24.10 and 24.15 cm3. Titration 1 (24.70 cm3) differs by 0.50 cm3 from the rest, far outside 0.20 cm3, so it is not concordant and must be rejected; it is a rough titration only.	1 mark: all four titres correct (24.20 / 24.70 / 24.10 / 24.15) 1 mark: 24.70 cm3 rejected, difference from the rest is more than 0.20 cm3	2
1	(b)	The concordant titres are 24.20, 24.10 and 24.15 cm3; the mean is (24.20+24.10+24.15)/3 = 24.15 cm3.	1 mark: only the three concordant titres used 1 mark: mean = 24.15 cm3 to 2 d.p.	2
1	(c)	n(KMnO4) = 0.0200 x 24.15 / 1000 = 4.83 x 10-4 mol.	1 mark: 4.83 x 10-4 mol (accept 4.8 x 10-4)	1
1	(d)	MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O. The ratio is 1:5, so n(Fe2+) = 5 x 4.83 x 10-4 = 2.415 x 10-3 mol.	1 mark: correct coefficients (5Fe2+, 8H+, 4H2O) 1 mark: 2.415 x 10-3 mol	2
1	(e)	n(Fe2+) = 2.415 x 10-3 x 10 = 2.415 x 10-2 mol; Mr = 9.45 / 2.415 x 10-2 = 391.3.	1 mark: x10 to give 2.415 x 10-2 mol 1 mark: Mr = 391 (accept 391.3)	2
1	(f)	m(Fe) = 2.415 x 10-2 x 55.8 = 1.348 g (1.35 g); percentage = 1.348 / 9.45 x 100 = 14.26%.	1 mark: mass of iron 1.35 g 1 mark: 14.26% (accept 14.3%)	2
1	(g)	Mr 391.3 differs from 392.1 by 0.2%, and 14.26% differs from 14.23% by 0.03 percentage points; both are within experimental error, so the sample is essentially pure. Improvement: stand the conical flask on a white tile so the first permanent pale pink end-point is easier to see, and add the last drops one at a time with constant swirling to avoid overshooting.	1 mark: comparison of Mr (0.2% difference) 1 mark: comparison of iron percentage (0.03 points) 1 mark: sensible improvement with justification	3
2	(a)	1. A colourless solution forms. 2. With a little NaOH a white precipitate forms; it dissolves in excess to give a colourless solution; on warming a gas is given off that turns damp red litmus paper blue. 3. With BaCl2(aq) a white precipitate forms that does not dissolve in excess dilute HCl. 4. With a little ammonia a white precipitate forms; it does not dissolve in excess ammonia.	1 mark each: colourless solution; white ppt soluble in excess with NH3 gas; white ppt insoluble in acid; white ppt insoluble in excess ammonia	4
2	(b)	NH4+: warming with alkali gives a gas that turns damp red litmus blue (ammonia). Al3+: NaOH gives a white precipitate that dissolves in excess NaOH (amphoteric) but not in excess ammonia. SO4(2-): BaCl2(aq) gives a white precipitate that is insoluble in excess dilute HCl, unlike a carbonate or sulfite.	1 mark: NH4+ with evidence 1 mark: Al3+ with evidence 1 mark: SO4(2-) with evidence	3
2	(c)	NH4 = 18.0; Al = 27.0; SO4 = 32.1 + 64.0 = 96.1, x2 = 192.2; 12H2O = 12 x 18.0 = 216.0. Mr = 18.0 + 27.0 + 192.2 + 216.0 = 453.2. Percentage water = 216.0 / 453.2 x 100 = 47.66%.	1 mark: correct parts for SO4 and 12H2O 1 mark: Mr = 453.2 1 mark: 47.66% (accept 47.7%)	3
2	(d)	Both give a white precipitate with a little ammonia. In excess ammonia, Zn2+ forms the soluble complex [Zn(NH3)4]2+ and the precipitate dissolves, whereas Al3+ does not form a stable ammine complex so its precipitate remains. A precipitate that is insoluble in excess ammonia therefore identifies Al3+.	1 mark: Zn2+ dissolves as a soluble complex 1 mark: Al3+ precipitate stays, so the two are distinguished	2
3	(a)	Maximum temperature 26.7 C, initial 20.0 C, so the rise is DT = 26.7 – 20.0 = 6.7 C. The maximum is the moment the reaction has just finished, before much heat is lost; the later readings fall as heat is lost to the surroundings, so the last reading (26.4 C) would underestimate DT.	1 mark: 26.7 C 1 mark: DT = 6.7 C 1 mark: maximum is before heat loss, last reading underestimates DT	3
3	(b)	Total volume = 25.0 + 25.0 = 50.0 cm3, mass = 50.0 x 1.00 = 50.0 g. q = mcDT = 50.0 x 4.18 x 6.7 = 1,400 J (1.40 kJ).	1 mark: mass 50.0 g from the total volume 1 mark: q = mcDT substituted correctly 1 mark: 1,400 J (1.40 kJ)	3
3	(c)	n(HCl) = 1.00 x 25.0 / 1000 = 0.0250 mol and n(NaOH) is also 0.0250 mol, so 0.0250 mol of water is formed. DH = -q/n = -1,400 / 0.0250 = -56,000 J/mol = -56.0 kJ/mol.	1 mark: 0.0250 mol of water 1 mark: q/n = 56,000 J/mol 1 mark: -56.0 kJ/mol with the negative sign explained	3
3	(d)	(1) Heat is lost to the cup and the surroundings, so the measured DT is too small. (2) The heat capacity of the cup is ignored and the density and specific heat capacity of the solution are taken as those of pure water, so q is underestimated.	1 mark: heat loss 1 mark: heat capacity of the cup ignored / properties taken as water	2
3	(e)	Improvements: use a polystyrene cup with a lid and pass the thermometer through the lid to cut heat loss and evaporation; plot temperature against time and extrapolate back to the moment of mixing (cooling correction), or use more concentrated solutions so DT is larger and the relative error smaller. The magnitude is smaller because some heat is lost between mixing and reading, so DT is too small and \|DH\| is underestimated; the Data Booklet value assumes perfect insulation.	1 mark: lid / better insulation 1 mark: cooling correction or larger DT 1 mark: heat loss makes DT and \|DH\| too small	3

Topic breakdown

Question	Topic	Marks
1	Titration: purity of the iron(II) salt FA 1 using potassium manganate(VII)	14
2	Qualitative analysis: identifying the ions in the unknown salt FA 4	12
3	Calorimetry: enthalpy change of neutralisation of sodium hydroxide with hydrochloric acid	14

Common lost marks

Plotting V rather than ln V, which gives a curve and a gradient that is not 1/(RC).
Omitting the unit s-1 for the gradient, or putting ln V into C = 1/(R x gradient) as if it were a voltage.
Writing only “measure the frequency” in the planning question without saying how resonance is obtained.
Counting only the 5% resistor tolerance and forgetting the 3% uncertainty in reading the gradient.

How to use this paper Sit the paper closed-book in 75 minutes, then mark point by point with the answer booklet; redo every lost mark and note whether it was the formula, the substitution or the unit.

Syllabus reference Format follows the CIE 9701 Chemistry syllabus (2025-2027) Paper 3: 2 h, 40 marks, advanced practical skills.

Frequently asked

What does Paper 3 test?
Paper 3 is the advanced practical skills paper: 2 hours, 40 marks. Question 1 is a titration with calculations (moles, Mr, percentage by mass), question 2 is the qualitative analysis of an unknown salt, and question 3 is the handling and evaluation of data from a calorimetry or rates experiment.

Can I sit it if I have not done the experiment myself?
Yes. Paper 3 tests practical technique and data handling rather than hands-on experience, but you must be fluent with burette readings and concordant titres, the coefficients of ionic equations, q = mcDT and error analysis.

Where are marks most often lost?
Titres must be quoted to 2 decimal places and only concordant readings averaged; ionic equations need correct coefficients and state symbols; calorimetry needs the total mass of solution; error analysis needs a specific source, not just “human error”.

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