S4 Chemistry Practice Paper: Mole Calculations, Concentration And Titration (12 Questions, 100 Marks)

S4 Chemistry Practice Paper: Mole Calculations, Concentration and Titration (12 questions, 100 marks)

2026-11-08

S4 Chemistry: Moles, Concentration and Titration

S4 Chemistry: moles, Avogadro constant, molar concentration, dilution and titration, plus molar gas volume and sample purity. 12 questions, 100 marks, 75 minutes, answers included.

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Section A: Moles and Mass Calculations (4 questions, 33 marks)

Moles: always show n = m / M with the relative atomic masses used; every answer needs units, and volumes must be converted to dm3 first.

#	Question	Answer and marking notes
1	Calculate the number of moles in 8.0 g of copper(II) oxide, CuO. (Cu = 64, O = 16)	M(CuO) = 64 + 16 = 80 g/mol; n = 8.0 / 80 = 0.10 mol. Marking: correct molar mass (1 mark), use of n = m / M (1 mark), answer with unit (1 mark).
2	Calculate the number of moles in 4.4 g of carbon dioxide, CO2, and the number of molecules it contains. (C = 12, O = 16, Avogadro constant L = 6.02 x 10^23 /mol)	M(CO2) = 12 + 2 x 16 = 44 g/mol; n = 4.4 / 44 = 0.10 mol; number of molecules = 0.10 x 6.02 x 10^23 = 6.02 x 10^22. Marking: molar mass, number of moles and number of particles each carry marks.
3	2.0 g of sodium hydroxide, NaOH, is dissolved in water to make 250 cm3 of solution. Calculate the molar concentration of the solution. (Na = 23, O = 16, H = 1)	M(NaOH) = 23 + 16 + 1 = 40 g/mol; n = 2.0 / 40 = 0.050 mol; V = 250 cm3 = 0.250 dm3; c = 0.050 / 0.250 = 0.20 mol/dm3. Marking: converting cm3 to dm3 (1 mark); the rest is mole and concentration work.
4	12.0 g of magnesium reacts completely with excess dilute hydrochloric acid: Mg + 2HCl -> MgCl2 + H2. Find (a) the number of moles of magnesium; (b) the number of moles of hydrogen produced; (c) the volume of hydrogen at room temperature and pressure (molar volume 24.0 dm3/mol). (Mg = 24)	(a) n(Mg) = 12.0 / 24 = 0.50 mol; (b) mole ratio 1:1, so n(H2) = 0.50 mol; (c) V = 0.50 x 24.0 = 12.0 dm3. Marking: stating the mole ratio (1 mark); assuming the ratio without an equation loses marks.

Section B: Solution Concentration, Dilution and Preparation (4 questions, 33 marks)

Dilution and preparation: use c1V1 = c2V2; include water of crystallisation in the molar mass when preparing a solution.

#	Question	Answer and marking notes
5	How many grams of CuSO4.5H2O are needed to prepare 500 cm3 of 0.50 mol/dm3 copper(II) sulphate solution? (Cu = 64, S = 32, O = 16, H = 1)	M(CuSO4) = 64 + 32 + 4 x 16 = 160 g/mol; M(CuSO4.5H2O) = 160 + 5 x 18 = 250 g/mol; n = 0.50 x 0.500 = 0.25 mol; m = 0.25 x 250 = 62.5 g. Marking: ignoring the water of crystallisation (using 160) earns only partial marks.
6	100 cm3 of 2.0 mol/dm3 hydrochloric acid is diluted with water to 500 cm3. Calculate the molar concentration of the diluted solution.	n = c1V1 = 2.0 x 0.100 = 0.20 mol; c2 = 0.20 / 0.500 = 0.40 mol/dm3. The same result follows from c1V1 = c2V2. Marking: the key idea is that the number of moles of solute is unchanged.
7	How many moles of sodium ions, Na+, are present in 25.0 cm3 of 0.20 mol/dm3 sodium carbonate solution? What is the mass of sodium carbonate in this volume? (Na = 23, C = 12, O = 16)	n(Na2CO3) = 0.20 x 0.0250 = 5.0 x 10^-3 mol; n(Na+) = 2 x 5.0 x 10^-3 = 1.0 x 10^-2 mol; M(Na2CO3) = 2 x 23 + 12 + 3 x 16 = 106 g/mol; m = 5.0 x 10^-3 x 106 = 0.53 g. Marking: the factor of 2 for Na+ is required.
8	5.85 g of sodium chloride is dissolved in water to make 200 cm3 of solution. Calculate its molar concentration. 25.0 cm3 of this solution is then diluted to 100 cm3; find the final concentration. (Na = 23, Cl = 35.5)	n(NaCl) = 5.85 / 58.5 = 0.100 mol; c = 0.100 / 0.200 = 0.50 mol/dm3; after dilution c2 = 0.50 x 25.0 / 100 = 0.125 mol/dm3. Marking: both concentrations carry marks; show the volume ratio.

Section C: Titration and Molar Gas Volume (4 questions, 34 marks)

Titration and gases: fix the mole ratio from the equation; use 24.0 dm3/mol at RTP; purity = mass of pure substance / mass of sample x 100%.

#	Question	Answer and marking notes
9	0.100 mol/dm3 hydrochloric acid is used to titrate 25.0 cm3 of sodium hydroxide solution of unknown concentration; 20.0 cm3 of the acid is used to reach the end point. Calculate the molar concentration of the sodium hydroxide solution. (NaOH + HCl -> NaCl + H2O)	n(HCl) = 0.100 x 0.0200 = 2.0 x 10^-3 mol; mole ratio 1:1, so n(NaOH) = 2.0 x 10^-3 mol; c = 2.0 x 10^-3 / 0.0250 = 0.080 mol/dm3. Marking: only the acid actually used in the titration counts.
10	25.0 cm3 of 0.050 mol/dm3 sodium carbonate solution reacts with 0.100 mol/dm3 hydrochloric acid: Na2CO3 + 2HCl -> 2NaCl + H2O + CO2. Calculate the volume of acid, in cm3, needed for complete reaction.	n(Na2CO3) = 0.050 x 0.0250 = 1.25 x 10^-3 mol; mole ratio 1:2, so n(HCl) = 2.50 x 10^-3 mol; V = 2.50 x 10^-3 / 0.100 = 0.0250 dm3 = 25.0 cm3. Marking: forgetting the factor of 2 is the common error; give the answer in cm3.
11	4.0 g of an impure calcium carbonate sample reacts with excess dilute hydrochloric acid and gives 0.60 dm3 of carbon dioxide at room temperature and pressure (molar volume 24.0 dm3/mol). Find (a) the number of moles of carbon dioxide; (b) the mass of calcium carbonate in the sample; (c) the percentage purity of the sample. (Ca = 40, C = 12, O = 16)	(a) n(CO2) = 0.60 / 24.0 = 0.025 mol; (b) mole ratio 1:1, M(CaCO3) = 40 + 12 + 48 = 100 g/mol, m = 0.025 x 100 = 2.50 g; (c) purity = 2.50 / 4.0 x 100% = 62.5%. Marking: the denominator of purity is the mass of the sample.
12	8.0 g of iron(III) oxide, Fe2O3, is completely reduced by excess carbon monoxide: Fe2O3 + 3CO -> 2Fe + 3CO2. Find (a) the number of moles of Fe2O3; (b) the mass of iron produced; (c) the volume of carbon dioxide produced at room temperature and pressure. (Fe = 56, O = 16, C = 12)	(a) M(Fe2O3) = 2 x 56 + 3 x 16 = 160 g/mol, n = 8.0 / 160 = 0.050 mol; (b) mole ratio 1:2, n(Fe) = 0.100 mol, m = 0.100 x 56 = 5.6 g; (c) mole ratio 1:3, n(CO2) = 0.150 mol, V = 0.150 x 24.0 = 3.6 dm3. Marking: all three mole ratios must be stated.

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