CAIE A-Level Chemistry 9701 Paper 2 Structured Practice Paper | 10 questions, 60 marks, 1 hour 15 minutes

2026-10-31

CAIE A-Level Chemistry 9701 Paper 2

CAIE A-Level Chemistry 9701 Paper 2 practice paper: 10 questions, 60 marks, 1 hour 15 minutes, covering titration, empirical formula, gas volume, Hess law, yield, bonding and equilibrium, with full worked answers.

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Paper details　Level: A-Level AS | Subject: Chemistry | Time: 1 hour 15 minutes | Marks: 60 | Questions: 10

Section A: structured questions (36 marks)

Structured questions: show your working, give units and use appropriate significant figures; answer every part of each question. Total 60 marks.

#	Question	Answer and marking notes
1	(Titration) 25.0 cm³ of sodium hydroxide solution is titrated with 0.150 mol dm⁻³ hydrochloric acid; 20.0 cm³ of the acid is required to reach the end point. (a) Calculate the amount, in mol, of hydrochloric acid used. (b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. (6 marks)	Answer: (a) n(HCl) = 0.150 × 20.0 ÷ 1000 = 0.00300 mol, i.e. 3.00 × 10⁻³ mol to 3 s.f. (b) NaOH reacts with HCl in a 1 : 1 ratio, so n(NaOH) = 0.00300 mol and c = 0.00300 ÷ 0.0250 = 0.120 mol dm⁻³. (6 marks)
2	(Empirical formula) An organic compound contains 40.0%% carbon, 6.7%% hydrogen and 53.3%% oxygen by mass. Deduce its empirical formula and state what additional information is needed to find the molecular formula. (6 marks)	Answer: Amount ratio C : H : O = 40.0 ÷ 12.0 : 6.7 ÷ 1.0 : 53.3 ÷ 16.0 = 3.33 : 6.70 : 3.33; dividing by the smallest value 3.33 gives 1 : 2.01 : 1, so the empirical formula is CH₂O (Mr = 30.0). The relative molecular mass must also be known; Mr ÷ 30.0 gives the multiplier. (6 marks)
3	(Molar gas volume) 0.243 g of magnesium reacts with excess dilute hydrochloric acid, giving hydrogen at room temperature and pressure. Take the molar gas volume as 24 dm³ mol⁻¹ and Ar(Mg) = 24.3. (a) Calculate the amount of magnesium. (b) Calculate the volume of hydrogen produced, in cm³. (6 marks)	Answer: (a) n(Mg) = 0.243 ÷ 24.3 = 0.010 mol. (b) From Mg + 2HCl → MgCl₂ + H₂, n(H₂) = n(Mg) = 0.010 mol, so V = 0.010 × 24 = 0.240 dm³ = 240 cm³. (6 marks)
4	(Hess law) Standard enthalpies of formation are CH₄(g) = -75, CO₂(g) = -394 and H₂O(l) = -286 kJ mol⁻¹. Use Hess law to calculate the standard enthalpy change of CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), and state whether the reaction is exothermic or endothermic. (6 marks)	Answer: ΔH = Σ ΔHf(products) − Σ ΔHf(reactants) = [-394 + 2 × (-286)] − (-75) = -891 kJ mol⁻¹. ΔH is negative, so the reaction is exothermic. (6 marks)
5	(Percentage yield) 4.86 g of magnesium is burned completely in oxygen; the theoretical mass of magnesium oxide is 8.06 g (Mr(MgO) = 40.3) but only 6.85 g is obtained. (a) Calculate the percentage yield. (b) Give two reasons why the yield is below 100%. (6 marks)	Answer: (a) n(Mg) = 4.86 ÷ 24.3 = 0.200 mol, so n(MgO) = 0.200 mol and m = 0.200 × 40.3 = 8.06 g; yield = 6.85 ÷ 8.06 × 100% = 85%. (b) Some MgO is lost as smoke; magnesium also reacts with nitrogen in the air to form magnesium nitride; a little product remains in the container during transfer. (6 marks)
6	(Atom economy) Calcium carbonate decomposes on heating: CaCO₃ → CaO + CO₂, with Mr(CaCO₃) = 100.1 and Mr(CaO) = 56.1. (a) Calculate the atom economy for the formation of CaO. (b) Explain why a high atom economy does not mean a high actual yield. (6 marks)	Answer: (a) Atom economy = 56.1 ÷ 100.1 × 100% = 56.0%. (b) Atom economy compares theoretical masses in the equation and is a property of the reaction design; the actual yield depends on incomplete reaction, side reactions, loss of material and separation losses, so the two are independent. (6 marks)

Section B: extended questions (24 marks)

Structured questions: show your working, give units and use appropriate significant figures; answer every part of each question. Total 60 marks.

#	Question	Answer and marking notes
7	(Bonding and structure) Magnesium oxide melts at 2852 °C, whereas carbon dioxide is a gas at room temperature. Explain the large difference in their melting and boiling points in terms of bonding and structure. (6 marks)	Answer: MgO is a giant ionic lattice of Mg²⁺ and O²⁻ ions held by ionic bonds, so melting requires overcoming many strong electrostatic attractions and the melting point is very high. CO₂ consists of simple molecules held together only by weak van der Waals (London dispersion) forces, so it boils at a low temperature. The C=O covalent bonds within a molecule are strong but are not broken on melting or boiling. (6 marks)
8	(Rates of reaction) Explain how raising the temperature affects the rate of a reaction, referring to the distribution of molecular energies, and explain how a catalyst increases the rate. (6 marks)	Answer: Raising the temperature increases the average kinetic energy of the molecules, so the Maxwell–Boltzmann distribution shifts to the right and becomes flatter; the fraction of molecules with energy greater than the activation energy rises sharply, so there are more successful collisions per unit time and the rate increases. A catalyst provides an alternative route of lower activation energy, so a larger fraction of collisions is successful; the enthalpy change is unchanged. (6 marks)
9	(Equilibrium constant) At 700 K, a 2.00 dm³ vessel at equilibrium contains 0.20 mol H₂, 0.20 mol I₂ and 1.60 mol HI for H₂(g) + I₂(g) ⇌ 2HI(g). (a) Calculate the concentration of each species. (b) Calculate Kc and state its units. (6 marks)	Answer: (a) [H₂] = 0.20 ÷ 2.00 = 0.10 mol dm⁻³, [I₂] = 0.20 ÷ 2.00 = 0.10 mol dm⁻³ and [HI] = 1.60 ÷ 2.00 = 0.80 mol dm⁻³. (b) Kc = [HI]² ÷ ([H₂][I₂]) = 0.80² ÷ (0.10 × 0.10) = 64.0; the powers of concentration in the numerator and denominator are equal, so Kc has no units. (6 marks)
10	(Organic reaction) Propene reacts with hydrogen bromide to give two products. (a) Name the type of reaction and give the structure of the major product. (b) Explain why the major product is formed in the greater amount, naming the intermediate. (6 marks)	Answer: (a) Electrophilic addition; the major product is 2-bromopropane, CH₃CHBrCH₃. (b) The reaction proceeds through a carbocation intermediate: H⁺ adds to the terminal carbon of the double bond to give the more stable secondary carbocation CH₃C⁺HCH₃ rather than the primary carbocation. The secondary ion is stabilised by the inductive effect of two alkyl groups and by hyperconjugation, so it forms faster and is longer lived, and 2-bromopropane predominates. (6 marks)

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