HKDSE Chemistry Practice Paper with Answers | 14 questions, 60 minutes, 100 marks

2026-10-28

Practice paper · HKDSE Chemistry

HKDSE Chemistry practice paper: 10 MC and 4 structured questions, 60 minutes, 100 marks, covering moles, titration, rates, redox and electrolysis, with worked answers and printable PDFs.

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Paper details　Level: S6 (HKDSE) | Subject: Chemistry | Time: 60 minutes | Total: 100 marks | Questions: 14

Section A: Multiple choice (20 marks)

1. Two marks each. Choose the best answer. (Ar: H=1, C=12, O=16, Mg=24, S=32, Cl=35.5, Cu=64)

#	Question	Answer and marking notes
1	What is the relative molecular mass of H₂SO₄?	Mr = 2(1) + 32 + 4(16) = 98. (2 marks)
2	How many moles of magnesium atoms are there in 24 g of magnesium?	n = m/Ar = 24/24 = 1 mol. (2 marks)
3	How many moles of CO₂ are there in 11 g of carbon dioxide?	Mr(CO₂) = 44; n = 11/44 = 0.25 mol. (2 marks)
4	0.50 mol of NaOH is dissolved in water and made up to 250 cm³. What is the concentration?	c = n/V = 0.50/0.250 = 2.0 mol dm⁻³. (2 marks)
5	What volume of 0.10 M HCl is needed to neutralise 25.0 cm³ of 0.20 M NaOH?	n(NaOH) = 0.0250 × 0.20 = 0.0050 mol; V(HCl) = 0.0050/0.10 = 0.050 dm³ = 50.0 cm³. (2 marks)
6	What is the pH of 0.010 M hydrochloric acid?	pH = -log(0.010) = 2.0. (2 marks)
7	What is the pH of 0.010 M sodium hydroxide solution?	pOH = -log(0.010) = 2.0; pH = 14 – 2.0 = 12.0. (2 marks)
8	What is the oxidation number of sulfur in the sulfate ion, SO₄²⁻?	Let x be the oxidation number of S: x + 4(-2) = -2, so x = +6. (2 marks)
9	In a reaction the concentration of a reactant falls from 1.20 to 0.20 mol dm⁻³ in 20 s. What is the average rate?	Rate = (1.20 – 0.20)/20 = 0.050 mol dm⁻³ s⁻¹. (2 marks)
10	Chlorine has two isotopes, ³⁵Cl and ³⁷Cl, with relative abundances of 75% and 25%. What is the relative atomic mass of chlorine?	Ar = 35 × 0.75 + 37 × 0.25 = 35.5. (2 marks)

Section B: Structured questions (80 marks)

2. Answer all questions. Show your working and units.

#	Question	Answer and marking notes
11	12.0 g of magnesium burns completely in oxygen to form magnesium oxide (2Mg + O₂ → 2MgO). (a) How many moles of magnesium are burned? (b) What mass of magnesium oxide is formed? (c) What volume of oxygen is needed at room conditions (molar volume = 24 dm³ mol⁻¹)? (20 marks)	(a) n(Mg) = 12.0/24 = 0.50 mol. (6 marks) (b) n(MgO) = n(Mg) = 0.50 mol; m = 0.50 × (24 + 16) = 20.0 g. (7 marks) (c) n(O₂) = 0.50/2 = 0.25 mol; V = 0.25 × 24 = 6.0 dm³. (7 marks)
12	25.0 cm³ of sodium hydroxide solution of unknown concentration is titrated with 0.10 M hydrochloric acid; 20.0 cm³ of the acid is needed to reach the end point (NaOH + HCl → NaCl + H₂O). (a) How many moles of HCl are used? (b) What is the concentration of the sodium hydroxide solution? (c) What colour change marks the end point if methyl orange is used? (20 marks)	(a) n(HCl) = 20.0/1000 × 0.10 = 0.0020 mol. (6 marks) (b) n(NaOH) = n(HCl) = 0.0020 mol; c = 0.0020/(25.0/1000) = 0.080 mol dm⁻³. (7 marks) (c) Yellow to orange: methyl orange is yellow in alkali and red-orange in acid, so the end point is taken at orange. (7 marks)
13	A strip of magnesium is added to excess dilute hydrochloric acid and the hydrogen ion concentration is measured: it falls from 1.20 to 0.20 mol dm⁻³ in 20 s (Mg + 2HCl → MgCl₂ + H₂). (a) What is the average rate of decrease of hydrogen ion concentration? (b) Explain why the curve flattens out. (c) State two ways to speed up the reaction, each with a reason. (20 marks)	(a) Rate = (1.20 – 0.20)/20 = 0.050 mol dm⁻³ s⁻¹. (6 marks) (b) As the reaction proceeds the acid becomes more dilute, so collisions are less frequent; the magnesium surface is also used up, reducing the surface area. Both make the rate fall, so the curve flattens. (7 marks) (c) Any two: raise the temperature, because particles have more kinetic energy and more collisions exceed the activation energy; use more concentrated acid, because there are more particles per unit volume and collisions are more frequent; cut the magnesium into small pieces, because the larger surface area exposes more particles to collision. (3.5 marks each, 7 marks)
14	A current of 2.0 A is passed through copper(II) sulfate solution for 965 s, with copper electrodes at both sides (Cu²⁺ + 2e⁻ → Cu; 1 mol of electrons = 96500 C). (a) What charge passes? (b) What mass of copper is deposited at the cathode? (c) What happens at the anode, and why does the colour of the solution hardly change? (20 marks)	(a) Q = It = 2.0 × 965 = 1930 C. (6 marks) (b) n(e⁻) = 1930/96500 = 0.0200 mol; n(Cu) = 0.0200/2 = 0.0100 mol; m = 0.0100 × 64 = 0.64 g. (7 marks) (c) Copper at the anode dissolves: Cu → Cu²⁺ + 2e⁻. For every mole of copper dissolved at the anode one mole is deposited at the cathode, so the Cu²⁺ concentration is almost unchanged and the blue colour stays. (7 marks)

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