中文（繁體）
中文（简体）
English
HKDSE Chemistry Paper 1B Structured-question Mock Paper (10 questions, 100 marks)
HKDSE Chemistry
HKDSE Chemistry Paper 1B structured-question mock paper: 10 questions, 100 marks, 2.5 hours, covering moles, titration, rates, equilibrium, electrolysis and organic.
Download Question paper PDF | Answer key PDF
Paper details Level: HKDSE | Subject: Chemistry | Time: 2 hours 30 minutes | Marks: 100 | Questions: 10
Section A (50 marks) — Questions 1 to 5
Structured questions: show full working, substitutions and units; balance every equation and give state symbols.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) 25.0 g of calcium carbonate (CaCO₃, Mr = 100.1) is heated strongly in a crucible until it fully decomposes. Write the balanced equation (with state symbols) and calculate the number of moles of calcium oxide formed. (b) Calculate the mass of calcium oxide formed and the volume of carbon dioxide released at room temperature and pressure (molar volume 24.0 dm³ mol⁻¹). (c) If only 12.0 g of calcium oxide is actually obtained, calculate the percentage yield and suggest one reason for the lower yield. |
(a) CaCO₃(s) → CaO(s) + CO₂(g). n(CaCO₃) = 25.0 / 100.1 = 0.2498 mol, so n(CaO) = 0.2498 mol. [2 marks: equation 1, moles 1] (b) m(CaO) = 0.2498 × 56.1 = 14.01 g; V(CO₂) = 0.2498 × 24.0 = 5.99 dm³. [3 marks] (c) Percentage yield = 12.0 / 14.01 × 100% = 85.6%. Reason: incomplete heating so some CaCO₃ is undecomposed / powder lost on transfer / crucible left open so some CaO is carried away. [3 marks] |
| 2 | (a) 25.0 cm³ of sodium hydroxide solution of unknown concentration is titrated with 0.0500 M sulfuric acid; 20.0 cm³ is used to reach the end point. Write the balanced ionic equation and calculate the number of moles of NaOH. (b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. (c) If the batch of solution is 250.0 cm³ in total, calculate the mass of NaOH in it; also calculate the volume of 0.100 M hydrochloric acid needed to titrate 25.0 cm³ of the same solution. |
(a) H⁺ + OH⁻ → H₂O (or H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O). n(H₂SO₄) = 0.0500 × 0.0200 = 1.00e-03 mol, n(NaOH) = 2 × 1.00e-03 = 2.00e-03 mol. [3 marks] (b) c(NaOH) = 2.00e-03 / 0.0250 = 0.0800 mol dm⁻³. [3 marks] (c) m(NaOH) = 0.0800 × 0.2500 × 40.0 = 0.800 g; V(HCl) = 2.00e-03 / 0.100 = 0.0200 dm³ = 20.0 cm³. [4 marks] |
| 3 | (a) 0.0600 g of magnesium ribbon is added to excess dilute hydrochloric acid and 56.0 cm³ of hydrogen is collected in 40 seconds (at room temperature and pressure, molar volume 24.0 dm³ mol⁻¹). Write the balanced equation and calculate the number of moles of magnesium. (b) Calculate the theoretical volume of hydrogen (in cm³) and the percentage yield. (c) Calculate the average rate of hydrogen production over the 40 seconds (cm³ s⁻¹) and explain why the rate falls later in the reaction. |
(a) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). n(Mg) = 0.0600 / 24.3 = 2.4691e-03 mol. [3 marks] (b) n(H₂) = 2.4691e-03 mol; V = 2.4691e-03 × 24.0 × 1000 = 59.3 cm³; percentage yield = 56.0 / 59.3 × 100% = 94.5%. [4 marks] (c) Average rate = 56.0 / 40.0 = 1.40 cm³ s⁻¹. Later the acid is less concentrated and the surface area of magnesium is smaller (covered by MgCl₂), so the frequency of effective collisions decreases. [3 marks] |
| 4 | Calculate the pH of each of the following solutions (assume complete ionisation, 25 °C): (a) 0.0100 M hydrochloric acid; (b) 0.0100 M sodium hydroxide; (c) 0.00500 M sulfuric acid; (d) solution (a) diluted 1,000 times; (e) the mixture of 25.0 cm³ of (a) with 25.0 cm³ of (b). |
(a) [H⁺] = 0.0100 M, pH = 2.00. [2 marks] (b) [OH⁻] = 0.0100 M, pOH = 2.00, pH = 12.00. [2 marks] (c) Sulfuric acid is diprotic, [H⁺] = 2 × 0.00500 = 0.0100 M, pH = 2.00. [2 marks] (d) [H⁺] = 1.00 × 10⁻⁵ M, pH = 5.00. [2 marks] (e) Equal moles of acid and alkali (2.50 × 10⁻⁴ mol each), complete neutralisation, pH = 7.00. [2 marks] |
| 5 | An aluminium cup holding 200 g of water is heated by burning 0.200 g of ethanol (C₂H₅OH, Mr = 46.0) with a spirit burner; the water temperature rises by 6.0 °C. Specific heat capacity of water = 4.18 J g⁻¹ K⁻¹. (a) Calculate the heat energy absorbed by the water (in J and kJ). (b) Calculate the enthalpy change of combustion of ethanol, ΔH, in kJ mol⁻¹. (c) The literature value is −1,366 kJ mol⁻¹. Calculate the experimental value as a percentage of the literature value, explain the difference and suggest two improvements. |
(a) q = mcΔT = 200 × 4.18 × 6.0 = 5016 J = 5.02 kJ. [4 marks] (b) n(C₂H₅OH) = 0.200 / 46.0 = 4.3478e-03 mol; ΔH = −5.02 / 4.3478e-03 = -1154 kJ mol⁻¹. [4 marks] (c) Percentage = 1154 / 1,366 × 100% = 84.5%. Heat is lost to the air and the aluminium cup, the ethanol burns incompletely and some heat heats the container; improvements: use a draught shield, use an insulated calorimeter, ensure complete combustion and heat more slowly. [4 marks] |
Section B (50 marks) — Questions 6 to 10
Structured questions: show full working, substitutions and units; balance every equation and give state symbols.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | In a 2.00 dm³ vessel at 450 °C, 0.500 mol of hydrogen is mixed with 0.500 mol of iodine vapour. At equilibrium for H₂(g) + I₂(g) ⇌ 2HI(g), 0.200 mol of hydrogen remains. (a) Calculate the concentration of each substance at equilibrium. (b) Calculate the equilibrium constant Kc at 450 °C. (c) Calculate the percentage conversion of hydrogen and explain the effect of compressing the vessel to 1.00 dm³ on the equilibrium concentration of HI. |
(a) [H₂]₀ = 0.500 / 2.00 = 0.250 M, [H₂]eq = 0.200 / 2.00 = 0.100 M; change −0.150 M, so [I₂]eq = 0.100 M and [HI]eq = 0.300 M. [3 marks] (b) Kc = [HI]² / ([H₂][I₂]) = 0.300² / (0.100 × 0.100) = 9.00. [4 marks] (c) Percentage conversion = (0.250 − 0.100) / 0.250 × 100% = 60.0%. After compression to 1.00 dm³ every concentration doubles, but the number of gas moles is the same on both sides (Δn = 0), so Kc is unchanged and the position of equilibrium does not shift; the equilibrium concentration of HI doubles to about 0.6 M. [3 marks] |
| 7 | A copper(II) sulfate solution is electrolysed with a current of 2.00 A for 30.0 minutes (Faraday constant = 96,500 C mol⁻¹). (a) Calculate the quantity of electricity passed and the number of moles of electrons. (b) Calculate the mass of copper deposited at the cathode (Ar(Cu) = 63.5). (c) Calculate the number of moles and the volume at room temperature and pressure of oxygen produced at the anode, and explain the change in pH of the solution during electrolysis. |
(a) Q = 2.00 × 1,800 = 3600 C; n(e⁻) = 3600 / 96,500 = 3.7306e-02 mol. [3 marks] (b) Cu²⁺ + 2e⁻ → Cu, n(Cu) = 3.7306e-02 / 2 = 1.8653e-02 mol, m = 1.8653e-02 × 63.5 = 1.18 g. [3 marks] (c) 2H₂O → O₂ + 4H⁺ + 4e⁻, n(O₂) = 3.7306e-02 / 4 = 9.3264e-03 mol, V = 9.3264e-03 × 24.0 × 1000 = 224 cm³. H⁺ is produced at the anode while Cu²⁺ is removed at the cathode, so the pH falls (the blue colour also fades). [4 marks] |
| 8 | (a) A compound contains 40.0% carbon, 6.70% hydrogen and 53.3% oxygen by mass; its relative molecular mass is 180. Find its empirical and molecular formulae. (b) 0.100 mol of ethanol is heated under reflux with excess ethanoic acid and 6.60 g of ester is obtained. Write the equation, name the product and calculate the percentage yield (Mr of ester = 88.0). |
(a) Mole ratio C : H : O = 40.0/12.0 : 6.70/1.0 : 53.3/16.0 = 3.33 : 6.70 : 3.33 = 1 : 2 : 1, empirical formula CH₂O (Mr = 30.0). Number of units = 180 / 30.0 = 6, so the molecular formula is C₆H₁₂O₆. [5 marks] (b) CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O; the product is ethyl ethanoate. Theoretical mass = 0.100 × 88.0 = 8.80 g; percentage yield = 6.60 / 8.80 × 100% = 75.0%. [5 marks] |
| 9 | 25.0 cm³ of an iron(II) solution is titrated with acidified potassium permanganate solution and 20.0 cm³ of 0.0200 M KMnO₄ is used to reach the end point. (a) State the oxidation number of manganese in MnO₄⁻ and in Mn²⁺, and the number of electrons gained per MnO₄⁻ ion. (b) Calculate the number of moles of Fe²⁺. (c) Calculate the concentration of the iron(II) solution and the mass of iron in 250.0 cm³ of it (Ar(Fe) = 55.8). |
(a) The oxidation number of Mn is +7 in MnO₄⁻ and +2 in Mn²⁺, so each MnO₄⁻ ion gains 5 electrons (it is reduced). [2 marks] (b) n(MnO₄⁻) = 0.0200 × 0.0200 = 4.00e-04 mol; MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, n(Fe²⁺) = 5 × 4.00e-04 = 2.00e-03 mol. [3 marks] (c) c(Fe²⁺) = 2.00e-03 / 0.0250 = 0.0800 mol dm⁻³; m(Fe) = 0.0800 × 0.2500 × 55.8 = 1.116 g. [5 marks] |
| 10 | At room temperature and pressure, 3.20 g of a gaseous compound of carbon, hydrogen and oxygen occupies 2.40 dm³ (molar volume 24.0 dm³ mol⁻¹). Elemental analysis gives C 37.5%, H 12.5% and O 50.0% by mass. (a) Calculate the number of moles and the relative molecular mass of the gas. (b) Find its empirical and molecular formulae. (c) Draw one possible structural formula and explain why it is completely miscible with water. |
(a) n = 2.40 / 24.0 = 0.100 mol; Mr = 3.20 / 0.100 = 32.0. [3 marks] (b) Mole ratio = 37.5/12.0 : 12.5/1.0 : 50.0/16.0 = 3.125 : 12.5 : 3.125 = 1 : 4 : 1, empirical formula CH₄O (Mr = 32.0), the same as Mr 32.0, so the molecular formula is also CH₄O. [4 marks] (c) Structural formula CH₃—OH (methanol); the –OH group forms hydrogen bonds with water molecules, so it is completely miscible with water. [3 marks] |
A TutorZone practice paper; original questions and passages, not publisher material.
