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HKDSE Chemistry Paper 1B Calculations and Practical Data Practice Paper (10 Questions, 100 Marks)
HKDSE Chemistry Paper 1B Calculations
HKDSE Chemistry Paper 1B calculations and practical data practice paper: 10 original questions worth 100 marks in 60 minutes, covering mole and mass conversions.
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Paper details Level: S6 (DSE) | Subject: Chemistry | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Matter and Chemical Calculations (50 marks)
Questions 1-5
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Magnesium burns completely in oxygen. 4.8 g of magnesium (Ar = 24) is heated in excess oxygen until the reaction is complete. (a) Calculate the number of moles of magnesium. (3 marks) (b) Calculate the mass of magnesium oxide formed (Mr = 40). (3 marks) (c) Calculate the volume of oxygen required at room temperature and pressure (molar volume = 24 dm3 mol-1). (4 marks) |
Answer: (a) 4.8 / 24 = 0.200 mol. (b) 2Mg + O2 -> 2MgO, ratio 1:1, so 0.200 x 40 = 8 g. (c) Moles of O2 = 0.200 / 2 = 0.100 mol, volume = 0.100 x 24 = 2.4 dm3. Marking notes: divide mass by relative atomic mass first; the equation fixes the 2:1 and 2:2 ratios; gas volume must be multiplied by the molar volume. |
| 2 | A 25 cm3 portion of 0.200 mol dm-3 sodium hydroxide solution is diluted to 250 cm3 with water. 25 cm3 of the diluted solution is then titrated with 0.100 mol dm-3 sulphuric acid. (a) Calculate the number of moles of sodium hydroxide before dilution. (2 marks) (b) Calculate the concentration of the diluted solution. (3 marks) (c) Calculate the volume of sulphuric acid required in cm3. (5 marks) |
Answer: (a) 0.200 x 0.025 = 0.0050 mol. (b) 0.0050 / 0.250 = 0.0200 mol dm-3. (c) Moles in the sample = 0.0200 x 0.025 = 0.000500 mol; H2SO4 + 2NaOH -> Na2SO4 + 2H2O, so acid = 0.000250 mol; volume = 0.000250 / 0.100 = 0.0025 dm3 = 2.5 cm3. Marking notes: dilution keeps moles constant and changes only concentration; the sample is one tenth of the total; the 1:2 acid to alkali ratio doubles the answer if omitted. |
| 3 | 25 g of calcium carbonate (Mr = 100) is strongly heated and cools to give 12.6 g of calcium oxide. (a) Calculate the number of moles of calcium carbonate. (2 marks) (b) Calculate the theoretical mass of calcium oxide (Mr = 56). (4 marks) (c) Calculate the percentage yield of the reaction. (4 marks) |
Answer: (a) 25 / 100 = 0.250 mol. (b) CaCO3 -> CaO + CO2, ratio 1:1, theoretical mass = 0.250 x 56 = 14 g. (c) 12.6 / 14 x 100% = 90%. Marking notes: the theoretical mass comes from the mole ratio, not from comparing masses; the theoretical value is the denominator of the yield; a yield below 100% is normal. |
| 4 | Burning 2 g of methanol (Mr = 32) raises the temperature of 500 g of water from 22.0 C to 43 C (specific heat capacity of water = 4.2 J g-1 K-1). (a) Calculate the heat absorbed by the water in J. (3 marks) (b) Calculate the number of moles of methanol burned. (3 marks) (c) Calculate the enthalpy change of combustion per mole in kJ mol-1 and state its sign. (4 marks) |
Answer: (a) 500 x 4.2 x 21 = 44,100 J = 44.1 kJ. (b) 2 / 32 = 0.0625 mol. (c) 44.1 / 0.0625 = 705.6 kJ mol-1, and since combustion releases heat, dH = -705.6 kJ mol-1. Marking notes: m in q = mcDT is the mass of water; dividing by moles gives the value per mole; exothermic reactions have negative enthalpy change. |
| 5 | 0.65 g of zinc (Ar = 65) is added to excess dilute hydrochloric acid and 240 cm3 of gas is collected in 80 s at room temperature and pressure (molar volume = 24 dm3 mol-1). (a) Calculate the number of moles of zinc. (3 marks) (b) Calculate the theoretical volume of hydrogen in cm3. (3 marks) (c) Calculate the mean rate of gas production in cm3 s-1 over the first 80 s. (4 marks) |
Answer: (a) 0.65 / 65 = 0.010 mol. (b) Zn + 2HCl -> ZnCl2 + H2, ratio 1:1, so 0.010 x 24 = 0.240 dm3 = 240 cm3. (c) 240 / 80 = 3 cm3 s-1. Marking notes: a rate is a change in volume divided by time, not the total volume; 24 dm3 is 24000 cm3; the rate falls as the acid is used up. |
Section B Equilibrium, Acids and Bases, and Electrochemistry (50 marks)
Questions 6-10
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | In a 2 dm3 vessel hydrogen and iodine vapour reach equilibrium: H2(g) + I2(g) <=> 2HI(g). At equilibrium the vessel contains 0.40 mol H2, 0.40 mol I2 and 3.2 mol HI. (a) Calculate the equilibrium concentration of each species. (3 marks) (b) Calculate the equilibrium constant Kc. (4 marks) (c) State the effect of a higher pressure and of adding a catalyst on Kc. (3 marks) |
Answer: (a) [H2] = 0.20, [I2] = 0.20, [HI] = 1.60 mol dm-3. (b) Kc = [HI]2 / ([H2][I2]) = 1.602 / (0.20 x 0.20) = 64. (c) Both sides have the same number of gas moles (Dn = 0), so pressure does not change Kc; a catalyst only speeds up the approach to equilibrium and leaves Kc unchanged. Marking notes: concentrations are moles divided by vessel volume; the coefficient 2 on HI must be squared; only temperature changes Kc. |
| 7 | At 25 C: (a) calculate the pH of 0.010 mol dm-3 hydrochloric acid (3 marks); (b) calculate the pH of 0.010 mol dm-3 sodium hydroxide solution (3 marks); (c) 10 cm3 of the acid is mixed with 30 cm3 of distilled water, calculate the pH of the mixture. (4 marks) | Answer: (a) [H+] = 0.010 mol dm-3, pH = 2. (b) [OH-] = 0.010, pOH = 2, pH = 14 – 2 = 12. (c) [H+] after mixing = 0.010 x 10 / 40 = 0.0025 mol dm-3, pH = 2.60. Marking notes: strong acids and alkalis are fully ionised; dilution is handled with conservation of moles; one pH unit is a tenfold change in concentration. |
| 8 | A current of 0.50 A is passed through copper(II) sulphate solution for 30 minutes (Faraday constant = 96500 C mol-1). (a) Calculate the quantity of charge passed in C. (2 marks) (b) Calculate the number of moles of electrons transferred. (4 marks) (c) Calculate the mass of copper deposited at the cathode (Ar = 63.5). (4 marks) |
Answer: (a) Q = It = 0.50 x 1800 = 900 C. (b) 900 / 96500 = 0.00933 mol. (c) Cu2+ + 2e- -> Cu, so copper = 0.00933 / 2 = 0.00466 mol and mass = 0.00466 x 63.5 = 0.2961 g. Marking notes: charge is current times seconds; divide by the Faraday constant for electron moles; the factor of 2 for Cu2+ must not be omitted. |
| 9 | 0.44 g of carbon dioxide (Mr = 44) is sealed in a container at room temperature and pressure (molar volume = 24 dm3 mol-1, Avogadro constant = 6.02 x 1023 mol-1). (a) Calculate the number of moles of carbon dioxide. (3 marks) (b) Calculate its volume in dm3. (3 marks) (c) Calculate the number of molecules present. (4 marks) |
Answer: (a) 0.44 / 44 = 0.010 mol. (b) 0.010 x 24 = 0.24 dm3. (c) 0.010 x 6.02 x 1023 = 6.02 x 1021 molecules. Marking notes: moles link mass, volume and particle number; index arithmetic must be handled carefully because 10-2 mol gives 1021 molecules. |
| 10 | 2.50 g of hydrated copper(II) sulphate (CuSO4.nH2O) is heated to constant mass, leaving 1.60 g of anhydrous copper(II) sulphate (Mr = 160, Mr of H2O = 18). (a) Calculate the mass of water lost. (2 marks) (b) Calculate the number of moles of anhydrous salt and of water. (4 marks) (c) Determine the value of n. (4 marks) |
Answer: (a) 2.50 – 1.60 = 0.90 g. (b) Salt = 1.60 / 160 = 0.010 mol; water = 0.90 / 18 = 0.050 mol. (c) n = 0.050 / 0.010 = 5, so the formula is CuSO4.5H2O. Marking notes: the water mass is the loss on heating; the mole ratio uses the anhydrous salt as the reference; overheating decomposes the salt and inflates n. |
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