中文（繁體）
中文（简体）
English
HKDSE Chemistry Paper 1B Structured Questions Practice Paper (6 Questions, 50 Marks, 60 Minutes)
HKDSE Chemistry Paper 1B
HKDSE Chemistry Paper 1B structured question practice paper: 6 original questions in 60 minutes on mole calculations, acid-base titration, enthalpy of neutralisation.
Download Question paper PDF | Answer key PDF
Paper details Level: S6 (DSE) | Subject: Chemistry | Time: 60 minutes | Marks: 50 | Questions: 6
Section A Inorganic and physical chemistry (25 marks)
Questions 1-3 (structured, marks per part as shown)
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Calcium carbonate reacts with hydrochloric acid: CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l). 25 g of calcium carbonate (relative molecular mass 100.09) is added to excess 1.00 mol dm^-3 hydrochloric acid and reacts completely. (a) Calculate the number of moles of calcium carbonate. (2 marks) (b) Calculate the volume of carbon dioxide produced at room conditions (molar volume 24.0 dm3 mol^-1), in dm3. (3 marks) (c) Calculate the volume of 1.00 mol dm^-3 hydrochloric acid required, in cm3. (3 marks) |
(a) n = 25 / 100.09 = 0.2498 mol (1 mark for the working, 1 mark for the answer). (b) n(CO2) = n(CaCO3) = 0.2498 mol; V = 0.2498 x 24.0 = 5.99 dm3 (1 mark ratio, 1 mark substitution, 1 mark answer). (c) n(HCl) = 2 x 0.2498 = 0.4996 mol; V = 0.4996 / 1.00 x 1000 = 499.6 cm3 (1 mark ratio, 1 mark division, 1 mark conversion). |
| 2 | 25.0 cm3 of sodium hydroxide solution of unknown concentration is titrated with 0.100 mol dm^-3 hydrochloric acid using phenolphthalein. The titres are 23.40, 23.60 and 23.50 cm3. (a) Find the mean titre. (3 marks) (b) Calculate the number of moles of hydrochloric acid used. (1 mark) (c) Calculate the molar concentration of the sodium hydroxide solution. (2 marks) (d) Express the mass concentration of the solution in g dm^-3 (relative molecular mass of NaOH = 40.00). (3 marks) |
(a) Mean = (23.40 + 23.60 + 23.50) / 3 = 23.50 cm3 (1 mark readings, 1 mark mean, 1 mark unit or significant figures). (b) n(HCl) = 0.100 x 23.50 / 1000 = 0.00235 mol. (c) n(NaOH) = n(HCl); [NaOH] = 0.00235 / 0.0250 = 0.0940 mol dm^-3. (d) Mass concentration = 0.0940 x 40.00 = 3.76 g dm^-3. |
| 3 | Calorimetry of neutralisation: 50.0 cm3 of 1.00 mol dm^-3 hydrochloric acid is mixed with 50.0 cm3 of 1.00 mol dm^-3 sodium hydroxide in a polystyrene cup (H+ + OH- -> H2O). The temperature rises by 6.5 K. Assume the mixture has density 1.00 g cm^-3 and specific heat capacity 4.18 J g^-1 K^-1, and that no heat is lost. (a) Calculate the heat released, in kJ. (3 marks) (b) Calculate the number of moles of water formed. (2 marks) (c) Find the enthalpy change of neutralisation in kJ mol^-1, with its sign. (3 marks) |
(a) q = mcDT = 100 x 4.18 x 6.5 = 2717 J = 2.72 kJ (1 mark m, 1 mark substitution, 1 mark answer). (b) Both reagents are limiting with 0.0500 mol, so n(H2O) = 0.0500 mol (1 mark working, 1 mark answer). (c) DH = -q / n = -2.72 / 0.0500 = -54.3 kJ mol^-1; the sign is negative because heat is released (1 mark division, 1 mark sign, 1 mark answer). |
Section B Organic chemistry, electrochemistry and equilibrium (25 marks)
Questions 4-6 (structured, marks per part as shown)
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | Compound X contains carbon, hydrogen and oxygen only. Complete combustion of 0.290 g of X gives 0.660 g of carbon dioxide and 0.270 g of water. (Relative atomic masses: C = 12.0, H = 1.0, O = 16.0; CO2 = 44.0, H2O = 18.0) (a) Calculate the masses of carbon and hydrogen in X. (3 marks) (b) Calculate the mass and the number of moles of oxygen in X. (3 marks) (c) Deduce the empirical formula of X. (3 marks) |
(a) n(CO2) = 0.660 / 44.0 = 0.0150 mol, m(C) = 0.0150 x 12.0 = 0.180 g; n(H2O) = 0.270 / 18.0 = 0.0150 mol, m(H) = 2 x 0.0150 x 1.0 = 0.030 g (1 mark each). (b) m(O) = 0.290 – 0.180 – 0.030 = 0.080 g; n(O) = 0.080 / 16.0 = 0.00500 mol (1 mark each). (c) C : H : O = 0.0150 : 0.0300 : 0.00500 = 3 : 6 : 1, so the empirical formula is C3H6O (1 mark ratio, 1 mark simplest ratio, 1 mark formula). |
| 5 | Copper(II) sulphate solution is electrolysed with a constant current of 2 A for 30 minutes using platinum electrodes. (Faraday constant 96500 C mol^-1; relative atomic mass of Cu = 63.5) (a) Calculate the quantity of charge passed, in C. (2 marks) (b) Calculate the number of moles of electrons passed. (2 marks) (c) Calculate the number of moles of copper deposited at the cathode. (2 marks) (d) Calculate the mass of copper deposited, in g. (2 marks) |
(a) Q = It = 2 x 1800 = 3600 C (1 mark substitution, 1 mark answer). (b) n(e-) = 3600 / 96500 = 0.03731 mol (1 mark division, 1 mark answer). (c) Cu2+ + 2e- -> Cu, so n(Cu) = 0.03731 / 2 = 0.01865 mol (1 mark ratio, 1 mark answer). (d) m = 0.01865 x 63.5 = 1.184 g (1 mark multiplication, 1 mark answer). |
| 6 | At 25 C, a 0.100 mol dm^-3 solution of ethanoic acid (CH3COOH) has Ka = 1.8 x 10^-5 mol dm^-3. (a) Write the Ka expression for ethanoic acid. (2 marks) (b) Calculate the hydrogen ion concentration and the pH of the solution. (4 marks) (c) Calculate the degree of dissociation of the acid as a percentage. (2 marks) |
(a) Ka = [H+][CH3COO-] / [CH3COOH] (1 mark numerator, 1 mark denominator). (b) Let [H+] = x, then x^2 / (0.100 – x) = approx x^2 / 0.100 = 1.8 x 10^-5, so x = 0.00134 mol dm^-3 (1 mark assumption, 1 mark square root, 1 mark concentration); pH = 2.87 (1 mark). (c) Degree of dissociation = 0.00134 / 0.100 x 100 per cent = 1.34 per cent (1 mark working, 1 mark answer). |
Find a tutor — freeTutor Registration →
A TutorZone practice paper; original questions and passages, not publisher material.
