IGCSE Chemistry 0620 Paper 4 Practice Paper: Moles, Electrolysis and Acids (100 marks, 60 min)

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 100 marks | Parts: 30

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (18 marks) – Moles and chemical calculations

(a) Calculate the relative molecular mass of hydrated copper(II) sulfate, CuSO4.5H2O (Cu = 64, S = 32, O = 16, H = 1). (3 marks)

(b) 25.0 cm3 of 0.200 mol/dm3 sodium hydroxide solution is exactly neutralised by 20.0 cm3 of dilute sulfuric acid. Calculate the concentration of the acid. (4 marks)

(c) An experiment needs 0.050 mol of sodium chloride. Calculate the mass required (Na = 23, Cl = 35.5). (4 marks)

(d) Using the theoretical mass from the previous part, a student obtains only 2.40 g of sodium chloride. Calculate the percentage yield to one decimal place. (4 marks)

(e) 0.10 mol of magnesium reacts with 0.15 mol of dilute hydrochloric acid: Mg + 2HCl -> MgCl2 + H2. Identify the limiting reactant and calculate the volume of hydrogen produced (molar volume 24 dm3/mol). (3 marks)

Question 2 (17 marks) – Electrolysis

(a) A current of 2.0 A passes through copper(II) sulfate solution for 1800 s. Calculate the quantity of charge in coulombs. (3 marks)

(b) Copper is deposited at the cathode. Calculate the mass of copper formed (Cu = 64, Faraday constant 96500 C/mol). (4 marks)

(c) With the same charge, oxygen is released at the anode. Calculate the volume of oxygen at room temperature and pressure (molar volume 24 dm3/mol), giving your answer in cm3. (4 marks)

(d) When copper plates are used as both electrodes in copper(II) sulfate solution, the mass of the anode slowly decreases. Explain this observation. (3 marks)

(e) State the products formed at the cathode and at the anode when molten lead(II) bromide is electrolysed with inert electrodes. (3 marks)

Question 3 (17 marks) – Acids, bases and salts

(a) Hydrochloric acid of concentration 0.010 mol/dm3 is fully ionised. Calculate the pH of the solution. (3 marks)

(b) 10.0 cm3 of 1.0 mol/dm3 hydrochloric acid is diluted with water to a total volume of 250 cm3. Calculate the new concentration. (4 marks)

(c) Copper(II) oxide reacts with dilute sulfuric acid to make 0.050 mol of copper(II) sulfate. Calculate the mass of copper(II) oxide needed (Cu = 64, O = 16). (4 marks)

(d) Describe a test to show that a solution contains sulfate ions, including the expected observation. (3 marks)

(e) Explain why an indicator is added in an acid-base titration and state one rule for choosing it. (3 marks)

Question 4 (16 marks) – The Periodic Table and chemical bonding

(a) Write the electron arrangement of a chlorine atom (atomic number 17) and state the group it belongs to. (3 marks)

(b) Magnesium reacts with oxygen to form magnesium oxide. Describe how the ions are formed and give the electron arrangement of each ion. (4 marks)

(c) Calculate the volume of 0.25 mol of carbon dioxide at room temperature and pressure (molar volume 24 dm3/mol) and its mass (C = 12, O = 16). (4 marks)

(d) Explain why graphite conducts electricity. (2 marks)

(e) Explain why the reactivity of Group I metals increases down the group. (3 marks)

Question 5 (16 marks) – Organic chemistry and polymers

(a) An alkane molecule contains six carbon atoms. Write its molecular formula and calculate its relative molecular mass (C = 12, H = 1). (3 marks)

(b) 0.10 mol of this alkane is burned completely. Calculate the volume of carbon dioxide formed (molar volume 24 dm3/mol). (4 marks)

(c) 0.20 mol of ethene (C2H4) undergoes addition polymerisation and is fully converted to poly(ethene). Calculate the mass of polymer formed (C = 12, H = 1). (4 marks)

(d) Give a simple test to show that a gas is an alkene, including the expected observation. (2 marks)

(e) State two purposes of cracking in industry. (3 marks)

Question 6 (16 marks) – Carbonates and the atmosphere

(a) 0.20 mol of calcium carbonate is completely decomposed by heat (CaCO3 -> CaO + CO2). Calculate the volume of carbon dioxide formed (molar volume 24 dm3/mol). (3 marks)

(b) From the same reaction, calculate the mass of calcium oxide formed (Ca = 40, O = 16). (4 marks)

(c) Explain how carbon dioxide contributes to the greenhouse effect. (4 marks)

(d) Give the reagent used to test for carbon dioxide and the positive result. (2 marks)

(e) Calculate the percentage by mass of nitrogen in ammonium nitrate, NH4NO3 (N = 14, H = 1, O = 16, one decimal place). (3 marks)

Answers and marking notes

Question 1 – Moles and chemical calculations

(a) 64 + 32 + 4 x 16 + 5 x 18 = 64 + 32 + 64 + 90 = 250 (3; lose 2 for omitting the water of crystallisation). [3 marks]

(b) n(NaOH) = 0.200 x 25.0 / 1000 = 5.00e-03 mol (1); 2NaOH + H2SO4 so n(H2SO4) = 2.50e-03 mol (1); c = 2.50e-03 / 0.0200 = 0.125 mol/dm3 (2). [4 marks]

(c) Mr(NaCl) = 23 + 35.5 = 58.5 (2); m = 0.050 x 58.5 = 2.925 g (2). [4 marks]

(d) Yield = 2.40 / 2.925 x 100 = 82.1% (4; lose 1 if the result is left as a decimal fraction). [4 marks]

(e) 0.10 mol of magnesium needs 0.20 mol of acid but only 0.15 mol is present, so the acid is limiting (1); n(H2) = 0.15 / 2 = 0.075 mol (1); V = 0.075 x 24 = 1.8 dm3 (1). [3 marks]

Question 2 – Electrolysis

(a) Q = I x t = 2.0 x 1800 = 3600.0 C (3; lose 1 for inverting the multiplication). [3 marks]

(b) n(e-) = 3600.0 / 96500 = 0.0373 mol (1); Cu2+ + 2e- -> Cu so n(Cu) = 0.0187 mol (2); m = 0.0187 x 64 = 1.19 g (1). [4 marks]

(c) 4OH- -> O2 + 2H2O + 4e- so n(O2) = 0.0373 / 4 = 0.00933 mol (2); V = 0.00933 x 24000 = 224 cm3 (2). [4 marks]

(d) Copper atoms at the anode lose electrons and pass into solution as copper ions (2); the anode is therefore oxidised and loses mass (1). [3 marks]

(e) Lead (Pb) at the cathode (1.5); bromine (Br2) at the anode (1.5). [3 marks]

Question 3 – Acids, bases and salts

(a) [H+] = 0.010 mol/dm3 (1); pH = -log(0.010) = 2 (2). [3 marks]

(b) n(HCl) = 1.0 x 10.0 / 1000 = 0.010 mol (2); c = 0.010 / 0.250 = 0.040 mol/dm3 (2). [4 marks]

(c) Mr(CuO) = 64 + 16 = 80 (2); m = 0.050 x 80 = 4.0 g (2). [4 marks]

(d) Add dilute hydrochloric acid and then barium chloride solution (1); a white precipitate forms (2). [3 marks]

(e) The indicator changes colour at the end point so neutralisation can be seen (2); its colour-change range should lie close to the pH at the end point (1). [3 marks]

Question 4 – The Periodic Table and chemical bonding

(a) 2, 8, 7 (2); Group VII (1). [3 marks]

(b) Magnesium loses two electrons to form Mg2+ (1); oxygen gains two electrons to form O2- (1); Mg2+ is 2, 8 and O2- is 2, 8 (2). [4 marks]

(c) V = 0.25 x 24 = 6.0 dm3 (2); Mr = 44 so m = 0.25 x 44 = 11.0 g (2). [4 marks]

(d) Each carbon atom bonds to only three others, leaving one delocalised electron free to move within the layer (2). [2 marks]

(e) Atomic radius increases (1); the outer electron is further from the nucleus so the attraction is weaker (1); the electron is therefore lost more easily (1). [3 marks]

Question 5 – Organic chemistry and polymers

(a) C6H14 (1); Mr = 6 x 12 + 14 = 86 (2). [3 marks]

(b) Each molecule has six carbon atoms so n(CO2) = 0.10 x 6 = 0.60 mol (2); V = 0.60 x 24 = 14.4 dm3 (2). [4 marks]

(c) Mr(C2H4) = 28 (2); addition polymerisation loses no atoms so m = 0.20 x 28 = 5.6 g (2). [4 marks]

(d) Bubble the gas through bromine water; the orange-brown colour is decolourised (2). [2 marks]

(e) To convert long-chain alkanes into shorter, more useful alkanes (1.5); to produce alkenes as feedstock for polymers (1.5). [3 marks]

Question 6 – Carbonates and the atmosphere

(a) n(CO2) = 0.20 mol (1); V = 0.20 x 24 = 4.8 dm3 (2). [3 marks]

(b) Mr(CaO) = 40 + 16 = 56 (2); m = 0.20 x 56 = 11.2 g (2). [4 marks]

(c) Carbon dioxide absorbs infrared radiation emitted by the Earth surface (2); the energy is trapped in the lower atmosphere and the average temperature rises (2). [4 marks]

(d) Bubble through limewater; the limewater turns milky (2). [2 marks]

(e) Mr = 2 x 14 + 4 x 1 + 3 x 16 = 80 (1); nitrogen contributes 28 so 28 / 80 x 100 = 35.0% (2). [3 marks]

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