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Answers:LEVEL: ASCHEMISTRY – Atoms, ions and molecules1. Formulae, equations and calculating moles (15 minutes)Data required for this question:A r (H) = 1.01; A r (C) = 12.0; A r (O) = 16.0, A r (Na) = 23.0, A r (Cl) = 35.5,L = 6.02 x 10 23 mol −1(a) Use the Avogadro constant, L, and the appropriate relative atomic masses to determine thenumber of chlorine atoms in each of the following.(i) 71.0 g of chlorine gas.(1 mark)Amount of Cl is 71.0g ÷ 35.5 = 2.00 molSo number of Cl atoms is thus 2.00 × 6.02 × 10 23 = 1.204 × 10 24(ii) 15.4 g of tetrachloromethane.(2 marks)Formula for tetrachloromethane is CCl 4Mr(CCl 4 ) is 12.0 + 4 × 35.5 = 154So 15.4 g CCl 4 is 0.1 mol containing 0.4 mol Cl or 0.4 × L = 2.408 × 10 23 atoms(b) Calculate the mass of anhydrous sodium carbonate, Na 2 CO 3 , needed to make exactly500 cm 3 of aqueous sodium carbonate of concentration 0.050 mol dm −3 .(3 marks)Mr(Na 2 CO 3 ) is 2 × 23.0 + 12.0 + 3 × 16.0 = 106.0Mass of sodium carbonate needed is 0.05 × 106.0 = 5.3 g for 1000 cm 3So mass needed for 500 cm 3 is 5.3 ÷ 2 = 2.65 g(c) (i) Write a balanced equation for formation of aqueous sodium sulphate from solid sodiumcarbonate and aqueous sulphuric acid.(1 mark)Na 2 CO 3 (s) + H 2 SO 4 (aq) → Na 2 SO 4 (aq) + 2H 2 O(l)(ii) Calculate the volume of aqueous sulphuric acid of concentration 0.25 mol dm –3 needed toreact with 25.00 cm 3 of aqueous sodium carbonate of concentration 0.05 mol dm –3 to produce aneutral solution of sodium sulphate.(2 marks)25.00 cm 3 × 0.05 mol dm −3 = V cm 3 × 0.25 mol dm −3So V must be 25.00 × 0.05 ÷ 0.25 = 5.00 cm 3 of aqueous sulphuric acidCopyright © Pearson Education Limited 20011

Answers:LEVEL: ASCHEMISTRY – Atoms, ions and molecules(d) The equation for the manufacture of ethanol by direct hydration of ethene isC 2 H 4 + H 2 O → C 2 H 5 OH(i) Calculate the minimum mass of steam needed to convert one tonne of ethene to ethanol.(3 marks)M r (C 2 H 4 ) is 2 × 12.0 + 4 × 1.01 = 28.04M r (H 2 O) is 2 × 1.01 + 16.0 = 18.02So mass of steam needed per tonne ethene is 18.02 ÷ 28.04 = 0.6427 tonne(ii) Calculate the maximum mass of ethanol that could be formed from one tonne of ethene.(3 marks)M r (C 2 H 5 OH) is 2 × 12.0 + 5 × 1.01 + 16.0 + 1 × 1.01 = 46.06So mass of ethanol per tonne ethene is 46.06 ÷ 28.04 = 1.643 tonne(Total marks 15)Copyright © Pearson Education Limited 20012