Answers:
LEVEL: AS
CHEMISTRY – Atoms, ions and molecules
1. Formulae, equations and calculating moles (15 minutes)
Data required for this question:
Ar(H) = 1.01; Ar(C) = 12.0; Ar(O) = 16.0, Ar(Na) = 23.0, Ar(Cl) = 35.5,
L = 6.02 x 1023 mol−1
(a) Use the Avogadro constant, L, and the appropriate relative atomic masses to determine the
number 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 mol
So number of Cl atoms is thus 2.00 × 6.02 × 1023 = 1.204 × 1024
(ii) 15.4 g of tetrachloromethane.
(2 marks)
Formula for tetrachloromethane is CCl4
Mr(CCl4) is 12.0 + 4 × 35.5 = 154
So 15.4 g CCl4 is 0.1 mol containing 0.4 mol Cl or 0.4 × L = 2.408 × 1023 atoms
(b) Calculate the mass of anhydrous sodium carbonate, Na2CO3, needed to make exactly
500 cm3 of aqueous sodium carbonate of concentration 0.050 mol dm−3.
(3 marks)
Mr(Na2CO3) is 2 × 23.0 + 12.0 + 3 × 16.0 = 106.0
Mass of sodium carbonate needed is 0.05 × 106.0 = 5.3 g for 1000 cm3
So mass needed for 500 cm3 is 5.3 ÷ 2 = 2.65 g
(c) (i) Write a balanced equation for formation of aqueous sodium sulphate from solid sodium
carbonate and aqueous sulphuric acid.
(1 mark)
Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
(ii) Calculate the volume of aqueous sulphuric acid of concentration 0.25 mol dm–3 needed to
react with 25.00 cm3 of aqueous sodium carbonate of concentration 0.05 mol dm–3 to produce a
neutral solution of sodium sulphate.
(2 marks)
25.00 cm3 × 0.05 mol dm−3 = V cm3 × 0.25 mol dm−3
So V must be 25.00 × 0.05 ÷ 0.25 = 5.00 cm3 of aqueous sulphuric acid
Copyright © Pearson Education Limited 2001
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Answers:
LEVEL: AS
CHEMISTRY – Atoms, ions and molecules
(d) The equation for the manufacture of ethanol by direct hydration of ethene is
C2H4 + H2O → C2H5OH
(i) Calculate the minimum mass of steam needed to convert one tonne of ethene to ethanol.
(3 marks)
Mr(C2H4) is 2 × 12.0 + 4 × 1.01 = 28.04
Mr(H2O) is 2 × 1.01 + 16.0 = 18.02
So 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)
Mr(C2H5OH) is 2 × 12.0 + 5 × 1.01 + 16.0 + 1 × 1.01 = 46.06
So mass of ethanol per tonne ethene is 46.06 ÷ 28.04 = 1.643 tonne
(Total marks 15)
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