In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0⁰C and atmospheric pressure. Calculate (i) the mass fraction and (ii) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 and water. Basis 1 m3 of water = 1000 kg Volume of carbon dioxide added = 3 m3 From the gas equation, pV = nRT 1 x 3 = n x 0.08206 x 273 n = 0.134 mole. Molecular weight of carbon dioxide = 44 And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg

Correct Answer: 5.9 × 10-3, 2.41 × 10-3
(i) Mass fraction of carbon dioxide in drink = 5.9 / (1000 + 5.9) = 5.9 × 10-3 (ii) Mole fraction of carbon dioxide in drink = 0.134 / (1000/18 + 0.134) = 2.41 × 10-3.