Addition of 4.0 g of a non-volatile solute to 54.0 g water results in a lowering of vapour pressure — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Let the molar mass of the solute be M. Moles of solute n = 4.0 / M. Moles of water N = 54.0 / 18 = 3.0. The relative lowering of vapour pressure is given by: (P^o - P) / P^o = n / (n + N) = 0.01. Thus, n / (n + 3) = 0.01 => n = 0.01n + 0.03 => 0.99n = 0.03 => 0.99 * (4.0 / M) = 0.03 => 3.96 / M = 0.03 => M = 3.96 / 0.03 = 132 g/mol.