A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is _____. (Nearest integer) [Given: boiling point of the pure solvent = 319.5 K, K_b of the solvent = 0.3 K kg mol^ -1 ]
A. A
From the relative lowering of vapour pressure, we have: P^0 - P_s P_s = n_B n_A where n_B is the number of moles of solute and n_A is the number of moles of solvent. Substituting the given values: 760 - 750 750 = n_B n_A 10 750 = n_B n_A n_A = 75 n_B The elevation in boiling point is given by: T_b = T_b - T_b^0 = 320 - 319.5 = 0.5 K We also know that T_b = K_b m, where m is the molality of the solution. m = n_B Mass of solvent in kg = n_B 0.04 Substituting the values into the elevation in boiling point equation: 0.5 = 0.3 n_B 0.04 n_B = 0.5 0.04 0.3 = 0.02 0.3 = 1 15 mol Now, substituting n_B back to find the number of moles of the solvent n_A: n_A = 75 1 15 = 5 mol Answer: 5
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