A current of dry air was passed through a series of bulbs containing 1.25 g of a solute (molar mass — Solutions and Colligative Properties Chemistry Question
Question
A current of dry air was passed through a series of bulbs containing 1.25 g of a solute $\text{A}_2\text{B}$ (molar mass = 90 g/mol) in 49 g of water and then through pure water. The loss in weight of the former series of bulbs was 0.98 g and in the later series 0.01 g. The percentage dissociation of solute is
💡 Solution & Explanation
By the Ostwald-Walker method, the relative lowering of vapour pressure is $(P^\circ - P) / P = (\text{loss in pure solvent}) / (\text{loss in solution}) = 0.01 / 0.98 = 1/98$. From Raoult's law, $(P^\circ - P) / P = n_{eff} / N$. Moles of water $N = 49 / 18 = 2.722 \text{ mol}$. $n_{eff} = N \times (1/98) = (49/18) \times (1/98) = 1 / 36 \approx 0.02778 \text{ mol}$. Actual moles of solute $n_{actual} = 1.25 / 90 = 0.01389 \text{ mol}$. The van't Hoff factor $i = n_{eff} / n_{actual} = (1/36) / (1.25/90) = 2.0$. For $\text{A}_2\text{B} \rightarrow 2\text{A}^+ + \text{B}^{2-}$ ($n=3$), $i = 1 + 2\alpha \implies 2.0 = 1 + 2\alpha \implies 2\alpha = 1.0 \implies \alpha = 0.50$. The percentage dissociation is 50%. Therefore, correct answer is 0050.