A solution containing 30 g of a non-volatile solute exactly in 90 g water has a vapour pressure of 2 — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Let $n$ be moles of solute. Initially, $N_1 = 90/18 = 5 \text{ mol}$. $(P^\circ - 2.8)/2.8 = n/5 \Rightarrow P^\circ = 2.8 + 0.56n$. After adding 18 g water, $N_2 = 108/18 = 6 \text{ mol}$. $(P^\circ - 2.9)/2.9 = n/6 \Rightarrow P^\circ = 2.9 + 0.4833n$. Equating the two $P^\circ$ values gives $2.8 + 0.56n = 2.9 + 0.4833n \Rightarrow 0.0767n = 0.1 \Rightarrow n = 1.304$. Molar mass $M = 30 / 1.304 = 23 \text{ g/mol}$. Therefore, correct answer is A.