Vapour pressure of solution containing 6 g of a non-volatile solute in 180 g water is 20 torr. If 1 — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Let $n$ be the moles of solute. Initially, moles of water $N_1 = 180/18 = 10 \text{ mol}$. Relative lowering: $(P^\circ - 20) / 20 = n / 10 \implies P^\circ = 20 + 2n$. After adding 1 mole water, $N_2 = 11 \text{ mol}$, and $P = 20.02 \text{ torr}$. $(P^\circ - 20.02) / 20.02 = n / 11 \implies P^\circ = 20.02 + 1.82n$. Equating $P^\circ$: $20 + 2n = 20.02 + 1.82n \implies 0.18n = 0.02 \implies n = 1/9 \text{ mol}$. Molar mass $M = \text{mass} / n = 6 / (1/9) = 54 \text{ g/mol}$. Therefore, correct answer is 0054.