The vapour pressure of water at room temperature is lowered by 5% on dissolving a non-volatile solut — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Relative lowering is 5%, so $P = 0.95 P^\circ$. $\frac{P^\circ - P}{P} = \frac{n}{N} \implies \frac{0.05}{0.95} = \frac{n}{N} \implies \frac{n}{N} = \frac{1}{19}$. Molality $m = (n / (N \cdot M_{\text{solvent}})) \times 1000 = (1/19) \times (1000 / 18) = 1000 / 342 \approx 2.92$ mol/kg. Therefore, correct answer is D.