At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_b=0.52$, the boiling point of this solution will be:

Solution & Explanation

\textbf{Answer:} (A) 101 °C \textbf{Solution:} Step 1 — Use relative lowering of vapour pressure to find the molar mass. At 100 °C the vapour pressure of pure water is 760 mm. $\frac{p^{\circ }-p}{p^{\circ }}=\frac{n_{solute}}{n_{solvent}}$ $\frac{760-732}{760}=\frac{6.5/M}{100/18}$ $\frac{28}{760}=\frac{6.5/M}{5.56}$, giving $M\approx 32$ g/mol. Step 2 — Molality: $m=\frac{6.5/32}{0.100}=\frac{0.203}{0.100}=2.03$ mol/kg. Step 3 — Elevation of boiling point: $\Delta T_b=K_{b}m=0.52\times 2.03\approx 1.05$ K. Boiling point $=100+1.05\approx 101$ °C.