Question

$10\, g$ of non-volatile solute is dissolved in $180 \,g$ of $H _{2}$ Oresulting in lowering of vapour pressure by $0.5\%$. Determine the boiling point of solution if $K_{b}$ of water is $0.52\, K\, kg\, mol ^{-1}$

• A. $100.01^{\circ} \,C$
• B. $100.15^{\circ}\, C$
• C. $100.23^{\circ} \,C$
• D. $100.32^{\circ} \,C$

Answer 1

Answer: (B)

$p=p^{\circ} x_{1}$
$99.50=100 \frac{n_{1}}{n_{1}+n_{2}}$
$\frac{n_{2}}{n_{1}}=\frac{1}{191}$
$\Delta T_{b}=k_{b} m=K_{b} \frac{n_{2}}{n_{1}} \times \frac{1000}{m_{1}}$
$=0.52 \times \frac{1}{191} \times \frac{1000}{18}=0.15$
$T_{b}=10015^{\circ} C$