Q52 / 200 · Manipal (MET) 2013 · Chemistry

10 g of non-volatile solute is dissolved in 180 g of $\mathrm{H}_2 \mathrm{Oresulting}$ in lowering of vapour pressure by $0.5 \%$. Determine the boiling point of solution if $K_b$ of water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$.

  • A
    $100.01^{\circ} \mathrm{C}$
  • Correct answer
    B
    $100.15^{\circ} \mathrm{C}$
  • C
    $100.23^{\circ} \mathrm{C}$
  • D
    $100.32^{\circ} \mathrm{C}$

Solution

$p=p^{\circ} x_1$ $99.50=100 \frac{n_1}{n_1+n_2}$ $\frac{n_2}{n_1}=\frac{1}{191}$ $\Delta T_b=k_b m=K_b \frac{n_2}{n_1} \times \frac{1000}{m_1}$ $\begin{aligned} & =0.52 \times \frac{1}{191} \times \frac{1000}{18}=0.15 \\ T_b & =100.15^{\circ} \mathrm{C}\end{aligned}$

Hence, the correct answer is option: 'B'.

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