At  ${100}^0\mathrm{C}$   the vapour pressure of a solution of  $6.5\mathrm{g}$ of a solute in $100\mathrm{g}$ water is $732\mathrm{mm}$. If ${\mathrm{k}}_{\mathrm{b}}=0.52$ the boiling point of this solution will be

$\text{ Given }:{\mathrm{W}}_{\mathrm{B}}=6.5 \mathrm{g},{\mathrm{W}}_{\mathrm{A}}=100 \mathrm{g},{\mathrm{p}}_{\mathrm{s}}=732 \mathrm{mm},{\mathrm{K}}_{\mathrm{b}}=0.52$

${\mathrm{T}}_{\mathrm{b}}°=100°\mathrm{C},\mathrm{p}°=760 \mathrm{mm}$

$\frac{\mathrm{p}°-{\mathrm{p}}_{\mathrm{s}}}{\mathrm{p}°}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\Rightarrow \frac{760-732}{760}=\frac{{\mathrm{n}}_2}{100/18}$

$\Rightarrow {\mathrm{n}}_2=\frac{28\times 100}{760\times 18}=0.2046 \mathrm{moles}$

${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\times \mathrm{m}$

${\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}°={\mathrm{K}}_{\mathrm{b}}\times \frac{{\mathrm{n}}_2\times 1000}{{\mathrm{W}}_{\mathrm{A}}\left(\mathrm{g}\right)}$

$\begin{array}{l}{\mathrm{T}}_{\mathrm{b}}-100°\mathrm{C}=\frac{0.52\times 0.2046\times 1000}{100}=1.06\\ {\mathrm{T}}_{\mathrm{b}}=100+1.06=101.06°\mathrm{C}\end{array}$