Given,
Mass of solute $\left(w_{B}\right)=10 \,g$
Molar mass of solute $\left(M_{B}\right)=M_{B}$
Mass of solvent $\left(w_{A}\right)=360 \,g$
Relative lowering in vapour pressure of solution
$=5 \times 10^{-3}$
Molar mass of water $\left(M_{A}\right)=18 \,g \,mol ^{-1}$
$\because \frac{\Delta p}{p^{\circ}}=$ Relative lower in vapour-pressure of solution.
$\frac{\Delta p}{p^{\circ}}=\chi_{B} $
$\Rightarrow \frac{n_{B}}{n_{A}+n_{B}}=5 \times 10^{-3}$
where, $n_{A}$ and $n_{B}$ are number of moles of solvent
(A) and solute $(B)$ respectively.
$n_{A}=\frac{360}{18}=20$
$n_{B}=\frac{w_{B}}{M_{B}}=\frac{10}{M_{B}}$
$ \because 5 \times 10^{-3}=\chi_{B}=\frac{\frac{10}{M_{B}}}{20+\frac{10}{M_{B}}}$
$5 \times 10^{-3}=\frac{\frac{10}{M_{B}}}{\frac{20 M_{B}+10}{M_{B}}}$
$5 \times 10^{-3}=\frac{10}{20\, M_{B}+10}$
$\left(20\, M_{B}+10\right) 5 \times 10^{-3} =10 $ or,
$20\, M_{B}+10 =\frac{10}{5} \times 10^{3} $
$20 M_{B}+10 =2000 $
$\Rightarrow 20 \,M_{B}=1990 $
$ M_{B} =\frac{1990}{20}=99.5 \,g \,mol ^{-1}$