Vapour pressure of pure water at the boiling point \(\left(P^{\circ}\right)\) \(=1.013 \mathrm{bar}\)
Vapour pressure of solution \(\left(P_{\mathrm{s}}\right)=1.004 \mathrm{bar}\)
Mass of non-volatile solute \(\left(w_2\right)=2 \%=2 \mathrm{~g}\)
Molar mass of solvent, water \(\left(M_1\right)=18 \mathrm{~g}\)
Mass of solvent \(\left(w_1\right)=98 \mathrm{~g}\)
Mass of solution \(=100 \mathrm{~g}\)
Applying Raoult's Law for dilute solutions, \(\frac{P^{\circ}-P_s}{P^{\circ}}=\frac{n_2}{n_1+n_2} \simeq \frac{n_2}{n_1}\)
[Dilute solution being \(2 \%\) ]
\(\begin{aligned}
&\frac{P^{\circ}-P_s}{P^{\circ}}=\frac{n_2}{n_1}=\frac{w_2 / M_2}{w_1 / M_1} \\
&\frac{(1.013-(1.004)}{(1.013)}=\frac{2 \times 18}{M_2 \times 98} \\
&\therefore M_2=\frac{2 \times 18}{98 \times 0.009} \times 1.013=41.35 \mathrm{~g} \mathrm{~mol}^{-1} .
\end{aligned}\)