Vapour pressure of pure water at the boiling point $(P^{0}) = 1 \text{ atm}$ $= 1.013 \text{ bar}$Vapour pressure of solution $(P_s) = 1.004 \text{ bar}$Let mass of solution = $100 \text{ g}$ , then, mass of solute = ( $w_2$ ) = $2 \text{ g}$Mass of solvent $(w_1) = 100 - 2 = 98 \text{ g}$Applying Raoult’s law for dilute solution, $\frac{P^{0}-P_{s}}{P^{0}}$ $= \frac{n_{2}}{n_{1}+n_{2}}$ $= \frac{n_{2}}{n_{1}}$ $= \frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1}}$ $= \frac{w_{2}}{M_{2}} \times \frac{M_{1}}{w_{1}}$ $\frac{1.013-1.004}{1.013}$ $= \frac{2 \text{ g}}{M_{2}} \times \frac{18 \text{ g mol}^{-1}}{98 \text{ g}}$or, $M_{2} = \frac{2 \times 18}{98} \times \frac{1.013}{0.009} \text{ g mol}^{-1}$ $= 41.35 \text{ g mol}^{-1}$