An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

Vapour pressure of pure water at boiling point = 1 atm = 1.013 bar

Vapour pressure of solution $\left(p_S\right)$=1.004 bar

Let mass of solution = 100g

Mass of solute = 2g

Mass of solvent = 100-2=98 g

$\frac{p^o-p_s}{p^o}=\frac{n_2}{n_2+n_2}=\frac{n_2}{n_1}=\frac{W_2/M_2}{W_1/M_1}$

$(\because n_2<<<n_1)$

$\frac{1.013-1.004}{1.013}=\frac{2}{M_2}X\frac{18}{98}$ or $M_2=\frac{2X18}{98}X\frac{1.013}{0.009}=41.35g$ $mo{l}^{-1}$