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

Here,

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar

Vapour pressure of pure water at normal boiling point $\left(p_{1}^{0}\right)=1.013$ bar

Mass of solute, (w2) = 2 g

Mass of solvent (water), (w1) = 98 g

Molar mass of solvent (water), (M1) = 18 g mol−1

According to Raoult’s law,

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$

$\Rightarrow \frac{1.013-1.004}{1.013}=\frac{2 \times 18}{M_{2} \times 98}$

$\Rightarrow \frac{0.009}{1.013}=\frac{2 \times 18}{M_{2} \times 98}$

$\Rightarrow M_{2}=\frac{1.013 \times 2 \times 18}{0.009 \times 98}$

= 41.35 g mol−1

Hence, the molar mass of the solute is 41.35 g mol−1.