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 molar mass of the solute?

(N/A) Given:
Vapour pressure of the solution at normal boiling point $(p_{1}) = 1.004 \ bar$
Vapour pressure of pure water at normal boiling point $(p_{1}^{\circ}) = 1.013 \ bar$
Mass of solute $(w_{2}) = 2 \ g$
Mass of solvent (water) $(w_{1}) = 98 \ g$
Molar mass of solvent (water) $(M_{1}) = 18 \ g \ mol^{-1}$
According to Raoult's law for relative lowering of vapour pressure:
$\frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
Substituting the values:
$\frac{1.013 - 1.004}{1.013} = \frac{2 \times 18}{M_{2} \times 98}$
$\frac{0.009}{1.013} = \frac{36}{M_{2} \times 98}$
$M_{2} = \frac{36 \times 1.013}{0.009 \times 98}$
$M_{2} = \frac{36.468}{0.882} \approx 41.35 \ g \ mol^{-1}$
Thus,the molar mass of the solute is $41.35 \ g \ mol^{-1}$.