Question

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

Answer 1

Here,

Vapour pressure of the solution at normal boiling point $(p_1)=1.004\ bar$

Vapour pressure of pure water at normal boiling point $(p_1^0)=1.013\ bar$

Mass of solute,$(w_2) = 2 \ g$

Mass of solvent (water),$(w_1) = 100 - 2 = 98 \ g$

Molar mass of solvent (water), $(M_1) = 18\ g \ mol ^{- 1}$

According to Raoult's law,

$\frac{(p_1^o - p1)} {p_1^o } = \frac{ (w_2 \times M_1 )} {(M_2 \times w_1 )}$ $\implies \frac{(1.013 - 1.004)} {1.013 }= \frac{(2 \times 18)}{(M_2 \times 98 )}$ $\implies M_2=41.35\ g\ mol^{-1}$