Question

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?

Answer 1

The correct answer is: $41.35 g mol^{- 1}.$
Here,
Vapour pressure of the solution at normal boiling point $(ρ_1) = 1.004 bar$
Vapour pressure of pure water at normal boiling point $(ρ^0_1)=1.013 bar$
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,
$\frac{(ρ^0_1-ρ_1)}{ρ^0_1} = \frac{(w_2\times M_1)}{(M_2 \times w_1)}$
$=\frac{(1.013-1.004)}{1.013} = \frac{(2\times 18)}{(M_2\times 98)}$
$\frac{0.009}{1.013} =\frac{(2\times18)}{(M_2\times 98)}$
$M_2= \frac{1.013\times2\times19}{0.009\times98}$
$= 41.35 g mol^{- 1 }$
Hence, the molar mass of the solute is $41.35 g mol^{- 1}.$