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 \(41.35 g mol^{- 1}.\)