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# 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?

$\begin{array}{1 1}41.35gmol^{-1}\\31.35gmol^{-1}\\21.35gmol^{-1}\\61.35gmol^{-1}\end{array}$

Vapour pressure of pure water at the boiling point
$(P_0)=$1 atm = 1.013 bar
Vapour pressure of solution $P_s=1.004$ bar
Mass of solute =$W_2$=2g
Mass of solution =100g
Mass of solvent =98g
Applying Raoult's law for dilute solution
$\large\frac{P_0-P_s}{P_0}=\frac{n_2}{n_1+n_2}=\frac{n_2}{n_1}=\frac{\Large\frac{W_2}{M_2}}{\Large\frac{W_1}{M_1}}$
$\large\frac{P_0-P_s}{P_0}=\frac{W_2}{M_2}\times \frac{M_1}{W_1}$
$\large\frac{1.013-1.004}{1.013}=\frac{2g}{M_2}\times \frac{188gmol^{-1}}{98g}$
(or)
$M_2=\large\frac{2\times 18\times 1.013}{98\times 0.009}$
$\;\;\;\;\;\;=41.35 g mol^{-1}$