A solution containing 10 g per ${\mathrm{dm}}^3$ of urea $\text{ (molecular mass }=60 \mathrm{g}{\text{ mol }}^{-}\text{) }$ is isotonic with a 5 % solution of a non - volatile solute whose molar mass is :

For isotonic solution :

Osmotic pressure of urea 

= Osmotic pressure of non-volatile solute

(i) For urea, 

$\begin{array}{l}\mathrm{M}=60 \mathrm{g} {\mathrm{mol}}^{-},\\ \mathrm{W}=10 \mathrm{g},\\ \mathrm{V}=1 {\mathrm{dm}}^3\\ \therefore {\mathrm{\pi }}_1=\frac{\mathrm{WRT}}{\mathrm{MV}}=\frac{10\mathrm{RT}}{60\times 1}\end{array}$

(ii) For non - volatile solute,

$\begin{array}{l}\mathrm{M}=?,\mathrm{W}=5 \mathrm{g};\\ \mathrm{V}=\frac{100}{1000} {\mathrm{dm}}^3=0.1{\mathrm{dm}}^3\\ \therefore {\mathrm{\pi }}_2=\frac{\mathrm{WRT}}{\mathrm{MV}}=\frac{5\mathrm{RT}}{\mathrm{M}\times 0.1}\end{array}$

Equating (1) and (2), we get :

$\begin{array}{l}\frac{10\mathrm{RT}}{60}=\frac{5\mathrm{RT}}{0.1\mathrm{M}}\\ \mathrm{M}=\frac{5\mathrm{RT} \times 60}{10\mathrm{RT} \times 0.1}=300 \mathrm{g} {\mathrm{mol}}^{-}.\end{array}$