The boiling point of water($100°\mathrm{C}$) becomes $100.25°\mathrm{C}$, If 3 grams of a nonvolatile solute is dissolved in 200ml of water. The molecular weight of solute is (${\mathrm{K}}_{\mathrm{b}}$for water is 0.6 K Kg / mole)

First boiling point of water = $100°C$

Final  boiling point of water = $100.52°C$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{s}} = 3\mathrm{g} , {\mathrm{W}}_0=200\mathrm{g} , {\mathrm{K}}_{\mathrm{b}}=0.6\mathrm{K} \mathrm{Kg} {\mathrm{mol}}^{-1} \\ ∆{\mathrm{T}}_{\mathrm{b}}=100.52-100 =0.52°\mathrm{C} \\ ∆{\mathrm{T}}_{\mathrm{b}}= {\mathrm{K}}_{\mathrm{b}}\mathrm{m}={\mathrm{K}}_{\mathrm{b}}\frac{{\mathrm{W}}_{\mathrm{s}}}{{\mathrm{M}}_{\mathrm{s}}}\times \frac{1000}{{\mathrm{W}}_{\mathrm{o}}} \\ {\mathrm{M}}_{\mathrm{s}} =\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_{\mathrm{s}}\times 1000}{∆{\mathrm{T}}_{\mathrm{b}}\times {\mathrm{W}}_{\mathrm{o}}}=\frac{0.6\times 3\times 1000}{0.52\times 200}=\frac{1800}{104}=17.3 {\mathrm{gmol}}^{-1} \end{gathered}$$

Where , ${\mathrm{W}}_{\mathrm{s}}=\mathrm{weight} \mathrm{of} \mathrm{solute} , {\mathrm{W}}_0=\mathrm{weight} \mathrm{of} \mathrm{solvent}$