We have ${\text{M}}_2=\frac{{\text{K}}_{\text{f}}\times {\text{w}}_2\times 1000}{\Delta {\text{T}}_{\text{f}}\times {\text{w}}_1}$

when ${\text{w}}_1=$ weight of solvent = 50 g

          ${\text{w}}_2=$ weight of solute = 1.00 g

         ${\text{K}}_{\text{f}}=$ molal depression constant

                $={\text{5.12 K kg mol}}^{-1}$

Substituting the values of various terms involved in the above equation, we get

${\text{M}}_2=\frac{\text{5.12}{\text{ K kg mol}}^{-1}\times \text{1.00 g}\times 1000{\text{ g kg}}^{-1}}{\text{0.40}\times 50\text{ g}}$

         $=256{\text{ g mol}}^{-1}$

Thus, molar mass of the solute = 256 g/​mol