A solution of 2.5 g of a non - volatile solute dissolved in 100 g of benzene boiled at 0.42⁰C higher than the boiling point of pure benzene. Then the molecular mass of the substance (Molal elevation constant of benzene is $2.67 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}) \mathrm{is}$

The formula of elevation of boiling point of a solution is

$∆{\mathrm{T}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_1\times 1000}{{\mathrm{M}}_1\times {\mathrm{W}}_2}$

Here ${\mathrm{K}}_{\mathrm{b}}$ is molal boiling point elevation constant.

${\mathrm{M}}_1=$molar mass of non-volatile solute

${\mathrm{W}}_1=$ weight of non-volatile solute

${\mathrm{W}}_2=$weight of solvent

$0.42=2.67\times \frac{2.5}{{\mathrm{M}}_1}\times \frac{1000}{100}$

$\therefore {\mathrm{M}}_1=\frac{2.67\times 25}{0.42}=158.9$