The boiling point of benzene is 353.23 K. When 1.80 g of a non - volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Then the molar mass of the solute $\left({\mathrm{K}}_{\mathrm{b}}\mathrm{for} \mathrm{benzene} \mathrm{is} 2.53 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\right) is$

The given information is as follows below.

${\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}=353.23 \mathrm{K}: {\mathrm{T}}_{\mathrm{b}}=354.11\mathrm{K}$

Elevation in boiling point of solution,

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}=354.11-353.23$

$∆{\mathrm{T}}_{\mathrm{b}}=0.88\mathrm{k}$

Weight of non-volatile solute, w =1.80g

Molar mass of solute, M.W =?

Weight of solvent (benzene), W =90g

Molal elevation constant for benzene, ${\mathrm{K}}_{\mathrm{b}}=2.53 \mathrm{K}.\mathrm{kg} {\mathrm{mol}}^{-1}$

Use the following formula to find the molar mass of non-volatile solute.

$0.88=2.53\times \frac{1.8}{\mathrm{m}}\times \frac{1000}{90}$

$\therefore \mathrm{M}.\mathrm{W} \mathrm{of} \mathrm{solute}=\frac{20\times 2.53}{0.88}=57.5\mathrm{g}$