Question

(ii) 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 pointisraised to 354.11 k. Calculate the molar mass ofthe solute. Kb for benzene is 2.53 k kgmol~1.

Answer 1

Hope this's right and it helps you well enough

Answer 2

Answer: 57.5 g/mol

Explanation: Elevation in boiling point :

$\Delta T_b=k_b\times m$

or $\Delta T_b=k_b\times \frac{\text{given mass of solute}}{{\text {Molar mass}}\times \text{ Mass of solvent in kg}}}$

where,

$\Delta T_b$ = change in boiling point  = (354.11-353.23) K= 0.88 K

$k_b$ = boiling point constant

m = molality

$0.88K=2.53Kkgmol^{-1}\times \frac{1.80g}{M\times 0.09kg}$

$M=57.5g/mol$