Question

The boiling point of benzene is 353.23 k when 1.80 g of sulphur

Answer 1

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