Question

45. The b.p. of a solution of 5 g of sulphur in 100 g of carbon disulphide is ${0.476}^0$ above that of pure solvent. Determine the moles of sulphur in the solvent. The b.p. of pure carbon disulphide is ${46.30}^{0}C$ and its heat of vaporisation is 84.1 calories per gram.

Answer 1

$\Delta T_h$ = ${0.476}^{0}C$
$K_h=2.4kkg/mal$
$M=32n$
$m=\frac{5\times 1000}{32n\times 100}$
$\Delta T_b=K_b\times m$
$0.476=2.4\times \frac{5\times 1000}{32n\times 100}$
$=\frac{24}{32}\times 5$
$=\frac{13}{4n}$
$n=\frac{15}{4\times 4.76}=\frac{15}{1.9}$
=$8$