Question

The boiling point of a solution of 5 g of sulphur in 100 g of carbon disulphide is 0.474°C above that of pure
solvent. Determine the molecular formula of sulphur in this solvent. The boiling point of pure carbon disulphide
is 47°C and its heat of vaporisation is 84 calories per gram.​

Answer 1

Answer: 246 g

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$\Delta T_b=k_b\times \frac{\text{ Moles of solute}\times 1000}{\text{ Mass of solvent in Kg}}$

where,

$\Delta T_b$ = change in boiling point

$k_b$ = boiling point constant

m = molality

$T_{solution}-T_{solvent}=2.34\times \frac{\text{ given mass}\times 1000}{\text {Mass of solvent in g}\times {\text {Molar mass of solute}}}$

$0.474=2.34\times \frac{5g\times 1000}{100g\times {\text {Molar mass of sulphur}}}$

${\text {Molar mass of sulphur}}=246g$