CBSE Class 12 · 2002 · All India Compartment · SolutionsShort Answer

A solution of 3.8 g of sulphur in 100 g of CS

_{2}

(boiling point 46.30°C) boils at 46.66°C. What is the formula of the sulphur molecule in this solution? (Atomic mass of sulphur = 32 g mol

^{- 1}

, K

_{b}

for CS

_{2}

= 2.40 K kg mol

^{- 1}

)

Answer: S₈

Solution

Elevation observed:

Δ T_{b} = 46.66 - 46.30 = 0.36 K

Molality:

m = \frac{Δ T _{b}}{K _{b}} = \frac{0.36}{2.40} = 0.15 mol kg^{- 1}

Moles of sulphur in 100 g CS

_{2}

0.15 \times 0.1 = 0.015

mol.
Molar mass:

M = \frac{3.8}{0.015} = 253.3 g mol^{- 1}

Number of S atoms per molecule:

x = \frac{253.3}{32} \approx 8

The molecular formula is S_{8}

CBSE Class 12 Board · 2002 · All India Compartment · Subjective