JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Solutions

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When 3.00 g of substance 'X' is dissolved in 100 g of CCl₄, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is ______ g mol⁻¹. (nearest integer) [Given K_b for CCl₄ is 5.0 K kg mol⁻¹]

Answer: 250

Solution

Elevation of the boiling point gives the molality of the solution directly:

m = \frac{Δ T _{b}}{K _{b}} = \frac{0.60}{5.0} = 0.12 mol k g^{- 1}

The solvent weighs

100 g = 0.100 kg

, so it holds

0.12 \times 0.100 = 0.012

mol of X. That amount of substance weighs 3.00 g, hence

M = \frac{3.00}{0.012} = 250 g mo l^{- 1}

The molar mass of X is

250 g mo l^{- 1}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer

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