The vapour pressure of acetone at $20^{\circ} \mathrm{C}$ is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at $20^{\circ} \mathrm{C}$, its vapour pressure was 183 torr. The molar mass $(\mathrm{g} \mathrm{mol}^{-1})$ of the substance is:

JEE Main · 2015 · 04 Apr 2015 · Offline · Solutions

The vapour pressure of acetone at

2 0^{\circ} C

is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at

2 0^{\circ} C

, its vapour pressure was 183 torr. The molar mass

(g mol^{- 1})

of the substance is:

(A)64✓ Correct
(B)128
(C)488
(D)32

Answer: (A)

Solution

A non-volatile solute lowers the vapour pressure in proportion to its mole fraction, which for a dilute solution is

\frac{p ^{0} - p}{p ^{0}} = \frac{n _{solute}}{n _{solvent}}

. Here

\frac{185 - 183}{185} = \frac{1.2/ M}{100/58}

. Solving,

M = \frac{1.2 \times 58 \times 185}{100 \times 2} \approx 64

g/mol.

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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