JEE Main 2019 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

At a certain temperature the pure-liquid vapour pressures of two completely miscible liquids A and B are 120 kPa and 40 kPa respectively. They form…

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At a certain temperature the pure-liquid vapour pressures of two completely miscible liquids A and B are

120\ \mathrm{kPa}

and

40\ \mathrm{kPa}

respectively. They form an ideal solution that boils at this temperature under an external pressure of

70\ \mathrm{kPa}

. The mole fraction of A in the vapour in equilibrium with the boiling liquid is

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Answer: (A) $0.64$

Boiling occurs when the total vapour pressure equals

70\ \mathrm{kPa}

P = x_A(120)+(1-x_A)(40) = 40+80x_A = 70

, so

x_A = 0.375

. By Dalton's law,

y_A = \dfrac{p_A}{P} = \dfrac{0.375\times120}{70} = \dfrac{45}{70} = 0.643

. The liquid composition

0.375

is the trap answer; the vapour is richer in the more volatile component A.

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake