The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K on mixing the two liquids, the sum of their initial volume is equal ot the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture, The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

Solutions PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q21 with Solution

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Correct answer: B

Since on mixing, the sum of initial volumes equals the final volume, the solution behaves as an ideal solution.

So, Raoult’s law is applicable: $p_A = x_A p_A^0, \qquad p_B = x_B p_B^0$ where:

For component $A$ : $p_A = x_A p_A^0 = 0.5 \times 400 = 200\ \text{mmHg}$

For component $B$ : $p_B = x_B p_B^0 = 0.5 \times 600 = 300\ \text{mmHg}$

$P = p_A + p_B = 200 + 300 = 500\ \text{mmHg}$

The mole fraction of a component in vapour phase is: $y_i = \frac{p_i}{P}$

So,

For $A$ : $y_A = \frac{200}{500} = 0.4$

For $B$ : $y_B = \frac{300}{500} = 0.6$

We get:

This matches Option B.

Stored correct answer: B

Our derived answer: B

So, they agree.