JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Solutions

The vapour pressures of pure liquids

A

and

B

are 400 and 600 mm Hg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to 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:

(A) $500 mmHg, 0.4, 0.6$ ✓ Correct
(B) $500 mmHg, 0.5, 0.5$
(C) $450 mmHg, 0.5, 0.5$
(D) $450 mmHg, 0.4, 0.6$

Answer: (A)

Solution

Treat the mixture as an ideal solution obeying Raoult's law. With

X_{A} = X_{B} = 0.5

P_{A}^{\circ} = 400

and

P_{B}^{\circ} = 600

mm Hg, the total vapour pressure is

P = X_{A} P_{A}^{\circ} + X_{B} P_{B}^{\circ} = 0.5 \times 400 + 0.5 \times 600 = 500

mm Hg. The partial pressure of A is

0.5 \times 400 = 200

mm Hg, so its vapour mole fraction is

Y_{A} = \frac{200}{500} = 0.4

and that of B is

Y_{B} = 1 - 0.4 = 0.6

. The solution thus shows

500

mm Hg with vapour mole fractions

0.4

and

0.6

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Solutions

Solution

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