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:

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Chemistry

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Raoult Law and Vapour Pressure

Question

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:

500 mm Hg, 0.4, 0.6

500 mm Hg, 0.5, 0.5

450 mm Hg, 0.5, 0.5

450 mm Hg, 0.4, 0.6

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P_total = X_A P_A⁰ + X_B P_B⁰ = 0.5 × 400 × 0.5 × 600 = 500 mmHg

Now, mole fraction of A in vapour

$Y_{A} = \frac{P_{A}}{P_{total}} = \frac{0 .5 \times 400}{500} = 0 .4$ and mole fraction of B in vapour

Y_B = 1 – 0.4 = 0.6