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 volumes is equal to the volume of the final mixture. The mole fractional of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the…

Question

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 volumes is equal to the volume of the final mixture. The mole fractional of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are

Aicharya · Accepted Answer

Let X A and X B be the mole fraction of liquid A and B in the mixture. Given, P A ° = 400 m m H g , P B ° = 600 m m H g P T o t a l = X A . P A ° + X B . P B ° = 0.5 × 400 + 0.5 × 600 = 500 m m H g Now, mole fraction of A in vapour, Y A = P A P t o t a l = 0.5 × 400 500 = 0.4 Hence [ P A = X A P A ° ] and mole fraction of B in vapour, Y B = 1 - 0.4 = 0.6 ~

The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two…

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