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 :

Options

A. 450 mmHg, 0.4, 0.6
B. 500 mmHg, 0.5, 0.5
C. 450 mmHg, 0.5, 0.5
D. 500mmHg, 0.4, 0.6

Answer

D

Explanation

According to Raoult.s law.
` p = p_(A)^(@) x_(A) + p_(B)^(@) x_(B)`
If `x_(B) = 0.5,` then `x_(A) = 1- 0.5 = 0.5`
` p = 400 xx 0.5 + 600 xx 0.5`
` = 200 + 300 = 500`
In Vapour phase,
`y_(A) = (p_(x)^(@) x_(A))/(p) = (0.5 xx 400)/(500) = 0.4`
`y_(B) = (p_(A)^(@) x_(B))/(p) = (0.5 xx 600)/(500)=0.6`