A mixture of two miscible liquids A and B is distilled under equilibrium conditions at 1 atm pressure. The mole fraction of A in solution and vapour phase are 0.30 and 0.60 respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure A to that of pure B.

Options

A. 4
B. 3.5
C. 2.5
D. 1.85

Answer

B

Explanation

In solution, `x_(A)=0.30, x_(B)=0.70`
In vapour phase, `x'_(A)=0.60, x'_(B)=0.40`
`x'_(A)=0.60=(p_(A))/(p)=(p_(A))/(p_(A)+p_(B))=(0.30 p_(A)^(@))/(0.30p_(A)^(@)+0.70p_(B)^(@))`
`x'_(B)=0.40=(p_(B))/(p)=(p_(B))/(p_(A)+p_(B))=(0.70)/(0.30p_(A)^(@)+0.70p_(B)^(@))`
`(x'_(A))/(x'_(B))=(0.60)/(0.40)=(0.30p_(A)^(@))/(0.70 p_(B)^(@))`
`(p_(A)^(@))/(p_(B)^(@))=(0.60xx0.70)/(0.40xx0.30)=(7)/(2)=3.5`