The liquids X and Y form an ideal solution at 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively -

Options

A. 200 and 300
B. 300 and 400
C. 400 and 600
D. 500 and 600

Explanation

`P_(T)=P_(X)^(@)X_(x)+P_(Y)^(@)X_(Y)`
`X_(x)=` mol fraction of X
`X_(Y)=` mol fraction of Y
`550=P_(X)^(@)(1/(1+3))+P_(Y)^(@)(3/(1+3))`
`550(4)=P_(X)^(@)+3P_(4)^(@)`………1
`:.560=P_(X)^(@)(1/(1+4))+P_(Y)^(@)(4/(1+4))`
`:.560(S)=P_(x)^(@)+4P_(Y)^(@)`.........2
By solving 1 and 2
We get `P_(x)^(@)=400m Hg" ", " "P_(Y)^(@)=600 mm Hg`