Two liquids X and Y from an ideal solution at`300K`, Vapour pressure of the Solution containing `1`mol of X and `3`mol of Y is `550`mmHg . At the same temperature, if `1` mol of Y is further added to this solution ,vapour pressur of the solutions increases by `100`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`

Answer

C

Explanation

`P_(T)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)`
`550=P_(A)^(@)xx(1)/(4)+P_(B)^(@)xx(3)/(4)`
Thus, `P_(A)^(@)+3P_(B)^(@)=2200` …(i)
when `1` mole of Y is further added to the solution, vapour pressure of a solution becomes `560`mmHg.
`560=P_(A)^(@)xx(1)/(5)+P_(B)^(@)xx(4)/(5)`
Thus, `P_(A)^(@)+4P_(B)^(@)=2800`...(ii)
on subtracting eq.(ii)by Eq.(i) we get
`P_(B)^(@)=2800-2000rArr P_(B)^(@)=600`
putting the value of `P_(B)^(@)` in Eq.(i)
`P_(A)^(@)+3xx600=2200`
`P_(A)^(@)=2200-1800=400`