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

3

Explanation

Accoding to Raoult's law
`P_(T)=P_(X)^(0) chi_(X)+P_(Y)^(0) chi_(Y)`
`550 mmHg =P_(X)^(0) (1/(1+3))+P_(Y)^(@)(3/(1+3))`
`=P_(X)^(0)/4+(3 P_(Y)^(0))/4`
or `4(550) mmHg=P_(X)^(0)+3P_(Y)^(0)`
When 1 mol of Y is further added to the same solution, vapour pressure of the solution increases by `10 mmHg`. Thus
`(550+10) mmHg=P_(X)^(0)(1/(1+4))+P_(Y)^(0)(4/(1+4))`
`=P_(X)^(0)/5+(4P_(Y)^(0))/5`
or
`5(560) mmHg =P_(X)^(0)+4P_(Y)^(0)(2)`
By solving (1) and (2) simultaneously, we get
`P_(X)^(0)=400 mmHg`
`P_(Y)^(0)=600 mmHg`