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

Options

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

Answer

A

Explanation

Mole fraction of `X= (1)/(1+3) = (1)/(4)`, Mole fraction of `Y= -(3)/(1+3) = (3)/(4)`
Total pressure = partial pressure of `X xx` Mole fraction of X+ Partial pressure of Y `xx` mole fraction of Y.
`550 = (P_(s)^(@))/(4) + (3P_(y)^(@))/(4)` ..(1)
`550 xx 4= P_(X)^(@) + 3P_(Y)^(@)`
When one mole of Y is added mole fraction of X becomes `1//(1+4)` = 1/5 and mole fraction of Y becomes 4/(1+4)= 4/5. Now, the total pressure given as
`560 =(p_(x)^(@))/(5) + (4p_(y)^(@))/(5)` (2)
`560 xx 5= p_(x)^(@) + 4p_(y)^(@)`
Solving Eqs (1) and (2), we get `p_(x)^(@)= 400mm Hg and p_(y)^(@)= 600mmHg`