Two liquids X and Y form an ideal solution. At 300K, 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 mm Hg. 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 300

Answer

C

Explanation

According to Raoult .s law .
`p = p_(X)^(@) x_(X) + p_(Y)^(@) x_(Y)`
Initially , p = 550 mm Hg `x_(X) = (1)/(4), x_(Y) = (3)/(4)`
` 550 = p_(X)^(@) (1)/(4) + p_(Y)^(@) (3)/(4)`
On adding 1 mole of Y
p = 560 mm , `Hg , x_(X) = (1)/(5), x_(Y) = (4)/(5)`
`560 = p_(X)^(@) (1)/(5) + p_(Y)^(@).(4)/(5)`
Solving eq. (i) and (ii) , `p_(X)^(@) = 400 mm Hg`
`p_(Y)^(@) = 600 mm Hg`