Two 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 Yis further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) 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^@ x^x x + P^@y^xy`
where, `P_T`= Total pressure
`p^@x` = Vapour pressure of X in pure state
`p^@Y`= Vapour pressure of Y in pure state
`x_X` = Mole fraction of `X= 1/4`
`x_Y` = Mole fraction of Y = `3/4`
(i) When `T = 300 K, P_T = 550 mm Hg`
`:. 550= P_x^@(1/4) +P_Y^@(3/4)`
`implies P_x^@ + 3P_Y^@ = 2200` ..... (1)
(ii) When at T = 300 K, 1 mole of Y is added,
`P_T=(550 + 10) mm Hg`
`:. x_x = 1//5 and x_y = 4//5`
`implies P_x^@(1/5) +P_Y^@(4/5)`
or `P_x^@ + 4P_Y^@ = 2800`
On solving equations (1) and (2), we get
`P_x^@ = 600 mm Hg and P_x^@ = 400 mm Hg`