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 600

Answer

C

Explanation

`P_(T) =P_(X)^(@)X_(X) + P_(Y)^(@)X_(Y)`
`X_(X) =` mole fraction of X
`X_(Y) = ` mol fraction of Y
`:. 550 =P_(X)^(@) ((1)/(1+3)) + P_(Y)^(@) ((3)/(1+3))`
`= (P_(X)^(@))/(4) = ( 3P_(Y)^(@))/(4)` ….(1)
Further 1 mol of Y is added and total pressure increases by 10 mm Hg.
`:. 550 + 10 = P_(X)^(@) ((1)/(1+4)) + P_(Y)^(@) ((4)/(1+4)) `
`:. 560 (5) =P_(X)^(@) + 4P_(Y)^(@)` ....(2)
By solving ( 1) and (2) , we get
`P_(X)^(@) =400 mm Hg, P_(Y)^(@) = 600 mmHg `