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

`x_(A)=(1)/(4)=0.25,x_(y)=0.75`
`P_("total")=x_(X)xxp_(X)^(@)+x_(Y)xxp_(Y)^(@)`
i.e., `550=0.25p_(X)^(@)+0.75p_(Y)^(@)`
`"or "2200=p_(X)^(@)+3p_(Y)^(@)" ...(i)"`
After adding 1 mol of Y,
`x_(X)=(1)/(5)=0.20, x_(Y)=0.80`
`therefore" "560=0.20p_(X)^(@)+0.80p_(Y)^(@)`
`"or "2800 = p_(X)^(@)+4p_(Y)^(@)" ...(ii)"`
Solving the two equations, we get
`p_(Y)^(@)=600mm, p_(X)^(@)=400mm`