Liquids `X` and `Y` form an ideal solution. The vapour pressure of `X` and `Y` at `100^(@)C` are `300` and `100 mm` of `Hg`, respectively. Suppose that a solution composed of `1 mol` of `X` and `1 mol` of `Y` at `100^(@)C` is collected and condensed. This condensate is then heated at `100@C` and vapour is again condensed to form a liquid `A`. What is the mole fraction of `X` in `A` ?

Explanation

Vapour presure due to vapours above tha solution
`P_(T)=X_(A)P_(A)^(@)+X_(B)P_(B)^(@)`
`P_(T)=300X_(A)+100 X_(B)`
It is given that in vapour pahse each of A and B are one mole each. Hence of them have mole fraction`0.5` in vaour phase.
`0.5=(300X_(A))/(P_(T))" "...(1)`
`0.5=(100X_(B))/(P_(T))" "...(2)`
And `X_(A)+X_(B)=1 " "...(3)`
Solving Eqs (1) ,(2) and (3) we get
`X_(A)=0.25 and X_(B)=0.75`
`P_(T)=300xx0.25+100xx0.75=75=150` mm
After condensation of vapours,
In condenstate (1)
`X_(A)=0.5`
`X_(B)=0.05`
`P_(T)=0.5xx300+0.5xx100`
`=150+50=200mm`
Mole fraction of A and B in vapour phase of condenstate
`Y_(A)=(P_(A))/(P)=0.75`
`V_(B)=1-0.75=0.25`
When the vapiours of the condenstate (1) will again be vapourizex in condenstate (2) liquid L
`X_(A)=0.75X_(B)=0.25`, where `X_(A).. and X_(B)..` are mole fractions of A and B in liquid L.
`P_(T)..=300xx0.75+100xx0.25=2556+25-=250 mm`
and mole fraction of A in vapour phase of the condenstate (2) is given by
`Y_(A)..=(225)/(250)=0.9`