At `25^(@)C, 1 mol` of A having a vapour pressure of 100 torr and 1 mole of B having a vapour pressure of 300 torr were mixed. The vapour at equilibrium is removed, condensed and the condensate is heated back to `25^(@)C`. The vapour now formed are again removed, recondensed and analyzed. what is the mole fraction of A in this condensate?

Answer

`0.1`

Explanation

`P_(A)^(@) = 100, P_(B)^(@) = 300, X_(A) = X_(B) = (1)/(2)`
`P_(T) = 200`
`Y_(A) = (100 xx(1)/(2))/(200) =(1)/(4)`
On condensation `X_(A) = (1)/(4),X_(B) = (3)/(4)`
`P_(T) = 100 xx (1)/(4) +300 xx (3)/(4) = 250`
`Y_(A) = (25)/(250) = 0.1`
on further condensation
`X_(A) = 0.1`