Solution of two volatile liquids `x` and `y` obey Raoult's law. At a certain temperature it is found that when the total pressure above a given solution is `400 mm` of `Hg`, the mole fraction of `x` in the vapour is `0.45` and in the liquid is `0.65`. What are the vapour pressures of two pure liquids at the given temperature?

Explanation

`chi_(x)` in liquid `= 0.65`
`chi_(y)` in liquid `= 0.35`
`p_(x) = chi_(x)p_(x)@`
`p_(y)= chi_(y)p_(y)@` , `p_(y)=0.35 p_(y) `, `p_(x) = 0.65 p_(x) `
`p_("total") = p_(x) + p_(y) = 400`
`chi_(x)^(v) = 0.45 = (p_(x))/(p_("total"))`
`0.45 = 0.65 xx (p_(x)@)/(400)`
`:. p_(x)@ = 276.9 mm`
`0.65 xx 276.9 + 0.35p_(y)@ = 400`
`p_(y)@ = 628.6 mm`