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Solution of two volatile liquids x and y...

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?

Text Solution

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The given data are
`chi_(A) = 0.60 , chi_(A)^(V) = 0.40 , P_(t) = 600` torr
`p_(A) = ? ,p_(B) = ?`
Using equation `chi_(i)^(V) = (p_(i)//p_("total"))`
`chi_(A)^(V) = p_(A)/P = (chi_(A)p_(A))/P`
or `p_(A)@ = (chi_(A)^Vp_(A)^@)/chi_(A) = ((0.40) (600)) /(0.60) = 400` torr
Similarly, `p_(B)@= (chi_(B)^(V)P)/chi_(B) = ((0.60)(600))/(0.40) = 900` torr
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