At `20^@C`, the vapour pressure of pure liquid `A` is `22 mm Hg` and that of pure liquid `B` is `75 mm Hg`. What is the composition of the solution of these two components that has vapour pressure of `48.5 mm Hg` at this temperature?

Answer

`x_(A) = x_(B) = 0.5` equal moles of A and B .

Explanation

Let `x_(A)` and `x_(B)` are the mole fractions of liquids A and B in the solution , then .
`p_(A) = p_(A)^(@)x_(A) = 22 x_(A)`
` p_(B) = p_(B)^(@) x_(B) = p_(B)^(@) (1-x_(A)) = 75(1-x_(A))`
`p_(A) + p_(B) = 22x_(A) + 75(1-x_(A)) = 48.5`
`therefore " "x_(A) = 0.5`
Thus , the solution contains equal moles of both liquids A and B .