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At 20^@C, the vapour pressure of pure li...

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?

Text Solution

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The given values are
`p_(A)@ =22 mm Hg , p_(B)@ =75 mm Hg , p_("total") = 48.5 mm Hg`
Let `chi_(A)` and ` chi_(B)` are the mole fractions of liquids `A` and `B` , respectively, in solution , then
`p_(A) = p_(A)@chi_(A) = 22 chi_(A)`
`p_(B) = p_(B)@chi_(B) = p_(B)@(1-chi_(A)) = 75(1-chi_(A))`
`p_("total") = p_(A) + p_(B) = 22chi_(A) + 75(1 - chi_(A)) = 48.5`
`:. chi_(A) = 0.5` and `chi_(B) = 1 - chi(A) = 0.5`
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