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Two liquids Aand B have vapour pressures...

Two liquids `A`and `B` have vapour pressures in the ratio of `p_(A)@,p_(B)@` = 1 : 2 at a certain temperature. Suppose we have an ideal solution of `A` and `B` in the mole fraction ratio `A:B = 1:2`. What would be the mole fraction of `A` in the vapour in equilibrium with the solution at a given temperature?
a.`0.25` , b.`0.2` , c.`0.5` d.`0.33`

Text Solution

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The vapour pressures `P_(A)^(@) and P_(B)^(@)` Of the two liquids A and B are in the ratio 1:2 The mole fractions `X_(A) and x_(B)` of the two liquids A and B are also in the ratio 1:2.
Partial vapour pressure of `A(P_(A)^('))=P_(A)^(@)x_(A)`
Partial vapour pressure of `B (P_(B)^('))=P_(A)^(@)x_(B)=2P_(A)^(@)xx2x_(A)`
`P_(A)^(')/P_(B)^(')=(P_(A)^(@)x_(A))/(4P_(A)^(@)x_(A))=1/4or P_(B)^(@)=4P_(A)^(')`
Mole fraction of A in the vapour phase may be given as: `x_(A)^(')=P_(A)^(')/(P_(A)^(')+P_(B)^('))=P_(A)^(')/(P_(A)^(')+4P_(A)^('))=1/5=0.2`
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