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The mole fraction of component A in vapo...

The mole fraction of component `A` in vapour phase is `chi_(1)` and mole fraction of component `A` in liquid mixture is `chi_(2) (P_(A)^(@)`=vapour pressure of pure `A`,`P_(B)^(@)`=vapour pressure of pure `B`).Then total vapour pressure of the liquid mixture is

A

`p_(A)^@(chi_(2))/(chi_(1))`

B

`p_(A)^@(chi_(1))/(chi_(2))`

C

`p_(B)^@(chi_(1))/(chi_(2))`

D

`p_(B)@^(chi_(2))/(chi_(1))`

Text Solution

Verified by Experts

The correct Answer is:
a
`p_(A) = p_(A)^@ chi_(2)`, vapour pressure of `A`.
Mole fraction of `A` in vapour = `p_(A)/p_("total")`
`chi_(1) = (p_(A)^@chi_(2))/p`
`p_("total")= (p_(A)^@chi_(2))/(chi_(1))`
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