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Mole fraction of component A in vapour p...

Mole fraction of component `A` in vapour phase is `chi_(1)` and that of component `A` in liquid mixture is `chi_2`, then (`p_(A)^@`)= vapour pressure of pure A, `p_(B)^@` = vapour pressure of pure B), the total vapour pressure of liquid mixture is

A

`(P_(A)^(@).X_(2))/(X_(1))`

B

`(P_(A)^(@).X_(1))/(X_(2))`

C

(c ) `(P_(B)^(@).X_(1))/(X_(2))`

D

(d) `(P_(B)^(@).X_(2))/(X_(1))`

Text Solution

Verified by Experts

The correct Answer is:
A
`P'_(A) = P_(A)^(@).X_(A)` and `P'_(A) = P_(M).x'_(A)` where `X_(A)` and `X'_(A)` are mole fraction of `A` in liquid and vapour phase respectively.
`:. P_(A)^(@).X_(2) = P_(M).X_(1)`
`:. P_(M) = (P_(A)^(@)X_(2))/(X_(1))`
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