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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)^(@)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
`chi_(A)^(V)=(P_(A)^(@)chi_(A))/(P_("total"))=chi_(1)(P_(A)^(@)chi_(2))/(P_("total")) ` or `P_("total")=P_(A)^(@)(chi_(2)/chi_(1))`
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