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

Options

A. `p_(A)^(0) (x_(2))/(x_(1))`
B. `p_(A)^(0)(x_(1))/(x_(2))`
C. `p_(B)^(0)(x_(1))/(x_(2))`
D. `p_(B)^(0) (x_(2))/(x_(1))`

Answer

A

Explanation

`p_(A)=p_(A)^(0)x_(2)`, vapour pressure of 'A'
Mole fraction of A in vapour `= (p_(A))/(p_("total"))`
`x_(1)=(p_(A)^(0)x_(2))/(p)`
`p=(p_(A)^(0)x_(2))/(x_(1))`]