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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 isA. `(P_(A)^(@)chi_(2))/chi_(1)`B. `(P_(A)^(@)chi_(1))/chi_(2)`C. `(P_(A)^(@)chi_(1))/chi_(2)`D. `(P_(B)^(@)chi_(2))/chi_(1)` |
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Answer» Correct Answer - A `chi_(1)=(P_(A)^(@)chi_(A))/(P_(Total))=(P_(A)^(@)chi_(2))/(P_(Total))` `:. (P_(Total)=(P_(A)^(@)chi_(2))/chi_(1)` |
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