If `P^(@)` and `P_(S)` are the vapour pressure of the solvent and its solution respectively and `x_(1)` and `x_(2)` are the mole fraction of the solvent and solute respectively, then

Options

A. `p_(A)^(@)//p_(B)^(@)`
B. `p_(B)^(@)//p_(A)^(@)`
C. `p_(A)^(@) - p_(B)^(@)`
D. `p_(B)^(@) - p_(A)^(@)`

Answer

B

Explanation

Mole fraction of component A in vapour phase .
`y_(A) = (p_(A))/(p_("tota")) = (p_(A)^(@) x_(A))/(p_(A)^(@) x_(A) + p_(B)^(@) x_(B)) = (p_(A)^(@) x_(A))/(p_(A)^(@) x_(A)^(@) + p_(B)^(@) (1-x_(A)))`
` =(p_(A)^(@) x_(A))/(x_(A)(p_(A)^(@) - p_(B)^(@)) + p_(B)^(@))`
`(1)/(y_(A)) = (x_(A) (P_(A)^(@) - p_(B)^(@)) + p_(B)^(@))/(x_(A)p_(A)^(@))`
` = (p_(A)^(@) - p_(B)^(@))/(p_(A)^(@)) + (p_(B)^(@))/(x_(A)p_(A)^(@))`
`therefore ` Plot of `(1)/(y_(A))` against `(1)/(p_(A)^(@))` give straight line with slope ` = (p_(B)^(@))/(p_(A)^(@))`