Derive the relationship between the relative lowering of Vapour pressure and the mole fractions of the non volatile solute

Explanation

According to Raoult.s Law
`p_(A) = p_(A)^(@) x_(A)`
where `p_(A) `= Vapour pressure of solution,
`x_(A)` = Mole fraction of solvent
`p_(A)^(@)` = Vapour pressure of solvent,
`x_(B)` = Mole fraction of solute
`p_(A) = p_(A)^(@) (1 - x_(B))`
`(p_(A))/( p_(A)^(@)) = 1 - x_(B)`
`x_(B) = 1- (p_(A))/( p_(A)^(@))`
`x_(B) = (p_(A)^(@) - p_(A))/( p_(A)^(@))`
Or `(p_(A)^(@) - P_(A))/( p_(A)^(@)) = x_(B)`
where `p_(A)^(@) - p_(A)` = Lowering in V.P.
`(p_(A)^(@) - p_(A))/( p_(A)^(@)) `= Relative lowering in V.P.
Thus , Relative lowering of V.P. = Mole fraction of non-volatile
As relative lowering in vapour pressure depend upon the mole fraction of the solute in the solution, it is a colligative property .