Define relative lowering of vapour pressure.

Explanation

According to Raoult.s law,
`P _("solution") prop x _(A), ` where `x _(A) =` mole fraction of solvent.
`P_("solution") =k x _(A),` where k = proportionality constnat
For a pure solvent,
Vapour pressure `= P^(@), x _(A) =1`
`therefore P _("solvent")^(@)= k xx 1 =k`
Substituting `P _("solvent"^(@)` in Raoult.s law
`P_("solution") =P_("solvent")^(@) x _(A)`
Relative lowering vapour pressure `= (P_("solvent)^(0) -P_("solution"))/(P _("solvent")^(0))`
Substituting `P _("solution")` as `P^(@) x _(A)` in the above equation
Relative lowering of vapour pressure=` (P ^(@) - P^(@) x _(A))/(P ^(@))= (P^(@) (1- x _(A)))/(P ^(@)) =1 - x _(A)`
Relative lowering of vapour pressure `=x _(B)`
`because x _(A) + x _(B) =1,` where `x _(B)=` mole fraction of solute.
It is clear that the relative lowering of vapour pressure depends only on the mole fraction of the solute `(x _(B))` and is independent of its nature. Therefore relative lowering of vapour pressure is a colligative property.