Show that relative lowering in vapour pressure is a colligative property

Explanation

It is the ratio of lowering in vapour pressure to vapour pressure of pure solvent. The relative lowering in vapour pressure of solution containing a nonvolatile solute is equal to the mole fraction of solute in the solution.
Let us consider `pa^@` and `P_A` is the vapour pressure of solvent at pure state and vapour pressure of solvent. `X_B` is the mole fraction of solute. So those are related as.
where, `(P_Å-P_A)/(P_A)=X_B` = relative lowering of vapour pressure
`(P_Å-P_A)/(P_Å) = (eta_B)/(eta_A+eta_B)`
for dilute solutions , `eta_B lt lt eta_A` . Hence
`(P_Å-P_A)/(P_Å) = (eta_B)/(eta_A)`
or `(P_Å-P_A)/(P_Å)= (W_B xxM_A)/(M_BxxW_A)`
`M_B = W_B/W_AxxM_A xx(P_A)/((P_Å-P_A))`
Above expression is used to find the molecular weight of an unknown solute dissolved in a given solvent. Where. `W_B and W_A` = mass of Solute and solvent respectively. `M_B and M_A` = molecular weight of solute and solvent respectively.