Show that relative lowering in vapour pressure is a colligative property

Explanation

Addition of a non-volatile solute to solvent lowers the vapour pressure of the solution. Particles of solute occupy some surface area. Lesser surface araa is availeble for the particles of solvent to evaporate. Hence, vapour pressure of the solution lowers. According to Raoult.s law, pressure of a solution `(p_(A))` is the product of vapour pressure of pure solvent `(p_(A)^(@))` and its mole fraction `(x_(A))`.
Thus, `p_(A)=p_(A)^(@)x_(A)` ...(i)
`because` For a binary solution `x_(A)+x_(B)=1orx_(A)=1-x_(B)`
Now equation (i) becomes `p_(A)=p_(A)^(@)(1-x_(B))`
`p_(A)=p_(A)^(@)-p_(A)^(@)x_(B)orp_(A)^(@)x_(B)=p_(A)^(@)-p_(A)orx_(B)=(p_(A)^(@)-p_(A))/(p_(A)^(@))`
Hence, relative lowering of vapour pressure is equal to the mole fraction of the solute. So, relative lowering of vapour pressure depends upon the mole fraction of solute. Therefore, it is a colligative property.