**Derivation of Raoult’s Law**

The vapour pressure of a liquid is governed by the ‘escaping tendency’ of the molecules in the liquid and the number of molecules per unit volume. If a solute is dissolved, the number of solvent molecules per unit volume decreases, since in the solution solute molecules will also be present. If we assume that the intermolecular forces between the solvent molecules remain unchanged, which is true for dilute solutions, then the number of solvent molecules leaving unit area of the solution surface per unit time will be less than the number leaving the surface of the pure solvent under the same conditions. Hence the vapour pressure of the solution will be lower than that of the solvent because the number of molecules per unit volume of the solvent in the vapour phase over liquid solvent will be more than the number of solvent molecules over the solution. The vapour pressure p of the solution will be proportional to the ratio of the number of solvent molecules, N, in solution to the total number of molecules in the solution, (n + N). Thus

**p = K [N / (n + N)] = K x _{N}**

where K is the proportionality constant and x_{N} is the mole fraction of the solvent. For the pure solvent x_{N} = I and p becomes equal to p^{0}. Therefore, K is equal to p^{0}. Thus,

p = P^{0} x_{N}.

which is one form of Raoult’s law.

*Raoult’s law states: The vapor pressure of an ideal solution is dependent on the vapor pressure of each chemical component and the mole fraction of the component present in the solution.*

It has, however, been assumed that the vapour of the solvent above the solution and the pure liquid behave like an ideal gas. Further assumptions have been made regarding the constancy of the intermolecular forces of the liquid in presence of the solute.