Derive the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid.

Raoult’s Law for solutions containing volatile liquids can be stated as follows:

For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

If P_i is the partial vapour pressure of i^th component in a solution and χ_i is its mole fraction in the solution, then by Raoult’s Law we have:

P_i αχ_i

P_i = constant × χ_{i …}……… (1)

If we consider the component in its pure state, then

P_i = P_i° and χ_i = 1 …….. (2)

Where

P_i° = vapour pressure of i^th component in pure state

Substituting the values of (2) in (1)

P_i° = constant × 1

P_i° = constant

Therefore the equation (1) becomes:

P_i = P_i° × χ_i

Raoul’s Law can also be stated as follows:

For solution of volatile liquids, the partial vapour pressure of each component in the solution at a particular temperature is equal to the product of the vapour pressure of each component in pure state and its mole fraction in the solution.

Let us assume a binary solution of two volatile liquids A and B in which the mole fraction of the liquid A be χ_A and that of the liquid B be χ_B. P_A be the vapour pressure of the liquid A and P_A° be the vapour pressure of the liquid A in pure state. P_B be the vapour pressure of the liquid B and P_B° be the vapour pressure of the liquid B in pure state.

According to Raoult’s Law, we have:

P_A = P_A° × χ_A

P_B = P_B° × χ_B

Therefore the total vapour pressure P of the solution is given by,

P = P_A° × χ_A + P_B° × χ_B

Raoult’s Law is only applicable when the volatile liquids are completely miscible. If the mixture of volatile liquids does not form a solution, the law is not applicable.