Question

Derive the equation of Raoult’s law for binary solution containing non-volatile solute.

Answer 1

Statement of Raoult’s Law:

Raoult’s law states that the partial vapor pressure of a volatile solvent in a solution is directly proportional to its mole fraction in the solution.

For a binary solution with:

Solvent (A) – Volatile

Solute (B) – Non-volatile

Since the solute is non-volatile, only the solvent contributes to the vapor pressure.

Raoult’s law Expression:

According to Raoult’s Law, the vapor pressure of the solvent in the solution is:

`P_A = X_AP_A^0`

where:

P_A = Vapor pressure of the solvent in the solution

`P_A^0` = Vapor pressure of the pure solvent

X_A ​= Mole fraction of the solvent (X_A = 1 − X_B)

Thus,

`P_A = (1 - X_B)P_A^0`

`P_A = P_A - X_BP_A^0`

Lowering of Vapor Pressure:

The lowering of vapor pressure due to the non-volatile solute is:

`ΔP = P_A^0 − P_A​`

Substituting `P_A = P_A^0 - X_BP_A^0`

`ΔP = P_A^0 − (P_A^0 - X_BP_A^0)`

`ΔP = X_BP_A^0`

Relative Lowering of Vapor Pressure:

Dividing both sides by `P_A^0`,

`(ΔP )/(P_A^0) = X_B`

Since the mole fraction of solute (X_B​) is

`X_B = (n_B)/(n_A​ + n_B​)​`

where n_A​ and n_B​ are the number of moles of solvent and solute, respectively, we get:

`(ΔP )/(P_A^0) = (n_B)/(n_A​ + n_B​)​`

For a dilute solution, where n_B ≪ n_A​, we approximate,

`(ΔP )/(P_A^0) ≈ (n_B)/(n_A)​`

The relative lowering of vapor pressure is directly proportional to the mole fraction of the non-volatile solute.