Question

State and explain Raoult's law for the solution of non-volatile solutes.

Answer 1

Raoult's law: The solvent's vapour pressure above the solution is calculated by multiplying its mole fraction in the solution by the solvent's vapour pressure when it is pure.

Thus, `P_1 = P_1^0x_1`

For a binary solution containing one solute,

x₁ = 1 − x₂

∴ `P_1 = P_1^0(1 - x_2)`

`P_1 = P_1^0 - P_1^0 x_2`

or `p_1^0 - p_1 = P_1^0x_2`

But `p_1^0 - p_1` = ΔP (Lowering of vapour pressure)

∴ ΔP = ` P_1^0 x_2`

Thus ΔP ∝ x₂,

where, x₂ = Number of solute particles.

Hence, ΔP (Lowering of vapour pressure) is a colligative property.