# Blog

# Raoult’s Law of Volatile Mixture

- April 10, 2017
- Posted by: admin
- Category: eHow

# Raoult’s Law of Volatile Mixture

Raoult’s law states partial pressure of a component in a solution is equal to its mole fraction multiplied by the vapor pressure of that component in pure state.

P = P^{0}.X

P- V.P of component in the solution, P^{0}– V.P of pure solvent, X-mole fraction of the component.

The total vapor pressure of a solution for two individual components A and B can be given by

P = P_{A} + P_{B} = P^{0}_{A}.X_{A} + P^{0}_{B}.X_{B}

The sum of mole fraction of all the components in a solution is unity.

X_{A} + X_{B} = 1

P = P^{0}_{A}.(1-X_{B}) + P^{0}_{B}.X_{B} = (P_{B}^{0} – P^{0}_{A})X_{B} + P^{0}_{A}

**Problem:**

Vapor pressure of pure “A” is 70 mm of Hg at 25˚C. It forms an ideal solution with “B” in which mole fraction of A is 0.8. If the vapor pressure of the solution is 84mm of Hg at 25˚C, the vapor pressure of pure “B” at 25˚C is:

**Solution:**

The mole fraction of B = 1 – Xa = 1- 0.8 = 0.2

According to Raoult’s law, P = P_{A} + P_{B} = P_{A}^{0}X_{A} + P_{B}^{0}X_{B}

84 = 0.8*70 + 0.2*P^{0}_{B}

P^{0}_{B} = 140 mmHg