Unit – III A – 02 |

## Raoult’s Law:

### Statement:

The partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.

#### Explanation:

- Let us consider a solution containing two volatile components say A
_{1}and A_{2}, with mole fractions x_{1}and x_{2}respectively. Let p_{1}^{0}and p_{2}^{0 }be the vapour pressures of the pure components A_{1}and A_{2}respectively, then by Raoult’s law

p_{1} = p_{1}^{0 }x_{1} and p_{2} = p_{2}^{0} x_{2}

- The total vapour pressure of the solutions of two volatile components is the sum of partial vapour pressures of the two components.

- The solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution. If a solution does not obey Raoul’s law are called non-ideal solutions.

### Raoult’s Law for a Solution of Non-Volatile Solute:

- Let us consider a solution containing two volatile component A
_{1}and non-volatile component A_{2}, with mole fractionsx_{1}and x_{1}respectively. - Let p
_{1}^{0}and p_{2}^{0 }be the vapour pressures of the pure components A_{1}and A_{2}respectively. Now component A_{2}is non-volatile, hence it will not contribute to vapour pressure. Thus p_{2}^{0 }= 0. We have

- Thus vapour pressure of a solution of non-volatile solute is the product of vapour pressure p
_{1}^{0}of pure solvent and mole fraction x_{1}of the solvent, which is Raoult’s law. - The equation shows that vapour pressure of the solution p < p
_{1}^{0}, i.e. there is lowering of the vapour pressure of the solution.

The lowering of vapour pressure is given by

- In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.

Now, the relative lowering of vapour pressure is given by

- This relation proves that the lowering of vapour pressure is colligative property because it depends on the concentration of non-volatile solute.

### Raoult’s Law as Special Case of Henry’s Law:

- By Raoult’s law, we have

- By Henry’s law, we have

- If we compare the two equations for Raoult’s law and Henry’s law, it can be seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant K
_{H}differs from p_{1}^{0}. Thus, Raoult’s law becomes a special case of Henry’s law in which K_{H}is equal to p_{1}^{0}.

### Relation Between Molar Mass of Solute and Lowering of Vapour Pressure:

Let W_{2} g of the solute of molar mass M_{2} be dissolved in W_{1} g of the solvent of molar mass M_{1}. The number of moles of solvent and solute are given by

The mole fraction of solute is given by

Now for dilute solutions n_{2} << n_{1}. Hence n_{2} can be neglected.

This is the relation between the molar mass of solute and lowering of vapour pressure.

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