# Explain Raoult's law for volatile and non-volatile solute.

## Raoult’s law for Volatile solutes

Partial vapour pressure of a solvent in a solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.

P_{Solution} = X_{Solvent} × P_{Solvent}

where, P_{Solution} = vapour pressure of the solution

X_{Solvent} = mole fraction of the solvent

P_{Solvent} = vapour pressure of the pure solvent

### Raoult's Law Calculator

P_{Solution}:

For two component system-

P_{1} = X_{1} × P'_{1}

and P_{2} = X_{2} × P'_{2}

According to Dalton's law of partial pressures, the total pressure( P_{total}) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as-

P_{total} = P_{1} + P_{2}

P_{total} = X_{1} × P'_{1} + X_{2} × P'_{2}

or, P_{total} = (1 − X_{2}) × P'_{1} + X_{2} × P'_{2}

or, P_{total} = P'_{1} − (X_{2} × P'_{1}) + X_{2} × P'_{2}

or, P_{total} = P'_{1} − (P'_{2} − P'_{1}) X_{2}

From Raoult's law, it is evident that as the mole fraction of a component reduces, its partial pressure also reduces in the vapour phase. The graph of pressure vs mole fraction for component 1 and component 2 are shown below.

## Raoult’s law for Non-Volatile solutes

Relative lowering of vapour pressure is equal to the mole fraction of the solute is Raoult's law for non-volatile solutes.

If a dilute solution is prepared by dissolving n mol of a solute in N mol solvent then, the vapour pressure of solvent above the solution is proportional to mole fraction of solvent.

P ∝ N/(N + N)

or, P = K N/(n + N) -----Eq:1

where, K is proportionality constant.

For pure solvent, n = 0

so, P_{o} = K N/N =K -----Eq:2

From Eq:1 and Eq:2, we have-

P/P_{o} = N/(n + N)

or, 1 − (P/P_{o}) = 1 − [N / (n + N)]

or, (P_{o} − P) / P_{o} = (n + N −N) / (n + N)

or, (P_{o} − P) / P_{o} = n / (n + N)

Thus, we can say that the relative lowering of vapour pressure is equal to the mole fraction of the solute.

### Calculate the vapour pressure lowering caused by the addition of 100 g of sucrose (mol mass = 342) to 1000 g of water if the vapour pressure of pure water at 25°C is 23.8 mm Hg.

## Answer

**Hints:**

Lowering of vapour pressure(ΔP) = 23.8 × (0.292 / 55.792) = 0.125mm Hg.