Q.The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
a. 24 mm Hg
b. 32 mm Hg
c. 48 mm Hg
d. 12 mm Hg

Raoult's Law: From Intuition to Precision

Imagine you have a beaker of pure water at room temperature. Some water molecules at the surface have enough energy to escape into the air above — that's vapour pressure. Now dissolve some sugar in that water. The sugar molecules take up space at the surface, blocking some water molecules from escaping. Fewer water molecules can leave the liquid per second, so the vapour pressure drops.

That's the core intuition: a non-volatile solute lowers the solvent's vapour pressure simply by getting in the way.

But what if both components can evaporate — say, a mixture of benzene and toluene? Then both kinds of molecules crowd the surface, and both contribute to the total vapour pressure. The question becomes: how much does each contribute?

The Precise Statement

For a solution of volatile liquids, Raoult's Law says:

where:

• $p_i$ = partial vapour pressure of component $i$ above the solution
• $x_i$ = mole fraction of component $i$ in the liquid solution
• $p_i^*$ = vapour pressure of pure component $i$ at the same temperature

The law applies to each volatile component separately. The total vapour pressure above the solution is simply the sum:

What This Means Physically

The mole fraction $x_i$ tells you the fraction of molecules at the surface that are of type $i$. If half the molecules in the liquid are benzene ($x_{\text{benzene}} = 0.5$), then roughly half the surface sites are occupied by benzene molecules. So the rate at which benzene escapes should be about half the rate from pure benzene — hence $p_{\text{benzene}} = 0.5 \times p_{\text{benzene}}^*$.

This is a linear relationship: plot $p_i$ against $x_i$, and you get a straight line from the origin (when $x_i = 0$, $p_i = 0$) up to $p_i^*$ (when $x_i = 1$, pure component).

A Concrete Example

Suppose you mix 2 moles of benzene ($p^* = 100$ mm Hg) with 3 moles of toluene ($p^* = 40$ mm Hg) at 25°C.

Mole fractions:

• $x_{\text{benzene}} = \frac{2}{2+3} = 0.4$
• $x_{\text{toluene}} = \frac{3}{5} = 0.6$

Partial pressures:

• $p_{\text{benzene}} = 0.4 \times 100 = 40$ mm Hg
• $p_{\text{toluene}} = 0.6 \times 40 = 24$ mm Hg

Total vapour pressure: $40 + 24 = 64$ mm Hg

Notice: the total pressure is not a simple average of the pure pressures. It's a weighted average, with mole fractions as weights.

Why This Matters

Raoult's Law is the foundation for understanding distillation. Because the vapour above a solution is richer in the more volatile component (the one with higher $p^*$), you can separate liquids by boiling and condensing — that's fractional distillation. The law also explains why adding salt raises the boiling point of water (boiling point elevation) and lowers its freezing point — both are direct consequences of the vapour pressure being reduced.

Raoult's Law: $p_i = x_i p_i^*$ — the partial pressure of each volatile component is proportional to its mole fraction in the liquid.

Because it involves a direct formula application, Raoult's Law is a recurring numerical topic in CBSE Class 12 board papers as well as JEE Main and NEET — students commonly search for "Raoult's Law formula and derivation" or "Raoult's Law class 12 chemistry solved examples", and this concept sits squarely within the NCERT-aligned Solutions unit.