Raoult's law

Dissolve a pinch of sugar in water and the water's own vapour pressure quietly drops, even though sugar itself never evaporates. Dissolve one volatile liquid in another — benzene in toluene, say — and each liquid still contributes its own share of vapour, but less than it would pure, in exact proportion to how much of it is actually there. Both observations are the same law wearing two costumes. Raoult's law says the partial vapour pressure a component contributes to a solution equals its vapour pressure when pure, scaled down by its mole fraction in the liquid. It is the single composition rule that opens the entire Solutions unit: from it fall the vapour-pressure diagrams that separate ideal from non-ideal mixtures, the principle behind fractional distillation, and — once the solute stops evaporating altogether — the whole family of colligative properties, starting with the relative lowering of vapour pressure worked out below.

The core insight

Each component's partial vapour pressure above a solution is its pure vapour pressure times its mole fraction in the liquid, $p_i = p_i^0 x_i$ , and summing over all components gives the solution's total vapour pressure.

The core idea

Two volatile liquids. When both components of a solution are volatile, each escapes into the vapour in proportion to how much of it sits in the liquid. For a binary mixture of components 1 and 2:

p_1 = p_1^0\, x_1, \qquad p_2 = p_2^0\, x_2

where: $p_1^0, p_2^0$ are the vapour pressures of the pure liquids at that temperature, and $x_1, x_2$ are their mole fractions in the liquid solution ( $x_1 + x_2 = 1$ ). Dalton's law of partial pressures then hands us the total vapour pressure above the solution:

p_{total} = p_1^0 x_1 + p_2^0 x_2

A straight line in composition. Substituting $x_2 = 1 - x_1$ turns this into $p_{total} = p_2^0 + (p_1^0 - p_2^0)x_1$ — a straight line when total vapour pressure is plotted against $x_1$ , running from $p_2^0$ at $x_1 = 0$ (pure component 2) to $p_1^0$ at $x_1 = 1$ (pure component 1). This straight line is the graphical definition of an ideal solution; real mixtures that bulge above it show positive deviation and those that dip below show negative deviation, the subject of the next node in this unit.

One non-volatile solute. If component 2 is a non-volatile solute (sugar, a salt, a non-volatile organic compound) it puts no vapour of its own into the gas phase, so $p_2 = 0$ and the solution's entire vapour pressure comes from the solvent alone:

p_{solution} = p_1^0\, x_1 = p_1^0(1 - x_2)

where: $x_2$ is the mole fraction of the dissolved solute. Rearranging gives the relative lowering of vapour pressure, the first colligative property and a direct consequence of Raoult's law:

\frac{p_1^0 - p_{solution}}{p_1^0} = x_2

Vapour is richer in the more volatile liquid. Because the component with the higher $p^0$ escapes more eagerly, the vapour above a binary solution is always enriched in the more volatile liquid relative to the liquid phase itself. Repeating vaporisation-condensation steps concentrates that enrichment further — the working principle of fractional distillation for a non-azeotropic mixture. Raoult's law can also be seen as a special case of Henry's law for the solvent, with the Henry's-law constant $K_H$ equal to $p_1^0$ : the same proportionality that governs a dissolved gas governs the solvent itself when it is nearly pure.

Figure

Vapour pressure plotted against mole fraction of component one showing two straight lines meeting the pure component axes at p one naught and p two naught

Derivation ARelative lowering of vapour pressure from Raoult's law

Starting only from Raoult's law applied to a non-volatile solute, show that the fractional drop in a solvent's vapour pressure upon dissolving a solute equals the solute's mole fraction, and use that relation to check the numbers of a worked measurement so the connection between the composition rule and the measurable colligative property is airtight.

Step 1. Write Raoult's law for the solvent alone. Since the solute is non-volatile, only the solvent (component 1) contributes vapour: $p_{solution} = p_1^0\, x_1$ . This is just $p_1 = p_1^0 x_1$ with the subscript dropped because it is the only vapour present.

Step 2. Replace $x_1$ using $x_1 + x_2 = 1$ , so $x_1 = 1 - x_2$ . Substituting: $p_{solution} = p_1^0(1 - x_2) = p_1^0 - p_1^0 x_2$ .

Step 3. Isolate the drop in vapour pressure. Rearranging gives $p_1^0 - p_{solution} = p_1^0 x_2$ , and dividing both sides by $p_1^0$ : $\dfrac{p_1^0 - p_{solution}}{p_1^0} = x_2$ . The left side is the relative lowering — the fraction by which the vapour pressure fell — and it equals the solute's mole fraction exactly, with no approximation.

Step 4. Sanity-check against numbers. For pure water $p_1^0 = 23.8\ \text{mm Hg}$ lowered to $p_{solution} = 23.0\ \text{mm Hg}$ by a non-volatile solute, the relation predicts $x_2 = (23.8-23.0)/23.8 = 0.8/23.8 = 0.0336$ . This is exactly the value used in the worked example below, confirming the derivation reproduces the composition-based form of Raoult's law used throughout this unit.

Worked example

At 25 degrees Celsius, pure benzene has vapour pressure $150\ \text{mm Hg}$ and pure toluene has vapour pressure $50\ \text{mm Hg}$ . A solution has mole fraction of benzene $x_{benzene} = 0.6$ . Find the partial and total vapour pressures, and then find the mole fraction of a non-volatile solute that would lower pure water's vapour pressure from $23.8\ \text{mm Hg}$ to $23.0\ \text{mm Hg}$ .

Step 1. Mole fraction of toluene: $x_{toluene} = 1 - 0.6 = 0.4$ .

Step 2. Apply Raoult's law to each volatile component: $p_{benzene} = 150 \times 0.6 = 90\ \text{mm Hg}$ and $p_{toluene} = 50 \times 0.4 = 20\ \text{mm Hg}$ .

Step 3. Total vapour pressure by Dalton's law: $p_{total} = 90 + 20 = 110\ \text{mm Hg}$ .

Step 4. Now the non-volatile case: apply the relative-lowering form, $x_2 = \dfrac{23.8 - 23.0}{23.8} = \dfrac{0.8}{23.8} = 0.0336$ .

The benzene-toluene solution has $p_{benzene} = 90\ \text{mm Hg}$ , $p_{toluene} = 20\ \text{mm Hg}$ , $p_{total} = 110\ \text{mm Hg}$ , and the non-volatile solute's mole fraction is about $0.034$ .

Competitive stretch

If the same water solution contained $90\ \text{g}$ of water (5 mol) and the solute's mole fraction is $0.0336$ , the moles of solute follow from $x_2 = n_2/(n_1+n_2)$ . Since $x_2$ is small, $n_2 \approx x_2 \times n_1 = 0.0336 \times 5 = 0.168\ \text{mol}$ — the same relative-lowering data, one algebra step from a molar-mass determination if the solute's mass were also given.

Common traps

Using mass fraction, molarity, or molality in place of mole fraction — Raoult's law is defined strictly in terms of $x_i$ .
Forgetting that a non-volatile solute contributes $p_2 = 0$ , so only the solvent's partial pressure survives in $p_{solution}$ .
Confusing $p_1^0$ (vapour pressure of the pure liquid) with $p_1$ (its partial pressure inside the solution) — they are linked by $x_1$ , not interchangeable.
Treating Raoult's law as exact for every solution — it is the ideal-solution benchmark; real mixtures deviate positively or negatively from it.

Ask the tutor: want to see how the same vapour-pressure diagram bends for solutions showing positive or negative deviation from Raoult's law? Ask the AI tutor to sketch both cases and explain what causes each.

Why does a non-volatile solute lower the vapour pressure at all if it never evaporates itself?

Because the solute occupies part of the liquid surface and dilutes the solvent, fewer solvent molecules per unit surface area have enough energy in the right place to escape into the vapour. The solute contributes nothing to the vapour phase, so the entire vapour pressure comes from the solvent term $p_1^0 x_1$ , and since $x_1$ is now less than $1$ , the vapour pressure must be lower than that of the pure solvent.

Is Raoult's law the same as Henry's law?

They describe the same underlying proportionality but are usually applied in opposite regimes. Raoult's law works well for a component present in large amount, close to pure, such as the solvent in a dilute solution. Henry's law is used for a component present in small amount, such as a dissolved gas, where the proportionality constant $K_H$ differs from the pure vapour pressure. For an ideal solution the two coincide, with $K_H = p_1^0$ for the solvent.

Why does fractional distillation work if Raoult's law just gives one vapour pressure for the whole solution?

Raoult's law gives the partial pressure of each component separately, $p_1^0 x_1$ and $p_2^0 x_2$ , and the more volatile component (higher $p^0$ ) always contributes a larger share of the vapour than its share in the liquid. So the vapour composition differs from the liquid composition, richer in the more volatile part. Condensing that vapour and re-evaporating it repeats the enrichment, which is exactly how fractional distillation separates two volatile liquids.

Raoult's law

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads