Vapour Pressure of Liquid Solutions

Leave a glass of pure water and a glass of sugar syrup side by side and, at the same temperature, the syrup evaporates more slowly — its vapour pressure is lower. That single observation, made precise, is Raoult's law, and it is the doorway into every colligative-property calculation in this chapter. Sugar molecules sitting at the liquid surface physically occupy space that solvent molecules would otherwise use to escape into the vapour, so fewer solvent molecules leave per second and the equilibrium vapour pressure drops in exact proportion to how much of the surface is "solvent." This article builds that proportionality into a formula, shows how it generalises when both liquids in a solution are volatile, and connects it to Henry's law so the whole Solutions unit reads as one connected idea rather than three separate rules to memorise.

The core insight

The vapour pressure of a solvent above a solution equals its mole fraction times the pure solvent's vapour pressure, $p_1 = x_1 p_1^0$ , and this single proportionality — Raoult's law — underlies every colligative property in the chapter.

The core idea

Why vapour pressure drops at all. A liquid's vapour pressure comes from molecules at the surface escaping into the gas phase until escape and return balance. Dissolve a non-volatile solute and some of that surface is now occupied by solute particles that do not evaporate, so the rate of solvent escape falls while the rate of return (condensation) is essentially unchanged for the diminished solvent vapour — a new, lower equilibrium vapour pressure is reached.

Raoult's law for a non-volatile solute. For a solution of a non-volatile solute in a volatile solvent, the vapour pressure of the solvent above the solution is directly proportional to its mole fraction in the liquid:

p_1 = x_1\, p_1^0

where: $p_1$ is the vapour pressure of the solvent above the solution, $p_1^0$ is the vapour pressure of the pure solvent at the same temperature, and $x_1$ is the mole fraction of the solvent. Because the solute is non-volatile it adds nothing of its own to the vapour, so $p_1$ is also the total vapour pressure of the solution.

Relative lowering of vapour pressure. Since mole fractions sum to one, $x_1 = 1 - x_2$ , where $x_2$ is the solute's mole fraction, so a short rearrangement of Raoult's law gives the fraction by which vapour pressure has fallen:

\dfrac{p_1^0 - p_1}{p_1^0} = x_2

This quantity is a colligative property — it depends only on how many solute particles are present relative to solvent particles, never on what the solute chemically is. That is why it can be used later to find an unknown solute's molar mass from a vapour-pressure measurement.

Two volatile liquids together. When both components of a solution are volatile — say two miscible liquids A and B — each has its own vapour pressure and each obeys Raoult's law independently, contributing its own partial pressure to the total:

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

Every point on a plot of vapour pressure against composition for such an ideal mixture lies on a straight line between the two pure vapour pressures, because each partial pressure rises linearly with that component's own mole fraction.

Figure

Two bars comparing vapour pressure of pure water and a glucose solution, the solution bar shorter to show Raoult's law lowering

Derivation ADeriving the relative lowering of vapour pressure from Raoult's law

The goal is to show, in one short algebraic step from Raoult's law for a non-volatile solute, exactly why the fraction $(p_1^0 - p_1)/p_1^0$ equals the solute's mole fraction $x_2$ — the relation that turns a measured vapour-pressure drop directly into a colligative property usable for molar-mass determination.

Step 1. Start from Raoult's law for the solvent: $p_1 = x_1 p_1^0$ . This is the statement that solvent vapour pressure scales with the fraction of the liquid surface that is still solvent.

Step 2. Subtract both sides from $p_1^0$ : $p_1^0 - p_1 = p_1^0 - x_1 p_1^0 = p_1^0(1 - x_1)$ . This isolates the actual drop in vapour pressure caused by adding the solute.

Step 3. Use $x_1 + x_2 = 1$ (only two components, solvent and solute) to write $1 - x_1 = x_2$ , so $p_1^0 - p_1 = p_1^0\, x_2$ .

Step 4. Divide both sides by $p_1^0$ to get the relative (fractional) lowering: $\dfrac{p_1^0 - p_1}{p_1^0} = x_2$ . This confirms the drop, expressed as a fraction of the pure solvent's vapour pressure, equals the solute's mole fraction exactly — no approximation was used, only the two-component mole-fraction identity.

Worked example

Problem: The vapour pressure of pure water at 298 K is $23.8$ mm Hg. Calculate the vapour pressure of a solution containing 18 g of glucose ( $M = 180\ \text{g mol}^{-1}$ ) dissolved in 90 g of water ( $M = 18\ \text{g mol}^{-1}$ ), and verify the answer using the relative-lowering relation.

Step 1. Find moles of each component: $n_{glucose} = 18/180 = 0.10\ \text{mol}$ and $n_{water} = 90/18 = 5.0\ \text{mol}$ .

Step 2. Find the mole fraction of water (the solvent): $x_1 = \dfrac{5.0}{5.0 + 0.10} = \dfrac{5.0}{5.10} = 0.9804$ .

Step 3. Apply Raoult's law directly: $p_1 = x_1\, p_1^0 = 0.9804 \times 23.8 = 23.34\ \text{mm Hg}$ .

Step 4. Cross-check with relative lowering: $x_2 = 0.10/5.10 = 0.0196$ , so the drop is $p_1^0 - p_1 = 0.0196 \times 23.8 = 0.466\ \text{mm Hg}$ , giving $p_1 = 23.8 - 0.466 = 23.33\ \text{mm Hg}$ . Both routes agree up to rounding. The vapour pressure of the glucose solution is 23.34 mm Hg, only slightly below the pure solvent's 23.8 mm Hg because the solute is dilute.

Competitive stretch

For a solution of two volatile liquids with $p_A^0 = 200$ mm Hg, $p_B^0 = 400$ mm Hg and $x_A = 0.6$ : $x_B = 1 - 0.6 = 0.4$ , so $p_A = 0.6 \times 200 = 120$ mm Hg, $p_B = 0.4 \times 400 = 160$ mm Hg, and $p_{total} = 120 + 160 = 280$ mm Hg. This is the two-volatile-liquid extension of the same law used above.

Common traps

Using the solute's mole fraction instead of the solvent's mole fraction in $p_1 = x_1 p_1^0$ — the formula always uses the mole fraction of the component whose vapour pressure is being calculated.
Forgetting that a non-volatile solute contributes nothing to the vapour phase; do not add a "solute vapour pressure" term for glucose, sucrose, or similar solutes.
Reporting the relative lowering $(p_1^0-p_1)/p_1^0 = x_2$ itself as if it were the vapour pressure — it is a dimensionless fraction (or the drop when multiplied by $p_1^0$ ), not $p_1$ .
Applying the single-solute formula $p_1 = x_1 p_1^0$ to a two-volatile-liquid mixture, where instead both components contribute and the total is the sum $p_A + p_B$ .

Ask the tutor: Want to see how the relative lowering of vapour pressure is used to calculate an unknown solute's molar mass from experimental data — ask the AI tutor to walk through that extension.

Why does dissolving a non-volatile solute lower the vapour pressure at all — nothing is being heated or cooled?

Vapour pressure comes from solvent molecules at the liquid surface escaping into the gas phase; equilibrium is reached when escape and return balance. When a non-volatile solute is dissolved, some of that surface is occupied by solute particles that never evaporate, so fewer solvent molecules leave per second while the return rate for the reduced vapour stays roughly matched — the system settles at a new, lower equilibrium vapour pressure. No temperature change is involved; it is a surface-occupancy effect tied purely to the mole fraction of solute present.

Is the relative lowering of vapour pressure the same thing as the vapour pressure itself?

No, and mixing the two up is the most common exam slip on this topic. The relative lowering, $(p_1^0-p_1)/p_1^0 = x_2$ , is a fraction (a pure number between 0 and 1) telling you what proportion of the pure vapour pressure has been lost. To get the actual solution vapour pressure $p_1$ , you must either use $p_1 = x_1 p_1^0$ directly, or multiply the fraction by $p_1^0$ to get the drop and subtract that from $p_1^0$ . Reporting $x_2$ itself as "the vapour pressure" throws away the units and the physical meaning.

How exactly is Raoult's law for a solvent a special case of Henry's law?

Henry's law states $p = K_H x$ for a gas dissolved in a liquid, where $K_H$ is a substance-specific constant. Raoult's law for the solvent states $p_1 = p_1^0 x_1$ , which has the identical mathematical form with $K_H$ replaced by $p_1^0$ , the solvent's own pure vapour pressure. So Raoult's law is exactly what Henry's law becomes when the "Henry's law constant" for a component happens to equal that component's pure vapour pressure — which is the case for the major component (the solvent) of a solution, especially a dilute one where the solvent behaves almost as if it were still pure.

Vapour Pressure of Liquid Solutions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads