Raoult's Law

Open a bottle of wine and pour a glass. The alcohol and water inside are constantly evaporating and condensing, and the vapour pressure above that liquid is lower than pure water's and lower than pure ethanol's alone. Raoult's law explains why, using nothing more than mole fractions and pure-component vapour pressures. This law is the foundation for understanding how solutions behave. It underlies boiling point elevation, freezing point depression, osmotic pressure, and distillation. Every one of those topics starts from the same simple proportionality. This article builds Raoult's law from the equality of chemical potentials at equilibrium. You will see the equation stated precisely. Then you will see it derived from first principles. Two worked numerical examples follow, along with common pitfalls and a look at the chemist who discovered it experimentally, decades before anyone could explain it thermodynamically.

Key takeaways

Raoult's law: the partial vapour pressure of a solvent equals its mole fraction times the pure solvent's vapour pressure.
It holds exactly only for ideal solutions, where solute and solvent molecules interact almost identically.
Real solutions obey it closely only when the solute is dilute or chemically similar to the solvent.
It follows directly from equating the chemical potential of a liquid to its vapour at equilibrium.
Deviations (positive or negative) signal that solute–solvent interactions differ from solvent–solvent interactions.
Key idea: at equilibrium, lowering a solvent's mole fraction lowers its escaping tendency, and hence its vapour pressure, in direct proportion.

The core insight

Raoult's law is a direct consequence of chemical equilibrium: because the chemical potential of a solvent in an ideal mixture decreases logarithmically with its mole fraction, the vapour pressure it can sustain in equilibrium must decrease by the same factor, $x$ , giving $p_A = x_A p_A^*$ .

The physical picture

Picture a liquid solvent A in a closed container, with its vapour above it. Molecules of A constantly leave the liquid surface and constantly return from the vapour. At equilibrium, the two rates match, and the vapour pressure above pure A settles at some value $p_A^*$ , the pure-component vapour pressure. Now dissolve a solute B into A. Some of the surface that used to hold A molecules now holds B molecules instead. Roughly speaking, if B occupies a fraction $x_B$ of the surface (equal to its mole fraction, if the two molecules are similar in size and behavior), then only a fraction $x_A = 1 - x_B$ of the surface can now offer up A molecules to the vapour. Fewer escaping A molecules per unit time means a lower equilibrium vapour pressure of A, and the effect is directly proportional to $x_A$ . This is a statistical dilution argument, but it becomes rigorous once cast in terms of chemical potential. In a mixture, the solvent's chemical potential is lower than in the pure liquid, because mixing increases the entropy accessible to the system. The vapour above the mixture must match that lowered chemical potential, and matching it point-for-point forces the vapour pressure of A to scale linearly with $x_A$ .

Figure

Partial vapour pressure vs. mole fraction for an ideal binary mixture of A and B

Figure

Molecular picture: solute molecules block part of the surface, reducing solvent escape

The core relations

p_A = x_A \, p_A^{*}

where: $p_A$ is the partial vapour pressure of solvent A above the solution, $x_A$ is its mole fraction in the liquid, and $p_A^{*}$ is the vapour pressure of pure A at the same temperature.

p_{\text{total}} = x_A p_A^{*} + x_B p_B^{*}

where: for an ideal binary liquid mixture, the total vapour pressure is the sum of both partial pressures, each obeying Raoult's law.

\mu_A(\text{soln}) = \mu_A^{*} + RT \ln x_A

where: $\mu_A^{*}$ is the chemical potential of pure liquid A, $R$ is the gas constant, and $T$ is the absolute temperature — this is the thermodynamic statement that Raoult's law encodes.

Derivation AFrom equal chemical potentials to Raoult's law

Goal. Show that if the solvent's chemical potential in an ideal liquid mixture is $\mu_A^{*} + RT\ln x_A$ , then equilibrium with its vapour requires $p_A = x_A p_A^{*}$ .

Step 1. Write the equilibrium condition. At a given temperature $T$ , liquid A in the mixture is in equilibrium with A vapour above it. Equilibrium means the chemical potential of A is the same in both phases:

\mu_A(\text{liquid, mixture}) = \mu_A(\text{vapour})

Step 2. Express the vapour-phase chemical potential. Treating the vapour as a perfect gas at partial pressure $p_A$ ,

\mu_A(\text{vapour}) = \mu_A^{\ominus}(g) + RT \ln\!\left(\frac{p_A}{p^{\ominus}}\right)

where $\mu_A^{\ominus}(g)$ is the standard chemical potential of gaseous A at reference pressure $p^{\ominus}$ .

Step 3. Express the liquid-phase chemical potential in the ideal mixture. From the definition of an ideal solution (built on the entropy of mixing of indistinguishable-sized molecules with no extra enthalpy of mixing),

\mu_A(\text{liquid, mixture}) = \mu_A^{*}(l) + RT \ln x_A

Here $\mu_A^{*}(l)$ is the chemical potential of pure liquid A at the same $T$ and total pressure.

Step 4. Apply the same equilibrium condition to pure A. For pure liquid A in equilibrium with its own vapour at pressure $p_A^{*}$ ,

\mu_A^{*}(l) = \mu_A^{\ominus}(g) + RT \ln\!\left(\frac{p_A^{*}}{p^{\ominus}}\right)

Step 5. Substitute Step 4 into Step 3, then equate to Step 2.

\mu_A^{\ominus}(g) + RT \ln\!\left(\frac{p_A^{*}}{p^{\ominus}}\right) + RT \ln x_A = \mu_A^{\ominus}(g) + RT \ln\!\left(\frac{p_A}{p^{\ominus}}\right)

Step 6. Cancel $\mu_A^{\ominus}(g)$ and combine logarithms.

RT \ln\!\left(\frac{p_A^{*} \, x_A}{p^{\ominus}}\right) = RT \ln\!\left(\frac{p_A}{p^{\ominus}}\right)

Step 7. Take the exponential of both sides. The arguments of equal logarithms must be equal, so

p_A = x_A \, p_A^{*}

This is Raoult's law. It holds exactly whenever the mixture is genuinely ideal — that is, whenever the entropy of mixing is the only thermodynamic consequence of mixing, with no extra enthalpy or volume change. $\blacksquare$

Worked examples

Example 1 — Foundational: vapour pressure of a simple two-component ideal mixture. Liquid benzene (A) and toluene (B) form a close-to-ideal mixture — they are structurally similar aromatics with comparable intermolecular forces. At 25 °C, pure benzene has vapour pressure $p_A^{*} = 12.7$ kPa and pure toluene has $p_B^{*} = 3.8$ kPa. Suppose you prepare a mixture with 0.60 mol benzene and 0.40 mol toluene. Step 1: mole fractions.

x_A = \frac{0.60}{0.60+0.40} = 0.60, \qquad x_B = 0.40

Step 2: apply Raoult's law to each component.

p_A = x_A p_A^{*} = (0.60)(12.7 \text{ kPa}) = 7.62 \text{ kPa}

p_B = x_B p_B^{*} = (0.40)(3.8 \text{ kPa}) = 1.52 \text{ kPa}

Step 3: total vapour pressure.

p_{\text{total}} = 7.62 + 1.52 = 9.14 \text{ kPa}

Sanity check: this total lies between the two pure-component vapour pressures (3.8 kPa and 12.7 kPa), as it must for any ideal mixture — a weighted average can never exceed the largest weight-one value or fall below the smallest. Step 4 (bonus): vapour-phase composition. By Dalton's law, the mole fraction of benzene in the vapour is

y_A = \frac{p_A}{p_{\text{total}}} = \frac{7.62}{9.14} = 0.834

Notice the vapour is richer in benzene (0.834) than the liquid (0.60) — because benzene is more volatile. This enrichment is exactly the principle exploited in fractional distillation.

Example 2 — Applied: vapour pressure lowering by a nonvolatile solute, and back-calculating a molar mass. A chemist dissolves 5.00 g of an unknown nonvolatile, non-dissociating organic solid in 100.0 g of water at 25 °C. Pure water's vapour pressure at 25 °C is $p_{\text{water}}^{*} = 3.169$ kPa. The measured vapour pressure of the solution is 3.116 kPa. Find the molar mass of the solute. Step 1: because the solute is nonvolatile, only water contributes to the vapour, and Raoult's law gives

p_{\text{water}} = x_{\text{water}} \, p_{\text{water}}^{*}

Solve for the mole fraction of water:

x_{\text{water}} = \frac{p_{\text{water}}}{p_{\text{water}}^{*}} = \frac{3.116}{3.169} = 0.9833

Step 2: convert to moles of water.

n_{\text{water}} = \frac{100.0 \text{ g}}{18.02 \text{ g mol}^{-1}} = 5.549 \text{ mol}

Step 3: use the mole fraction to find moles of solute. Since

x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{solute}}}

rearrange:

n_{\text{water}} + n_{\text{solute}} = \frac{n_{\text{water}}}{x_{\text{water}}} = \frac{5.549}{0.9833} = 5.643 \text{ mol}

n_{\text{solute}} = 5.643 - 5.549 = 0.094 \text{ mol}

Step 4: molar mass.

M = \frac{5.00 \text{ g}}{0.094 \text{ mol}} = 53 \text{ g mol}^{-1}

Sanity check: this is a physically reasonable molar mass for a small organic molecule (comparable to, say, acrylonitrile at 53.06 g/mol). The mole fraction of solute (0.017) is small, consistent with the dilute-solution assumption implicit in treating water as behaving ideally near $x_A = 1$ .

Limiting cases & common traps

Situation	What happens
$x_A \to 1$ (dilute solute)	Raoult's law becomes exact for the solvent; this is the origin of colligative properties
$x_A \to 0$ (solvent is the minor component)	The solvent instead typically obeys Henry's law, $p_A = x_A K_A$ , with $K_A \neq p_A^{*}$
Similar A and B (e.g., benzene/toluene, hexane/heptane)	Nearly ideal across the whole composition range
Strong A–B attraction (e.g., acetone/chloroform, H-bonding)	Negative deviation: $p_{\text{total}}$ below the ideal line
Weak A–B attraction relative to A–A, B–B (e.g., ethanol/hexane)	Positive deviation: $p_{\text{total}}$ above the ideal line, can produce an azeotrope
Nonvolatile solute	Only the solvent contributes vapour pressure; solute mole fraction directly measures vapour-pressure lowering

The chemist behind it

François-Marie Raoult was a French chemist working in Grenoble in the 1880s, decades before Gibbs's chemical potential framework was widely applied to solutions by chemists. Raoult discovered his proportionality purely empirically — by measuring freezing-point depressions and vapour pressures of many solutions and noticing the same simple pattern recur across wildly different solutes and solvents. Only later, once thermodynamicists connected his empirical rule to the chemical potential of an ideal mixture, did the law acquire the rigorous derivation shown above. This is a common story in physical chemistry: the phenomenology arrives first, the mechanism follows once the right theoretical tools exist.

Origins · 1 minFrançois-Marie Raoult and how this was discovered

François-Marie Raoult

1830-1901 · French chemist who established the vapour-pressure law and colligative-property relations from careful cryoscopic and vapour-pressure measurements

Ask the tutor: How does Raoult's law extend to predict boiling point elevation and freezing point depression quantitatively?

Does Raoult's law apply to solids or only to liquids in solution

Raoult's law is usually applied to a volatile liquid component in a liquid mixture, but the same reasoning extends to any condensed phase in equilibrium with its vapour, including a solid solvent below its melting point (relevant to freezing-point depression). The essential requirement is that the component obeying the law behaves ideally, meaning it experiences a similar molecular environment whether surrounded by its own kind or by the other component. This is most reliably true for the major, or solvent, component in a dilute solution.

Why do real solutions deviate from Raoult's law

Deviations arise because real molecules interact differently with each other than the ideal-mixture assumption requires. If A–B attractions are stronger than the average of A–A and B–B attractions, molecules resist escaping into the vapour, and pressure falls below the ideal line (negative deviation). If A–B attractions are weaker, molecules escape more easily, giving positive deviation. Chloroform and acetone hydrogen-bond strongly together and show negative deviation; ethanol and hexane repel each other's escaping tendency and show positive deviation, sometimes forming an azeotrope.

How is Raoult's law related to Henry's law

Both describe partial vapour pressure as proportional to mole fraction, but they apply in opposite composition limits. Raoult's law, $p = xp^{*}$ , describes a component near $x=1$ , referenced to its own pure vapour pressure. Henry's law, $p = xK$ , describes a dilute solute near $x=0$ , referenced to an empirical constant $K$ that reflects its unusual environment surrounded almost entirely by solvent. A component can obey Raoult's law as solvent and Henry's law as solute in the very same mixture.

Can Raoult's law be used to find the molar mass of an unknown solute

Yes, this is one of its classic applications, shown in the second worked example above. By measuring how much a nonvolatile solute lowers a solvent's vapour pressure, you can extract the solute's mole fraction directly from Raoult's law. Combined with the known mass of solvent and solute, that mole fraction gives the number of moles of solute, and hence its molar mass. This method works best for dilute solutions, where deviations from ideality are smallest.

Why is Raoult's law considered the starting point for colligative properties

Boiling point elevation, freezing point depression, and osmotic pressure all trace back to the same cause: a nonvolatile solute lowers the solvent's chemical potential, exactly as Raoult's law describes for its vapour pressure. Each colligative property is a different experimental consequence of that same lowered chemical potential, so all of them depend on the solute's mole fraction (or molality) rather than its chemical identity. Raoult's law is the cleanest, most direct expression of that underlying principle.