Vapour-Pressure Lowering

Dissolve sugar in water and the liquid becomes harder to boil away. The solvent molecules escape into the vapour phase more slowly than they would from pure solvent. This single fact — that a non-volatile solute lowers the solvent's vapour pressure — is the seed from which every colligative property grows. Vapour-pressure lowering matters because it is the most direct experimental window into solution thermodynamics. It is measured with barometric simplicity, yet it encodes the chemical potential of the solvent, the entropy of mixing, and the mole fraction of whatever is dissolved. Boiling-point elevation, freezing-point depression, and osmotic pressure are all downstream consequences of the same equation. This article builds Raoult's law into a quantitative statement about vapour-pressure lowering. You will see the derivation from chemical potentials. Then you will see two fully worked numerical examples. Finally you will see where the simple law breaks down and what replaces it in real solutions.

Key takeaways

A non-volatile solute lowers the solvent's vapour pressure in proportion to its mole fraction.
The governing relation is Raoult's law: $p_A = x_A p_A^*$ , applied to the solvent only.
The lowering itself is $\Delta p = x_B p_A^*$ , where $x_B$ is the solute mole fraction.
The effect is colligative: it depends on how many solute particles are present, not their identity.
Real solutions deviate from ideality; activities and activity coefficients correct for this.
Key idea: Diluting the solvent with solute molecules lowers its escaping tendency, and hence its equilibrium vapour pressure, in direct proportion to the solute's mole fraction.

The core insight

Adding a non-volatile solute dilutes the solvent at the liquid surface and in the bulk, lowering the solvent's chemical potential and therefore its equilibrium vapour pressure by an amount proportional to the mole fraction of solute present, regardless of the solute's chemical nature.

The physical picture

Picture a closed container holding pure liquid solvent A in equilibrium with its vapour. Molecules leave the liquid surface and return at equal rates; the vapour pressure $p_A^*$ is the pressure at which these rates balance. Now dissolve a non-volatile solute B into the liquid. Solute particles occupy some of the surface and bulk positions that solvent molecules used to occupy. Fewer solvent molecules sit at the interface per unit area, so fewer escape per unit time into the vapour. The condensation rate from the vapour, however, is essentially unchanged at low vapour density. A new, lower equilibrium pressure $p_A$ is reached. Crucially, this is an entropy effect, not an energy effect, provided the solution is ideal. Mixing solvent with solute increases the entropy of the liquid phase relative to pure solvent, without changing the vapour phase. The chemical potential of the liquid solvent drops below that of the pure liquid by $RT\ln x_A$ , and vapour pressure must drop to match. Because the effect originates in a counting (entropy) argument, it applies to any non-volatile solute at a given mole fraction — one mole of sucrose lowers vapour pressure by the same fraction as one mole of urea, at the same total moles of solvent. This is what makes vapour-pressure lowering a colligative property, tying it directly to the general framework established for colligative properties.

Figure

Solvent escape from pure liquid versus solution: fewer surface solvent molecules lower the escaping tendency

The core relations

p_A = x_A p_A^{*}

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

\Delta p = p_A^{*} - p_A = x_B\, p_A^{*}

where: $\Delta p$ is the vapour-pressure lowering, $x_B = 1 - x_A$ the total mole fraction of all non-volatile solute species.

\frac{\Delta p}{p_A^{*}} = x_B

where: the relative lowering equals the solute mole fraction exactly, independent of solvent identity, for an ideal solution.

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

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

Derivation AFrom equal chemical potentials to Raoult's law

Goal: show that equilibrium between liquid solution and vapour forces $p_A = x_A p_A^{*}$ , and hence $\Delta p = x_B p_A^{*}$ .

Step 1. Equilibrium condition. At equilibrium the solvent's chemical potential must be equal in both phases,

\mu_A(\mathrm{l}) = \mu_A(\mathrm{g}).

This holds for the pure liquid too, giving a reference state to compare against.

Step 2. Chemical potential of the vapour. Treating solvent vapour as an ideal gas,

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

For the pure liquid in equilibrium with its own vapour at pressure $p_A^{*}$ ,

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

Step 3. Chemical potential of the ideal solution. Mixing solvent and solute ideally contributes an entropy-of-mixing term only,

\mu_A(\mathrm{l}) = \mu_A^{*}(\mathrm{l}) + RT\ln x_A,

valid because the solute is non-volatile (it does not appear in the vapour) and the solution is ideal (no enthalpy of mixing, no volume change).

Step 4. Combine the three chemical-potential expressions. Setting $\mu_A(\mathrm{l}) = \mu_A(\mathrm{g})$ and substituting Steps 2 and 3,

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

Subtract the pure-liquid equilibrium expression from Step 2 term by term,

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

Step 5. Simplify. Dividing by $RT$ and exponentiating both sides,

x_A = \frac{p_A}{p_A^{*}} \quad\Longrightarrow\quad p_A = x_A p_A^{*}.

This is Raoult's law for the solvent.

Step 6. Extract the lowering. Since $x_A = 1 - x_B$ ,

p_A = (1-x_B)p_A^{*} = p_A^{*} - x_B p_A^{*}.

Rearranging,

\boxed{\;p_A^{*} - p_A = \Delta p = x_B\, p_A^{*}\;}

The vapour-pressure lowering is directly proportional to solute mole fraction and to the pure solvent's vapour pressure — no other property of the solute enters, which is exactly the colligative signature.

Worked examples

Example 1 — Foundational: sucrose in water. Dissolve 68.4 g of sucrose ( $M = 342.3\ \mathrm{g\,mol^{-1}}$ ) in 500.0 g of water ( $M = 18.02\ \mathrm{g\,mol^{-1}}$ ) at 25 °C, where $p_{\mathrm{H_2O}}^{*} = 3.170\ \mathrm{kPa}$ . Find the vapour pressure of the solution and the lowering. Step 1 — moles.

n_{\text{sucrose}} = \frac{68.4\ \mathrm{g}}{342.3\ \mathrm{g\,mol^{-1}}} = 0.1998\ \mathrm{mol}

n_{\text{water}} = \frac{500.0\ \mathrm{g}}{18.02\ \mathrm{g\,mol^{-1}}} = 27.75\ \mathrm{mol}

Step 2 — mole fractions.

x_B = \frac{0.1998}{0.1998 + 27.75} = \frac{0.1998}{27.95} = 7.148\times10^{-3}

x_A = 1 - 7.148\times10^{-3} = 0.99285

Step 3 — vapour pressure of the solution.

p_A = x_A p_A^{*} = (0.99285)(3.170\ \mathrm{kPa}) = 3.1473\ \mathrm{kPa}

Step 4 — lowering.

\Delta p = x_B p_A^{*} = (7.148\times10^{-3})(3.170\ \mathrm{kPa}) = 0.02266\ \mathrm{kPa} = 22.7\ \mathrm{Pa}

Sanity check: $p_A + \Delta p = 3.1473 + 0.0227 = 3.170\ \mathrm{kPa}$ , matching $p_A^{*}$ exactly. A lowering of 22.7 Pa out of 3170 Pa (0.7%) is small, consistent with a dilute solution ( $x_B \ll 1$ ).

Example 2 — Applied, harder: back-calculating molar mass from a measured lowering, with a dissociating solute check. A chemist dissolves 4.50 g of an unknown non-volatile, non-electrolyte solute in 100.0 g of benzene at 25 °C. Pure benzene has $p^{*} = 12.7\ \mathrm{kPa}$ at this temperature ( $M_{\text{benzene}} = 78.11\ \mathrm{g\,mol^{-1}}$ ). The measured vapour pressure of the solution is 12.30 kPa. Determine the molar mass of the solute. Step 1 — find $\Delta p$ and $x_B$ .

\Delta p = 12.7 - 12.30 = 0.40\ \mathrm{kPa}

x_B = \frac{\Delta p}{p^{*}} = \frac{0.40}{12.7} = 0.03150

Step 2 — moles of benzene.

n_{\text{benzene}} = \frac{100.0\ \mathrm{g}}{78.11\ \mathrm{g\,mol^{-1}}} = 1.2802\ \mathrm{mol}

Step 3 — solve for moles of solute. Since $x_B = n_B/(n_B+n_A)$ and $x_B$ is small,

n_B = \frac{x_B\, n_A}{1-x_B} = \frac{(0.03150)(1.2802)}{1 - 0.03150} = \frac{0.04033}{0.9685} = 0.04164\ \mathrm{mol}

Step 4 — molar mass.

M_B = \frac{4.50\ \mathrm{g}}{0.04164\ \mathrm{mol}} = 108.1\ \mathrm{g\,mol^{-1}}

Sanity check and discussion: 108 g/mol is close to the molar mass of phenol (94.1 g/mol) or naphthalene (128.2 g/mol) — a plausible order of magnitude for a small organic non-electrolyte, confirming the calculation is physically sensible. Note the method's key assumption: the solute must not dissociate or associate in benzene. If the same mass of a salt were used instead, each formula unit could release two or three particles into solution, inflating $x_B$ for a given mass and yielding a falsely low apparent molar mass — this is why the van 't Hoff factor $i$ must be included for ionic solutes, with $\Delta p = i x_B p^{*}$ replacing the simple form.

Limiting cases & common traps

Situation	What happens	Why
Very dilute solution ( $x_B \to 0$ )	$\Delta p \approx x_B p_A^{*}$ holds to excellent accuracy	Raoult's law is exact in this limit (ideal-dilute solvent)
Concentrated / non-ideal solution	$\Delta p \neq x_B p_A^{*}$ ; must use activity $a_A = \gamma_A x_A$	Solvent–solute interactions deviate from ideal mixing
Volatile "solute"	Simple lowering formula fails; use Raoult's law for both components	The solute contributes its own partial pressure to total $p$
Ionic solute (electrolyte)	$\Delta p = i\, x_B\, p_A^{*}$ with $i > 1$	Each formula unit dissociates into multiple particles
Two immiscible solvents / azeotropes	Raoult's law breaks down qualitatively	Strong deviations from ideality dominate the vapour composition

The chemist behind it

François-Marie Raoult spent decades in Grenoble performing meticulous vapour-pressure and freezing-point measurements on dozens of solvent–solute combinations. Working with primitive manometry by today's standards, he established empirically, in the 1880s, that the relative lowering of vapour pressure equals the mole fraction of solute — a law announced before the modern chemical-potential formalism existed to explain it. His freezing-point data, gathered in parallel, became one of the first reliable methods for determining molecular weights in solution, directly anticipating the boiling-point elevation and freezing-point depression relations built on this same foundation.

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

François-Marie Raoult

1830–1901 · French chemist whose cryoscopic and vapour-pressure measurements established the law bearing his name

Ask the tutor: How does this same chemical-potential argument lead to the exact formula for boiling-point elevation?

Does this law apply to volatile solutes too?

Not directly, in the simple $\Delta p = x_B p_A^{*}$ form. That equation assumes the solute contributes zero vapour pressure of its own. For a volatile solute, both components obey Raoult's law separately, $p_A = x_A p_A^{*}$ and $p_B = x_B p_B^{*}$ , and the total pressure is their sum, $p = p_A + p_B$ . The "lowering" concept as a single colligative number only makes clean sense for genuinely non-volatile solutes.

How is vapour-pressure lowering related to boiling-point elevation?

They share the identical thermodynamic root: the solvent's chemical potential is reduced by $RT\ln x_A$ in solution. Because the liquid needs a higher temperature to raise its vapour pressure back up to atmospheric pressure, the solution boils above the pure solvent's boiling point. Vapour-pressure lowering is the cause; boiling-point elevation is the observable temperature-domain consequence, quantified by the ebullioscopic constant.

Why does the derivation require an ideal solution?

The step $\mu_A(\mathrm{l}) = \mu_A^{*}(\mathrm{l}) + RT\ln x_A$ assumes no enthalpy change on mixing and no excess entropy beyond the ideal combinatorial term. Real solutions with strong solvent–solute interactions (hydrogen bonding, dipole effects) deviate from this. The fix is to replace $x_A$ with the activity $a_A = \gamma_A x_A$ , where $\gamma_A$ is an activity coefficient measured experimentally or estimated from a model like Margules or van Laar.

Can vapour-pressure lowering be used to find molar mass, like in Example 2?

Yes, and historically it was one of the first such methods, alongside cryoscopy. It requires a precise manometric or isopiestic measurement of $\Delta p$ , knowledge of both pure component vapour pressures, and confirmation that the solute is non-volatile and non-dissociating. Modern molar-mass determination favors mass spectrometry or osmometry for precision, but the vapour-pressure method remains a clean pedagogical demonstration of colligative reasoning.