Raoult's Law Calculator — Vapour Pressure of a Solution

Background.

This Raoult's law calculator works out the vapour pressure above a liquid solution: p_A = x_A × p*_A, where x_A is the liquid-phase mole fraction of a component and p*_A is the vapour pressure of that component when pure, at the same temperature. It handles three genuinely different situations — a non-volatile solute lowering a solvent's vapour pressure, the same case with the composition entered as mole amounts, and a binary mixture of two volatile liquids where the vapour composition differs from the liquid composition.

The intuition is a surface-area argument. Molecules escape a liquid from its surface. If a fraction 1 − x_A of the surface sites are occupied by something else, only the fraction x_A of the escape attempts are attempts by A, so A's escape rate — and therefore its equilibrium vapour pressure — falls in the same proportion. The consequences are everywhere: dissolving anything in water lowers water's vapour pressure, which raises its boiling point and depresses its freezing point, which is why salt melts ice, why antifreeze works, and why a sugar syrup boils above 100 °C.

Raoult's law is isothermal and the page has no temperature field, which is a convention you must respect rather than an omission. Every pressure in the equation — the pure-component values you enter and the answer that comes out — belongs to one single temperature, and it is your job to supply pure-component vapour pressures for the temperature you care about. Mixing a p* for one component at 25 °C with a p* for the other at 80 °C produces a number with no physical meaning. The NIST Chemistry WebBook publishes Antoine parameters for most common compounds, in the form log₁₀(p/bar) = A − B/(T + C) with T in kelvin, and that is the standard way to generate the two numbers this page needs.

**The single most important caveat, and it belongs here rather than in an FAQ: this is an ideal-solution model, and real solutions are frequently not ideal.** Raoult's law is exact only in the limit x_A → 1, where a molecule of A is surrounded almost entirely by other A molecules. It stays accurate across the whole composition range only when the two components are chemically alike — benzene and toluene, hexane and heptane, two adjacent members of a homologous series. When A–B attractions are weaker than the A–A and B–B attractions, molecules escape more easily than predicted and the real vapour pressure sits above the Raoult line (a positive deviation, as in ethanol–water); when A–B attractions are stronger, it sits below (a negative deviation, as in acetone–chloroform). Large deviations produce azeotropes, mixtures that boil without changing composition and cannot be separated by simple distillation — the reason ethanol from a fermentation column stops at about 95.6 percent by mass. Formally the exact statement is p_A = x_A f_A p*_A with an activity coefficient f_A that Raoult's law sets to 1; IUPAC defines that coefficient as the one 'referenced to Raoult's law' precisely because it measures the departure from this page's assumption. Treat the numbers here as a first estimate, and check them against measured vapour–liquid equilibrium data before designing anything.

Three more conventions. Composition is a **liquid-phase mole fraction**, dimensionless and in the range 0 < x ≤ 1, not a percentage, a mass fraction or a molarity. Pressures are absolute, and the units only have to be consistent — enter both pure-component values in Torr and the answer comes out in Torr. Vapour-pressure lowering is reported as a **positive number**: it is defined as p*_A − p_A, and because x_A can never exceed 1, it can never be negative. Finally, the solver carries every step at 30-digit precision and rounds once, at the end, to ten decimal places; round your own answer to the precision of your pure-component vapour pressures, which for tabulated Antoine fits is rarely better than three significant figures.

What is raoult's law calculator (vapour pressure of solutions)?

Raoult's law states that the partial pressure of a component above an ideal liquid solution equals its liquid-phase mole fraction multiplied by the vapour pressure of that component in its pure state at the same temperature: p_A = x_A p*_A. It was established by François-Marie Raoult in the 1880s and it is a limiting law — exact in the limit x_A → 1, approximate elsewhere. The IUPAC Compendium of Chemical Terminology gives the rigorous form under its combined entry for Henry's and Raoult's laws: 'For the solvent (A) the relationship is called Raoult's law, and the proportionality factor is the fugacity of the pure solvent: p_A = p̃*_A a_A', where a_A is the activity of the solvent. Setting the activity equal to the mole fraction, a_A = x_A, is exactly the ideal-solution assumption this calculator makes; the IUPAC Green Book names the correction factor for that assumption the 'activity coefficient referenced to Raoult's law', f_B = a_B / x_B, which is 1 for an ideal solution. Two results follow. For a non-volatile solute, the solute contributes nothing to the vapour, the total vapour pressure above the solution is x_A p*_A, and the relative lowering Δp/p*_A equals 1 − x_A — the solute's mole fraction, independent of the solute's chemical identity. That independence makes vapour-pressure lowering a colligative property, in the same family as boiling-point elevation, freezing-point depression and osmotic pressure. For a mixture of two volatile components, both obey their own Raoult term and the total pressure is x_A p*_A + x_B p*_B, a straight line between the two pure-component values. The vapour above such a mixture is not the same composition as the liquid: by Dalton's law y_A = x_A p*_A / (x_A p*_A + x_B p*_B), which is larger than x_A whenever A is the more volatile component. Repeating that enrichment step is what a distillation column does. Raoult's law is the solvent-side counterpart of Henry's law, which governs a dilute solute in the opposite limit x → 0; the two are different limiting laws for the same physical system and neither is a special case of the other.

How to use this calculator.

1. 01Pick the situation. 'Two volatile components' if both liquids evaporate; either non-volatile option if the solute is a sugar, a salt, a polymer or anything else with negligible vapour pressure.
2. 02Decide the temperature you are working at, and look up the pure-component saturation vapour pressures AT THAT TEMPERATURE. The NIST Chemistry WebBook gives Antoine parameters in the form log₁₀(p/bar) = A − B/(T + C), T in kelvin. Both values must be for the same temperature.
3. 03Enter the liquid composition as a mole fraction between 0 and 1 — 50 percent is 0.5. Or choose the 'mole amounts' mode and enter moles directly.
4. 04For an ionic solute, count particles rather than formula units: 1 mol of NaCl gives close to 2 mol of ions in dilute solution, and 1 mol of CaCl₂ close to 3. Colligative properties count dissolved particles.
5. 05Read the total vapour pressure, then the two partial pressures and the lowering. The relative lowering should come out equal to 1 − x_A; if it does not, an input is wrong.
6. 06In the binary mode, compare the vapour mole fraction y_A with the liquid mole fraction x_A. The gap between them is the separation a single equilibrium stage of distillation would achieve.
7. 07Before trusting the number for anything real, check whether your two components are chemically similar. If they hydrogen-bond to each other, or if one is polar and the other is not, look for measured vapour–liquid equilibrium data instead.

The formula.

For each volatile component, Raoult's law gives p_i = x_i p*_i. Summing over the components gives the total vapour pressure, and dividing one partial pressure by that total gives the vapour composition through Dalton's law:

p_A = x_A p*_A p_B = x_B p*_B = (1 − x_A) p*_B (zero for a non-volatile solute) P = p_A + p_B y_A = p_A / P Δp = p*_A − p_A = (1 − x_A) p*_A (the lowering, always ≥ 0) Δp / p*_A = 1 − x_A (the relative lowering)

The last identity is worth pausing on, because it is the classic colligative statement: the fractional amount by which the solvent's vapour pressure drops equals the mole fraction of solute, and it does not matter at all what the solute is. Sucrose, urea and a polymer at the same mole fraction lower water's vapour pressure by exactly the same fraction. What does matter is how many particles the solute produces: sodium chloride dissociates into two ions, so one mole of NaCl behaves close to two moles of solute, which is why de-icing salt is effective at modest mass loadings.

Rounding stage: FINAL ONLY. Every multiplication, subtraction and division above is carried at 30-digit Decimal.js working precision and a single rounding to ten decimal places is applied at the return boundary. Nothing is rounded in between, which is why the cross-mode identities hold to the tenth decimal place: entering 10 mol solvent and 1 mol solute in the mole-amount mode and entering x_A = 0.909090909… in the fraction mode give the same vapour pressure.

Directional behaviour, read off the equations rather than from memory. In the worked example below, benzene (p* = 12.677 kPa) is about 3.3 times as volatile as toluene (p* = 3.7915 kPa) at 25 °C. An equimolar liquid therefore gives partial pressures of 6.33850 kPa and 1.89575 kPa, a total of 8.23425 kPa, and a vapour that is 0.7698 benzene by mole — much richer in benzene than the 0.5 liquid. Increasing x_A always increases both p_A and y_A; y_A equals x_A only when the two pure-component vapour pressures are equal, in which case no separation is possible at all. This is the correct direction: the vapour is enriched in the MORE volatile component, the one with the HIGHER pure vapour pressure and therefore the LOWER boiling point.

Invalid domain. The mole fraction must satisfy 0 < x_A ≤ 1: exactly 1 is the pure component and is allowed, exactly 0 is rejected because there would be no A left to evaporate and the vapour composition y_A would be 0/0. Values above 1 are rejected, which catches a percentage entered by mistake. Pure-component vapour pressures must be strictly positive; a zero would make the relative lowering 0/0. In the mole-amount mode, solvent moles must be positive and solute moles must be zero or positive — zero solute is allowed and returns the pure solvent. Every rejection names the field rather than returning a NaN.

A worked example.

Frequently asked questions.

References& sources.

1. [1]International Union of Pure and Applied Chemistry. Compendium of Chemical Terminology (the 'Gold Book'), 5th edition, online version 5.0.0 (2025), entry 'Henry's law' (H02783), which also defines Raoult's law: 'For the solvent (A) the relationship is called Raoult's law, and the proportionality factor is the fugacity of the pure solvent, p̃*_A: p_A = p̃*_A a_A.' Source document: Pure Appl. Chem. 1984, 56, 567 (Physicochemical quantities and units in clinical chemistry, Recommendations 1983), p. 571. Independent standards body; open access; retrieved and quoted 2026-07-29. The alias R05135 'Raoult's law' resolves to this entry.
2. [2]International Union of Pure and Applied Chemistry. Quantities, Units and Symbols in Physical Chemistry (the 'Green Book'), 3rd edition, 2nd printing 2012, p. 59: 'activity coefficient referenced to Raoult's law f, f_B = a_B / x_B, SI unit 1', with the note that the quantity 'applies to pure phases, substances in mixtures, or solvents'. This is the formal definition of the departure from the ideal-solution assumption this calculator makes (f = 1). Consulted as a SECOND authority independent of the Gold Book entry above; agrees. Open access PDF, text extracted and read locally; retrieved 2026-07-29.
3. [3]National Institute of Standards and Technology. NIST Chemistry WebBook, SRD 69 — benzene (CAS 71-43-2), Antoine equation parameters, log₁₀(p/bar) = A − B/(T + C) with T in K. Willingham, Taylor, Pignocco & Rossini (1945): A = 4.01814, B = 1203.835, C = −53.226, valid 287.70–354.07 K, giving 12.677 kPa (95.08 Torr) at 298.15 K — the p*_A used in the worked example. Note the page also lists an independent short-range fit (Deshpande & Pandya 1967, 297.9–318.0 K) that gives 12.16 kPa at the same temperature, 4.1 % lower; see the dossier for why the Willingham fit was chosen. Open access; retrieved 2026-07-29.
4. [4]National Institute of Standards and Technology. NIST Chemistry WebBook, SRD 69 — toluene (CAS 108-88-3), Antoine equation parameters. Pitzer & Scott (1943): A = 4.14157, B = 1377.578, C = −50.507, valid 273–323 K, giving 3.7915 kPa (28.44 Torr) at 298.15 K — the p*_B used in the worked example. Cross-checked against the independent fit of Besley & Bottomley (1974), A = 4.23679, B = 1426.448, C = −45.957, valid 273.13–297.89 K, which gives 3.80727 kPa — agreement to 0.42 %. Open access; retrieved 2026-07-29.
5. [5]National Institute of Standards and Technology. NIST Chemistry WebBook, SRD 69 — Thermophysical Properties of Fluid Systems, water saturation table, computed from the IAPWS Formulation 1995 (Wagner & Pruss, J. Phys. Chem. Ref. Data 31, 387, 2002). Saturation pressure of water at 25.0000 °C = 3.169 93 kPa, the value quoted in the vapour-pressure-lowering FAQ and used as a reference point in the input hints. Open access; value retrieved directly from the data endpoint 2026-07-29.