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Raoult's Law

P=x P0P = x \, P^{0}

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Worked example: x = 0.90, P0 = 3.17 kPa → P = 2.853 kPa. Press Try an example to run it live, then adjust anything.

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Clausius–Clapeyron Equation (Two-Point Form)Gas Density from Molar MassPartial Pressure from Mole FractionCompressibility Factor (Z = PV/nRT)
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See also

Antoine Equation (Vapour Pressure): Both use vapour pressure over solution.Clausius–Clapeyron Equation (Two-Point Form): Both use vapour pressure over solution.
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Raoult's Law explained

P0Px

Molecules can only evaporate from the liquid's surface, and dissolving a non-volatile solute means some of that surface is occupied by particles that cannot leave. François-Marie Raoult found in the 1880s that the effect is exactly proportional: the solvent's vapor pressure drops to its mole fraction times the pure-solvent value. Pure water at 25 °C exerts 3.17 kPa; a solution where water's mole fraction is 0.90 (say, 1 mol of glucose per 9 mol of water) exerts P = 0.90 × 3.17 = 2.85 kPa.

This vapor-pressure lowering is the root of all colligative properties: boiling-point elevation and freezing-point depression both follow from it. Real solutions obey Raoult's law best when dilute and when solute–solvent interactions resemble solvent–solvent ones; strong deviations from the straight line are the chemist's first diagnostic that a mixture is far from ideal, like the ethanol–water pair that refuses to distill past 95%.

Raoult's Law formula

P=x P0P = x \, P^{0}
Where
  • PP= Vapour pressure over solution (kPa)
  • xx= Mole fraction of solvent
  • P0P^{0}= Vapour pressure of pure solvent (kPa)
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Colligative properties

4 formulasBoiling-point elevation, freezing-point depression, osmotic pressure and Raoult's law — the four effects that count particles, not identity.

Revised 16 October 2026
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formula.expert. (2026, October 16). Raoult's Law Formula & Calculator. https://formula.expert/solvers/raoults-law

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