Raoult's Law

P = x \, P^{0}

Fluid

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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Raoult's Law explained

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 \, P^{0}

Where

$P$ = Vapor pressure over solution (kPa)
$x$ = Mole fraction of solvent
$P^{0}$ = Vapor pressure of pure solvent (kPa)

Missing one of these? Work it out first, then come back

Vapor pressure over solution — Ideal Gas Law, Gas Density from Molar Mass
Mole fraction of solvent — Mole Fraction, Partial Pressure from Mole Fraction
Vapor pressure of pure solvent — Ideal Gas Law, Gas Density from Molar Mass

Part of this formula set

Colligative properties

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