Raoult's Law Calculator (Vapour Pressure of a Mixture)

How to use the Raoult’s law calculator

Enter the mole fraction of component A in the liquid and the vapour pressures of the pure components at the mixture temperature. The calculator returns each partial pressure, the total vapour pressure and the composition of the vapour. For a solvent with a non-volatile solute, set P°_B to 0.

Raoult’s law formula

P_A = x_A P°_A  ·  P_B = (1 − x_A) P°_B  ·  P = P_A + P_B  ·  y_A = P_A ÷ P

Example

A benzene (A) – toluene (B) liquid with x_A = 0.4 at 80 °C, where P°_benzene = 101 kPa and P°_toluene = 38.8 kPa:

• P_A = 0.4 × 101 = 40.40 kPa
• P_B = 0.6 × 38.8 = 23.28 kPa
• P = 40.40 + 23.28 = 63.68 kPa
• y_A = 40.40 ÷ 63.68 = 0.6344 — the vapour is 63% benzene although the liquid is only 40%.

Benzene–toluene at 80 °C

x benzene (liquid)	P benzene	P toluene	Total P	y benzene (vapour)
0.0	0.0 kPa	38.8 kPa	38.8 kPa	0.000
0.1	10.1 kPa	34.9 kPa	45.0 kPa	0.224
0.2	20.2 kPa	31.0 kPa	51.2 kPa	0.394
0.3	30.3 kPa	27.2 kPa	57.5 kPa	0.527
0.4	40.4 kPa	23.3 kPa	63.7 kPa	0.634
0.5	50.5 kPa	19.4 kPa	69.9 kPa	0.722
0.6	60.6 kPa	15.5 kPa	76.1 kPa	0.796
0.7	70.7 kPa	11.6 kPa	82.3 kPa	0.859
0.8	80.8 kPa	7.8 kPa	88.6 kPa	0.912
0.9	90.9 kPa	3.9 kPa	94.8 kPa	0.959
1.0	101.0 kPa	0.0 kPa	101.0 kPa	1.000

Pure-component vapour pressures (Antoine equation)

Temperature	Benzene	Toluene
20 °C	10.0 kPa	2.9 kPa
40 °C	24.4 kPa	7.9 kPa
60 °C	52.2 kPa	18.5 kPa
80 °C	101.0 kPa	38.8 kPa
100 °C	180.1 kPa	74.2 kPa

Calculated with the Antoine equation (log₁₀P[mmHg] = A − B ÷ (C + T[°C])), benzene A = 6.90565, B = 1211.033, C = 220.79; toluene A = 6.95464, B = 1344.8, C = 219.482.

Notes

• Benzene–toluene is close to ideal; most real mixtures deviate somewhat.
• In the vapour, the partial pressures add up by Dalton’s law.

Frequently asked questions

What is Raoult's law?

The partial vapour pressure of a component in an ideal solution equals its mole fraction in the liquid times the vapour pressure of the pure component — Pᵢ = xᵢ P°ᵢ. The total pressure is the sum of the partial pressures.

How does Raoult's law explain vapour pressure lowering?

With a non-volatile solute (P°_B = 0), P = x_solvent × P°_solvent. Dissolving a solute lowers the solvent's mole fraction, so the vapour pressure drops — which is why solutions boil at a higher temperature.

Why is the vapour richer in the more volatile component?

The component with the higher pure vapour pressure contributes more partial pressure per mole, so y_A = x_A P°_A ÷ P is larger than x_A. Distillation exploits this difference.

When does Raoult's law not apply?

For non-ideal mixtures with strong attraction or repulsion between unlike molecules (ethanol–water, acetone–chloroform). These show positive or negative deviations and can form azeotropes. Dilute dissolved gases follow Henry's law instead.

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