Dissolving a non-volatile solute in a liquid lowers its vapour pressure, because fewer solvent molecules sit at the surface ready to escape. Raoult’s law, found by François-Marie Raoult in the 1880s, puts a number on it: the vapour pressure of the solution equals the vapour pressure of the pure solvent multiplied by the solvent’s mole fraction. Add enough sugar to water that 1 molecule in 10 is sugar and the vapour pressure drops by 10%.
This vapour-pressure lowering is a colligative property – it depends on how many solute particles there are, not on what they are – and it is behind the boiling-point elevation and freezing-point depression of solutions. Salts that split into ions count every ion, which the van ‘t Hoff factor i takes care of.
When both liquids are volatile, each contributes its own partial pressure, and the vapour above an ideal mixture is richer in the more volatile component. That is the principle of fractional distillation. The chart plots vapour pressure against composition – a straight line for an ideal solution.
How to use the Raoult’s Law Calculator
- I know
- Vapour pressure of the pure solvent (p°)
- Moles of solute
- Moles of solvent
- Mass of solute
- Molar mass of solute
- Mass of solvent
- Molar mass of solvent
- Van ‘t Hoff factor i (particles per formula unit)
- Mole fraction of the solvent
- Vapour pressure of the solution
- Vapour pressure of pure liquid A
- Vapour pressure of pure liquid B
- Mole fraction of A in the liquid
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Pick what you know. The calculator turns moles or masses into the mole fraction of the solvent (multiplying the solute moles by the van ‘t Hoff factor), multiplies by the pure-solvent vapour pressure, and reports the lowering. For two volatile liquids it adds the two partial pressures and finds the vapour composition from Dalton’s law.
Worked example
Sugar in water: 50 g of sucrose (342.30 g/mol, 0.1461 mol) in 100 g of water (5.551 mol) gives a solvent mole fraction of 5.551 ÷ 5.697 = 0.9744. At 25 °C, p = 0.9744 × 23.76 = 23.15 mmHg, a lowering of 0.61 mmHg (2.56%). Two liquids: a benzene–toluene mixture with x(benzene) = 0.4 has partial pressures 0.4 × 95.1 = 38.04 mmHg and 0.6 × 28.4 = 17.04 mmHg; the total is 55.08 mmHg and the vapour is 38.04 ÷ 55.08 = 69.1% benzene.
Vapour pressure of water
| Temperature | mmHg | kPa |
|---|---|---|
| 0 °C | 4.58 | 0.611 |
| 20 °C | 17.54 | 2.339 |
| 25 °C | 23.76 | 3.169 |
| 37 °C | 47.1 | 6.28 |
| 50 °C | 92.6 | 12.35 |
| 100 °C | 760 | 101.3 |
Tips
- Use the final volume of the solution, not the volume of solvent added, for molarity.
- For dilutions, C₁V₁ = C₂V₂ only holds when both concentrations use the same unit.
- Add acid to water, never water to acid.
- Everything is calculated in your browser.
Common mistakes to avoid
- Mixing w/w, w/v and v/v percentages.
- Forgetting that pH is logarithmic – one unit is a tenfold change in [H⁺].
- Using pKa where Ka is asked for (or the other way round).
Formula
p(solution) = x(solvent) × p°(solvent)Δp = p° − p = x(solute) × p°Two volatile liquids: p = x_A·p°_A + x_B·p°_B, y_A = x_A·p°_A ÷ pWorked example
Inputs — I know: Moles of solute and solvent; Vapour pressure of the pure solvent (p°): 23.76 mmHg (Torr); Moles of solute: 0.5 mol; Moles of solvent: 5 mol; Van 't Hoff factor i (particles per formula unit): 1.
Results — Vapour pressure of the solution = 21.6; Vapour pressure of the pure solvent (p°) = 23.76; Total vapour pressure = 21.6; Vapour pressure lowering (Δp) = 2.16; Relative lowering = 9.091 %; Mole fraction of solvent = 0.90909.
Key terms
- Vapour pressure
- Pressure of a vapour in equilibrium with its liquid at a given temperature.
- Mole fraction
- Moles of one component divided by total moles.
- Colligative property
- A property that depends on the number of solute particles, not their kind.
- Ideal solution
- A mixture that obeys Raoult's law at every composition.
- Van 't Hoff factor (i)
- Number of particles one formula unit forms in solution.
Frequently asked questions
What does Raoult's law say?
The vapour pressure of a solution equals the mole fraction of the solvent times the vapour pressure of the pure solvent: p = x·p°.
What is the vapour pressure of water at 25 °C?
About 23.76 mmHg (3.17 kPa); at 100 °C it is 760 mmHg, which is why water boils there at sea level.
What is the vapour pressure of a solution with 0.25 mole fraction of solvent and p° = 12 mmHg?
0.25 × 12 = 3 mmHg.
How do salts affect vapour pressure?
Each ion counts. 1 mol of NaCl gives about 2 mol of particles (i ≈ 2), so it lowers the vapour pressure about twice as much as 1 mol of sugar.
When does Raoult's law fail?
For concentrated solutions and for mixtures whose molecules attract each other more or less strongly than in the pure liquids – these show positive or negative deviations (and can form azeotropes).
What is the vapour composition over benzene–toluene?
For x(benzene) = 0.4 at 25 °C (p° = 95.1 and 28.4 mmHg), the vapour holds 69% benzene, which is why distillation separates them.
Is Raoult's law the same as Henry's law?
No. Raoult's law applies to the solvent (the major component); Henry's law describes a dilute dissolved gas or solute using its own constant.
What formula does the Raoult's Law Calculator use?
p(solution) = x(solvent) × p°(solvent).