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Raoult's Law Calculator

Find the vapor pressure lowering of a solution.

How it is calculated

Raoult's law says that adding a non-volatile solute to a solvent lowers the solvent's vapour pressure in proportion to the solvent's share of the mixture: P = x × P0. Here P0 is the vapour pressure of the pure solvent at the temperature of interest, and x is the mole fraction of the solvent, a number from 0 to 1, found as moles of solvent / (moles of solvent + moles of solute). The tool also gives the vapour pressure lowering, ΔP = P0 − P. Results are in the same pressure unit as P0 (mmHg, kPa, atm) and rounded to two decimals. x must lie between 0 and 1. The law is exact only for ideal solutions; strong interactions between solute and solvent cause deviations.

Worked example

A solvent has a pure vapour pressure of 23.8 mmHg at the working temperature, and its mole fraction in the solution is 0.9. Vapour pressure above the solution: P = 0.9 × 23.8 = 21.42 mmHg. Lowering: ΔP = 23.8 − 21.42 = 2.38 mmHg. This equals the solute's mole fraction (0.1) times 23.8, so a 10% solute content lowers the vapour pressure by 10% of its pure value.

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Common mistakes

  • Using the solute's mole fraction as x instead of the solvent's, which gives the lowering rather than the vapour pressure.
  • Entering a mass or mole percentage, such as 90, instead of a fraction (0.9), or using mass fraction where a mole fraction is required.
  • Using the vapour pressure of the pure solvent at the wrong temperature; vapour pressure changes strongly with temperature.
  • Applying the law to a solute that is itself volatile, where each component's partial pressure must be added together.
  • Expecting accurate results from strongly non-ideal mixtures, such as ethanol and water, where observed pressure differs from the straight-line prediction.

About this calculator

Raoult's law states that the vapor pressure of a solvent above an ideal solution equals the vapor pressure of the pure solvent multiplied by its mole fraction in the solution: P = x·P₀. Because adding a solute always lowers the solvent's mole fraction below 1, dissolving something in a liquid always lowers its vapor pressure — one of the four classic colligative properties, alongside boiling-point elevation, freezing-point depression, and osmotic pressure.

Chemists and chemical engineers use Raoult's law to predict vapor-liquid equilibrium in distillation columns, to model how antifreeze or dissolved salts change a solution's evaporation behavior, and as the baseline "ideal" case against which real (non-ideal) mixtures with positive or negative deviations are measured.

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Comments (1)

  • Editorial team

    The law holds for ideal solutions, and it counts dissolved particles, not formula units. One mole of NaCl in 9 mol of water releases roughly 2 mol of ions if it dissociates fully, so the solvent mole fraction is about 9 ÷ 11 = 0.82 rather than 0.90. Real salt solutions deviate from this. Use the pure-solvent pressure at the same temperature as the solution. Which solvent and solute pair are you working with?

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