Raoult's Law Calculator

How to use the Raoult's Law Calculator

Enter solvent vapour pressure, moles of solvent and solute to apply Raoult's Law.

1. Choose vapour pressure or boiling point modeTab 1 applies Raoult's Law to find solution vapour pressure. Tab 2 applies the boiling point elevation formula ΔTb = i × Kb × m.
2. Enter pure solvent vapour pressureFor water at 25°C this is 3.169 kPa. Furthermore, values at other temperatures can be found from the Antoine equation or steam tables.
3. Enter moles of solvent and soluteUse moles, not masses. Convert mass to moles using molar mass if needed.
4. Set the van't Hoff factorNon-electrolytes (sugar, urea) have i = 1. NaCl gives i = 2, CaCl₂ gives i = 3. Furthermore, actual i values are slightly lower than ideal for concentrated solutions.
5. Click CalculateMole fraction, vapour pressure depression and total vapour pressure appear together. Moreover, the boiling point tab shows molality and the new boiling point above 100°C for water.

Options and variants explained

Key ebullioscopic constants for common solvents.

The formula explained

Raoult's Law is a colligative property — it depends on the number of solute particles, not their identity. Consequently, adding any non-volatile solute reduces vapour pressure by the same amount for the same mole fraction — regardless of the solute's chemical identity. The van't Hoff factor corrects for electrolyte dissociation, which increases the effective particle count.

Worked example: 0.1 mol NaCl in 1 mol water, vapour pressure at 25°C

NaCl dissociates into Na⁺ and Cl⁻, so i = 2. Effective solute moles = 0.1 × 2 = 0.2. X_water = 1 / (1 + 0.2) = 0.8333. P = 3.169 × 0.8333 = 2.640 kPa. Vapour pressure depression ΔP = 3.169 − 2.640 = 0.529 kPa.

Boiling point elevation: molality = 0.1 mol ÷ (1 mol × 0.018 kg/mol) = 5.56 mol/kg. ΔTb = 2 × 0.512 × 5.56 = 5.69°C. New boiling point: 100 + 5.69 = 105.69°C. Furthermore, this is why salt water boils above 100°C. Kitchen quantities produce only about 0.5°C elevation in practice.

The colligative nature of vapour pressure depression

Vapour pressure depression is a colligative property. It depends only on the number of solute particles — not their chemical identity. Moreover, 0.1 mol of glucose and 0.1 mol of urea (both i = 1) produce identical vapour pressure depression in water.

What is Raoult's Law?

Raoult's Law states that the partial vapour pressure of a solution component equals the pure-component vapour pressure multiplied by its mole fraction. Furthermore, it was formulated by French chemist François-Marie Raoult in the 1880s based on experimental observations of solutions.

Raoult's Law applies ideally when solute-solvent interactions resemble pure-component interactions — as in benzene-toluene mixtures. Moreover, real solutions deviate either positively (weaker cross-interactions) or negatively (stronger cross-interactions).

Colligative properties — vapour pressure lowering, boiling point elevation, freezing point depression and osmotic pressure — all depend on the mole fraction of solute particles. Furthermore, they can all be derived from the chemical potential framework that underlies Raoult's Law.

Why Raoult's Law matters in chemistry and engineering

Distillation — separating components of a liquid mixture by boiling — depends on Raoult's Law for ideal mixtures. Vapour composition at a given temperature is determined by component vapour pressures and mole fractions. Furthermore, column design uses Raoult's Law to calculate the theoretical plates needed for a given separation.

Osmotic pressure — the pressure needed to prevent osmosis through a semipermeable membrane — is directly related to mole fraction and is used in water purification (reverse osmosis) and drug delivery. Moreover, colligative properties are used to determine unknown molar masses by measuring boiling point elevation or freezing point depression.

Anti-freeze solutions work through freezing point depression, another colligative property related to Raoult's Law. Adding ethylene glycol to water raises the boiling point and lowers the freezing point simultaneously. Furthermore, the optimal concentration is calculated using the same mole-fraction approach.

Common Raoult's Law mistakes

Using moles of solute rather than mole fraction in the formula is the most common error. The formula requires the mole fraction of the solvent — moles of solvent divided by total moles. Furthermore, forgetting to include solute moles in the denominator gives a mole fraction of 1 for all solutions.

Forgetting the van't Hoff factor for electrolytes underestimates the effect. NaCl produces 2 particles per formula unit; CaCl₂ produces 3. Moreover, the effective particle count drives all colligative effects, so omitting i gives an answer that can be 2–3 times too small for strong electrolytes.

Applying Raoult's Law to volatile solutes requires modification — Dalton's Law of partial pressures must be included because the solute also contributes to total vapour pressure. Furthermore, for miscible volatile liquid mixtures, both components contribute to total pressure using their respective Raoult's Law terms.

Tips for Raoult's Law calculations

Convert mass to moles before applying the formula. Divide mass in grams by molar mass in g/mol. Furthermore, for electrolytes, identify the complete dissociation products and determine i from the number of ions produced.

For aqueous solutions at 25°C, the pure water vapour pressure (3.169 kPa) is a standard reference. For other temperatures, use Antoine equation constants or steam table values. Moreover, the vapour pressure of water at 100°C is exactly 101.325 kPa (1 atm) by the definition of the normal boiling point.

Check your answer with the vapour pressure depression formula: ΔP/P° = X_solute (for dilute solutions). This simplified form shows that fractional vapour pressure lowering equals the solute mole fraction. Furthermore, this is Raoult's Law simplified for the special case of dilute solutions.

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