Raoult's Law Calculations — Formula, Worked Examples and Deviations
Raoult's law is the foundation of the whole Solutions chapter. Every colligative property you will meet later — boiling point elevation, freezing point depression, osmotic pressure — is a consequence of the one idea in this law: a dissolved solute lowers the vapour pressure of the solvent. Get Raoult's law right and the rest of the chapter follows. This guide gives the two forms of the law, three fully worked calculations, and the mistakes that turn an easy numerical into a lost mark.
The law in one line
For a solution of two volatile liquids, the partial vapour pressure of each component is its mole fraction in the liquid multiplied by its vapour pressure when pure:
ptotal = pA + pB = xA pA° + xB pB°
What each symbol means:
| Symbol | Meaning | Unit |
|---|---|---|
| pA | partial vapour pressure of A above the solution | any pressure unit (kPa, bar, mm Hg) |
| pA° | vapour pressure of pure A at the same temperature | same unit as pA |
| xA | mole fraction of A in the liquid | no unit (0 to 1) |
The mole fractions of all components in the liquid add to 1, so xB = 1 − xA for a two-component mixture.
The second form — a non-volatile solute
If the solute cannot evaporate (sugar, urea, a salt), it contributes no vapour of its own. Only the solvent evaporates, and its mole fraction is now less than 1:
Relative lowering: (p° − p) ÷ p° = xsolute
Read that second line carefully — it is the most useful result in the chapter. The fractional drop in vapour pressure equals the mole fraction of the solute. It does not depend on what the solute is, only on how many particles of it there are. That is exactly what "colligative" means.
Worked example 1 — two volatile liquids
Problem: 2.0 mol of benzene is mixed with 3.0 mol of toluene at 25 °C. Take the vapour pressures of the pure liquids as p°(benzene) = 12.7 kPa and p°(toluene) = 3.8 kPa. Find the total vapour pressure and the composition of the vapour.
Step 1 — mole fractions in the liquid. Total = 2.0 + 3.0 = 5.0 mol
x(benzene) = 2.0 ÷ 5.0 = 0.40 x(toluene) = 3.0 ÷ 5.0 = 0.60
Step 2 — partial pressures.
p(benzene) = 0.40 × 12.7 = 5.08 kPa
p(toluene) = 0.60 × 3.8 = 2.28 kPa
Step 3 — total. ptotal = 5.08 + 2.28 = 7.36 kPa
Step 4 — composition of the vapour (Dalton's law: mole fraction in the
vapour = partial pressure ÷ total pressure):
y(benzene) = 5.08 ÷ 7.36 = 0.690
y(toluene) = 2.28 ÷ 7.36 = 0.310
Check: 0.690 + 0.310 = 1.000 ✓
The liquid was only 40% benzene but the vapour is 69% benzene. The vapour is always richer in the more volatile component, and that single fact is the whole basis of fractional distillation.
Worked example 2 — molar mass from relative lowering
Problem: 18.0 g of a non-volatile, non-electrolyte solute is dissolved in 200 g of water at 25 °C. The vapour pressure falls from 3.170 kPa to 3.126 kPa. Find the molar mass of the solute.
Step 1 — relative lowering.
(p° − p) ÷ p° = (3.170 − 3.126) ÷ 3.170 = 0.044 ÷ 3.170 = 0.013880
Step 2 — moles of water. M(H₂O) = 18.02 g/mol
n₁ = 200 ÷ 18.02 = 11.099 mol
Step 3 — the dilute approximation. For a dilute solution
x₂ = n₂/(n₁ + n₂) ≈ n₂/n₁, so
n₂ ≈ 0.013880 × 11.099 = 0.15406 mol
M₂ = 18.0 ÷ 0.15406 = 116.8 g/mol
Step 4 — the exact route, as a cross-check. Without the approximation,
x₂ = n₂/(n₁ + n₂) = 0.013880 gives
n₂(1 − 0.013880) = 0.013880 × 11.099 = 0.154059
n₂ = 0.154059 ÷ 0.986120 = 0.156228 mol
M₂ = 18.0 ÷ 0.156228 = 115.2 g/mol
The approximation is out by about 1.4%. That is acceptable in most board questions, but if a question says "exact" or the solution is concentrated, use the full expression. Show which one you used — examiners accept either when the working is visible.
Worked example 3 — predicting the vapour pressure of a salt solution
Problem: 5.85 g of NaCl is dissolved in 500 g of water at 25 °C, where p°(water) = 3.170 kPa. Estimate the vapour pressure of the solution, assuming NaCl is fully dissociated.
Step 1 — moles of NaCl. M(NaCl) = 22.990 + 35.45 = 58.44 g/mol
n = 5.85 ÷ 58.44 = 0.1001 mol
Step 2 — moles of particles. NaCl → Na⁺ + Cl⁻, so each formula
unit gives 2 particles (van 't Hoff factor i = 2 for complete dissociation):
n(particles) = 2 × 0.1001 = 0.2002 mol
Step 3 — moles of water. n₁ = 500 ÷ 18.02 = 27.747 mol
Step 4 — mole fraction of solute particles.
x₂ = 0.2002 ÷ (27.747 + 0.2002) = 0.2002 ÷ 27.9472 = 0.007163
Step 5 — vapour pressure.
p = p°(1 − x₂) = 3.170 × (1 − 0.007163) = 3.170 × 0.992837 =
3.147 kPa
Lowering = 3.170 − 3.147 = 0.023 kPa
Ignoring the dissociation would have given exactly half this lowering. For any ionic solute you must count particles, not formula units.
Ideal and non-ideal solutions
A solution is called ideal when it obeys Raoult's law at every composition. That happens when the A–B attractions in the mixture are about as strong as the A–A and B–B attractions in the pure liquids — benzene with toluene, or hexane with heptane, are close to ideal. Real mixtures often deviate.
| Positive deviation | Negative deviation | |
|---|---|---|
| Observed ptotal | higher than Raoult predicts | lower than Raoult predicts |
| A–B interaction | weaker than A–A and B–B | stronger than A–A and B–B |
| ΔHmixing | positive (absorbs heat) | negative (releases heat) |
| ΔVmixing | positive (slight expansion) | negative (slight contraction) |
| Typical pair | ethanol + water; acetone + carbon disulphide | chloroform + acetone; nitric acid + water |
| Azeotrope formed | minimum boiling | maximum boiling |
Chloroform and acetone show negative deviation because a hydrogen bond forms between the chloroform hydrogen and the acetone oxygen — a new attraction that did not exist in either pure liquid, so fewer molecules escape into the vapour.
Common mistakes that cost marks
- Using mass fraction or percentage instead of mole fraction. Raoult's law counts particles. Convert every mass to moles first.
- Using the vapour mole fraction in the law. x in p = x·p° is the composition of the liquid. The vapour composition is what you calculate afterwards.
- Forgetting the van 't Hoff factor for ionic solutes. NaCl gives 2 particles, CaCl₂ gives 3, K₂SO₄ gives 3.
- Applying the non-volatile form to a volatile solute. If the solute evaporates, it has its own p° term and the total pressure goes up, not down.
- Mixing pressure units. p and p° must be in the same unit; the ratio then cancels the unit, which is why relative lowering has no unit.
- Using the dilute approximation on a concentrated solution. Above roughly 5 mol% solute the error grows quickly.
- Assuming p° is a constant. It is strongly temperature dependent, so every p° in a problem must be at the temperature stated.
Where Raoult's law appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Total vapour pressure of a binary mixture; relative lowering; molar mass of a solute |
| JEE/NEET | Vapour composition, deviations, azeotrope identification |
| IIT-JAM / CUET-PG | Ideal versus non-ideal behaviour, ΔH and ΔV of mixing |
| GATE / CSIR-NET | Activity and activity coefficients as the correction to Raoult's law |
Every Raoult's law problem starts with molar masses. You cannot get a mole fraction without them, and a wrong molar mass ruins an otherwise perfect answer. The Molar Mass & Composition tool takes any formula — C₆H₆, C₇H₈, NaCl, C₁₂H₂₂O₁₁ — and returns the molar mass with an element-wise breakdown.
Open the Molar Mass & Composition Calculator →The Solutions chapter carries steady marks in the board paper and feeds straight into JEE and NEET. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and through online classes across India, with weekly numerical practice built in: abcchemistry.in.