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Colligative Properties Calculator
In short: Colligative properties calculator: freezing-point depression, boiling-point elevation, osmotic pressure and relative lowering of vapour pressure, with the van 't Hoff factor and molar mass determination. It is free, works in your web browser and needs no download.
Choose a property, pick a solvent to fill in Kb or Kf, and leave the unknown blank. Work from masses (to find the molar mass of a solute) or from molality. Osmotic pressure accepts atm, bar, kPa or mmHg. A separate tab converts between the van 't Hoff factor and the degree of dissociation or association.
Leave the value you want to find blank.
Leave the value you want to find blank.
Fill in either α or i.
i = observed colligative property ÷ value expected with no dissociation = normal molar mass ÷ observed molar mass.
Colligative property formulas
| Property | Formula | Molar mass of solute |
|---|---|---|
| Boiling-point elevation | ΔTb = i·Kb·m | M₂ = i·Kb·w₂·1000 / (ΔTb·w₁) |
| Freezing-point depression | ΔTf = i·Kf·m | M₂ = i·Kf·w₂·1000 / (ΔTf·w₁) |
| Osmotic pressure | π = i·M·R·T | M₂ = i·w₂·R·T / (π·V) |
| Relative lowering of vapour pressure | (P° − P)/P° = i·n₂/(n₁ + i·n₂) ≈ i·n₂/n₁ | M₂ ≈ i·w₂·M₁·P° / (w₁·(P° − P)) |
m = molality (mol solute per kg solvent), w₂ = mass of solute in g, w₁ = mass of solvent in g, M = molarity in mol/L, V in L, T in K and R = 0.08206 L·atm mol⁻¹ K⁻¹ (or 0.08314 L·bar mol⁻¹ K⁻¹). ΔT has the same size in K and °C.
van 't Hoff factor i
dissociation (1 unit → n ions): i = 1 + α(n − 1) association (n molecules → 1 unit): i = 1 − α(1 − 1/n)
| Solute | Ideal i |
|---|---|
| Glucose, sucrose, urea (non-electrolytes) | 1 |
| NaCl, KCl, KNO₃, MgSO₄ | 2 |
| CaCl₂, Na₂SO₄, K₂SO₄, Ba(NO₃)₂ | 3 |
| AlCl₃, Na₃PO₄, K₃[Fe(CN)₆] | 4 |
| K₄[Fe(CN)₆], Al₂(SO₄)₃ | 5 |
| Acetic or benzoic acid fully dimerised in benzene | 0.5 |
Real electrolytes give slightly smaller i values because of ion pairing (0.1 m NaCl has i ≈ 1.87). Exam problems usually assume the ideal value unless α is given.
Solvent constants
| Solvent | Normal bp (°C) | Kb (K·kg/mol) | Freezing point (°C) | Kf (K·kg/mol) |
|---|---|---|---|---|
| Water | 100 | 0.512 | 0 | 1.86 |
| Benzene | 80.1 | 2.53 | 5.5 | 5.12 |
| Cyclohexane | 80.7 | 2.79 | 6.5 | 20 |
| Camphor | 204 | — | 179.8 | 37.7 |
| Ethanol | 78.4 | 1.22 | −114.1 | 1.99 |
| Chloroform | 61.2 | 3.63 | −63.5 | 4.68 |
| Acetic acid | 117.9 | 3.07 | 16.6 | 3.9 |
| Naphthalene | 218 | 5.8 | 80.2 | 6.94 |
Typical textbook values. Some books use Kf = 40 for camphor, 4.9 for benzene (NCERT) or 20.2 for cyclohexane; type your course's value into the K box.
Worked examples (JEE / NEET style)
Freezing point of salt water
0.1 m NaCl in water, i = 2, Kf = 1.86 K kg mol⁻¹.
ΔTf = 2 × 1.86 × 0.1
ΔTf = 0.372 K, so it freezes at −0.372 °C
Boiling point of a glucose solution
18 g glucose (M = 180 g/mol) in 1.0 kg water. Kb = 0.512.
m = 0.10 mol/kg; ΔTb = 1 × 0.512 × 0.10
ΔTb = 0.0512 K, bp = 100.051 °C
Molar mass from freezing point (NCERT)
1.00 g of a non-electrolyte in 50 g benzene lowers the freezing point by 0.40 K. Kf = 5.12.
M₂ = 5.12 × 1.00 × 1000 / (0.40 × 50)
M₂ = 256 g/mol
Association of benzoic acid (NCERT)
2 g benzoic acid (M = 122) in 25 g benzene gives ΔTf = 1.62 K, Kf = 4.9.
Observed M = 4.9 × 2 × 1000 / (25 × 1.62) = 242. i = 122/242 = 0.504. α = (1 − 0.504)/(1 − 1/2)
α = 0.99, so 99% dimerised
Degree of dissociation
0.10 m CaCl₂ freezes at −0.50 °C. Kf = 1.86.
i = 0.50 / (1.86 × 0.10) = 2.69. α = (i − 1)/(n − 1) = 1.69/2
α = 0.84 (84% dissociated)
Molar mass of a protein from π (NCERT)
1.26 g protein in 200 mL solution, π = 2.57 × 10⁻³ bar at 300 K.
M₂ = 1.26 × 0.08314 × 300 / (2.57 × 10⁻³ × 0.200)
M₂ ≈ 6.1 × 10⁴ g/mol
Relative lowering of vapour pressure (NCERT)
0.5 g of a non-volatile solid in 39.0 g benzene (M = 78) lowers P from 0.850 to 0.845 bar.
Dilute form: (0.850 − 0.845)/0.850 = (0.5/M₂)/(39/78)
M₂ = 170 g/mol (169 g/mol with the exact form)
Isotonic solutions
Blood has π ≈ 7.65 atm at 37 °C. What NaCl molarity is isotonic (i = 2)?
M = 7.65 / (2 × 0.08206 × 310.15)
M ≈ 0.150 mol/L (about 0.9% w/v)
Units and common mistakes
- Molality uses kg of solvent, not litres of solution. 18 g glucose in 500 g water is 0.20 m.
- Osmotic pressure uses molarity (mol per litre of solution), not molality.
- Forgetting i. Electrolytes give more particles. 0.1 m CaCl₂ depresses the freezing point about three times as much as 0.1 m glucose.
- R must match the pressure unit. Use 0.08206 L·atm, 0.08314 L·bar or 8.314 L·kPa mol⁻¹ K⁻¹.
- Temperature in kelvin for π. ΔT itself is the same in K and °C.
- Abnormal molar mass. Dissociation makes the observed molar mass too low (i greater than 1). Association makes it too high (i less than 1).
- Non-volatile solute only. The vapour pressure formula assumes the solute does not evaporate.
Drawing solutes and solvents? MolDraw is a free online chemical structure editor for molecules, reactions and lab diagrams.
Open MolDraw editorFrequently asked questions
What are colligative properties?
Properties that depend on the number of solute particles, not on what they are: vapour pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure.
How do I calculate freezing-point depression?
ΔTf = i·Kf·m. For 0.1 m NaCl in water, i = 2 and Kf = 1.86 K kg/mol, so ΔTf = 0.372 K and the solution freezes at −0.372 °C.
How do I find molar mass from freezing-point depression?
M₂ = i·Kf·w₂·1000 / (ΔTf·w₁), with masses in grams. 1.00 g solute in 50 g benzene with ΔTf = 0.40 K gives M₂ = 5.12 × 1.00 × 1000 / (0.40 × 50) = 256 g/mol.
What is the van 't Hoff factor?
The number of particles each formula unit gives in solution. Ideal values: glucose 1, NaCl 2, CaCl₂ 3. For partial dissociation i = 1 + α(n − 1); for association into n-mers i = 1 − α(1 − 1/n).
Why is the van 't Hoff factor of acetic acid in benzene about 0.5?
Acetic acid forms hydrogen-bonded dimers in non-polar solvents, so two molecules act as one particle. Complete dimerisation gives i = 0.5 and an apparent molar mass twice the true value.
How do I calculate osmotic pressure?
π = iMRT. 0.10 M glucose at 25 °C gives π = 1 × 0.10 × 0.08206 × 298.15 = 2.45 atm. Osmotic pressure is the best method for large molecules such as proteins because π is easy to measure even at very low concentration.
Which colligative property is best for finding the molar mass of a polymer?
Osmotic pressure. A protein of molar mass 60 000 g/mol at about 6 g/L lowers the freezing point by only about 2 × 10⁻⁴ K, but gives π of about 2.5 mbar, which is easy to measure at room temperature.
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