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🧊Colligative Properties Calculator - Freezing Point Depression, Boiling Point Elevation & Osmotic Pressure

Colligative properties depend only on how many solute particles are dissolved, not on what they are. The four equations are freezing point depression ΔTf = i·Kf·m and boiling point elevation ΔTb = i·Kb·m (m = molality in mol/kg), osmotic pressure Π = i·M·R·T (M = molarity) and Raoult’s law P = x(solvent)·P°. The van ’t Hoff factor i is the number of particles per formula unit: 1 for sugar or glucose, 2 for NaCl, 3 for CaCl₂, 5 for Al₂(SO₄)₃. A 1 mol/kg NaCl solution in water therefore has ΔTf = 2 × 1.86 × 1 = 3.72 °C and freezes near −3.72 °C, and a 0.1 M glucose solution at 25 °C has Π = 0.1 × 0.082057 × 298.15 ≈ 2.45 atm. The calculator has Kf, Kb, normal freezing and boiling points and 25 °C vapour pressures for water, benzene, cyclohexane, camphor, acetic acid, ethanol, carbon tetrachloride, chloroform, naphthalene, phenol, nitrobenzene, diethyl ether and acetone. Type a formula and it recognises strong electrolytes (salts, strong acids and bases), giving the ideal i and the molar mass, which you can override with a measured i. Each equation can be solved for any unknown: ΔT from concentration, the mass of solute needed for a target temperature (antifreeze, road salt), the molar mass of an unknown from measured ΔTf, ΔTb, Π or vapour pressure (the classic lab experiment, the Rast camphor method, polymer osmometry), or i from experimental data. A compare tab ranks several solutions by i·m to answer “which has the lowest freezing point” questions.

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How to use Colligative Properties Calculator

  1. 1Choose freezing point, boiling point, osmotic pressure, vapour pressure or compare, or click an example
  2. 2Pick the solvent, type the solute formula (i and molar mass are filled in), choose what to solve for and enter the known values
  3. 3Read the result with step-by-step working and copy the numbers you need

FAQ

How do I find the van ’t Hoff factor i?
Ideally i equals the number of ions one formula unit releases: 1 for non-electrolytes (glucose, sucrose, urea), 2 for NaCl or KNO₃, 3 for CaCl₂ or Na₂SO₄, 4 for FeCl₃ or K₃PO₄ and 5 for Al₂(SO₄)₃. Real solutions show ion pairing, so measured values are lower — about 1.9 for NaCl and 1.3 for MgSO₄ at 0.05 mol/kg — and the gap grows with concentration.
How do you calculate molar mass from freezing point depression?
Dissolve a mass w of solute in a mass W of solvent and measure ΔTf. Then m = ΔTf / (i·Kf), moles n = m × W (in kg), and molar mass M = w / n. For example, 2.00 g of an unknown in 25.0 g of benzene lowers the freezing point by 2.05 °C: m = 2.05 / 5.12 = 0.400 mol/kg, n = 0.0100 mol and M ≈ 200 g/mol. Camphor’s large Kf (about 37.7) makes it useful for tiny samples.
Why is osmotic pressure used for proteins and polymers?
Large molecules give very few particles per gram, so the freezing point drops by only thousandths of a degree, while the osmotic pressure is still easy to measure. A solution of 1.00 g of protein in 100 mL at 25 °C with Π = 1.85 mmHg gives c ≈ 9.95×10⁻⁵ mol/L and M ≈ 1.0×10⁵ g/mol.
Which solution has the lowest freezing point?
In the same solvent, compare the total particle concentration i·m: the larger it is, the lower the freezing point, the higher the boiling point and the larger the osmotic pressure. Among 0.25 m glucose (i·m = 0.25), 0.10 m NaCl (0.20), 0.08 m CaCl₂ (0.24) and 0.03 m Al₂(SO₄)₃ (0.15), the glucose solution freezes lowest. If a question gives measured i values, use those instead.