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Colligative properties calculator

Choose the solvent and the solute, type the masses or the molality and get the four colligative properties of the solution with every step; or, the other way round, the molar mass of an unknown solute from the measured freezing point, boiling point, osmotic pressure or vapor pressure.

What you can do

Guide: colligative properties

When a nonvolatile solute is dissolved, the solvent changes its behavior: it boils at a higher temperature, freezes at a lower one, its vapor pressure drops and, if a membrane separates it from the pure solvent, an osmotic pressure appears. These are the colligative properties, and they depend on how many particles of solute there are, not on which ones. This guide explains the four formulas, where the constants come from, and works through an example.

The four formulas (dilute, ideal solution)

b is the molality (moles of solute per kilogram of solvent), c the molarity, T the temperature in kelvin and R = 8.314 J/(mol·K). The van 't Hoff factor i counts the particles per formula unit: 1 for a nonelectrolyte such as glucose, 2 for NaCl, 3 for CaCl2 (somewhat less in reality, because of ion pairs).

Where Kb and Kf come from

They are not magic constants: phase-equilibrium thermodynamics gives them from pure-solvent data.

Kb = R · Tb² · M / ΔHvap     Kf = R · Tf² · M / ΔHfus

with the temperatures in kelvin, the solvent molar mass in kg/mol and the enthalpies of vaporization and fusion in J/mol. That is why a heavy solvent with a low enthalpy of fusion (such as camphor) has a huge Kf and is used to measure molar masses. The calculator does this calculation with the data of each solvent and shows their source; Kf only appears where the enthalpy of fusion agrees in two references.

Worked example: 10 g of glucose in 200 g of water

  1. Molality: glucose (C6H12O6) has M = 180.156 g/mol, so n = 10 / 180.156 = 0.05551 mol and b = 0.05551 / 0.200 kg = 0.2775 mol/kg. It is not an electrolyte: i = 1.
  2. Boiling: for water Kb = 0.5121 K·kg/mol; ΔTb = 0.5121 × 0.2775 = 0.1421 °C, so it boils at 99.974 + 0.1421 = 100.116 °C.
  3. Freezing: Kf = 1.86 K·kg/mol; ΔTf = 1.86 × 0.2775 = 0.5161 °C, so it freezes at −0.5161 °C.
  4. Vapor pressure at 25 °C: each kilogram of water holds 1000 / 18.015 = 55.51 mol of water, xwater = 55.51 / (55.51 + 0.2775) = 0.995025; with P° = 3.167 kPa, P = 3.152 kPa (15.76 Pa lower).
  5. Osmotic pressure at 25 °C: since the solution is dilute, c ≈ b · ρwater = 0.2767 mol/L and π = 276.7 mol/m³ × 8.314 × 298.15 K = 686 kPa, about 6.77 atm.

Notice the sizes: the boiling point barely rises a tenth of a degree, but the osmotic pressure is almost seven atmospheres. That is why osmometry is used to measure large molar masses (polymers, proteins), where ΔT would be too small to measure.

The molar mass of an unknown solute

The same formulas, the other way round: if 2.00 g of a solute in 100 g of water lower the freezing point by 0.40 °C, then b = 0.40 / 1.86 = 0.2151 mol/kg, n = 0.2151 × 0.100 = 0.02151 mol and M = 2.00 / 0.02151 = 92.98 g/mol. The calculator does it from the measured freezing point depression, boiling point elevation, osmotic pressure or vapor pressure.

How to use the calculator

  1. Choose what to compute: the properties of the solution or the molar mass of the solute.
  2. Choose the solvent (29 solvents, with their constants and sources) or "Other solvent" to use the constants your exercise gives.
  3. Type the solute (formula or molar mass; for salts, pick it from the list and the factor i comes by itself) and the masses of solute and solvent, or the molality.
  4. Read the results with every step and download the detailed solution as a PDF.

Common mistakes

Further reading

P. Atkins and J. de Paula, Physical Chemistry (Oxford University Press); R. Chang, Chemistry (McGraw-Hill). Simuveliq is not affiliated with these authors or publishers.

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