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Generated on: July 20, 2026 at 09:33 PM

Colligative Molar Mass Calculator

Inputs

Freezing point depression1.5 °C
Cryoscopic constant1.86 °C·kg/mol
van't Hoff factor1
Solute mass5 g
Solvent mass0.1 kg

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Chemistry

Colligative Molar Mass Calculator

Find the molar mass of an unknown solute from the freezing point depression it causes. Enter the measured ΔTf, the cryoscopic constant, and the masses of solute and solvent to get molality and molar mass.

Inputs

°C
°C·kg/mol

Results

Enter a value to see results.

\begin{aligned} M &= \dfrac{i \cdot K_f \cdot w}{\Delta T_f \cdot m_{solv}} \\ &= \dfrac{1 \times 1.86\,\text{°C·kg/mol} \times 5\,\text{g}}{1.5\,\text{°C} \times 0.1\,\text{kg}} \\ &= ?\,\text{g/mol} \end{aligned}

Details

\begin{aligned} b &= \dfrac{\Delta T_f}{i \cdot K_f} \\ &= \dfrac{1.5\,\text{°C}}{1 \times 1.86\,\text{°C·kg/mol}} \\ &= ?\,\text{mol/kg} \end{aligned}

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Molar mass from freezing point depression

When an unknown compound is dissolved in a solvent, it lowers the temperature at which that solvent freezes. That lowering — the freezing point depression — depends only on how many solute particles are present, not on what they are. Because the number of particles is tied to the number of moles, a careful measurement of the freezing point drop reveals the molar mass of the solute. This classic technique is called cryoscopy.

The formula

Freezing point depression follows a colligative law:

ΔTf=i⋅Kf⋅b\Delta T_f = i \cdot K_f \cdot b

where bb is the molality of the solution. Solving for molality and then for molar mass gives the chain this calculator uses:

b=ΔTfi⋅Kf,M=wb⋅msolv=i⋅Kf⋅wΔTf⋅msolvb = \frac{\Delta T_f}{i \cdot K_f}, \qquad M = \frac{w}{b \cdot m_{solv}} = \frac{i \cdot K_f \cdot w}{\Delta T_f \cdot m_{solv}}
SymbolQuantityUnit
ΔTfFreezing point depression°C
KfCryoscopic constant°C·kg/mol
ivan't Hoff factor—
wSolute massg
m_solvSolvent masskg
bMolalitymol/kg
MMolar massg/mol

Worked example

A student dissolves 5.00 g of an unknown molecular compound in 100 g (0.100 kg) of water and measures a freezing point of −0.31 °C, a depression of ΔTf = 0.31 °C. Water has Kf = 1.86 °C·kg/mol, and a molecular solute has i = 1.

First the molality:

b=0.311×1.86=0.167 mol/kgb = \frac{0.31}{1 \times 1.86} = 0.167\ \text{mol/kg}

Then the molar mass:

M=1×1.86×5.000.31×0.100=300 g/molM = \frac{1 \times 1.86 \times 5.00}{0.31 \times 0.100} = 300\ \text{g/mol}

The unknown has a molar mass of about 300 g/mol.

Why molality, not molarity

Colligative properties are governed by molality — moles of solute per kilogram of solvent — rather than molarity. Molality is defined by mass, which does not change with temperature, so it stays fixed as the solution is cooled toward freezing. Molarity depends on volume, and the solution contracts slightly as it cools, so it would drift during the measurement. Working in molality keeps the depression law exact.

The van't Hoff factor

The factor ii counts the particles each formula unit releases in solution:

SoluteDissociates intoi (ideal)
Sugar, ureaone molecule1
NaClNa⁺ + Cl⁻2
CaCl₂Ca²⁺ + 2 Cl⁻3

For identifying an unknown molecular compound, i = 1 is the normal assumption. If the compound actually dissociates and you leave i at 1, the calculated molar mass comes out too small by roughly the factor i.

Choosing a solvent

Because ΔTf is often only a fraction of a degree, the temperature reading is the limiting source of error. Solvents with a large cryoscopic constant produce a bigger, more measurable drop for the same molality:

SolventKf (°C·kg/mol)Normal fp (°C)
Water1.860
Benzene5.125.5
Camphor~40179

Camphor's very large constant made it the historical solvent of choice for molar-mass determination by the Rast method. Dilute solutions, a pure solvent, and a solute that neither associates nor dissociates unexpectedly all keep the result accurate.

Frequently Asked Questions (FAQ)

How do you find molar mass from freezing point depression?

Start from the freezing point depression law ΔTf = i · Kf · b, where b is the molality (moles of solute per kilogram of solvent). Rearrange for molality, b = ΔTf / (i · Kf). Because molality is the moles of solute divided by the solvent mass in kilograms, the moles of solute equal b times the solvent mass. Dividing the measured solute mass by those moles gives the molar mass.

Combined into one step: M = (i · Kf · w) / (ΔTf · m_solvent), where w is the solute mass in grams and m_solvent is the solvent mass in kilograms.

For example, 5 g of a nonelectrolyte dissolved in 0.100 kg of water lowers the freezing point by 0.31 °C. With Kf = 1.86 and i = 1, M = (1 × 1.86 × 5) / (0.31 × 0.100) = 300 g/mol.

Why does this use molality instead of molarity?

Colligative properties depend on molality — moles of solute per kilogram of solvent — rather than molarity — moles per litre of solution. Molality is defined by mass, which does not change with temperature, so it stays constant even as the solution is cooled toward its freezing point. Molarity would drift because the solution volume contracts as it cools. Using molality keeps ΔTf = i · Kf · b exact over the temperature range of the measurement.

What van't Hoff factor should I use?

The van't Hoff factor i is the number of independent particles each formula unit produces when it dissolves. Molecular solutes that do not dissociate — sugar, urea, most organic compounds — have i = 1, and this is the usual choice when you are identifying an unknown molecular compound. Salts split into ions: NaCl gives i ≈ 2 and CaCl₂ gives i ≈ 3, though real values fall slightly below the ideal count because of ion pairing. If you assume i = 1 for a compound that actually dissociates, the molar mass will come out too small by roughly the factor i.

How accurate is molar mass by cryoscopy?

The freezing point depression method — cryoscopy — works best for dilute solutions where ΔTf stays proportional to concentration. The main source of error is the temperature measurement itself: because ΔTf is often only a fraction of a degree, a small error in reading the depression translates into a large error in molar mass. Solvents with a large cryoscopic constant, such as camphor (Kf ≈ 40), magnify the temperature drop and were historically preferred for this reason. Concentrated solutions, association or dissociation of the solute, and impurities in the solvent all reduce accuracy.

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