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Freezing Point Depression Calculator

Find how much a solute lowers the freezing point.

Result

-5.58 °C
5.58 °C

How it is calculated

The tool applies ΔTf = i × Kf × m. Here i is the van't Hoff factor (particles per formula unit), Kf is the solvent's cryoscopic constant in °C·kg/mol (1.86 for water) and m is the molality in mol of solute per kg of solvent. It then reports the new freezing point as 0 − ΔTf, so it assumes the pure solvent freezes at 0 °C, which is correct for water. For another solvent, subtract ΔTf from its own freezing point yourself. If the van't Hoff factor is left empty or is zero or negative, the tool uses 1, as for a non-dissociating solute. Negative Kf or molality gives no result. Results are rounded to two decimals.

Worked example

A 1.5 mol/kg solution of table salt in water: Kf is 1.86 and NaCl splits into two ions, so i = 2. Multiply 2 × 1.86 × 1.5 = 5.58 °C of depression. The solution freezes at 0 − 5.58 = -5.58 °C. If you left the van't Hoff factor blank, the tool would use i = 1 and report only 2.79 °C.

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Common mistakes

  • Using molarity (mol per litre of solution) instead of molality (mol per kg of solvent). They differ noticeably in concentrated solutions.
  • Leaving the van't Hoff factor blank for salts. It then defaults to 1 and the depression is half of what NaCl really gives in the ideal model.
  • Applying the water value of Kf, 1.86, to a different solvent such as benzene.
  • Dividing the solute mass by the whole solution mass. The kilograms must be of the solvent only.
  • Reading the freezing point of a non-water solvent from the result. The tool subtracts from 0 °C.

About this calculator

This calculator finds freezing point depression using the colligative property formula ΔTf = i·Kf·m, where i is the van't Hoff factor (number of particles the solute dissociates into), Kf is the solvent's cryoscopic (molal freezing point depression) constant, and m is the solution's molality. Subtracting ΔTf from the pure solvent's freezing point gives the solution's actual freezing point.

Chemistry students use it in lab to determine an unknown solute's molar mass from measured freezing point depression, and it explains real-world uses like why salt lowers the freezing point of icy roads and why antifreeze (ethylene glycol) protects a car's radiator.

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Comments (1)

  • Editorial team

    Real solutions behave less ideally than the formula, especially above about 1 mol/kg, where ions pair up and the true van't Hoff factor drops below the textbook count. Salt on an icy road is a good example: the number here is the most you could expect, not a promise. The tool also takes 0 degrees as the solvent's freezing point, so it only suits water. Which solute and solvent do you want to try?

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