Freezing Point Depression Calculator — ΔTf = Kf × m × i

Salt on icy roads. Antifreeze in car radiators. The creamy texture of ice cream. All of these rely on the same chemistry: dissolved solutes lower the freezing point of a solvent. The CalcHub Freezing Point Depression Calculator quantifies exactly how much, using the same elegant formula that chemists have used for over a century.

The Formula

ΔTf = Kf × m × i

Where:

• ΔTf = freezing point depression (°C) — always a positive number


• Kf = cryoscopic constant of the solvent


• m = molality (mol of solute per kg of solvent)


• i = van't Hoff factor

New freezing point = normal freezing point − ΔTf

Kf Values for Common Solvents

Solvent	Normal Freezing Point (°C)	Kf (°C·kg/mol)
Water	0.0	1.853
Benzene	5.5	5.12
Acetic acid	16.6	3.90
Cyclohexane	6.6	20.0
Camphor	178.4	37.7

Camphor's enormous Kf makes it useful for molecular weight determination — even tiny solute amounts produce measurable freezing point changes.

How to Use the Calculator

1. Open CalcHub and navigate to the Freezing Point Depression Calculator.
2. Enter the solute molality in mol/kg.
3. Select or enter the Kf of your solvent.
4. Enter the van't Hoff factor (1 for non-electrolytes, 2 for NaCl, 3 for CaCl₂, etc.).
5. Get ΔTf and the new freezing point immediately.

Real-World Examples

Road salt on ice: Road salt is often CaCl₂ (i = 3). Applying 0.5 mol/kg to water: ΔTf = 1.853 × 0.5 × 3 = 2.78°C Effective down to about −3°C from this concentration, higher concentrations go lower. Car antifreeze (ethylene glycol, i = 1): At 6 mol/kg (a 50/50 water-glycol mix by volume): ΔTf = 1.853 × 6 × 1 ≈ −11°C below 0°C

Higher concentrations push this to −37°C or lower.

Application	Solute	Approx ΔTf
Road deicing	NaCl (0.5 m)	−0.9°C
Road deicing	CaCl₂ (0.5 m)	−2.8°C
Car antifreeze (50%)	Ethylene glycol	−37°C
Homemade ice cream	Salt in ice	−5 to −10°C

Why Salt Makes Ice Colder for Ice Cream

When you pack ice cream in a salt-ice mixture, the salt depresses the freezing point of the ice-water slurry to below 0°C. The mixture absorbs heat from the cream to keep melting, chilling the cream below 0°C and freezing it. It's not magic — it's colligative properties.

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Why is Kf for water (1.853) larger than Kb (0.512)?

It's just a property of water's specific molecular structure and thermodynamics at its freezing vs. boiling points. Different phase transitions have different sensitivities. Other solvents show different patterns — benzene has Kf = 5.12 and Kb = 2.53.

Does freezing point depression depend on what the solute is?

No — only on the number of particles dissolved per kg of solvent. That's the definition of a colligative property. 1 mol of glucose and 1 mol of urea in 1 kg of water produce identical freezing point depressions (assuming no dissociation).

How do I find the molar mass of an unknown substance using this?

Weigh a known mass of solute, dissolve it in a known mass of solvent, measure the freezing point change. Then: M = (Kf × mass_solute) / (ΔTf × mass_solvent). This is cryoscopy — a classical technique for molar mass determination.

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Related calculators: Boiling Point Calculator · Molarity Calculator · Concentration Converter