Freezing Point Depression
Calculate the freezing point depression of a solution from the cryoscopic constant, molality, and van't Hoff factor.
About this calculator
Freezing point depression is a colligative property — it depends on how many dissolved particles are floating in the solvent, not on what those particles chemically are. This calculator applies ΔTf = i × Kf × m: the cryoscopic constant Kf is a property of the solvent alone (water's is 1.86 °C·kg/mol), molality is moles of solute per kilogram of solvent, and the van't Hoff factor i accounts for how many particles each formula unit actually splits into once dissolved. That last factor is where most of the real-world complexity — and most input errors — live: table sugar stays as one intact molecule so i = 1, but NaCl dissociates into Na⁺ and Cl⁻ so i = 2, and CaCl₂ splits into three ions so i = 3. Using i = 1 for an ionic compound is the single most common mistake, and it understates the actual freezing point drop by exactly that dissociation factor.
The calculator subtracts the computed ΔTf from the solvent's normal freezing point to report the new, lower freezing point directly — useful for road-salting calculations, antifreeze mixtures, or lab solution prep where you need to know exactly how cold a solution can get before it solidifies. It also reports the "effective particle molality" (i × m), which is the number that actually drives the physics; two solutions with the same molality but different van't Hoff factors will freeze at different temperatures even though they have the same molar concentration of solute. This model assumes ideal, dilute-solution behavior — at high concentrations, ion pairing and other interactions mean the real i drifts below its theoretical value, so measured freezing points in concentrated solutions can come in less depressed than this straightforward calculation predicts.
Inputs
Results
Freezing Point Depression (ΔTf)
1.86 °C
New Freezing Point
-1.86 °C
How to Use This Calculator
- Enter the cryoscopic constant (Kf) for your solvent — water is 1.86 °C·kg/mol.
- Set molality (mol/kg) and van't Hoff factor (i).
- Enter the normal freezing point of the pure solvent.
- Review Freezing Point Depression (ΔTf) and New Freezing Point.
How the result changes with Cryoscopic Constant (Kf)
| Cryoscopic Constant (Kf) | Freezing Point Depression (ΔTf) | New Freezing Point |
|---|---|---|
| 0.93 | 0.93 °C | -0.93 °C |
| 1.4 | 1.4 °C | -1.4 °C |
| 2.79 | 2.79 °C | -2.79 °C |
| 4.65 | 4.65 °C | -4.65 °C |
What each input means
- Cryoscopic Constant (Kf)
- Freezing point depression constant for the solvent (water Kf = 1.86 °C·kg/mol)
- Molality
- Moles of solute per kilogram of solvent
- van't Hoff Factor (i)
- Number of particles per formula unit in solution (NaCl = 2, CaCl₂ = 3, sugar = 1)
- Normal Freezing Point
- Freezing point of the pure solvent (water = 0 °C)
How this is calculated
Formula
ΔTf = i × Kf × mWorked example, using the default values
- Identify Input Parameters4 parametersCryoscopic Constant (Kf) = 1.86, Molality = 1, van't Hoff Factor (i) = 1, Normal Freezing Point = 0 = 4 input(s) provided
- Calculate Freezing Point DepressionFreezing Point Depression1.86 = 1.86
- Calculate New Freezing PointNew Freezing Point-1.86 = -1.86
- Calculate Effective Particle MolalityEffective Particle Molality1 = 1
Engine last updated . Checked against 1 independently-derived test — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
How do I choose the right van't Hoff factor for my solute?
It's the number of particles one formula unit breaks into when it dissolves: molecular compounds that stay intact, like table sugar (sucrose), use i = 1, while ionic compounds dissociate — NaCl splits into Na⁺ and Cl⁻ for i = 2, MgCl₂ splits into three ions for i = 3, and so on. Using i = 1 for an ionic compound is the most common error and will understate the freezing point drop, since the calculator multiplies i directly into ΔTf = i × Kf × m.
Why does the same molality of two different solutes give different freezing points?
Freezing point depression is a colligative property, driven by the total number of dissolved particles, not the identity of the solute — the calculator's 'effective particle molality' output (i × m) is the number that actually determines ΔTf. Two solutes at identical molality but different van't Hoff factors have different effective particle molalities, so they freeze at different temperatures even though their molar concentrations match.
Where do I find the cryoscopic constant (Kf) for my solvent?
Kf is a fixed property of the solvent itself, not the solute — water's is 1.86 °C·kg/mol, which is the calculator's default, but other solvents like benzene or camphor have their own published Kf values you'd need to look up and enter in place of water's. Using water's Kf for a non-aqueous solution will give a physically meaningless result.
Why might my measured freezing point differ from what this calculator predicts?
This model assumes ideal, dilute-solution behavior where the van't Hoff factor is exactly its theoretical value. At higher concentrations, ions in solution start pairing up and interacting with each other, so the effective van't Hoff factor drifts below the theoretical number you'd enter here, and the real freezing point ends up less depressed than the straightforward i × Kf × m calculation predicts.
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