A freezing-point depression calculator that applies the standard colligative-property formula to find any one of four quantities — the depression in freezing point ΔTf, solution molality, the solvent’s cryoscopic constant Kf, or the van ‘t Hoff factor i — given the other three. Seven common solvents are pre-loaded with their literature Kf values and normal freezing points; a Custom option accepts any solvent. Every result comes with the new freezing point of the solution and a numbered step-by-step working trace.
The formula
Freezing-point depression is one of the four colligative properties — solution properties that depend only on the number of dissolved particles, not on what those particles are. The governing equation is:
ΔTf = i × Kf × m
- ΔTf — the decrease in freezing point (°C). The solution freezes at (Tf,pure − ΔTf).
- i — the van ‘t Hoff factor, counting the number of particles released per formula unit on dissolution. Non-electrolytes: i = 1. NaCl: i = 2. CaCl₂: i = 3.
- Kf — the cryoscopic (molal freezing-point depression) constant of the pure solvent (°C·kg/mol). This is a fixed physical constant for each solvent — water’s value is 1.853, camphor’s is 37.7.
- m — the molality of the solute (mol of solute per kg of pure solvent).
Rearranging gives three further solve-for forms, all available in the dropdown above:
- m = ΔTf / (i × Kf)
- Kf = ΔTf / (i × m)
- i = ΔTf / (Kf × m)
How it works
Select what you want to find, choose your solvent (or enter a custom Kf), fill in the three known quantities, and the calculator produces the answer plus a line-by-line arithmetic trace. Where the solvent is known, the new freezing point of the solution (pure-solvent f.p. minus ΔTf) is shown immediately below the result — useful for quickly checking whether a road-salt mixture will work at a given temperature, or confirming that an antifreeze concentration is safe for winter conditions.
Worked example
Problem: 11.7 g of NaCl (M = 58.44 g/mol) is dissolved in 500 g of water. What is the new freezing point of the solution?
- Moles of NaCl: n = 11.7 / 58.44 = 0.2002 mol
- Molality: m = 0.2002 / 0.500 = 0.4003 mol/kg
- Van ‘t Hoff factor for NaCl: i = 2 (Na⁺ + Cl⁻)
- Kf for water: 1.853 °C·kg/mol
- ΔTf = 2 × 1.853 × 0.4003 = 1.483 °C
- New freezing point: 0 − 1.483 = −1.483 °C
Enter these values in the calculator (select Water, set i = 2, m = 0.4003, solve for ΔTf) and the working section confirms every step. To reproduce the molality from scratch, use the linked molality calculator.
Cryoscopic constants reference table
| Solvent | Normal f.p. (°C) | Kf (°C·kg/mol) |
|---|---|---|
| Water (H₂O) | 0.0 | 1.853 |
| Benzene (C₆H₆) | 5.5 | 5.12 |
| Cyclohexane (C₆H₁₂) | 6.5 | 20.0 |
| Acetic acid (CH₃COOH) | 16.6 | 3.9 |
| Camphor (C₁₀H₁₆O) | 178.8 | 37.7 |
| Ethanol (C₂H₅OH) | −114.1 | 1.99 |
| Chloroform (CHCl₃) | −63.5 | 4.68 |
Camphor’s exceptionally large Kf makes it the solvent of choice in the Rast method for molar-mass determination: even a small mass of unknown compound produces a measurable depression, reducing experimental error.
Why colligative properties depend on particle count
When a solute dissolves, its particles occupy positions at the liquid surface and disrupt the regular arrangement that allows solvent molecules to escape into the solid phase. More dissolved particles mean a larger disruption and a lower equilibrium freezing temperature. Because the effect scales purely with how many particles are present (not their size, charge, or identity), the same formula works for sugar, salt, antifreeze, or any other non-volatile solute — only i and m change.
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