About the Freezing Point Depression Calculator
This freezing point depression calculator works out how much a dissolved solute lowers the freezing point of a solvent, using ΔTf = i·Kf·m. Enter the molality directly, or give the mass of solute, its molar mass and the mass of solvent, and it returns the depression and the new freezing point. It also runs the classic cryoscopy experiment in reverse: from a measured freezing-point drop it calculates the molar mass of an unknown solute.
It is built for chemistry students studying colligative properties, teachers preparing lab exercises and anyone curious about why salt melts ice or antifreeze protects engines. Common solvents — water, benzene, cyclohexane, camphor and more — are built in with their cryoscopic constants, or you can enter your own Kf.
The van ’t Hoff factor i is the number of particles each formula unit produces in solution: 1 for sugars and other non-electrolytes, about 2 for NaCl, 3 for CaCl₂. These are ideal values; real electrolytes form ion pairs and give somewhat smaller depressions, especially above about 0.1 mol/kg. The equation also assumes a dilute solution and a solute that does not freeze out with the solvent.
How to use the freezing point depression calculator
- 1Choose what you know: masses, molality, or a measured freezing point drop.
- 2Pick the solvent, or select Custom and enter its Kf and freezing point.
- 3Enter the solute and solvent amounts (or the molality, or the measured ΔTf).
- 4Set the van ’t Hoff factor — 1 for non-electrolytes, about 2 for NaCl.
- 5Read the freezing point depression, the new freezing point or the molar mass.
Formula and method
Freezing point depression is a colligative property: it depends on how many solute particles are dissolved, not what they are. The drop in freezing point equals the van ’t Hoff factor i times the solvent’s cryoscopic constant Kf times the molality m (moles of solute per kilogram of solvent). The new freezing point is the pure solvent’s freezing point minus ΔTf.
Molality is used rather than molarity because it does not change with temperature. Rearranging the same equation gives the molality from a measured ΔTf, and dividing the solute mass by moles (molality × kg of solvent) gives its molar mass — the basis of cryoscopic molar-mass determination. Solvents with large Kf, such as camphor, give bigger, easier-to-measure drops.
- ΔTf
- Freezing point depression, °C or K
- i
- Van ’t Hoff factor (particles per formula unit)
- Kf
- Cryoscopic constant of the solvent, K·kg/mol
- m
- Molality, mol solute per kg solvent
- M
- Molar mass of the solute, g/mol
Worked examples
10 g of glucose in 100 g of water
10 g ÷ 180.156 g/mol = 0.05551 mol in 0.100 kg of water is 0.5551 mol/kg. ΔTf = 1 × 1.86 × 0.5551 = 1.03 °C, so the solution freezes at about −1.03 °C.
1 mol/kg sodium chloride in water
NaCl splits into Na⁺ and Cl⁻, so i ≈ 2. ΔTf = 2 × 1.86 × 1 = 3.72 °C, giving an ideal freezing point of −3.72 °C (measured values are slightly smaller because of ion pairing).
Molar mass of an unknown in benzene
A 1.00 °C drop in benzene means m = 1.00 ÷ 5.12 = 0.1953 mol/kg, so 0.050 kg holds 0.009766 mol. 1.25 g ÷ 0.009766 mol = 128 g/mol — consistent with naphthalene (128.17 g/mol).
Frequently asked questions
What is the formula for freezing point depression?+
ΔTf = i × Kf × m, where i is the van ’t Hoff factor, Kf is the solvent’s cryoscopic constant and m is the molality in mol/kg. Subtract ΔTf from the pure solvent’s freezing point to get the new one.
What is the Kf of water?+
Water’s cryoscopic constant is 1.86 K·kg/mol (more precisely about 1.853). Each mole of dissolved particles per kilogram of water lowers the freezing point by roughly 1.86 °C.
Why does salt lower the freezing point of water more than sugar?+
Freezing point depression depends on the number of dissolved particles. Each NaCl unit splits into two ions (i ≈ 2), while sugar stays as one molecule (i = 1), so the same molality of salt has about twice the effect.
Why use molality instead of molarity?+
Molality is based on the mass of solvent, which does not change with temperature, while molarity depends on solution volume, which expands and contracts. Colligative property equations are derived in terms of molality.
How is freezing point depression used to find molar mass?+
Dissolve a weighed amount of the unknown in a known mass of solvent and measure ΔTf. Molality = ΔTf ÷ (i·Kf); moles = molality × kg of solvent; molar mass = grams ÷ moles.