Freezing point depression calculator with boiling point elevation
This freezing point depression calculator uses ΔTf = i · Kf · m and, for the boiling point, ΔTb = i · Kb · m with the constants from OpenStax Chemistry 2e. Example: a 1 m solution of a non-electrolyte in water freezes at −1.86 °C and boils at 100.51 °C.
Example: Solvent: water · Concentration as: molality m · Molality m 1 mol/kg · van ’t Hoff factor i 1 → Freezing and boiling point: Freezing point -1.86 °C, boiling point 100.51 °C. Source: OpenStax Chemistry 2e, 11.4 Colligative Properties: ΔT_b = K_b·m, ΔT_f = K_f·m, Table 11.2 (K_b, K_f of water, acetic acid, benzene, chloroform, nitrobenzene), Table 11.3 (van’t Hoff factors), Examples 11.7–11.9 and 11.13 (read 09.10.2026). Updated: .
How it is calculated
ΔTf = i × Kf × m. The freezing point of a solution drops in proportion to its molality m, the moles of dissolved particles per kilogram of solvent. Kf is a property of the solvent; for water it is 1.86 °C·kg/mol. The boiling point rises the same way with Kb: ΔTb = i × Kb × m, for water with 0.512. The van ’t Hoff factor i counts the particles per formula unit – 1 for sugar, 2 for NaCl.
Constants (OpenStax Table 11.2, at 1 atm)
| Solvent | Boiling point | Kb | Freezing point | Kf |
|---|---|---|---|---|
| water | 100.0 °C | 0.512 | 0.0 °C | 1.86 |
| acetic acid | 118.1 °C | 3.07 | 16.6 °C | 3.9 |
| benzene | 80.1 °C | 2.53 | 5.5 °C | 5.12 |
| chloroform | 61.26 °C | 3.63 | −63.5 °C | 4.68 |
| nitrobenzene | 210.9 °C | 5.24 | 5.67 °C | 8.1 |
Worked examples
| Solution | m | i | Freezing point | Boiling point |
|---|---|---|---|---|
| 1 m sugar in water | 1 | 1 | −1.86 °C | 100.51 °C |
| 0.33 m in benzene | 0.33 | 1 | 3.81 °C | 80.93 °C |
| 92.1 g I₂ in 800.0 g chloroform | 0.454 | 1 | −65.62 °C | 62.91 °C |
| 4.2 g NaCl in 125 g water | 0.575 | 2 | −2.14 °C | 100.59 °C |
OpenStax rounds these to 80.9 °C, 3.8 °C, 62.91 °C and −2.1 °C (the salt solution is a model of seawater). 1.0 g of glycerin (92.09 g/mol) in 47.8 g of water boils at 100.12 °C, and a 1.85 m solution in nitrobenzene freezes at −9.3 °C.
Electrolytes and the van ’t Hoff factor
Assuming complete dissociation, 1 m NaCl should lower the freezing point of water by 2 × 1.86 = 3.72 °C; OpenStax reports a measured 3.4 °C. Ions attract each other, so the effective number of particles is lower – Table 11.3 gives measured factors of 1.9 for NaCl, 1.3 for MgSO₄ and 2.7 for MgCl₂ at 0.050 m. Enter the measured value for a closer result. The molality itself can be worked out with the molality calculator and the molar mass of the solute with the molar mass calculator.
Frequently asked questions
What is the formula for freezing point depression?
ΔTf = i · Kf · m. For water Kf = 1.86 °C·kg/mol, so a 1 m sugar solution freezes 1.86 °C lower, at −1.86 °C.
How do I calculate the boiling point elevation?
ΔTb = i · Kb · m with Kb = 0.512 for water. 0.33 m in benzene (Kb 2.53) raises 80.1 °C by 0.83 °C to 80.93 °C.
What is the van ’t Hoff factor of NaCl?
Ideally 2 (Na⁺ and Cl⁻). Measured at 0.050 m it is 1.9, because the ions do not act fully independently.
How do I get molality from grams?
Moles of solute divided by kilograms of solvent: 4.2 g NaCl ÷ 58.44 g/mol = 0.072 mol in 0.125 kg water = 0.575 mol/kg. With i = 2 the solution freezes at −2.14 °C.
Why does salt melt ice on roads?
Dissolved salt lowers the freezing point of the water film. CaCl₂ gives three particles per formula unit, NaCl two, so CaCl₂ lowers it more per mole.
How exact is the result?
It is calculated for ideal, dilute solutions of non-volatile solutes. At higher concentrations and with electrolytes the real change is smaller than the estimate.
Sources and legal basis
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