Boiling Point Elevation and Freezing Point Depression Calculator
Choose boiling or freezing, pick the solvent, and enter the molality and van 't Hoff factor to get the temperature change and the new boiling or freezing point.
Updated
Report a problem
Formula
ΔT = i × K × m
ΔTis the change in boiling or freezing pointKisKb(boiling) orKf(freezing) in °C kg/mol, specific to the solventmis the molality of the solute in mol/kgiis the van 't Hoff factor (particles per formula unit)
New boiling point = pure boiling point + ΔTb. New freezing point = pure freezing point - ΔTf.
| Solvent | Kb | Kf | Boils at | Freezes at |
|---|---|---|---|---|
| Water | 0.512 | 1.86 | 100 °C | 0 °C |
| Ethanol | 1.22 | 1.99 | 78.37 °C | -114.1 °C |
| Benzene | 2.53 | 5.12 | 80.1 °C | 5.5 °C |
| Acetic acid | 3.07 | 3.90 | 118.1 °C | 16.6 °C |
| Cyclohexane | 2.79 | 20.0 | 80.7 °C | 6.5 °C |
| Chloroform | 3.63 | 4.68 | 61.2 °C | -63.5 °C |
Worked example
1 m NaCl in water (i = 2):
- boiling:
ΔT = 2 × 0.512 × 1 =1.024 °C, new boiling point 101.024 °C - freezing:
ΔT = 2 × 1.86 × 1 =3.72 °C, new freezing point -3.72 °C
Benzene, 0.5 m non-electrolyte (i = 1): ΔTf = 0.5 × 5.12 = 2.56 °C, so it freezes at 5.5 - 2.56 = 2.94 °C.
Limits
Kb and Kf are tabulated values from standard general-chemistry tables (for example OpenStax). The formula holds for dilute solutions; the calculator reports an error when the new freezing point would be below -273.15 C or the elevation would exceed 100 K.
Frequently asked questions
What is the boiling point elevation formula?
ΔTb = i × Kb × m, where m is the molality, Kb the solvent's ebullioscopic constant and i the van 't Hoff factor. For 1 m NaCl (i = 2) in water: 2 × 0.512 × 1 = 1.024 °C, so it boils at about 101.024 °C.
How do I find the freezing point of a salt solution?
Use ΔTf = i × Kf × m and subtract it from the pure freezing point. For 1 m NaCl in water: 2 × 1.86 × 1 = 3.72 °C, so it freezes at about -3.72 °C.
Why does the answer depend on the solvent?
Each solvent has its own Kb and Kf. Benzene has Kf = 5.12, so 0.5 m of a non-electrolyte lowers its freezing point by 0.5 × 5.12 = 2.56 °C, from 5.5 °C to 2.94 °C. The same solution in water would drop only 0.93 °C.
How accurate are the results?
The law is for ideal dilute solutions of non-volatile solutes, and the constants are tabulated values from standard general-chemistry tables (for example OpenStax) that differ slightly between sources. The formula holds for dilute solutions only, so the calculator reports an error if the freezing point would fall below -273.15 C or the elevation would exceed 100 K. Concentrated or strongly ionic solutions deviate, because the real i is below the ideal count.