What is Freezing Point Depression?
When a nonvolatile solute is dissolved in a volatile solvent, the freezing point of the resulting solution is always lower than that of the pure solvent. This phenomenon is called freezing point depression — one of the four colligative properties of solutions (along with boiling point elevation, vapor pressure lowering, and osmotic pressure).
The effect depends only on the number of dissolved particles, not on their chemical nature. That is why both sugar and salt lower the freezing point of water, but salt is more effective because it dissociates into two ions (Na⁺ and Cl⁻), effectively doubling the number of solute particles.
Freezing Point Depression Formula
The change in freezing point is described by:
Where:
- ΔTf — decrease in freezing point (°C or K)
- Kf — cryoscopic constant of the solvent (°C·kg/mol)
- m — molality of the solution (mol of solute per kg of solvent)
- i — van't Hoff factor (number of particles the solute produces per formula unit)
The new freezing point of the solution is:
How to Calculate Freezing Point Depression
- Choose your solvent and note its cryoscopic constant Kf and pure freezing point Tf°.
- Calculate molality: m = (mass of solute ÷ molar mass of solute) ÷ (mass of solvent in kg).
- Determine the van't Hoff factor i: 1 for nonelectrolytes, 2 for 1:1 electrolytes like NaCl, 3 for MgCl₂, etc.
- Apply the formula: ΔTf = Kf × m × i.
- Subtract ΔTf from the pure solvent's freezing point.
The Van't Hoff Factor
The van't Hoff factor i accounts for the degree of dissociation or association of the solute:
- Nonelectrolytes (glucose, sugar, urea): i = 1
- NaCl, KCl, KBr (fully dissociate into 2 ions): i ≈ 2
- MgCl₂, CaCl₂ (dissociate into 3 ions): i ≈ 3
- AlCl₃ (dissociates into 4 ions): i ≈ 4
- For real solutions, i may be a non-integer due to ion pairing.
Molal Freezing Point Depression Constants (Kf)
| Solvent | Kf (°C·kg/mol) | Freezing Point (°C) |
|---|---|---|
| Water | 1.853 | 0.0 |
| Benzene | 5.12 | 5.5 |
| Camphor | 39.7 | 179.8 |
| Acetic Acid | 3.9 | 16.6 |
| Nitrobenzene | 7.00 | 5.7 |
| Cyclohexane | 20.8 | 6.5 |
| Phenol | 7.27 | 40.9 |
| Naphthalene | 6.89 | 80.2 |
| Chloroform | 4.68 | −63.5 |
| Ethanol | 1.99 | −114.1 |
| Diethyl Ether | 1.79 | −116.3 |
Examples of Freezing Point Depression
Example 1 — Sugar in Water
Dissolving 10 g of glucose (M = 180.16 g/mol) in 100 g of water (Kf = 1.853 °C·kg/mol, i = 1):
- Molality: m = (10/180.16) / (100/1000) = 0.0555 / 0.1 = 0.5551 mol/kg
- ΔTf = 1.853 × 0.5551 × 1 = 1.0286 °C
- New freezing point: 0 − 1.0286 = −1.03 °C (28.15 °F)
Example 2 — Salt on Roads
Dissolving 58.44 g of NaCl (M = 58.44 g/mol, i = 2) in 1000 g of water:
- Molality: m = (58.44/58.44) / 1.0 = 1.00 mol/kg
- ΔTf = 1.853 × 1.00 × 2 = 3.706 °C
- New freezing point: 0 − 3.706 = −3.71 °C (25.32 °F)
FAQs
- Why does salt melt ice on roads?
- Salt (NaCl) dissolves in the thin film of water on ice and dissociates into Na⁺ and Cl⁻ ions (van't Hoff factor ≈ 2), significantly lowering the freezing point below 0 °C so the ice melts at typical winter temperatures.
- Does freezing point depression depend on the type of solute?
- No — it is a colligative property. It depends only on the number of solute particles per kilogram of solvent, not on the chemical identity of the solute.
- What is the cryoscopic constant?
- The cryoscopic constant Kf is a property of the solvent. It represents the decrease in freezing point when one mole of solute is dissolved in one kilogram of the solvent.
- Can I use this calculator for antifreeze concentrations?
- Yes. Ethylene glycol antifreeze uses the same principle. Enter the glycol as the solute with its molar mass (62.07 g/mol) and water as the solvent. Use i = 1 since glycol is a nonelectrolyte.
- What units does this calculator support?
- Metric (grams, °C) and US customary (ounces, °F). Molar mass is always in g/mol.