Boiling & Freezing Point
Find how a dissolved solute raises the boiling point and lowers the freezing point of water.
The solution freezes at -3.72 °C, below pure water’s 0 °C.
Freezing point drops by 3.72 °C.
Dissolving 1 m of table salt (NaCl) in water lowers its freezing point by ΔTf = i·Kf·m = 2 × 1.86 × 1 = 3.72 °C, so the solution freezes at −3.72 °C instead of 0 °C. The same solution raises the boiling point by ΔTb = 2 × 0.512 × 1 ≈ 1.0 °C, boiling near 101.0 °C.
What colligative properties are
Boiling-point elevation and freezing-point depression are colligative properties: they depend only on the number of dissolved particles, not on what those particles are. One mole of dissolved salt and one mole of dissolved sugar behave the same per particle — but salt splits into more particles, so it shifts the temperature more.
The size of the shift, ΔT, scales with three things: the van’t Hoff factor i (how many particles each formula unit releases), the solvent’s characteristic constant K, and the molality m of the solution. Boiling point goes up by ΔTb; freezing point comes down by ΔTf.
i = van’t Hoff factor · K = Kb (boiling) or Kf (freezing) · m = molality (mol/kg)
Worked example
Find the freezing and boiling points of 1 m NaCl in water (i = 2, Kf = 1.86 °C/m, Kb = 0.512 °C/m).
- 1 Find the van’t Hoff factor. NaCl dissociates into Na⁺ and Cl⁻, so each formula unit makes 2 particles: i = 2.
- 2 Pick the constant for water. Use Kf = 1.86 °C/m for freezing and Kb = 0.512 °C/m for boiling.
- 3 Multiply for the freezing shift. ΔTf = i · Kf · m = 2 × 1.86 × 1 = 3.72 °C. The freezing point drops, so it falls from 0 °C to −3.72 °C.
- 4 Multiply for the boiling shift. ΔTb = i · Kb · m = 2 × 0.512 × 1 = 1.024 °C. The boiling point rises from 100 °C to 101.02 °C.
Water constants and van’t Hoff factors
Kb and Kf are for water; the van’t Hoff factor i is the ideal particle count per formula unit.
| Quantity | Value | Meaning |
|---|---|---|
| Kb (water) | 0.512 °C/m | Boiling-point elevation per unit molality |
| Kf (water) | 1.86 °C/m | Freezing-point depression per unit molality |
| Normal bp / fp | 100 °C / 0 °C | Pure-water reference points |
| Sugar (sucrose), i | 1 | Dissolves without splitting into ions |
| NaCl, i | 2 | Splits into Na⁺ + Cl⁻ |
| CaCl₂, i | 3 | Splits into Ca²⁺ + 2 Cl⁻ |
Why it matters: salt on roads and antifreeze
Spreading salt on icy roads works because it lowers water’s freezing point — the brine stays liquid below 0 °C, so ice melts. CaCl₂ is even more effective than NaCl because it releases three particles per formula unit instead of two, giving a larger ΔTf for the same molality.
Engine antifreeze (ethylene glycol) uses the same idea from both directions: it depresses the freezing point so the coolant doesn’t freeze in winter and elevates the boiling point so it doesn’t boil over in summer. Because these effects count particles rather than identities, the chemistry is identical whether the solute is salt, sugar, or glycol — only i and m change the magnitude.