Freezing-point depression: antifreeze and de-icing explained

What freezing-point depression is

Pure water freezes at 0 °C. Dissolve enough salt or glycol in it and it stays liquid at −10 °C, −20 °C, or lower. Nothing about the water molecules changed — what changed is how hard it is for them to lock into ice.

Freezing-point depression is the lowering of a solvent's freezing point by dissolved solute. It is the colligative property behind antifreeze, road salt, and ice-cream churns, and it is the most commonly measured of the four colligative effects because freezing points are easy to pin down precisely.

Why solute lowers the freezing point

Freezing is an ordering process: liquid water molecules must arrange into the hexagonal lattice of ice. Dissolved particles — ions, sugar molecules, glycol — physically get in the way of that ordering. With the lattice disrupted, the temperature must fall further before the solid phase becomes thermodynamically favorable.

Equivalently, the solute lowers the liquid's vapor pressure while the solid ice's vapor pressure is unchanged; the freezing point is where the two vapor-pressure curves meet, and lowering the liquid curve moves that intersection to a lower temperature. Both pictures give the same formula.

The formula: ΔTf = i·Kf·m

The quantitative form mirrors boiling-point elevation:

$$\Delta T_f = i \cdot K_f \cdot m$$

• $\Delta T_f$: the freezing-point drop in °C (a positive number; subtract it from the pure solvent's freezing point).
• $i$: the van't Hoff factor (2 for NaCl, 3 for CaCl2, 1 for nonelectrolytes like glycol).
• $K_f$: the solvent's freezing-point depression constant — 1.86 °C·kg/mol for water.
• $m$: molality in mol/kg of solvent.

Note that $K_f$ is larger than $K_b$ for water (1.86 vs 0.512), so freezing points respond more dramatically than boiling points to the same concentration.

Worked example: antifreeze at 1.00 m

Calculate the freezing point of a 1.00 m aqueous ethylene glycol solution. Glycol does not dissociate, so $i = 1$; water's $K_f = 1.86$ °C·kg/mol.

1. Substitute: $\Delta T_f = 1 \times 1.86 \times 1.00 = 1.86$ °C.
2. Subtract from water's freezing point: $0.00 - 1.86 = -1.86$ °C.

The solution freezes at −1.86 °C. Scale up to real antifreeze: a 50/50 glycol-water mix is roughly 16 m in glycol, giving an ideal $\Delta T_f$ near 30 °C — real mixtures freeze around −37 °C once non-ideality is accounted for. The 1.00 m example shows the mechanism; the concentrated coolant shows its power.

For comparison, 1.00 m NaCl ($i \approx 2$) gives $\Delta T_f \approx 3.72$ °C and freezes near −3.7 °C — about twice the depression per mole, because each formula unit yields two particles.

De-icing roads and churning ice cream

Spread NaCl on an icy road and it dissolves into the thin water film on the ice, creating a concentrated brine whose freezing point sits well below 0 °C. Ice in contact with that brine is now above its (depressed) freezing point, so it melts — diluting the brine until a new equilibrium is reached.

The limit is solubility and temperature: NaCl's effectiveness fades near −10 °C, while CaCl2 ($i \approx 3$, plus an exothermic heat of solution) keeps working toward −25 °C. Ice-cream makers exploit the same physics deliberately: a salt-ice bath around the churn can reach −20 °C, freezing the cream mixture far faster than ice alone could.

Cryoscopy: weighing molecules with a thermometer

Freezing-point measurements run the formula in reverse to identify unknowns — a technique called cryoscopy. Dissolve a weighed sample of an unknown compound in a known mass of solvent, measure how far the freezing point drops, and solve:

$$m = \frac{\Delta T_f}{i \cdot K_f}$$

Then moles = $m \times$ kg of solvent, and molar mass = mass / moles. Because freezing points can be measured to hundredths of a degree with simple apparatus, cryoscopy was historically one of the most accessible routes to molecular weights, and it remains a standard teaching-lab experiment.

Errors to avoid

The classic errors: using molarity instead of molality, forgetting $i$ for electrolytes, and adding $\Delta T_f$ to the freezing point instead of subtracting it. $\Delta T_f$ is defined as a positive drop — the new freezing point is $T_f^\circ - \Delta T_f$.

Also keep the solute requirements in mind: it must dissolve (undissolved solid contributes nothing) and must not incorporate into the solvent's crystal. And remember the direction pairing one final time: solute depresses freezing points and elevates boiling points. Mixing those two directions up is the most common conceptual slip in the colligative unit.

Variables and units

Common mistakes

Real-world applications

• Automotive antifreeze: ethylene glycol keeps coolant liquid below 0 °C

• Road de-icing: salts melt ice by lowering water's freezing point

• Ice cream making: salt-ice baths reach well below 0 °C for churning

• Aviation de-icing fluids for aircraft wings

• Cryoscopy: determining molar masses of unknown compounds in the lab

• Refrigeration brines: calcium chloride solutions as low-temperature heat transfer fluids