Freezing Point Depression

Lowering a solvent's freezing point by adding solute

The Formula

ΔTf = i Kf m
Tf,solution = Tf,pure - ΔTf

Variables

  • ΔTf = freezing point lowering (C)
  • i = van't Hoff factor
  • Kf = cryoscopic constant (C·kg/mol)
  • m = molality (mol solute / kg solvent)

Typical Kf Values

  • Water: 1.86 C·kg/mol
  • Benzene: 5.12 C·kg/mol
  • Acetic acid: 3.90 C·kg/mol

Step-by-Step Example

Problem:

What is the freezing point of a solution made by dissolving 0.500 mol NaCl in 1.00 kg water? (Assume i = 1.9, Kf water = 1.86 C·kg/mol)

1) Molality

m = 0.500 mol / 1.00 kg = 0.500 m

2) Calculate ΔTf

ΔTf = i Kf m = 1.9 × 1.86 × 0.500 = 1.77 C

3) New freezing point

Tf,solution = 0.00 C - 1.77 C = -1.77 C

Answer:

The solution freezes at -1.77 C (ideal assumption).

Common Mistakes to Avoid

Using molarity

Use molality (mol/kg solvent), not molarity.

Ignoring i

Electrolytes dissociate; include the van't Hoff factor.

Wrong units

Keep Kf in C·kg/mol and m in mol/kg.

Non-ideal behavior

At high concentration, i is lower than the ideal integer.

Frequently Asked Questions

What is i for NaCl?

Ideal i = 2, but effective i is typically 1.8-1.9 at moderate concentration.

Why use molality?

Molality is temperature independent; molarity changes with thermal expansion.

Does this work for nonelectrolytes?

Yes, with i = 1 for nonelectrolytes like glucose.