Freezing Point Depression: Formula and Solved Example

What Is Freezing Point Depression?

Freezing point depression is the decrease in a solvent’s freezing temperature when a solute is dissolved in it. It is a colligative property because, for a dilute ideal solution, the effect depends on the number of dissolved particles rather than on their chemical identity.

Dissolved particles lower the chemical potential of the liquid solvent. Equilibrium between the liquid and solid phases is therefore reached at a lower temperature.

Other colligative properties include vapor-pressure lowering, boiling-point elevation, and osmotic pressure.

Freezing Point Depression Formula

The magnitude of the freezing point depression is

\[\Delta T_f = i K_f \, m\]

where:

  • $\Delta T_f$ is the decrease in freezing temperature;
  • $i$ is the van ’t Hoff factor, representing the effective number of dissolved particles produced by each solute unit;
  • $K_f$ is the cryoscopic or molal freezing-point-depression constant of the solvent;
  • $m$ is the molality of the solute.

For water,

\[K_f = 1.86\,\mathrm{K\,kg\,mol^{-1}}.\]

Molality is defined as

\[m = \frac{n_{\text{solute}}}{m_{\text{solvent}}(\mathrm{kg})}.\]

When Does the Formula Apply?

The relation $\Delta T_f=iK_fm$ is a dilute-solution approximation. It applies when:

  • the solute is effectively nonvolatile;
  • the solution is sufficiently dilute and behaves approximately ideally;
  • solvent and solute do not undergo a chemical reaction that changes the species present;
  • $K_f$ is the constant for the chosen solvent;
  • the factor $i$ represents the effective number of dissolved particles, including dissociation or association.

How to Calculate Freezing Point Depression: Solved Example

A solution is prepared by dissolving 10.0 g of NaCl in 200 g of water.

  1. Calculate the freezing point depression $\Delta T_f$.
  2. Assume complete dissociation of NaCl.
  3. Use $K_f(\text{H}_2O) = 1.86\, K\,kg\,mol^{-1}$.

Step-by-Step Solution

Step 1. Moles of solute
Molar mass NaCl = $58.44\, g\,mol^{-1}$
\(n = \frac{10.0}{58.44} = 0.171\, mol\)

Step 2. Molality
Mass of solvent = $200 g = 0.200 kg$
\(m = \frac{0.171}{0.200} = 0.855\, mol\,kg^{-1}\)

Step 3. van ’t Hoff factor
NaCl dissociates ideally into 2 ions ($Na^+, Cl^-$), so:
\(i = 2\)

Effective molality:
\(m_{\text{eff}} = i m = 2 \times 0.855 = 1.71\, mol\,kg^{-1}\)

Step 4. Freezing point depression
\(\Delta T_f = K_f m_{\text{eff}} = (1.86)(1.71) = 3.18\, K\)

So the new freezing point is:
\(T_f = 273.15 - 3.18 = 269.97\,K \;\approx -3.2^\circ C\)

Notes

  • The assumption of complete dissociation is an idealization; in real solutions the van ’t Hoff factor $i$ is slightly less than 2 due to ion pairing.
  • Colligative properties provide a powerful experimental tool to determine molar masses of solutes or to estimate their degree of dissociation.
  • This case illustrates why adding salt lowers the freezing point of water — the scientific basis of road de-icing in winter and of antifreeze mixtures in car engines.

← Back to Thermodynamics