Colligative Properties — Freezing Point Depression

Theoretical Background

  • Colligative properties depend only on the number of solute particles present in solution, not on their chemical identity.
  • Typical colligative properties: vapor-pressure lowering, boiling-point elevation, freezing-point depression, osmotic pressure.

  • For freezing point depression:
    \(\Delta T_f = i K_f \, m\) where:
    • $\Delta T_f$ = freezing point lowering
    • $i$ = van ’t Hoff factor (number of effective particles per solute unit)
    • $K_f$ = cryoscopic constant of the solvent
    • $m$ = molality of solute
  • Molality is defined as:
    \(m = \frac{n_{\text{solute}}}{m_{\text{solvent}}(kg)}\)

Exercise

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.