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.
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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.
- Calculate the freezing point depression $\Delta T_f$.
- Assume complete dissociation of NaCl.
- 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.