Colligative properties

Definition. A colligative property is a measurable property of a solution — its vapour pressure, freezing point, boiling point, or osmotic pressure — that depends only on the number of solute particles present (relative to solvent), not on their size, mass, or chemical nature. Equal moles of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) and urea ($\text{CH}_4\text{N}_2\text{O}$) dissolved in the same mass of water produce identical freezing-point depressions, even though the two molecules are structurally nothing alike.

The four properties, formula by formula. Relative lowering of vapour pressure compares a solution's vapour pressure $p_1$ to the pure solvent's $p_1^0$:

where: $p_1^0$ is the vapour pressure of the pure solvent, $p_1$ is the vapour pressure above the solution, and $x_2$ is the mole fraction of the (non-volatile) solute. Freezing-point depression and boiling-point elevation both scale with molality $m$:

where: $K_f$ and $K_b$ are the molal depression and elevation constants of the solvent (fixed physical constants — for water, $K_f = 1.86\ \text{K kg mol}^{-1}$ and $K_b = 0.512\ \text{K kg mol}^{-1}$), and $m$ is the molality of the solution (mol solute per kg solvent). Osmotic pressure scales with molarity:

where: $M$ is the molarity of the solution, $R$ is the universal gas constant, and $T$ is the absolute temperature.

Why one idea explains all four. A non-volatile solute occupies some of the solvent's surface, so fewer solvent molecules escape into the vapour at any given temperature — the vapour pressure curve of the solution sits below the pure solvent's curve at every temperature. A lower vapour-pressure curve crosses the solid-liquid equilibrium line at a lower temperature (this is freezing-point depression) and needs a higher temperature to reach atmospheric pressure and boil (boiling-point elevation). The same lowered escaping tendency is what drives solvent molecules across a semipermeable membrane toward the more concentrated solution — osmotic pressure. One cause, four symptoms.

When the plain formulae fail — the van't Hoff factor. All four formulae above assume the solute stays intact in solution, one dissolved formula unit giving exactly one solute particle. Electrolytes dissociate (NaCl gives $\text{Na}^+$ and $\text{Cl}^-$) and some solutes associate (carboxylic acids can pair into dimers), so the actual particle count differs from the nominal moles dissolved. The van't Hoff factor $i$ corrects for this: $\Delta T_f = i K_f m$, $\Delta T_b = i K_b m$, $\Pi = i M R T$, and $\dfrac{p_1^0 - p_1}{p_1^0} = i\, x_2$. For a solute that dissociates into $n$ ions with degree of dissociation $\alpha$, $i = 1 + (n-1)\alpha$; for full dissociation of NaCl, $i \approx 2$.

1 mole of glucose and 1 mole of sucrose are each dissolved separately in 1 kg of water. Find the freezing-point depression of each solution, given $K_f(\text{water}) = 1.86\ \text{K kg mol}^{-1}$, and then find the boiling-point elevation of the glucose solution given $K_b(\text{water}) = 0.512\ \text{K kg mol}^{-1}$.

Step 1. Identify the particle count. Glucose and sucrose are both non-electrolytes — neither dissociates nor associates in water — so each solution contains exactly 1 mole of independent solute particles per kg of water. Molality is identical for both: $m = 1\ \text{mol kg}^{-1}$.

Step 2. Apply $\Delta T_f = K_f m$ to both. $\Delta T_f = 1.86 \times 1 = 1.86\ \text{K}$ for glucose, and the identical $1.86\ \text{K}$ for sucrose — both solutions freeze $1.86\ ^\circ\text{C}$ below pure water, confirming that molecular identity plays no role once the particle count is fixed.

Step 3. Apply $\Delta T_b = K_b m$ to the same glucose solution, using its molality $m = 1\ \text{mol kg}^{-1}$ unchanged: $\Delta T_b = 0.512 \times 1 = 0.512\ \text{K}$. The glucose solution boils $0.512\ ^\circ\text{C}$ above pure water's $100\ ^\circ\text{C}$.

The freezing point of both solutions is $-1.86\ ^\circ\text{C}$, and the boiling point of the glucose solution is $100.512\ ^\circ\text{C}$.

Competitive stretch

A student instead dissolves 0.1 mol of NaCl and, separately, 0.1 mol of glucose, each in 1 kg of water. Which solution boils at a higher temperature, and by roughly what factor?

Step 1. Glucose does not dissociate, so it contributes exactly $0.1$ mol of particles per kg of water; its van't Hoff factor is $i_{\text{glucose}} \approx 1$.

Step 2. NaCl is a strong electrolyte and dissociates almost completely, $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$, so the same $0.1$ mol of formula units yields close to $0.2$ mol of ions per kg of water; its van't Hoff factor is $i_{\text{NaCl}} \approx 2$.

Step 3. Since $\Delta T_b = i K_b m$ with the same nominal $m = 0.1\ \text{mol kg}^{-1}$ for both, the boiling-point elevation scales directly with $i$. The NaCl solution boils at roughly double the elevation of the glucose solution — about $\Delta T_b \approx 2 \times 0.512 \times 0.1 = 0.102\ \text{K}$ versus $\approx 0.051\ \text{K}$ for glucose — purely because dissociation doubles the dissolved particle count.

• Thinking a colligative property depends on the mass or size of solute particles — it depends strictly on their number, never on identity, size, or mass.
• Applying the plain (no-$i$) formula to a known electrolyte and getting a molar mass that "doesn't match" — this mismatch is the very signal that the van't Hoff factor $i$ must be inserted.
• Mixing up concentration scales across the four formulae: mole fraction for vapour-pressure lowering, molality for $\Delta T_f$ and $\Delta T_b$, molarity for $\Pi$ — swapping any of these is one of the most common numerical slips.
• Forgetting all four formulae assume a dilute solution of a non-volatile solute and are derived, directly or indirectly, from Raoult's law — they are not universal laws valid at any concentration.

Ask the tutor: want a mixed problem set that makes you pick the right colligative formula and concentration scale for each scenario?