Colligative Properties and Molar Mass

Why "colligative." A dilute solution of a non-volatile, non-electrolyte solute (one mole of solute gives exactly one mole of dissolved particles, and the solute does not evaporate) shows four linked effects, all traceable to the same cause: dissolved solute particles get in the way of solvent molecules escaping into vapour, so the solution's vapour pressure is lower than the pure solvent's at every temperature.

1. Relative lowering of vapour pressure. For a solute mole fraction $x_2$ this lowering is

where $p_1^0$ is the pure solvent's vapour pressure, $p_1$ is the solution's vapour pressure, and $n_1$, $n_2$ are moles of solvent and solute.

2. Boiling point elevation. Because the solution's vapour pressure curve sits below the pure solvent's, it takes a higher temperature to push the vapour pressure back up to 1 atmosphere, so the solution boils hotter:

where: $\Delta T_b = T_b - T_b^0$ is the rise in boiling point, $m$ is molality (mol of solute per kg of solvent), and $K_b$ is the molal elevation constant of the solvent (units K kg mol⁻¹) — a fixed property of the solvent, never of the solute.

3. Freezing point depression. The same vapour-pressure argument, run at the freezing point instead, gives a lower freezing point for the solution:

where $\Delta T_f = T_f^0 - T_f$ and $K_f$ is the molal depression constant (cryoscopic constant) of the solvent.

4. Osmotic pressure. Osmosis is the spontaneous flow of solvent through a semipermeable membrane toward the more concentrated side; osmotic pressure $\Pi$ is the extra pressure needed on the solution side to stop that net flow. For dilute solutions it obeys a van't Hoff equation shaped exactly like the ideal gas law:

where $C$ is molar concentration (mol L⁻¹), $R$ is the gas constant, and $T$ is absolute temperature. The figure below shows why boiling point rises and freezing point falls together: the solution's vapour-pressure curve is uniformly shifted down, so it crosses the 1-atm line at a higher temperature and crosses the solid-liquid line at a lower one.

Getting to molar mass. Every equation above hides $n_2 = w_2/M_2$, the moles of solute expressed as mass over molar mass. Once $\Delta T_b$, $\Delta T_f$, or $\Pi$ is measured in the lab, the equation can be rearranged to solve for the one unknown, $M_2$.

Problem: A solution containing 6.0 g of a non-volatile, non-electrolyte solute dissolved in 1.0 L of solution has an osmotic pressure of 4.92 atm at 300 K. Calculate the molar mass of the solute. (Use $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$.)

Step 1. Write the van't Hoff equation and solve for concentration: $\Pi = CRT \implies C = \dfrac{\Pi}{RT} = \dfrac{4.92}{0.0821 \times 300}$.

Step 2. Compute the denominator and divide: $C = \dfrac{4.92}{24.63} = 0.1998 \approx 0.200\ \text{mol L}^{-1}$.

Step 3. Relate $C$ to molar mass using $n_2 = w_2/M_2$, with 6.0 g of solute in 1.0 L of solution: $C = \dfrac{w_2/M_2}{V} \implies 0.200 = \dfrac{6.0/M_2}{1.0}$.

Step 4. Solve for $M_2$: $M_2 = \dfrac{6.0}{0.200} = 30\ \text{g mol}^{-1}$.

The solute has a molar mass of 30 g mol⁻¹.

Competitive stretch

The same 6.0 g sample, if instead dissolved in 50 g of water, would raise the boiling point by $\Delta T_b = \dfrac{K_b w_2}{M_2\, w_1(\text{kg})} = \dfrac{0.52 \times 6.0}{30 \times 0.050} = 2.08\ \text{K}$ — a rise nearly twenty times larger than the 0.104 K seen for the 1.0 g, 100 g mol⁻¹ solute of the earlier boiling-point example. This is exactly why osmotic pressure, not $\Delta T_b$, is preferred for very large solute molar masses such as proteins and polymers: their tiny molar concentration at reasonable working masses gives an almost unmeasurable $\Delta T_b$, but $\Pi = CRT$ still produces a comfortably measurable pressure.

• Mixing up molality and molarity. $\Delta T_b$ and $\Delta T_f$ use molality $m$ (moles of solute per kilogram of solvent); $\Pi = CRT$ uses molarity $C$ (moles of solute per litre of solution) — using volume where mass is required, or vice versa, is the single most common slip.
• Sign confusion. $\Delta T_b = T_b - T_b^0$ is a rise; $\Delta T_f = T_f^0 - T_f$ is written the opposite way round so that it too comes out positive — plugging temperatures in mechanically without checking the direction flips the sign of the answer.
• Applying these formulas unmodified to electrolytes. NaCl, CaCl₂, and other electrolytes dissociate into more particles than moles dissolved, so plain $\Delta T_b = K_b m$ or $\Pi = CRT$ underestimate the effect; that correction (the van't Hoff factor) belongs to the abnormal molar mass topic that follows this one.
• Treating $K_b$ or $K_f$ as solute properties. Both constants belong to the solvent alone and never change with which solute is dissolved in it.

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