Colligative Properties: How Dissolved Particles Change a Solvent

Counting particles: molality and mole fraction

A colligative property depends on the ratio of solute particles to solvent particles. A mole of sucrose molecules and a mole of urea molecules dissolved in the same mass of water lower its freezing point by the same amount, even though their molar masses differ by a factor of almost six.

Because of this, the concentration units used here count particles. Mole fraction, x, is the moles of one component divided by the total moles. Molality, m, is moles of solute per kilogram of solvent. Molality is used instead of molarity because it is based on masses, so it does not change when the solution expands or contracts with temperature. For dilute aqueous solutions, molality and molarity are numerically close, but they are not the same quantity.

The equations below assume an ideal dilute solution with a nonvolatile solute, meaning one that contributes essentially nothing to the vapor above the solution.

Raoult's law and the drop in vapor pressure

Raoult's law states that the vapor pressure of the solvent above a solution equals its mole fraction times the vapor pressure of the pure solvent: P_solution = x_solvent × P°_solvent. Since the mole fractions add to 1, the drop in vapor pressure is ΔP = x_solute × P°_solvent.

Worked example 1. Find the vapor pressure at 25 °C of a solution of 100.0 g sucrose, C₁₂H₂₂O₁₁, in 250.0 g water. The vapor pressure of pure water at 25 °C is 23.8 torr (OpenStax Table 9.2). Using IUPAC standard atomic weights, M(sucrose) = 342.30 g/mol and M(H₂O) = 18.015 g/mol, so n(sucrose) = 0.2921 mol and n(H₂O) = 13.88 mol. x(H₂O) = 13.88 ÷ (13.88 + 0.2921) = 0.9794. P = 0.9794 × 23.8 torr = 23.3 torr.

The lowering, x(sucrose) × P° = 0.02062 × 23.8 torr = 0.491 torr, is small even though the solution is about 29 percent sugar by mass. Sucrose molecules are heavy, so 100 g of them is still only a small fraction of the molecules present.

Boiling-point elevation and freezing-point depression

Lowering the vapor pressure means the solution must be heated further before its vapor pressure reaches the outside pressure, so the boiling point rises. The solute also disrupts the formation of the solid solvent, so the freezing point falls. For dilute solutions both shifts are proportional to molality: ΔTb = Kb × m and ΔTf = Kf × m.

The constants belong to the solvent, not the solute. OpenStax Table 11.2 gives Kb = 0.512 °C·kg/mol and Kf = 1.86 °C·kg/mol for water, and Kb = 2.53 and Kf = 5.12 °C·kg/mol for benzene. Water's Kf is more than three times its Kb, so a given solute changes water's freezing point far more than its boiling point.

Worked example 2. A coolant is made by dissolving 250. g ethylene glycol, C₂H₆O₂ (62.07 g/mol), in 1.000 kg water. n = 250. g ÷ 62.07 g/mol = 4.028 mol, so m = 4.028 mol/kg. ΔTf = 1.86 × 4.028 = 7.49 °C, so the ideal freezing point is −7.49 °C. ΔTb = 0.512 × 4.028 = 2.06 °C, so the ideal boiling point is 102.1 °C. At this concentration the solution is no longer very dilute, so treat these as estimates, not measured values.

Electrolytes and the van't Hoff factor

An ionic solute releases more than one particle per formula unit. The van't Hoff factor, i, is the number of moles of particles in solution per mole of solute dissolved, and it multiplies each colligative equation: ΔTf = i × Kf × m. The ideal value is the number of ions in the formula: 2 for NaCl, 3 for CaCl₂ and 1 for a nonelectrolyte such as glucose.

Measured values are smaller, because oppositely charged ions attract one another and some form ion pairs that act as one particle. OpenStax Table 11.3 lists measured factors for 0.050 m solutions: 1.9 for NaCl, 2.7 for MgCl₂, 1.3 for MgSO₄ and 3.4 for FeCl₃, against ideal values of 2, 3, 2 and 4. In dilute solutions, the gap generally grows as concentration rises and as ion charges increase.

Worked example 3. Estimate the freezing point of 15.0 g CaCl₂ (110.98 g/mol) dissolved in 500.0 g water. n = 0.1352 mol and m = 0.1352 mol ÷ 0.5000 kg = 0.2703 mol/kg. With the ideal i = 3, ΔTf = 3 × 1.86 × 0.2703 = 1.51 °C, giving −1.51 °C. The real depression is smaller. If you use 2.7, the measured factor for the similar salt MgCl₂ at 0.050 m, the estimate becomes 1.36 °C. That figure is still only approximate, because this solution is more than five times as concentrated.

Using a freezing point to find a molar mass

If you know the mass of solute and solvent and measure ΔTf, you can work backward to the molar mass. The freezing-point method is preferred to the boiling-point method because Kf is larger than Kb for common solvents, and because a boiling point also shifts with the day's atmospheric pressure.

Worked example 4. Dissolving 1.50 g of an unknown nonvolatile nonelectrolyte in 25.0 g benzene lowers the freezing point by 1.62 °C. Molality: m = ΔTf ÷ Kf = 1.62 ÷ 5.12 = 0.3164 mol/kg. Moles of solute: 0.3164 mol/kg × 0.0250 kg = 7.910 × 10⁻³ mol. Molar mass: 1.50 g ÷ 7.910 × 10⁻³ mol = 190 g/mol, to three significant figures.

Benzene is a good solvent for this experiment because its Kf is nearly three times water's, so the same amount of solute gives a larger, easier-to-measure change. The method assumes the solute neither dissociates nor associates. Carboxylic acids, for example, form hydrogen-bonded pairs in benzene and give an apparent molar mass close to double the true value.

Osmotic pressure and very large molecules

When a semipermeable membrane separates a solution from pure solvent, solvent flows into the solution. The pressure that must be applied to the solution to stop this flow is the osmotic pressure, Π = i × M × R × T, where M is molarity, R = 0.08206 L·atm/(mol·K), and T is in kelvins.

Worked example 5. Intravenous saline is about 0.155 mol/L NaCl. At body temperature, 37.0 °C or 310.15 K, the ideal osmotic pressure is Π = 2 × 0.155 × 0.08206 × 310.15 = 7.89 atm. Using i = 2 overstates it slightly, because the real factor is a little below 2. A pressure that large is why solutions given into a vein must match the solute concentration of blood.

Worked example 6. A solution of 0.250 g of a protein in 50.0 mL of water has an osmotic pressure of 4.20 torr at 25.0 °C. Π = 4.20 ÷ 760 = 5.526 × 10⁻³ atm. M = Π ÷ (R × T) = 5.526 × 10⁻³ ÷ (0.08206 × 298.15) = 2.259 × 10⁻⁴ mol/L, so n = 1.129 × 10⁻⁵ mol in 0.0500 L, and the molar mass is 0.250 g ÷ 1.129 × 10⁻⁵ mol = 2.21 × 10⁴ g/mol. The same solution would freeze only about 4 × 10⁻⁴ °C below pure water, which is too small to measure well. That is why osmotic pressure, not freezing point, is used for polymers and proteins.

Mistakes that throw off colligative calculations

A quick check catches most errors: the solution should always boil higher and freeze lower than the pure solvent, and the shifts for dilute aqueous solutions are usually a few degrees at most.

• Using molarity in ΔT = K × m. These equations need molality, which uses kilograms of solvent, not liters of solution.
• Dividing by the mass of the solution instead of the mass of the solvent.
• Leaving out the van't Hoff factor for an ionic solute, or using it for a molecular solute such as sucrose or ethylene glycol.
• Reporting ΔTf as the freezing point. Subtract it from the pure solvent's freezing point.
• Using Celsius temperature in Π = iMRT. Convert to kelvins first.
• Taking Kf or Kb from the wrong solvent. They are properties of the solvent.