Four properties of dilute solutions depend only on the number of solute particles, not on what they are: vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure. Each can be used to find a molar mass.
A non-volatile solute lowers the vapour pressure of the solvent by a fraction equal to the solute’s mole fraction. That fraction depends on how many solute particles there are, not on what kind they are. Four measurable properties follow from this.
They are called colligative properties, from the Latin for “bound together”. In a dilute solution, each depends only on the number of solute particles. One mole of glucose and one mole of urea, each dissolved in 1 kg of water, lower its freezing point by the same amount.
Because each property counts particles, measuring it gives the number of moles of solute. If the mass of solute is known, its molar mass follows.
By Raoult’s law, the relative lowering of vapour pressure equals the mole fraction of the solute:
(p₁° − p₁) / p₁° = x₂
Measuring the vapour pressure of the pure solvent and of the solution gives x₂. From x₂ and the masses of solute and solvent, the molar mass of the solute can be calculated.
A liquid boils when its vapour pressure equals the external pressure. A solution has a lower vapour pressure than the pure solvent at every temperature. So it must be heated to a higher temperature before its vapour pressure reaches the external pressure. Its boiling point is higher. For a dilute solution, the rise ΔTb is proportional to the molality m:
ΔTb = Kb m
Kb is the boiling point elevation constant, a property of the solvent. For water, Kb = 0.52 K kg mol⁻¹. A solution of 1 mol of glucose in 1 kg of water boils at about 100.52°C.
A solution freezes when its vapour pressure equals the vapour pressure of the solid solvent. Because the solution’s vapour pressure is lowered, this happens at a lower temperature. The freezing point of the solution is lower than that of the pure solvent:
ΔTf = Kf m
Kf is the freezing point depression constant. For water, Kf = 1.86 K kg mol⁻¹. Salt spread on an icy road dissolves in the thin layer of water on the ice. The solution has a freezing point below 0°C, so if the road is at, say, −5°C, the ice melts. Ethylene glycol in a car’s radiator lowers the freezing point of the coolant in the same way.
Boiling-point elevation and freezing-point depression have the same cause. What is it, and why does one temperature rise while the other falls?
The cause is the lowering of the solvent’s vapour pressure by the solute.
Boiling point rises: boiling requires the vapour pressure to reach the external pressure. With the vapour pressure lowered, a higher temperature is needed to reach it.
Freezing point falls: freezing occurs where the vapour pressure of the liquid equals that of the solid. The solid is pure solvent, so its vapour pressure is unchanged. The lowered vapour pressure of the solution meets that of the solid at a lower temperature.
Both changes are proportional to the number of solute particles per kilogram of solvent, so both are colligative.
A semipermeable membrane lets solvent molecules pass but not solute particles. Cell membranes and parchment are examples. If a solution and pure solvent are separated by such a membrane, solvent flows through the membrane into the solution. This net flow of solvent, from lower to higher solute concentration, is osmosis.
The flow is caused by the difference in solvent concentration across the membrane: more solvent molecules strike the membrane per second from the pure-solvent side. Applying extra pressure to the solution increases the rate at which solvent molecules are pushed back through. The osmotic pressure, Π, is the extra pressure that must be applied to the solution to stop the net flow. For a dilute solution:
Π = C R T
C is the molar concentration of solute, R is the gas constant, 0.0821 L atm K⁻¹ mol⁻¹, and T is the temperature in kelvin. A 0.010 M solution at 300 K has Π = 0.010 × 0.0821 × 300 ≈ 0.25 atm. That pressure supports a column of water about 2.5 m high, which is easy to measure. The same solution would lower the freezing point of water by only about 0.019 K.
Why is osmotic pressure the preferred method for finding the molar mass of a protein?
The effect is large enough to measure. A protein has a molar mass of tens of thousands of g/mol, so a few grams per litre is only about 10⁻⁴ mol/L. The freezing-point depression would be about 0.0002 K, too small to measure accurately. The osmotic pressure would be about 0.0025 atm, which supports a water column of about 2.5 cm and can be read precisely.
It is measured at room temperature. Boiling-point measurements require heating to 100°C, at which many proteins lose their structure. Osmotic pressure can be measured without changing the protein.
1A colligative property of a dilute solution depends on…
2Which of these is NOT a colligative property?
3A non-volatile solute raises the boiling point of a solvent because…
4Freezing-point depression is given by ΔTf = ____ × m, where m is the molality.
5Why does salt melt ice on a road at −5°C?
6Osmosis is the net flow of…
7Fluid injected into the blood must be isotonic with blood because…
Next: solutes that split apart or join together, abnormal molar masses, and the van’t Hoff factor.