Lesson 5 · 30 min

Colligative properties: freezing, boiling and osmosis

The number of dissolved particles, not their kind, sets freezing-point depression, boiling-point elevation and osmotic pressure. See how antifreeze, road salt and seawater desalination use them — and why ions fall short of their full count.

In this lesson

Use Raoult’s law to explain why a solute lowers the solvent’s vapour pressure
Calculate freezing-point depression, boiling-point elevation and osmotic pressure, and find a molar mass by cryoscopy
Explain the van ’t Hoff factor and why measured values often fall short of the ion count
Explain antifreeze, road salt, intravenous drips and reverse-osmosis desalination with colligative properties

Before a snowstorm, road crews spread salt; car radiators are filled with ethylene-glycol antifreeze; castaways surrounded by seawater cannot drink a drop; and coastal plants force seawater through a thin membrane at 60 bar. What these share: some properties of a solution depend only on how many particles are dissolved, hardly on what they are. They are the colligative properties.

Predict first

Dissolve 1 mol of glucose in one kilogram of water, and 1 mol of NaCl in another. Which freezes at a lower temperature, and by how much?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

The same: equal amounts of solute NaCl — exactly twice as low (−3.72 °C) NaCl — close to, but less than, twice (about −3.4 °C) Glucose, because its molar mass is larger

The root: a lower solvent vapour pressure

Solute particles dilute the solvent: only part of the surface belongs to solvent molecules, so fewer can escape. For a non-volatile solute, Raoult’s law says the solvent’s vapour pressure above the solution equals the pure solvent’s vapour pressure times the solvent’s mole fraction. In a 1 mol/kg sucrose solution the mole fraction of water is 0.982, so its vapour pressure drops by 1.8 %.

With the whole vapour-pressure curve shifted down, two things follow: the vapour pressure reaches the outside pressure only at a higher temperature (boiling-point elevation), and the water in the solution is in equilibrium with pure ice only at a lower temperature (freezing-point depression).

p = x_\text{solvent}\,p^{*} \qquad \Delta T_\mathrm{b} = i\,K_\mathrm{b}\,m \qquad \Delta T_\mathrm{f} = i\,K_\mathrm{f}\,m \qquad \pi = i\,c\,R\,T

Raoult’s law and the three dilute-solution laws; they hold only for dilute solutions.

$m$: molality, mol solute per kg solvent
$K_\mathrm{f},\,K_\mathrm{b}$: water’s cryoscopic constant 1.86 and ebullioscopic constant 0.512 K kg/mol
$i$: van ’t Hoff factor: effective particles per formula unit
$c,\ \pi$: molar concentration (mol/L) and osmotic pressure

Worked example

A molar mass from freezing-point depression

2.50 g of an unknown non-electrolyte dissolved in 50.0 g of water lowers the freezing point by 0.517 K. Find its molar mass.

Osmosis and reverse osmosis

Separate pure water from a solution with a semipermeable membrane that passes only water, and water flows spontaneously into the solution: osmosis. The pressure you must apply to the solution side to stop it is the osmotic pressure $\pi$ . For seawater at 35 g of salt per kg, $\pi$ is about 27 bar — equivalent to a column of water 280 m tall.

Apply more than the osmotic pressure and water is squeezed out of the solution through the membrane: reverse osmosis. Desalination plants push seawater at about 60 bar, producing fresh water on one side and a more concentrated brine on the other. Thermodynamics demands at least $\pi \times V \approx 0.8$ kWh of work per cubic metre; modern plants use about 3 kWh/m³ in total.

Across a membrane: pure water | salt water

Osmosis and reverse osmosis

Salt content (as NaCl) · 35 g/kg Pressure applied to the salt side · 0 bar Temperature · 25 °C

Osmosis: water flows into the salt side

Osmotic pressure (from water activity)
van ’t Hoff π = icRT with i = 2
Applied pressure

$\pi = -\frac{RT}{V_{\ce{H2O}}}\ln a_{\ce{H2O}} \approx i\,c\,R\,T$

Osmotic pressure π: 28.3 bar (ideal 30.7)
Net driving pressure ΔP − π: −28.3 bar
Water flux (typical membrane): −28.3 L·m⁻²·h⁻¹
Equivalent water column: 289 m
Minimum energy (per m³ desalted): 0.79 kWh

Start with seawater: with no applied pressure, water flows into the salt side. Raise the applied pressure slowly past about 28 bar and the flow reverses — reverse osmosis begins. Then double the salt to 70 g/kg (the brine a plant discharges) and see how much more pressure it takes.

Concept check

Which of these 0.10 mol/kg aqueous solutions has the lowest freezing point?

Glucose NaCl CaCl₂ MgSO₄

Try the calculation

What is the osmotic pressure (bar) of 0.30 mol/L glucose at 37 °C? (R = 0.08314 L bar mol⁻¹ K⁻¹)

Key takeaways

Colligative properties depend only on the number of solute particles: vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure.
For dilute solutions ΔTf = iKf·m (water Kf = 1.86), ΔTb = iKb·m (Kb = 0.512) and π = icRT.
Electrolytes have effective i values below the ion count (ion atmospheres and ion pairs), most of all for doubly charged ions; in concentrated solutions hydration pushes i up again.
Pressure beyond the osmotic pressure drives reverse osmosis; seawater’s is about 27 bar and desalination plants run near 60 bar.

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Data and references

P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 5B (colligative properties).
K. S. Pitzer & G. Mayorga, “Thermodynamics of electrolytes. II”, J. Phys. Chem. 77, 2300 (1973); J. Solution Chem. 3, 539 (1974).
R. A. Robinson & R. H. Stokes, Electrolyte Solutions, 2nd ed. (Butterworths, 1959): osmotic coefficients.
R. J. Bodnar, “Revised equation and table for determining the freezing point depression of H2O–NaCl solutions”, Geochim. Cosmochim. Acta 57, 683 (1993).
M. Elimelech & W. A. Phillip, “The future of seawater desalination: energy, technology, and the environment”, Science 333, 712 (2011).