Solutions & Colligative Properties

Why salt melts icy roads, antifreeze protects an engine, and your cells don't burst — all from counting dissolved particles.

High schoolIntro Gen ChemUni Year 1
⏱️ About 18 min

Cities scatter salt on winter roads and antifreeze keeps a car engine from freezing solid. Both work for the same reason, and it has nothing to do with what the dissolved stuff IS — only how many particles it breaks into. That surprising idea is what makes colligative properties so powerful.

💡
The big idea: A colligative property depends only on the NUMBER of dissolved solute particles, not their identity. Dissolving a solute lowers a liquid's freezing point, raises its boiling point, and drives osmosis — all scaling with how many particles the solute contributes.
🎯 By the end, you'll be able to
  • Define a colligative property and the concept of molality
  • Calculate freezing-point depression and boiling-point elevation
  • Use the van't Hoff factor i to count particles from ionic compounds
  • Explain osmosis and osmotic pressure in everyday terms
📎 Helpful to know first

It's about counting, not identity

A colligative property is one that depends only on the number of dissolved particles in a solution, not on what kind they are. One mole of dissolved sugar and one mole of dissolved particles from any other non-splitting solute shift a solvent's behaviour by the same amount.

We measure concentration here in molality (m): moles of solute per kilogram of solvent. Molality is used instead of molarity because it doesn't change with temperature.

🔑 The van't Hoff factor: count the pieces
What matters is particles in solution, so a compound that splits up counts for more. The van't Hoff factor (i) is how many particles one formula unit releases. Sugar (glucose) doesn't dissociate, so i = 1. Table salt, \(\ce{NaCl -> Na+ + Cl-}\), gives two ions, so i = 2. Calcium chloride, \(\ce{CaCl2 -> Ca^2+ + 2Cl-}\), gives three, so i = 3.

Freezing-point depression

Dissolved particles get in the way of the solvent forming its neat solid structure, so the solution must be cooled further before it freezes. The freezing point drops by an amount set by the molality and a constant Kf that belongs to the solvent (for water, Kf = 1.86 °C/m).

\[ \Delta T_f = i\,K_f\,m \]
Freezing-point depression: how many degrees the freezing point drops. i = van't Hoff factor, Kf = freezing-point constant, m = molality.

Boiling-point elevation

Dissolved particles also make it harder for the solvent to escape as vapour, so the solution must be heated higher before it boils. The boiling point rises by a similar formula, with the solvent's boiling-point constant Kb (for water, Kb = 0.512 °C/m).

\[ \Delta T_b = i\,K_b\,m \]
Boiling-point elevation: how many degrees the boiling point rises above the pure solvent's value.
📝 Worked example: What is the freezing point of a 0.10 m aqueous NaCl solution? (Kf = 1.86 C/m for water; for NaCl, i = 2.)
  1. NaCl splits into Na⁺ and Cl⁻, so i = 2.
  2. ΔTₔ = i Kₔ m = 2 × 1.86 × 0.10 = 0.372 °C.
  3. Freezing point drops from 0 °C by 0.372, so it freezes at about −0.37 °C.
✓ About -0.37 C (a depression of 0.372 C).
✏️ Practice: A 0.50 m aqueous glucose solution (glucose does not dissociate, i = 1). By how many degrees Celsius is its freezing point depressed? (Kf = 1.86 C/m.)
C
Solution
  1. Glucose stays as whole molecules, so i = 1.
  2. ΔTₔ = i Kₔ m = 1 × 1.86 × 0.50.
  3. = 0.93 °C. (So the solution freezes at about −0.93 °C.)
✏️ Practice: A 1.0 m aqueous glucose solution (i = 1). By how many degrees Celsius is its boiling point elevated? (Kb = 0.512 C/m.)
C
Solution
  1. Glucose does not dissociate, so i = 1.
  2. ΔTₛ = i Kₛ m = 1 × 0.512 × 1.0.
  3. = 0.512 °C. (So the solution boils at about 100.51 °C.)

Osmosis: solvent flows toward the crowd

Put pure water and a sugar solution on opposite sides of a membrane that lets water through but not sugar. Water flows from the dilute side into the concentrated side, trying to even out the particle concentration. This is osmosis. The pressure you would have to apply to stop that flow is the osmotic pressure, and it too depends only on the particle concentration.

This is why a freshwater fish can't live in the sea, why over-salting a garden kills plants, and why an IV drip must match the salt concentration of blood — get it wrong and cells either shrivel or burst.

\[ \Pi = i\,M\,R\,T \]
Osmotic pressure: Pi = i * M * R * T, where M is molarity, R the gas constant and T the absolute temperature. Even small concentrations give large osmotic pressures.

Check your understanding

1. A colligative property depends on which of the following?
Colligative properties depend only on how many solute particles are dissolved, not on what they are. That's why the van't Hoff factor (how many pieces a solute splits into) matters so much.
2. Which 1 m aqueous solution has the LARGEST freezing-point depression?
Depression scales with i*Kf*m. At the same molality, CaCl2 releases the most particles (3 ions per formula unit), giving the biggest effect.
3. In osmosis, which way does the solvent (water) move?
Water moves from the dilute side (fewer solute particles) to the concentrated side (more particles), acting to equalise the concentrations across the membrane.
🔎 Related lessons
Previous: Phase Changes & Phase Diagrams Kinetic-Molecular Theory The Mole & Molar Mass
✅ Key takeaways
  • A colligative property depends on the NUMBER of dissolved particles, not their identity.
  • The van't Hoff factor i counts particles per formula unit (glucose 1, NaCl 2, CaCl2 3).
  • Freezing-point depression: delta-Tf = i*Kf*m (water Kf = 1.86 C/m).
  • Boiling-point elevation: delta-Tb = i*Kb*m (water Kb = 0.512 C/m).
  • Osmosis drives solvent from dilute to concentrated; osmotic pressure Pi = i*M*R*T.
➡️ You've now seen matter in every state and how dissolved particles bend its behaviour. That completes the States of Matter module — a foundation you'll lean on across thermochemistry, equilibrium and beyond.
Want to test yourself on this? Try the Chemistry practice test →