Freezing Point Depression and Boiling Point Elevation

Physical Chemistry · Solutions

Freezing Point Depression and Boiling Point Elevation

Two effects with one cause, and a pair of constants that belong to the solvent rather than to the solute.

BSc & MSc · Physical Chemistry · Concept

The short answer: Adding a non-volatile solute lowers the solvent’s vapour pressure, which raises the temperature needed to boil and lowers the temperature at which the solid appears. Both shifts are proportional to molality, with constants characteristic of the solvent alone.

The common cause

A non-volatile solute lowers the vapour pressure of the solvent, because some of the surface is occupied by solute and fewer solvent molecules escape. That single change produces both effects.

  • Boiling point rises because boiling requires the vapour pressure to reach atmospheric, and starting lower means a higher temperature is needed.
  • Freezing point falls because freezing requires the vapour pressures of solid and solution to be equal, and lowering the solution's vapour pressure moves that crossing point to a lower temperature.
Only the solvent freezes out, and that is why the freezing point continues to fall as freezing proceeds. As pure solvent crystallises, the remaining solution becomes more concentrated, so its freezing point drops further. A solution therefore has a freezing range rather than a sharp point — unlike a pure substance, which is why sharp melting is a purity test.

The equations

ΔTb = Kb m     ΔTf = Kf m

The constants depend only on the solvent, not on the solute — which is why they are tabulated per solvent and why the same solute produces different shifts in different solvents. The cryoscopic constant is generally the larger, which is one reason freezing point depression is preferred experimentally.

Molality is used rather than molarity because it is independent of temperature, and these experiments necessarily involve changing the temperature. Using molarity would introduce an error that grows with the size of the temperature change.

Determining molar mass

Measure the depression or elevation for a known mass of solute in a known mass of solvent, obtain the molality, and hence the moles of solute. Dividing the mass by the moles gives the molar mass.

M = (Kf × wsolute × 1000) / (ΔTf × wsolvent)

The method works well for small molecules. For macromolecules the effect is far too small to measure, which is why osmotic pressure is used instead — a comparison that appears frequently.

The van't Hoff factor

Colligative properties count particles, so a solute that dissociates gives a larger effect than its formula concentration predicts, and one that associates gives a smaller one.

BehaviouriEffect on the measured shift
No dissociation or association1As calculated
Complete dissociation into n ionsnn times larger
Partial dissociationBetween 1 and nIntermediate
DimerisationAbout 0.5Half as large

An apparent molar mass calculated without the factor will therefore be wrong — too small for a dissociating solute and too large for an associating one. Comparing apparent with true molar mass is the standard way such questions are framed, and the ratio gives the factor directly.

Practical points

  • Supercooling is common in freezing point measurements, so the true freezing point is obtained by extrapolation rather than by reading the lowest temperature observed.
  • The relationships are strictly valid only for dilute solutions, since they assume ideal behaviour.
  • The solute must be non-volatile for boiling point elevation, or it contributes its own vapour pressure and the treatment fails.

Frequently asked questions

Why does a solution not have a sharp freezing point?

Because pure solvent freezes out first, concentrating the remaining solution and lowering its freezing point further. The process therefore occurs over a range.

Why is molality used rather than molarity?

Because molality does not change with temperature, and these experiments involve temperature changes that would alter a volume-based concentration.

Why is freezing point depression preferred experimentally?

Because the cryoscopic constant is usually larger, giving a bigger and more accurately measurable effect, and because no heating is required so thermally sensitive solutes are unaffected.

What does an apparent molar mass smaller than the true value indicate?

Dissociation, since more particles produce a larger shift, which the calculation interprets as more moles and hence a smaller molar mass.

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