Lesson 5 of 10 · 7 min
Relative lowering of vapour pressure
NCERT §1.6; §1.6.1
Leave a cup of water and a cup of the coolant open side by side. The water level drops noticeably over days; the coolant barely changes. The glycol does not evaporate, so how does it slow the water down?
The lesson in notes
In short
Colligative properties are set by how many solute particles there are relative to all the particles present, whatever those particles are (Latin co, together, and ligare, to bind). There are four: the relative lowering of the solvent's vapour pressure, the rise in its boiling point, the fall in its freezing point, and the osmotic pressure of the solution.
All four come from the same cause: a non-volatile solute lowers the vapour pressure of the solvent.
For a non-volatile solute, the lowering is Δp₁ = p₁° − p₁ = x₂ p₁°, and the relative lowering (p₁° − p₁)/p₁° equals x₂, the mole fraction of the solute.
With several non-volatile solutes, the lowering depends on the sum of their mole fractions.
Since x₂ = n₂/(n₁ + n₂), and n₂ is much smaller than n₁ in a dilute solution, (p₁° − p₁)/p₁° ≈ n₂/n₁ = (w₂ × M₁)/(M₂ × w₁), which gives the molar mass M₂ of the solute from measured masses and pressures.