Lowering of vapour pressure and boiling point

Add sugar to water and the water evaporates a little more slowly: some of the surface is taken up by sugar particles, which do not evaporate. So a non-volatile solute lowers the vapour pressure of the solvent.

Properties like this depend only on the number of solute particles, not on what the particles are. They are called colligative properties ('bound together' with the number). There are four: lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Relative lowering of vapour pressure equals the mole fraction of the solute:

(p⁰ – p) ÷ p⁰ = x₂

For a dilute solution this becomes (p⁰ – p) ÷ p⁰ ≈ (w₂ × M₁) ÷ (M₂ × w₁), where w is a mass and M a molar mass (1 = solvent, 2 = solute). So the formula can be used to find the molar mass of a solute.

Elevation of boiling point. A liquid boils when its vapour pressure reaches the outside pressure (1.013 bar at the normal boiling point). With a solute present the vapour pressure is lower, so the solution must be heated more before it boils:

ΔT_b = K_b × m

m is the molality and K_b is the boiling point elevation constant of the solvent (for water, 0.52 K kg mol⁻¹). Put K_b in exactly as given; never change its units. Rearranged, the formula gives a molar mass: M₂ = (1000 × K_b × w₂) ÷ (ΔT_b × w₁), with w₁ in grams.

ΔT_b is only the rise. To give the new boiling point, add it to the solvent's own boiling point.