Question

What do you understand by lowering of vapour pressure and relative lowering of vapour pressure? Show that these are colligative properties.

Answer 1

When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution becomes lower than that of the pure solvent.

Δp = p° − p

Where:

p° = vapour pressure of pure solvent

p = vapour pressure of solution

Δp = lowering of vapour pressure

The ratio of the lowering of vapour pressure to the vapour pressure of pure solvent is called relative lowering of vapour pressure:

`(Delta p)/p^circ = (p^circ - p)/p^circ`

According to Raoult’s law for a dilute solution:

p = p° ⋅ χ_1​

Where χ_1​ = mole fraction of solvent

Thus,

p° − p = p°(1 − χ_1​​) = p°χ₂​

Where χ_2​ = mole fraction of solvent

So,

`(p^circ - p)/p^circ = chi_2 = n_2/(n_1 + n_2)`

For dilute solutions (n₂ ≪ n₁​):

`(Delta p)/p^circ = n_2/n_1`

Where:

n₂ = moles of solute

n₁ = moles of solvent