Question

A solute dimerises in water. The boiling point of a 2 molal solution of A is 100.52°C. The percentage association of A is ______. (Round off to the nearest integer).

[Use: K_b for water = 0.52 K kg mol⁻¹, Boiling point of water = 100°C]

Answer 1

A solute dimerises in water. The boiling point of a 2 molal solution of A is 100.52°C. The percentage association of A is 100.

Explanation:

Given: Molality m = 2 mol/kg

Boiling point of solution = 100.52°C

Boiling point of pure water = 100°C

Boiling point elevation ΔT_b = 0.52°C

K_b = 0.52 K kg mol⁻¹

Let the solute be A, which dimerises:

\[\ce{2A <=> A2}\]

Let degree of association = α

α moles dimerise → form `alpha/2` moles of A₂

Remaining moles of A = 1 − α

Total particles in solution (i) = `(1 - α) + alpha/2` 

= `1 - alpha/2`

ΔT_b = i K_b m

`0.52 = (1 - alpha/2) * 0.52 * 2`

`0.52 = 1.04 (1 - alpha/2)`

`0.52/1.04 = 1 - alpha/2`

`0.5 = alpha/2`

α = 1

So 100% of the association.