If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ____________ molal. (Rounded off to the nearest integer)
[K b = 0.52 K kg mol⁻¹]

Solution

Chemistry ma’am · 3 steps

AB splits into two ions, so first find the van't Hoff factor and only then use Δ T_b.

1. Van't Hoff factor. AB → A⁺ + B⁻ gives n = 2 particles. With degree of dissociation α = 0.75, i = 1 + α(n - 1) = 1 + 0.75 = 1.75.

   Δ T_b = i K_b m ⇒ m = (Δ T_b)/(i K_b)

2. Put in numbers. i K_b = 1.75 × 0.52 = 0.91 K kg mol⁻¹. So m = 2.5/0.91 = 2.75 mol kg⁻¹.
3. So. The molality is about 2.75 molal, which rounds to 3.

The common slip: 5. Leaving out the van't Hoff factor gives 2.5/0.52 = 4.8, which rounds to 5. Dissociation means more particles, so less solute is needed.

The idea to keep. Colligative properties count particles, not formula units. For an electrolyte, put i = 1 + α(n - 1) into the formula before you solve for molality.

Answer from the NTA final answer key. Solution by PrepWiser.