1.24 g of AX₂ (molar mass 124 g mol⁻¹) is dissolved in 1 kg of water to form a solution with boiling point of 100.015°C, while 25.4 g of AY₂ (molar mass 250 g mol⁻¹) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.
K b (H₂O) = 0.52 kg mol⁻¹
Which of the following is correct?

Solution

Chemistry ma’am · 3 steps

Work out the ΔTb each salt would give with no ionisation, then compare it with the measured one.

1. The test. With no ionisation, Δ T_b = K_b m. The ratio of the observed to this expected value is the van 't Hoff factor i. An AB₂ salt that ionises fully gives three ions, so i = 3.

   i = (Δ T_b (observed))/(K_b m)

   Salt	Molality m	K_b m (no ionisation)	Observed Δ T_b	i
   AX₂	(1.24/124)/1 = 0.01 mol kg⁻¹	0.52 × 0.01 = 0.0052 K	0.015 K	≈ 2.9, so 3
   AY₂	(25.4/250)/2 = 0.0508 mol kg⁻¹	0.52 × 0.0508 = 0.026 K	0.026 K	≈ 1

2. Read off i. For AX₂, i ≈ 3: it splits fully into A²⁺ and 2X⁻. For AY₂, i ≈ 1: the particle count matches the formula units, so it does not ionise.
3. So. AX₂ is fully ionised and AY₂ is completely unionised. That is option (d).

The idea to keep. In a colligative question about ionisation, find i as observed over expected. i = 1 means no ionisation. i equal to the number of ions per formula unit means full ionisation.

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