AB₂ is 10% dissociated in water to A²⁺ and B⁻. The boiling point of a 10.0 molal aqueous solution of AB₂ is __________°C. (Round off to the Nearest Integer).
[Given : Molal elevation constant of water K b = 0.5 K kg mol⁻¹ boiling point of pure water = 100°C]

Solution

Chemistry ma’am · 3 steps

Only 10% splits up, so find the real number of particles with i before you use Δ T_b = iK_bm.

1. Count the particles. AB₂ → A²⁺ + 2B⁻, so each formula unit that dissociates gives n = 3 ions. With α = 0.1, each mole of AB₂ leaves 0.9 mol undissociated and gives 0.1 + 0.2 = 0.3 mol of ions, a total of 1.2 mol of particles.

   i = 1 + α(n-1) = 1 + 0.1 × (3-1) = 1.2

2. Find the elevation. Δ T_b = iK_bm = 1.2 × 0.5 K kg mol⁻¹ × 10 mol kg⁻¹ = 6 K. A change of 6 K is the same as a change of 6 °C.
3. So. The boiling point is 100 + 6 = 106 °C, so the answer is 106.

The common slip: 6. Reporting only the elevation Δ T_b gives 6, not 106. Add it to the 100 °C boiling point of pure water.

The idea to keep. For partial dissociation into n ions, use i = 1 + α(n-1). Then add Δ T_b to the boiling point of the pure solvent, because the question asks for the boiling point.

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