A solution boils at 100.5°C (water at 100). If K_b = 0.52, molality is:
- 1.5 m
- 1 m (approx) using ΔT/K_b
- 2 m
- 0.5 m
Show answer and solution
Correct answer: B. 1 m (approx) using ΔT/K_b
We have ΔT_b = K_b × m → 0.5.
So ΔT_b = 0.52 × m → m ≈ 0.96 ≈ 1 m.
(Assuming non-electrolyte, i = 1.)