A solution boils at 100.5°C (water at 100). If K_b = 0.52, molality is:
A1.5 m
B1 m (approx) using ΔT/K_b
C2 m
D0.5 m
Answer & Solution
Correct answer: B. 1 m (approx) using ΔT/K_b
ΔT_b = K_b × m → 0.5 = 0.52 × m → m ≈ 0.96 ≈ 1 m. (Assuming non-electrolyte, i = 1.)
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