A 1.0 molal aqueous solution of electrolyte A2B3 is 60% ionised. What is the boiling point of this solution at 1 atm pressure? (Kb for water = 0.52 K kg mol⁻¹)

274.76 K
377 K
376.4 K
374.76 K

Correct answer: 374.76 K

Solution

A2B3 -> 2A⁺ + 3B²- gives n = 5 ions. Van't Hoff factor i = 1 + (5-1)(0.60) = 1 + 2.4 = 3.4. Boiling point elevation: delta_Tb = i * Kb * m = 3.4 * 0.52 * 1.0 = 1.768 K. Boiling point = 373 + 1.768 = 374.768 K ≈ 374.76 K.

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