Practice question

MCQ

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A solution of a non-volatile solute in water boils at 100.39°C at 1 atm. If the solute’s molality is 0.75 mol/kg, what is the value of Kb for water?

Choose one · Correct answer highlighted

0.5 K kg mol⁻¹ 0.48 K kg mol⁻¹ 0.55 K kg mol⁻¹ 0.52 K kg mol⁻¹ Correct

Δ Tb = Kb · m . 100.39 - 100 = 0.39 . 0.39 = Kb × 0.75 . Kb = (0.39/0.75) = 0.52 K kg mol⁻¹ .

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