Practice question
Question
A solution boils at 100.208°C at 1 atm. What is the molality if Kb = 0.52 K kg mol⁻¹ ?
Explanation
Δ Tb = Kb · m . 100.208 - 100 = 0.52 · m . m = (0.208/0.52) = 0.4 mol/kg .
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