Practice question
Question
A solution of 18 g of glucose (molar mass = 180 g/mol) in 1 kg of water freezes at -0.186°C. What is the Kf of water?
Explanation
Molality = (18 / 180/1) = 0.1 mol/kg . Δ Tf = Kf · m . 0.186 = Kf × 0.1 . Kf = (0.186/0.1) = 1.86 K kg mol⁻¹ .
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