Practice question
Question
A solution of 10 g of a non-volatile solute in 400 g of water has a boiling point elevation of 0.13°C. What is the molar mass of the solute? ( Kb = 0.52 K kg mol⁻¹ )
Explanation
Δ Tb = Kb · m . 0.13 = 0.52 × (moles/0.4) . Moles = (0.13 × 0.4/0.52) = 0.1 . Molar mass = (10/0.1) = 100 g/mol .
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