Practice question
Question
What is the boiling point elevation of a solution containing 5 g of glucose (molar mass = 180 g/mol) in 250 g of water? ( Kb = 0.52 K kg mol⁻¹ )
Explanation
Moles of glucose = (5/180) ≈ 0.0278 mol . Molality = (0.0278/0.25) ≈ 0.1112 mol/kg . Δ Tb = 0.52 × 0.1112 ≈ 0.0578 K .
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