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Moles of fructose = (10/180) ≈ 0.0556 mol . Molality = (0.0556/0.4) ≈ 0.139 mol/kg . Δ Tb = 0.52 × 0.139 ≈ 0.0723 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
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