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Moles of sucrose = (18/342) ≈ 0.0526 mol . Molality = (0.0526/0.4) ≈ 0.1315 mol/kg . Δ Tf = 1.86 × 0.1315 ≈ 0.2446 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Moles of sucrose = (12/342) ≈ 0.0351 mol . Mass of solvent = 300 g = 0.3 kg. Molality = (0.0351/0.3) ≈ 0.117 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
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