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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 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of glucose = (9/180) = 0.05 mol . Volume = 500 mL = 0.5 L. Molarity = (0.05/0.5) = 0.1 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
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 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Moles of glucose = (60/180) = 0.333 mol . Mass of solvent = 540 g = 0.54 kg. Molality = (0.333/0.54) ≈ 0.617 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Moles of glucose = (9/180) = 0.05 mol . Molality = (0.05/0.25) = 0.2 mol/kg . Δ Tf = 1.86 × 0.2 = 0.372 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Moles of glucose = 12/180 ≈ 0.0667 mol. Mass of water = 48 g = 0.048 kg. Molality = 0.0667 / 0.048 ≈ 1.39 m.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Importance and Nature of Chemistry and Properties of Matter
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