6 public questions tagged with this topic.
Moles = (20/40) = 0.5 . Δ Tf = Kf · m . 0.465 = 1.86 × (0.5/w) , w = (0.5 × 1.86/0.465) ≈ 2 kg = 2000 g . Total mass = 20 + 2000 = 2020 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Moles of solute = (12/60) = 0.2 mol . Molality = (0.2/0.2) =appendicitis = 1 mol/kg . Δ Tf = Kf · m = 1.86 × 1 = 1.86 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Δ Tf = i · Kf · m . 0.744 = i × 1.86 × 0.2 . i = (0.744/1.86 × 0.2) = 2 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Δ Tf = i · Kf · m . 0.744 = i × 1.86 × 0.2 . i = (0.744/1.86 × 0.2) = 2 . i = 1 + α (n - 1) , 2 = 1 + α (2 - 1) , α = 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
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 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
