#Kf

Colligative Properties - Relative Lowering and Elevation of Boiling Point

A solution freezes at -0.558°C. What is the molality if Kf = 1.86 K kg mol⁻¹ ?

Δ Tf = Kf · m . 0.558 = 1.86 · m . m = (0.558/1.86) = 0.3 mol/kg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point

View Details

Colligative Properties - Osmotic Pressure and Reverse Osmosis

A solution is prepared by dissolving 15 g of a non-volatile solute in 300 g of water. If the freezing point depression i

Δ Tf = Kf · m . 0.465 = 1.86 × (moles/0.3) . Moles = (0.465 × 0.3/1.86) ≈ 0.075 . Molar mass = (15/0.075) = 200 g/mol .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis

View Details

Colligative Properties - Depression of Freezing Point

A 0.1 molal NaCl solution has a van’t Hoff factor of 1.8. What is the freezing point depression? ( Kf = 1.86 K kg mol⁻¹

Δ Tf = i · Kf · m . Δ Tf = 1.8 × 1.86 × 0.1 ≈ 0.335 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

View Details

Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a solution containing 8 g of urea (molar mass = 60 g/mol) in 200 g of water? (

Moles of urea = (8/60) ≈ 0.1333 mol . Molality = (0.1333/0.2) ≈ 0.6665 mol/kg . Δ Tf = 1.86 × 0.6665 ≈ 1.24 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

View Details

Colligative Properties - Depression of Freezing Point

A solution freezes at -0.372°C. What is the molality of the solution if Kf = 1.86 K kg mol⁻¹ ?

Δ Tf = Kf · m . 0.372 = 1.86 · m . m = (0.372/1.86) = 0.2 mol/kg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

View Details

Colligative Properties - Depression of Freezing Point

A solution is made by mixing 40 g of a solute (molar mass = 80 g/mol) with water. If the freezing point depression is 0.

Moles = (40/80) = 0.5 . Δ Tf = Kf · m . 0.93 = 1.86 × (0.5/w) , w = (0.5 × 1.86/0.93) = 1 kg = 1000 g . Total mass = 40 + 1000 = 1040 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

View Details

Colligative Properties - Depression of Freezing Point

A 0.25 molal NaCl solution has a van’t Hoff factor of 1.9. What is the freezing point depression? ( Kf = 1.86 K kg mol⁻¹

Δ Tf = i · Kf · m . Δ Tf = 1.9 × 1.86 × 0.25 ≈ 0.8835 K .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point

View Details