#Kb

Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

A solution of a non-volatile solute in water boils at 100.39°C at 1 atm. If the solute’s molality is 0.75 mol/kg, what i

Δ Tb = Kb · m . 100.39 - 100 = 0.39 . 0.39 = Kb × 0.75 . Kb = (0.39/0.75) = 0.52 K kg mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

What is the boiling point elevation of a solution containing 12 g of glucose (molar mass = 180 g/mol) in 600 g of water?

Moles of glucose = (12/180) ≈ 0.0667 mol . Molality = (0.0667/0.6) ≈ 0.111 mol/kg . Δ Tb = 0.52 × 0.111 ≈ 0.0577 K .

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

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

What is the boiling point elevation of a 0.5 molal solution of a non-electrolyte in water? ( Kb = 0.52 K kg mol⁻¹ )

Δ Tb = Kb · m . Δ Tb = 0.52 × 0.5 = 0.26 K .

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

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

What is the boiling point elevation of a solution containing 6 g of urea (molar mass = 60 g/mol) in 500 g of water? ( Kb

Moles of urea = (6/60) = 0.1 mol . Molality = (0.1/0.5) = 0.2 mol/kg . Δ Tb = 0.52 × 0.2 = 0.104 K .

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

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

What is the boiling point elevation of a solution containing 5 g of glucose (molar mass = 180 g/mol) in 250 g of water?

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

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

A solution of 5 g of a solute (molar mass = 100 g/mol) in 200 g of water boils at 100.13°C at 1 atm. What is the van’t H

Moles = (5/100) = 0.05 . Molality = (0.05/0.2) = 0.25 mol/kg . Δ Tb = i · Kb · m . 0.13 = i × 0.52 × 0.25 , i = (0.13/0.52 × 0.25) = 1 .

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

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

A solution boils at 100.13°C at 1 atm. What is the molality if Kb = 0.52 K kg mol⁻¹ ?

Δ Tb = Kb · m . 100.13 - 100 = 0.52 · m . m = (0.13/0.52) = 0.25 mol/kg .

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

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

The van’t Hoff factor of a 0.08 m CaCl₂ solution is 2.6. What is the boiling point elevation? ( Kb = 0.52 K kg mol⁻¹ )

Δ Tb = i · Kb · m . Δ Tb = 2.6 × 0.52 × 0.08 ≈ 0.1082 K .

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

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Colligative Properties - Depression of Freezing Point

A solution is made by dissolving 15 g of a solute (molar mass = 30 g/mol) in water. If the boiling point elevation is 0.

Moles = (15/30) = 0.5 . Δ Tb = Kb · m . 0.26 = 0.52 × (0.5/w) , w = (0.5 × 0.52/0.26) = 1 kg = 1000 g .

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

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Colligative Properties - Depression of Freezing Point

A solution boils at 100.208°C at 1 atm. What is the molality if Kb = 0.52 K kg mol⁻¹ ?

Δ Tb = Kb · m . 100.208 - 100 = 0.52 · m . m = (0.208/0.52) = 0.4 mol/kg .

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

The Ka of HCN is 6.2 × 10⁻¹⁰ . What is the Kb of CN- at 298 K?

Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰) ≈ 1.61 × 10⁻⁵ .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

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Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

The Ka of HSO₄- is 1.2 × 10⁻² . What is the Kb of SO₄²⁻ at 298 K?

Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/1.2 × 10⁻²) = 8.33 × 10⁻¹³ .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

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