#boiling point elevation

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

A solution of a non-volatile solute in water has a vapor pressure of 22.4 mm Hg at a temperature where pure water’s vapo

Δ Tb = 100.208 - 100 = 0.208 K . Δ Tb = Kb · m . 0.208 = 0.52 · m , m = (0.208/0.52) = 0.4 mol/kg . Cross-check: xsolute = (24 - 22.4/24) = 0.0667 , consistent for dilute solution.

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

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Solubility - of Solids and Gases in Liquids Henry's Law

A 0.2 molal solution of a non-electrolyte in water has its boiling point elevated by 0.104°C. What is the value of Kb fo

Δ Tb = Kb · m . 0.104 = Kb × 0.2 . Kb = (0.104/0.2) = 0.52 K kg mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Solubility - of Solids and Gases in Liquids Henry's Law

A solution of a non-volatile solute in water boils at 100.52°C at 1 atm. What is the molality of the solution? ( Kb = 0.

Δ Tb = Kb · m . 100.52 - 100 = 0.52 · m . m = (0.52/0.52) = 1 mol/kg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Solubility - of Solids and Gases in Liquids Henry's Law

A solution boils at 100.26°C at 1 atm. What is the molality if Kb = 0.52 K kg mol⁻¹ ?

Δ Tb = Kb · m . 100.26 - 100 = 0.52 · m . m = (0.26/0.52) = 0.5 mol/kg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law

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Colligative Properties - Relative Lowering and Elevation of Boiling Point

The van’t Hoff factor for a 0.1 m solution of NaCl is approximately 1.87. What is the observed boiling point elevation i

Δ Tb = i · Kb · m . Δ Tb = 1.87 × 0.52 × 0.1 ≈ 0.097 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 10 g of fructose (molar mass = 180 g/mol) in 400 g of water

Moles of fructose = (10/180) ≈ 0.0556 mol . Molality = (0.0556/0.4) ≈ 0.139 mol/kg . Δ Tb = 0.52 × 0.139 ≈ 0.0723 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 10 g of a non-volatile solute in 400 g of water has a boiling point elevation of 0.13°C. What is the molar

Δ Tb = Kb · m . 0.13 = 0.52 × (moles/0.4) . Moles = (0.13 × 0.4/0.52) = 0.1 . Molar mass = (10/0.1) = 100 g/mol .

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 0.25 molal solution of a solute in water has a boiling point elevation of 0.26°C. If the solute dissociates into 2 ion

Δ Tb = i · Kb · m . 0.26 = i × 0.52 × 0.25 . i = (0.26/0.52 × 0.25) = 2 . i = 1 + α (n - 1) , 2 = 1 + α (2 - 1) , α = 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

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

The van’t Hoff factor of a 0.1 m K₂SO₄ solution is 2.4. What is the boiling point elevation? ( Kb = 0.52 K kg mol⁻¹ )

Δ Tb = i · Kb · m . Δ Tb = 2.4 × 0.52 × 0.1 = 0.1248 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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