#freezing point depression

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 23.7 mm Hg at a temperature where pure water’s vapo

Δ Tf = Kf · m . 0.372 = 1.86 · m . m = (0.372/1.86) = 0.2 mol/kg . Cross-check: xsolute = (25 - 23.7/25) = 0.052 , 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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Types of Solutions and Expressing Concentration

A solution of 10 g of a solute (molar mass = 40 g/mol) in 400 g of water has a freezing point of -0.93°C. What is the va

Moles = (10/40) = 0.25 . Molality = (0.25/0.4) = 0.625 mol/kg . Δ Tf = i · Kf · m . 0.93 = i × 1.86 × 0.625 , i = (0.93/1.86 × 0.625) ≈ 0.8 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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Types of Solutions and Expressing Concentration

A solution of 8 g of a solute (molar mass = 80 g/mol) in 500 g of water freezes at -0.372°C. What is the van’t Hoff fact

Moles = (8/80) = 0.1 . Molality = (0.1/0.5) = 0.2 mol/kg . Δ Tf = i · Kf · m . 0.372 = i × 1.86 × 0.2 , i = (0.372/1.86 × 0.2) = 1 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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Types of Solutions and Expressing Concentration

A solution is made by dissolving a non-volatile solute in 900 g of water. If the freezing point decreases by 0.93°C and

Δ Tf = Kf · m . 0.93 = 1.86 · m . m = (0.93/1.86) = 0.5 mol/kg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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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

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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

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Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a solution containing 15 g of mannitol (molar mass = 182 g/mol) in 500 g of wat

Moles of mannitol = (15/182) ≈ 0.0824 mol . Molality = (0.0824/0.5) ≈ 0.1648 mol/kg . Δ Tf = 1.86 × 0.1648 ≈ 0.3065 K .

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 0.2 molal Na₂SO₄ solution has a van’t Hoff factor of 2.5. What is the freezing point depression? ( Kf = 1.86 K kg mol⁻

Δ Tf = i · Kf · m . Δ Tf = 2.5 × 1.86 × 0.2 = 0.93 K .

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 is made by dissolving 20 g of a solute (molar mass = 40 g/mol) in water. If the freezing point depression is

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

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Colligative Properties - Depression of Freezing Point

What is the freezing point depression of a solution containing 12 g of a non-electrolyte solute (molar mass = 60 g/mol)

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

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Colligative Properties - Depression of Freezing Point

A 0.2 molal solution of NaBr in water has a freezing point depression of 0.744°C. What is the van’t Hoff factor? ( Kf =

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

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Colligative Properties - Depression of Freezing Point

A 0.2 molal solution of a solute in water has a freezing point depression of 0.744°C. If the solute dissociates into 2 i

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

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