#freezing point depression

CHEMISTRY

A solution of 1.8 g of benzoic acid ( C₆H₅COOH ) in 30 g of benzene shows a freezing point depression of 1.2 K. What is

Given: A solution of 1.8 g of benzoic acid ( C₆H₅COOH ) in 30 g of benzene shows a freezing point depression of 1.2 K. What is the van't Hoff factor if it dimerizes? ( K_f = 5.12 K kg/mol, Molar mass = 122 g/mol ) Formula: Molality = 1.8 / 122/0.03 = 0.492 mol/kg. Substitution & Calculation: Normal Δ T_f = 5.12 × 0.492 = 2.52 K . Observed Δ T_f = 1.2 K . i = 1.2/2.52 = 0.476 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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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 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 .