Colligative Properties - Relative Lowering and Elevation of Boiling Point

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

A gas dissolves in a solvent with a Henry’s law constant of 250 bar at 25°C. If the mole fraction of the gas doubles whe

Initial: p = 250 · x . New: 10 = KH' · 2x . Assume initial p = 5 bar , then x = (5/250) = 0.02 . New x = 0.04 , 10 = KH' × 0.04 , KH' = 250 bar . Recalculate: KH' = (10/0.04) = 250 bar (consistent).

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 gas has a Henry’s law constant of 250 bar. If its partial pressure is 5 bar and the solvent mass is 1 kg, what is the

p = KH · x . x = (5/250) = 0.02 . Molality ≈ x for dilute solutions, so m = 0.02 mol/kg . Moles = 0.02 × 1 = 0.02 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

What is the molarity of a solution prepared by dissolving 4 g of NaOH (molar mass = 40 g/mol) in 250 mL of solution?

Moles of NaOH = (4/40) = 0.1 mol . Volume = 250 mL = 0.25 L. Molarity = (0.1/0.25) = 0.4 M .

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 a gas in water has a mole fraction of 0.03 at a partial pressure of 4.5 bar. If the temperature increases

Henry’s law: p = KH · x . New p = 200 × 0.03 = 6 bar .

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 gas has a Henry’s law constant of 300 bar at 25°C. If 0.01 mole of the gas is dissolved in 500 g of water at 6 bar, wh

Molality = (moles/mass of solvent in kg) = (0.01/0.5) = 0.02 mol/kg . Cross-check: x = (6/300) = 0.02 , consistent with molality for dilute solutions.

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 mole fraction of methanol (molar mass = 32 g/mol) in a solution containing 16 g of methanol and 72 g of wate

Moles of methanol = (16/32) = 0.5 mol . Moles of water = (72/18) = 4 mol . Total moles = 0.5 + 4 = 4.5 . Mole fraction = (0.5/4.5) ≈ 0.111 .

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 is prepared with 46 g of ethanol (molar mass = 46 g/mol) and 54 g of water. If the mole fraction of ethanol b

Initial moles of ethanol = (46/46) = 1 . Moles of water = (54/18) = 3 . Let additional moles of ethanol = x . New mole fraction = (1 + x/1 + x + 3) = 0.4 . 1 + x = 0.4 (4 + x) , 1 + x = 1.6 + 0.4x , 0.6x = 0.6 , x = 1 . Mass added = 1 × 46 = 46 g .

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

How many grams of KCl (molar mass = 74.5 g/mol) are needed to prepare 250 mL of a 0.4 M solution?

Moles of KCl = 0.4 × 0.25 = 0.1 mol . Mass = 0.1 × 74.5 = 7.45 g .

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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