Types of Solutions and Expressing Concentration

Types of Solutions and Expressing Concentration

The van’t Hoff factor of a 0.05 m CaCl₂ solution is 2.7. What is the osmotic pressure at 27°C? ( R = 0.0821 L atm mol⁻¹

Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.05 M . Pi = 2.7 × 0.05 × 0.0821 × 300 ≈ 3.326 atm .

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 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

What is the mole fraction of a solute if its solution lowers the vapor pressure of water from 25 mm Hg to 24 mm Hg?

(p⁰ - p/p⁰) = xsolute . (25 - 24/25) = 0.04 .

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 containing 0.1 mole of a non-volatile solute in 1 kg of water has an osmotic pressure of 2.46 atm at 27°C. Wh

Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.1 M . 2.46 = i × 0.1 × 0.0821 × 300 . i = (2.46/0.1 × 0.0821 × 300) ≈ 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 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

The vapor pressure of a pure solvent is 40 mm Hg. What is the vapor pressure of a solution with a mole fraction of solut

Mole fraction of solvent = 1 - 0.25 = 0.75 . Vapor pressure = 40 × 0.75 = 30 mm Hg .

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

What is the mass percentage of a solution if 5 g of a solute is dissolved in 45 g of water?

Total mass = 5 g + 45 g = 50 g. Mass % = (5/50) × 100 = 10% .

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 two volatile liquids has vapor pressures of 180 mm Hg and 240 mm Hg for pure components. If the total vapo

P = P₁⁰ · x₁ + P₂⁰ · x₂ , where x₁ + x₂ = 1 . 216 = 180 (1 - x₂) + 240 x₂ . 216 = 180 - 180 x₂ + 240 x₂ . 216 - 180 = 60 x₂ , x₂ = (36/60) = 0.6 .

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

What is the mole fraction of a solute if the vapor pressure of a solution is 22 mm Hg and that of pure solvent is 25 mm

(p⁰ - p/p⁰) = xsolute . (25 - 22/25) = (3/25) = 0.12 .

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

How many grams of CaCl₂ (molar mass = 111 g/mol) are needed to prepare 600 mL of a 0.2 M solution?

Moles of CaCl₂ = 0.2 × 0.6 = 0.12 mol . Mass = 0.12 × 111 = 13.32 g .

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 reduces the vapor pressure of a solvent from 35 mm Hg to 33 mm Hg. What is the mole fraction of the solute?

(p⁰ - p/p⁰) = xsolute . (35 - 33/35) = (2/35) ≈ 0.0571 .

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 two volatile liquids has vapor pressures of 200 mm Hg and 300 mm Hg for pure components. If the mole fract

Mole fraction of second = 1 - 0.7 = 0.3 . Total vapor pressure = 200 × 0.7 + 300 × 0.3 = 140 + 90 = 230 mm Hg .

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

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