10 public questions tagged with this topic.
Lambdam° = λ°Ca²⁺ + 2 λ°Cl⁻ = 119 + 2 × 76.3 = 271.6 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Lambdam° = λ°Ca²⁺ + 2 λ°Cl⁻ = 119 + 2 × 76.3 = 271.6 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Lambdam° = λ°Ca²⁺ + 2λ°Cl⁻ . 258 = 119 + 2λ°Cl⁻ , 2λ°Cl⁻ = 139 , λ°Cl⁻ = 69.5 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Lambdam° = λ°Ca²⁺ + 2λ°Cl⁻ . 258 = 119 + 2λ°Cl⁻ , 2λ°Cl⁻ = 139 , λ°Cl⁻ = 69.5 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
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
Δ Tb = i · Kb · m . Δ Tb = 2.6 × 0.52 × 0.08 ≈ 0.1082 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Reaction: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. Moles of Ca(OH)₂ = 25/74 ≈ 0.3378 mol. 1 mol Ca(OH)₂ produces 1 mol CaCl₂; mass = 0.3378 × 111 ≈ 37.5 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
Reaction: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. Moles of CaCl₂ = 11.1/111 = 0.1 mol. 1 mol CaCl₂ from 1 mol Ca(OH)₂; mass = 0.1 × 74 = 7.4 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses
15 ppm = 15 g CaCl₂ in 10⁶ g solution. Volume = 1000 L. Moles = 15/111 ≈ 0.1351 mol; molarity = 0.1351/1000 ≈ 1.351 × 10⁻⁴ M.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Concentration Terms - Molarity Molality and Mole Fraction
15 ppm = 15 g CaCl₂ in 10⁶ g water. Moles = 15 / 111 ≈ 0.1351 mol. Mass of solvent = 10⁶ g = 1000 kg. Molality = 0.1351 / 1000 ≈ 1.351 × 10⁻⁴ m.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Concentration Terms - Molarity Molality and Mole Fraction
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