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Kohlrausch's Law - UNSOLVED PRACTICE SET

Class 12

Chapter: Electrochemistry | Topic: Kohlrauschs Law

Study Material.
Class 12

KOHLRAUSCH'S LAW - UNSOLVED PRACTICE SET

Topic: Kohlrauschs Law

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. Kohlrausch's law of independent migration of ions is applicable at:

  1. Any concentration
  2. Infinite dilution only
  3. High concentrations only
  4. Only for strong electrolytes

Q2. According to Kohlrausch's law, the limiting molar conductivity of an electrolyte is:

  1. The product of ionic conductivities
  2. The sum of the limiting molar conductivities of its cations and anions
  3. The difference between cation and anion conductivities
  4. Independent of ionic conductivities

Q3. The limiting molar conductivity of BaCl₂ can be expressed as:

  1. λ°(Ba²⁺) + λ°(Cl⁻)
  2. λ°(Ba²⁺) + 2λ°(Cl⁻)
  3. 2λ°(Ba²⁺) + λ°(Cl⁻)
  4. λ°(Ba²⁺) × λ°(Cl⁻)

Q4. Kohlrausch's law is particularly useful for determining:

  1. The conductivity of strong electrolytes at high concentration
  2. The limiting molar conductivity of weak electrolytes
  3. The pH of solutions
  4. The boiling point of solutions

Q5. The transport number of an ion is defined as:

  1. The fraction of total current carried by that ion
  2. The total current carried by all ions
  3. The velocity of the ion
  4. The charge on the ion

Q6. If the transport number of Na⁺ in NaCl is 0.4, then the transport number of Cl⁻ is:

  1. 0.4
  2. 0.6
  3. 1.0
  4. 0.2

Short Answer Questions

Q7. State Kohlrausch's law of independent migration of ions. Write its mathematical expression for a general electrolyte AₓBᵧ.

Q8. Given Λ°m(NaCl) = 126.4, Λ°m(HCl) = 425.9, and Λ°m(CH₃COONa) = 91.0 S·cm²·mol⁻¹, calculate Λ°m(CH₃COOH).

Q9. Explain why Kohlrausch's law cannot be applied directly to determine Λ°m of weak electrolytes by extrapolation.

Q10. The limiting ionic conductivity of Ag⁺ is 61.9 S·cm²·mol⁻¹ and that of NO₃⁻ is 71.4 S·cm²·mol⁻¹. Calculate Λ°m(AgNO₃) and predict its solubility if the measured conductivity of a saturated solution is 1.33 × 10⁻³ S·cm⁻¹.

Q11. Your teacher asks why H⁺ and OH⁻ have exceptionally high ionic mobilities compared to other ions. Explain the Grotthuss mechanism for H⁺.

Q12. Calculate the transport number of K⁺ in KCl if λ°(K⁺) = 73.5 and λ°(Cl⁻) = 76.3 S·cm²·mol⁻¹.

SECTION NAME

Q13. Discuss Kohlrausch's law in detail:

(a) Statement and mathematical formulation

(b) Theoretical basis — ions migrate independently at infinite dilution

(c) Limiting ionic molar conductivities (λ°) for common ions

(d) Verification of the law using experimental data

(e) Factors affecting ionic mobility: ionic size, solvation, charge, temperature

(f) Special cases: H⁺ and OH⁻ — Grotthuss mechanism and their anomalously high conductivities

Q14. Explain the applications of Kohlrausch's law:

(a) Determination of Λ°m for weak electrolytes (e.g., CH₃COOH, NH₄OH)

(b) Calculation of the degree of dissociation and the dissociation constant of weak electrolytes

(c) Determination of solubility and solubility product of sparingly soluble salts (e.g., AgCl, BaSO₄)

(d) Determination of the ionic product of water (Kw)

(e) Calculation of transport numbers of ions

Illustrate each application with a worked numerical example.

Q15. Kohlrausch's law underpins modern electrochemical research in India. Discuss:

(a) How Indian researchers use ionic conductivity data to design better electrolytes for lithium-ion batteries

(b) The application of Kohlrausch's law in developing solid oxide fuel cells (SOFCs) for clean energy

(c) How transport number measurements help optimise membrane performance in desalination plants

(d) The role of limiting ionic conductivities in understanding ion transport in biological systems (nerve impulses)

SECTION NAME

Q16. The molar conductivity at infinite dilution for NaCl, HCl, and CH₃COONa are 126.4, 425.9, and 91.0 S cm² mol⁻¹ respectively at 298 K. Calculate the molar conductivity at infinite dilution for acetic acid (CH₃COOH).


Also, if the measured molar conductivity of 0.05 M acetic acid solution is 39.05 S cm² mol⁻¹, calculate the degree of dissociation (α) of acetic acid in this solution. What does this value of α tell you about the nature of acetic acid as an electrolyte?

Q17. (a) The conductivity of a 0.001 M solution of CH₃COOH at 298 K is 4.95 × 10⁻⁵ S cm⁻¹. Calculate the molar conductivity of the solution.

(b) If the limiting molar conductivity of CH₃COOH is 390.5 S cm² mol⁻¹, calculate the degree of dissociation of acetic acid at this concentration.

(c) Calculate the dissociation constant (Ka) of acetic acid using the expression Ka = Cα² / (1−α).

Q18. The conductivity of 0.00241 M acetic acid solution is 7.896 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity. If Λ°m for acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?


(Given: λ°(H⁺) = 349.6 S cm² mol⁻¹, λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹)


Show all steps clearly and state the significance of the dissociation constant value you obtain.


Total: 30 Marks | Time: 40 mins

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