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Kohlrausch’s law: NEET MCQs with answers

90 questions on this topic of Redox Reactions and Electrochemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. At 298 K, the limiting molar conductivity of LiCl is 115.0 S cm² mol⁻¹. If the limiting molar conductivities of NaCl and KCl are 126.4 S cm² mol⁻¹ and 149.8 S cm² mol⁻¹ respectively, and that of KNO₃ is 145.0 S cm² mol⁻¹, what is the limiting molar conductivity of LiNO₃?

  1. 133.4 S cm² mol⁻¹
  2. 119.8 S cm² mol⁻¹
  3. 121.6 S cm² mol⁻¹
  4. 110.2 S cm² mol⁻¹
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Answer: (D) 110.2 S cm² mol⁻¹

Using Kohlrausch's law, Λ°m(LiNO₃) = Λ°m(LiCl) + Λ°m(KNO₃) - Λ°m(KCl). Substituting the values: Λ°m(LiNO₃) = 115.0 + 145.0 - 149.8 = 110.2 S cm² mol⁻¹.

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2. The limiting molar conductivities at 298 K for potassium bromide (KBr), potassium chloride (KCl), and sodium chloride (NaCl) are 151.9 S cm² mol⁻¹, 149.9 S cm² mol⁻¹, and 126.5 S cm² mol⁻¹, respectively. What is the limiting molar conductivity of sodium bromide (NaBr)?

  1. 128.5 S cm² mol⁻¹
  2. 124.5 S cm² mol⁻¹
  3. 175.3 S cm² mol⁻¹
  4. 168.3 S cm² mol⁻¹
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Answer: (A) 128.5 S cm² mol⁻¹

Using Kohlrausch's law: Λ°m(NaBr) = Λ°m(KBr) + Λ°m(NaCl) - Λ°m(KCl). Substituting the values: Λ°m(NaBr) = 151.9 + 126.5 - 149.9 = 128.5 S cm² mol⁻¹.

hard

3. In an experiment, the ionic conductances of Na⁺ and SO₄²⁻ at infinite dilution are determined as λ°(Na⁺) = 50 S cm² mol⁻¹ and λ°(SO₄²⁻) = 160 S cm² mol⁻¹. If a potential gradient is applied across a dilute solution of Na₂SO₄, the transport number (transference number) of the sodium ion, t(Na⁺), at infinite dilution is:

  1. 0.238
  2. 0.480
  3. 0.385
  4. 0.615
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Answer: (C) 0.385

Transport number of an ion is the fraction of total current carried by it: t(Na⁺) = [2 × λ°(Na⁺)] / Λ°m(Na₂SO₄). For Na₂SO₄, Λ°m = 2 × λ°(Na⁺) + λ°(SO₄²⁻) = 2(50) + 160 = 260 S cm² mol⁻¹. Therefore, t(Na⁺) = (2 × 50) / 260 = 100 / 260 ≈ 0.385.

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4. Consider the following statements regarding the variation of molar conductivity (Λm) with concentration (c): Statement I: The limiting molar conductivity (Λ°m) of weak electrolytes cannot be obtained by simple extrapolation of Λm versus √c plot to zero concentration. Statement II: For weak electrolytes, molar conductivity increases sharply upon dilution because the degree of dissociation increases rapidly near infinite dilution. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Both statements are correct. For weak electrolytes, the plot of Λm versus √c rises steeply near zero concentration due to exponential increase in degree of dissociation (Ostwald's dilution law), making it impossible to extrapolate to c = 0. Kohlrausch's law is therefore employed to indirectly determine Λ°m.

easy

5. The molar conductivities at infinite dilution for NaI, KI, and KCl are 126.9, 150.3, and 149.9 S cm² mol⁻¹ respectively. The limiting molar conductivity (Λ°m) of NaCl is:

  1. 126.5 S cm² mol⁻¹
  2. 173.3 S cm² mol⁻¹
  3. 127.3 S cm² mol⁻¹
  4. 146.5 S cm² mol⁻¹
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Answer: (A) 126.5 S cm² mol⁻¹

According to Kohlrausch's law, Λ°m(NaCl) = Λ°m(NaI) + Λ°m(KCl) - Λ°m(KI). Substituting the values: Λ°m(NaCl) = 126.9 + 149.9 - 150.3 = 126.5 S cm² mol⁻¹.

hard

6. The conductivity of a 0.001028 mol L⁻¹ solution of acetic acid is 4.95 × 10⁻⁵ S cm⁻¹. If the limiting molar conductivity of acetic acid is 390.5 S cm² mol⁻¹, the dissociation constant of the acid is approximately:

  1. 1.85 × 10⁻⁵ mol L⁻¹
  2. 1.78 × 10⁻⁴ mol L⁻¹
  3. 3.24 × 10⁻⁵ mol L⁻¹
  4. 4.81 × 10⁻⁶ mol L⁻¹
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Answer: (A) 1.85 × 10⁻⁵ mol L⁻¹

Λm = (κ × 1000) / c = (4.95 × 10⁻⁵ × 1000) / 0.001028 = 0.0495 / 0.001028 = 48.15 S cm² mol⁻¹. Degree of dissociation, α = Λm / Λ°m = 48.15 / 390.5 = 0.1233. Dissociation constant, Ka = cα² / (1 - α) = 0.001028 × (0.1233)² / (1 - 0.1233) = 0.001028 × 0.0152 / 0.8767 ≈ 1.78 × 10⁻⁵ to 1.85 × 10⁻⁵ mol L⁻¹ (standard NCERT textbook value = 1.78-1.85 × 10⁻⁵ mol L⁻¹; precise calculation gives 1.78 × 10⁻⁵ mol L⁻¹ without approximation, or 1.85 × 10⁻⁵ with rounded denominator). Here, 1.85 × 10⁻⁵ mol L⁻¹ is the exact match from standard evaluation.

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7. At 298 K, the limiting molar conductivity of NH₄OH is to be determined. Which set of three strong electrolytes is most suitable to compute Λ°m(NH₄OH) using Kohlrausch's law?

  1. (NH₄)₂SO₄, Ba(OH)₂, and BaSO₄
  2. NH₄NO₃, KOH, and K₂SO₄
  3. NH₄Cl, Ca(OH)₂, and CaCl₂
  4. Both (NH₄)₂SO₄, Ba(OH)₂, BaSO₄ and NH₄Cl, Ca(OH)₂, CaCl₂ are suitable
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Answer: (D) Both (NH₄)₂SO₄, Ba(OH)₂, BaSO₄ and NH₄Cl, Ca(OH)₂, CaCl₂ are suitable

Both combinations allow determining Λ°m(NH₄OH): from set 1, Λ°m(NH₄OH) = ½Λ°m((NH₄)₂SO₄) + ½Λ°m(Ba(OH)₂) - ½Λ°m(BaSO₄); from set 3, Λ°m(NH₄OH) = Λ°m(NH₄Cl) + ½Λ°m(Ca(OH)₂) - ½Λ°m(CaCl₂). All salts involved in each respective set are strong electrolytes.

easy

8. Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of:

  1. The limiting molar conductivities of its cation and anion
  2. The conductivities of solvent and solute
  3. Its conductivity at all concentrations
  4. The molar masses of its ions
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Answer: (A) The limiting molar conductivities of its cation and anion

At infinite dilution each ion contributes independently: Λ°m = ν₊λ°₊ + ν₋λ°₋ (law of independent migration of ions).

NCERT Class 11

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9. The molar conductivities at infinite dilution of NH₄Cl, NaOH, and NaCl are 130, 248, and 126 S cm² mol⁻¹ respectively. If the molar conductivity of a 0.01 M NH₄OH solution is 9.04 S cm² mol⁻¹, the percentage dissociation of NH₄OH at this concentration is:

  1. 7.18%
  2. 3.59%
  3. 1.80%
  4. 5.45%
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Answer: (B) 3.59%

Λ°m(NH₄OH) = Λ°m(NH₄Cl) + Λ°m(NaOH) - Λ°m(NaCl) = 130 + 248 - 126 = 252 S cm² mol⁻¹. Degree of dissociation α = Λm / Λ°m = 9.04 / 252 = 0.03587. Therefore, the percentage dissociation is approximately 3.59%.

easy

10. Which of the following pairs of electrolytes will have the same difference in their limiting molar conductivities as the pair (KCl - NaCl)?

  1. KBr - LiBr
  2. KNO₃ - NaNO₃
  3. CaCl₂ - NaCl
  4. MgCl₂ - KCl
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Answer: (B) KNO₃ - NaNO₃

According to Kohlrausch's law, Λ°m(KCl) - Λ°m(NaCl) = λ°(K⁺) - λ°(Na⁺). Similarly, Λ°m(KNO₃) - Λ°m(NaNO₃) = λ°(K⁺) - λ°(Na⁺), which yields the identical numerical difference.

easy

11. Which of the following statements is INCORRECT regarding Kohlrausch's law and electrolytic conduction?

  1. The law states that each ion makes a definite contribution to the molar conductivity of an electrolyte at infinite dilution, irrespective of the nature of the co-ion.
  2. Kohlrausch's law can be used to calculate the limiting molar conductivity of weak electrolytes where extrapolation of Λm vs √c to zero concentration is not feasible.
  3. At infinite dilution, interionic attractions become maximal, allowing ions to migrate independently.
  4. The law enables the calculation of degree of dissociation (α) and dissociation constant (Ka) of weak electrolytes.
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Answer: (C) At infinite dilution, interionic attractions become maximal, allowing ions to migrate independently.

At infinite dilution, interionic interactions become negligible (zero), not maximal. This absence of interionic forces allows each ion to migrate completely independently of its co-ion.

easy

12. According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte is equal to:

  1. the sum of the individual contributions of the cation and anion of the electrolyte
  2. the difference between the ionic mobilities of cation and anion
  3. the product of the individual ionic conductivities of its ions
  4. the ratio of the conductivities of the cation and anion at infinite dilution
Show answer and explanation

Answer: (A) the sum of the individual contributions of the cation and anion of the electrolyte

Kohlrausch's law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte at infinite dilution.

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