Conductance And Kohlrausch's Law
The molar conductivity of a 0.001 M acetic acid solution is 49.5 S cm² mol⁻¹. If the limiting molar conductivity of acetic acid is 390.5 S cm² mol⁻¹, the degree of dissociation (α) of acetic acid at this concentration is approximately:
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Solution
0.127
The degree of dissociation is given by α = Λm / Λ°m, where Λm is the molar conductivity at the given concentration and Λ°m is the limiting molar conductivity (at infinite dilution). Substituting: α = 49.5 / 390.5 ≈ 0.1268 ≈ 0.127. This relationship holds because molar conductivity increases with dilution for weak electrolytes, reaching a maximum (Λ°m) at infinite dilution where the electrolyte is completely dissociated. At finite concentrations, only a fraction α of the electrolyte dissociates, so the observed conductivity is proportionally reduced.
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About This Question
- Subject
- chemistry
- Chapter
- redox reactions and electrochemistry
- Topic
- conductance and kohlrausch's law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
0.127
The degree of dissociation is given by α = Λm / Λ°m, where Λm is the molar conductivity at the given concentration and Λ°m is the limiting molar conductivity (at infinite dilution). Substituting: α = 49.5 / 390.5 ≈ 0.1268 ≈ 0.127. This relationship holds because molar conductivity increases with dilution for weak electrolytes, reaching a maximum (Λ°m) at infinite dilution where the electrolyte is completely dissociated. At finite concentrations, only a fraction α of the electrolyte dissociates, so the observed conductivity is proportionally reduced.
This medium difficulty chemistry question is from the chapter redox reactions and electrochemistry, covering the topic of conductance and kohlrausch's law. It appeared in the 2025 exam.
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