Question:
The molar conductivity \((\Lambda_{m})\) acetic acid is \(78.1 \text{S cm}^{2} \text{mol}^{-1}\) . Its degree of dissociation \((\alpha)\) is \((\Lambda_{m}^{0})\) for acetic acid= 390.5 S \(cm^{2} \text{mol}^{-1}\)
The molar conductivity \((\Lambda_{m})\) acetic acid is \(78.1 \text{S cm}^{2} \text{mol}^{-1}\) . Its degree of dissociation \((\alpha)\) is \((\Lambda_{m}^{0})\) for acetic acid= 390.5 S \(cm^{2} \text{mol}^{-1}\)
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Kohlrausch's law: \(\Lambda_m^0\) for weak electrolyte = sum of ionic conductivities of cation and anion.
Updated On: Apr 24, 2026
- 0.12
- 0.40
- 0.02
- 0.20
- 0.04
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Degree of dissociation \(\alpha = \frac{\Lambda_m}{\Lambda_m^0}\) for weak electrolytes.
Step 2: Detailed Explanation:
Given \(\Lambda_m = 78.1 \text{ S cm}^2 \text{ mol}^{-1}\) and \(\Lambda_m^0 = 390.5 \text{ S cm}^2 \text{ mol}^{-1}\).
\(\alpha = \frac{78.1}{390.5} = 0.2\).
Step 3: Final Answer:
The degree of dissociation is 0.20.
Degree of dissociation \(\alpha = \frac{\Lambda_m}{\Lambda_m^0}\) for weak electrolytes.
Step 2: Detailed Explanation:
Given \(\Lambda_m = 78.1 \text{ S cm}^2 \text{ mol}^{-1}\) and \(\Lambda_m^0 = 390.5 \text{ S cm}^2 \text{ mol}^{-1}\).
\(\alpha = \frac{78.1}{390.5} = 0.2\).
Step 3: Final Answer:
The degree of dissociation is 0.20.
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