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Electrochemistry · Class 12 · CBSE · Science Chemistry

Limiting molar conductivity (Lambdam^0) for NaCl, HCl, and CH3COONa are 126.4, 425.9, and 91.0text{ S cm}^2text{ mol}^{-1} respectively. The limiting…

Practice question from Electrochemistry (Class 12 · CBSE · Science Chemistry), with the correct answer and a step-by-step solution.

medium · MCQ · Q-286CE8

Limiting molar conductivity (Λm0\Lambda_m^0) for \NaCl\NaCl, \HCl\HCl, and \CH3COONa\CH_3COONa are 126.4126.4, 425.9425.9, and 91.0 S cm2 mol−191.0\text{ S cm}^2\text{ mol}^{-1} respectively. The limiting molar conductivity for \CH3COOH\CH_3COOH is:

  1. A.390.5 S cm2 mol−1390.5\text{ S cm}^2\text{ mol}^{-1}Correct answer
  2. B.425.5 S cm2 mol−1425.5\text{ S cm}^2\text{ mol}^{-1}
  3. C.542.3 S cm2 mol−1542.3\text{ S cm}^2\text{ mol}^{-1}
  4. D.207.5 S cm2 mol−1207.5\text{ S cm}^2\text{ mol}^{-1}

Solution

According to Kohlrausch's law, Λm0(\CH3COOH)=Λm0(\CH3COONa)+Λm0(\HCl)−Λm0(\NaCl)\Lambda_m^0(\CH_3COOH) = \Lambda_m^0(\CH_3COONa) + \Lambda_m^0(\HCl) - \Lambda_m^0(\NaCl). Substituting the values: 91.0+425.9−126.4=390.5 S cm2 mol−191.0 + 425.9 - 126.4 = 390.5\text{ S cm}^2\text{ mol}^{-1}.

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