Question: 65. The molar conductance of \mathrm{NaCl}, \mathrm{HCl} and \mathrm{CH}_{3} \mathrm{COONa} at infinite dilution are 126.45 , 426.16 and 91.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} respectively. The molar conductance of \mathrm{CH}_{3} \mathrm{COOH} at infinite dilution is. Choose the right option for your answer.
(1) 201.28 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}
(2) 390.71 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}
(3) 698.28 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}
(4) 540.48 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}
Answer: Option (2)
Explanation:
The molar conductance of weak electrolytes at infinite dilution is calculated using Kohlrausch’s law of independent migration of ions.
According to Kohlrausch’s law, the molar conductance at infinite dilution is the sum of the individual ionic conductances.
For acetic acid, the relation is:
\Lambda^{\circ}(\mathrm{CH_3COOH}) = \Lambda^{\circ}(\mathrm{HCl}) + \Lambda^{\circ}(\mathrm{CH_3COONa}) - \Lambda^{\circ}(\mathrm{NaCl})Substituting the given values:
\Lambda^{\circ}(\mathrm{CH_3COOH}) = 426.16 + 91.0 - 126.45 \Lambda^{\circ}(\mathrm{CH_3COOH}) = 390.71 \ \mathrm{S \ cm^2 \ mol^{-1}}Hence, the correct answer is option (2).