(A) The molar conductivity at infinite dilution for acetic acid is given by Kohlrausch's law: $\Lambda_m^0(CH_3COOH) = \lambda_m^0(H^+) + \lambda_m^0(

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(A) The molar conductivity at infinite dilution for acetic acid is given by Kohlrausch's law: $\Lambda_m^0(CH_3COOH) = \lambda_m^0(H^+) + \lambda_m^0(CH_3COO^-) = 349 + 41 = 390 \ S \ cm^2 \ mol^{-1}$.Degree of dissociation $\alpha = \frac{\Lambda_m^c}{\Lambda_m^0} = \frac{7.8}{390} = 0.02$.The dissociation constant $K_a$ is given by $K_a = \frac{c \alpha^2}{1 - \alpha}$.Given $c = 0.04 \ M$,$K_a = \frac{0.04 \times (0.02)^2}{1 - 0.02} = \frac{0.04 \times 0.0004}{0.98} = \frac{0.000016}{0.98} \approx 1.63 \times 10^{-5}$.

Class 12 Chemistry Electrochemistry Mix Examples-Electrochemistry

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At $T(K)$,the molar conductivity of $0.04 \ M$ acetic acid is $7.8 \ S \ cm^2 \ mol^{-1}$. If the limiting molar conductivities of $H^{+}$ and $CH_3COO^{-}$ at $T(K)$ are $349$ and $41 \ S \ cm^2 \ mol^{-1}$ respectively,the dissociation constant of acetic acid is

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A
$1.63 \times 10^{-5}$
B
$8.33 \times 10^{-5}$
C
$1.63 \times 10^{-4}$
D
$8.33 \times 10^{-4}$

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At $T(K)$,the molar conductivity of $0.04 \ M$ acetic acid is $7.8 \ S \ cm^2 \ mol^{-1}$. If the limiting molar conductivities of $H^{+}$ and $CH_3COO^{-}$ at $T(K)$ are $349$ and $41 \ S \ cm^2 \ mol^{-1}$ respectively,the dissociation constant of acetic acid is

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At $300 \ K$,the conductivity of $0.01 \ mol \ dm^{-3}$ aqueous solution of acetic acid is $19.5 \times 10^{-5} \ S \ cm^{-1}$ and the limiting molar conductivity of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation of acetic acid is:

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Given below are two statements:
Statement $I$: For $KI$,molar conductivity increases steeply with dilution.
Statement $II$: For carbonic acid,molar conductivity increases slowly with dilution.
In the light of the above statements,choose the correct answer from the options given below: