The dissociation constant of n-butyric acid i | vedclass

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(B) The degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.6 	imes 10^{-5}}{0.01}} = \sqrt{1.6 	imes 10^{-3}

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$1.52 	imes 10^{-2} \ S \ m^{-1}$

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(B) The degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.6 	imes 10^{-5}}{0.01}} = \sqrt{1.6 	imes 10^{-3}} \approx 0.04$. Since $\alpha = \frac{\Lambda_m}{\Lambda_m^\infty}$,we have $\Lambda_m = \alpha 	imes \Lambda_m^\infty = 0.04 	imes 380 	imes 10^{-4} \ S \ m^2 \ mol^{-1} = 15.2 	imes 10^{-4} \ S \ m^2 \ mol^{-1}$. The molar conductivity $\Lambda_m$ is related to specific conductance $\kappa$ and molar concentration $C$ by $\Lambda_m = \frac{\kappa}{C}$. Here,$C = 0.01 \ M = 10 \ mol \ m^{-3}$. Therefore,$\kappa = \Lambda_m 	imes C = (15.2 	imes 10^{-4} \ S \ m^2 \ mol^{-1}) 	imes (10 \ mol \ m^{-3}) = 1.52 	imes 10^{-2} \ S \ m^{-1}$. Thus,the correct option is $B$.

The dissociation constant of $n$-butyric acid is $1.6 \times 10^{-5}$ and the molar conductivity at infinite dilution is $380 \times 10^{-4} \ S \ m^2 \ mol^{-1}$. The specific conductance of the $0.01 \ M$ acid solution is

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