The dissociation constant of acetic acid is 1 | vedclass

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(B) Given: $K_a = 1.6 	imes 10^{-5}$,$C = 0.01 \, M$,$\lambda_m^\infty = 380 	imes 10^{-4} \, S \, m^2 \, mol^{-1}$.Degree of dissociation $\alpha =

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

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(B) Given: $K_a = 1.6 	imes 10^{-5}$,$C = 0.01 \, M$,$\lambda_m^\infty = 380 	imes 10^{-4} \, S \, m^2 \, mol^{-1}$.Degree of dissociation $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.6 	imes 10^{-5}}{0.01}} = \sqrt{1.6 	imes 10^{-3}} = 0.04$.Molar conductance $\lambda_m = \alpha 	imes \lambda_m^\infty = 0.04 	imes 380 	imes 10^{-4} = 15.2 	imes 10^{-4} \, S \, m^2 \, mol^{-1}$.Specific conductance $\kappa = \lambda_m 	imes C$ (in $S \, m^{-1}$ units,$C$ must be in $mol \, m^{-3}$).$C = 0.01 \, mol \, L^{-1} = 0.01 	imes 10^3 \, mol \, m^{-3} = 10 \, mol \, m^{-3}$.$\kappa = (15.2 	imes 10^{-4} \, S \, m^2 \, mol^{-1}) 	imes (10 \, mol \, m^{-3}) = 15.2 	imes 10^{-3} \, S \, m^{-1} = 1.52 	imes 10^{-2} \, S \, m^{-1}$.

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

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