Conductivity of $0.00241 \, M$ acetic acid is $7.896 \times 10^{-5} \, S \, cm^{-1}$. Calculate its molar conductivity. If $\Lambda_m^o$ for acetic acid is $390.5 \, S \, cm^2 \, mol^{-1}$,what is its dissociation constant?

(N/A) Given,$\kappa = 7.896 \times 10^{-5} \, S \, cm^{-1}$,$c = 0.00241 \, mol \, L^{-1}$.
Molar conductivity,$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} = 32.76 \, S \, cm^2 \, mol^{-1}$.
Degree of dissociation,$\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{32.76}{390.5} = 0.084$.
Dissociation constant,$K_a = \frac{c \alpha^2}{1 - \alpha} = \frac{0.00241 \times (0.084)^2}{1 - 0.084} = \frac{0.00241 \times 0.007056}{0.916} = 1.86 \times 10^{-5} \, mol \, L^{-1}$.