Conductivity of $$0.00241\\ M$$ acetic acid is $$7.896 \ imes 10^{-5} S cm^{-1}$$. Calculate its molar conductivity. If $$\\wedge^{0}_{m}$$ for acetic acid is $$390.5\\ S\\ cm^{2} mol^{-1}$$, what is its dissociation constant?
Molar conductivity $$\\displaystyle \\Lambda$$ and specific conductivity $$\\displaystyle \\kappa$$ are related by the  following expression:$$\\displaystyle \\Lambda = \\dfrac {1000 \\kappa}{C}$$Here, $$C$$ is the molar concentration.$$\\displaystyle \\Lambda = \\dfrac {1000 \ imes 7.896 \ imes 10^{-5}}{0.00241} = 32.76 \\: S cm^2 /mol$$The relationship between the degree of dissociation $$\\displaystyle \\alpha$$, the molar conductivity at concentration $$C$$, $$\\displaystyle \\Lambda^C$$  and the molar conductivity at infinite dilution $$\\displaystyle \\Lambda ^o$$ is as follows:$$\\displaystyle \\alpha = \\dfrac {\\Lambda ^C}{\\Lambda ^0} $$$$\\displaystyle \\alpha = \\dfrac {32.76}{390.5} = 8.39 \ imes 10^{-2}$$The dissociation constant $$\\displaystyle K = \\dfrac {C \\alpha^2}{1 - \\alpha}$$$$\\displaystyle K = \\dfrac {0.00241 \ imes (8.39 \ imes 10^{-2})^2}{(1-8.39 \ imes 10^{-2})} = 1.86 \ imes 10^{-5}$$