Chemistry > Physical Chemistry > Electrochemistry

Calculate $\Lambda°_m$ for acetic acid and its degree of dissociation ($\alpha$) if its molar conductivity is 48.1 $\Omega^{-1}$ cm$^2$ mol$^{-1}$.

Given: $\Lambda°_m$ (HCl) = 426 $\Omega^{-1}$ cm$^2$ mol$^{-1}$ $\Lambda°_m$ (NaCl) = 126 $\Omega^{-1}$ cm$^2$ mol$^{-1}$ $\Lambda°_m$ (CH$_3$COONa) = 91 $\Omega^{-1}$ cm$^2$ mol$^{-1}$