Specific conductance of 0.1 M CH₃COOH at ${25}^{\circ }\mathrm{C}\text{ is }3.9\times {10}^{-4}{\mathrm{ohm}}^{-1}{\mathrm{cm}}^{-1}\text{. If }{\mathrm{\lambda }}^{\mathrm{\infty }}\left({\mathrm{H}}_3{\mathrm{O}}^{+}\right)$ $\text{ and }\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)\text{at }{25}^{\circ }\mathrm{C}\text{ are }349.0\text{ and }41.0$ ohm^–1 cm² mol^–1 respectively degree of ionization of CH₃COOH at the given concentration is

The specific conductance (κ) of 0.1 M acetc acid solution is 3.9  X 10 ⁻⁴ Ω ⁻¹ cm ⁻¹.

According to Kohlrausch law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the contributions of the molar conductivites of its ions.
Hence,$$\begin{gathered} {\wedge }_{C{H}_{3}COOH}^o= {\lambda }_{C{H}_{3}CO{O}^{-}}^o +\hspace{0.17em}\hspace{0.17em}{\lambda }_{H_3{O}^{+}}^o \\ = 41.0 +\hspace{0.17em}349.0 \\ =390 {\Omega }^{-1}c{m}^{2 }mo{l}^{-1} \end{gathered}$$

Degree of ionisation ($\alpha$) = $\frac{molar conductivity at given concentration }{Molar conductivity at infinite dilution }\frac{{\wedge }_m^c}{{\wedge }_{cm}^o}=\frac{3.9}{390}=0.01= 1\%$