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Following limiting molar conductivities are given as
$\lambda^0_{m(H_2SO_4)}=x$ S cm$^2$ mol$^{-1}$, $\lambda^0_{m(K_2SO_4)}=y$ S cm$^2$ mol$^{-1}$, $\lambda^0_{m(CH_3COOK)}=z$ S cm$^2$ mol$^{-1}$
$\lambda^0_m$ (in S cm$^2$ mol$^{-1}$) for $CH_3COOH$ will be:
Detailed Solution
By Kohlrausch's law, $\lambda^0_{CH_3COOH}=\lambda^0_{CH_3COO^-}+\lambda^0_{H^+}=\lambda^0_{CH_3COOK}+\frac{1}{2}\lambda^0_{H_2SO_4}-\frac{1}{2}\lambda^0_{K_2SO_4}=z+\frac{x}{2}-\frac{y}{2}=\frac{(x-y)}{2}+z$.