Following limiting molar conductivities are given as given as $$ \begin{aligned} & \lambda_{\mathrm{m}\left(\mathrm{H}_2 \mathrm{SO}_4\right)}^0=\mathrm{x} \mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{m}\left(\mathrm{K}_2 \mathrm{SO}_4\right)}^0=\mathrm{y} \mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-1} \\ & \end{aligned} $$ $\lambda_{\mathrm{m}\left(\mathrm{CH}_2 \mathrm{COOK}\right)}^0=\mathrm{Z} \mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-1}$

• A

  $x-y+2 z$
• B

  $x+y-z$
• C

  $x-y+z$
• Correct answer
  D

  $\frac{(x-y)}{2}+z$

Solution

Key Idea According to Kohlrausch's law, $$ \lambda_m^0 \text { for } A_x B_y=x \lambda_{(A)^{0+}}^0+y \lambda_{(B)^{x-}} $$ where, $\lambda_m^{\circ}=$ limiting molar conductivity of electrolyte and $\lambda_{(A)^{y+}}^0$ and $\lambda_{(B)^{x-}}^0$ are the limiting molar conductivities of cation $\left(A^{y+}\right)$ and anion $\left(B^{x-}\right)$, respectively.

Hence, the correct answer is option: 'D'.