Given below are 4 equations showing Molar conductivities at infinite dilution of various electrolytes. Which one of them represents the correct equation?

$$\begin{aligned} & \left(\Lambda_m^0\right)_{N a B r}-\left(\Lambda_m^0\right)_{N a C l}=\left(\Lambda_m^0\right)_{\mathrm{KCl}}-\left(\Lambda_m^0\right)_{\mathrm{KBr}} \\ & \left(\Lambda_m^0\right)_{\mathrm{HCl}}+\left(\Lambda_m^0\right)_{\mathrm{KOH}}-\left(\Lambda_m^0\right)_{\mathrm{KCl}}=\left(\Lambda_m^0\right)_{\mathrm{H}_2 \mathrm{O}} \\ & \left(\Lambda_m^0\right)_{\mathrm{KBr}}-\left(\Lambda_m^0\right)_{\mathrm{NaBr}}=\left(\Lambda_m^0\right)_{\mathrm{NaBr}}-\left(\Lambda_m^0\right)_{\mathrm{Nal}} \\ & \left(\Lambda_m^0\right)_{N H_4 \mathrm{Cl}}-\left(\Lambda_m^0\right)_{\mathrm{NH}_4 \mathrm{NO}_3}=\left(\Lambda_m^0\right)_{N H_4 \mathrm{Cl}}-\left(\Lambda_m^0\right)_{N H 4 B r} \end{aligned}$$

Assuming no change in volume, the time required to obtain solution of $$\mathrm{pH}=4$$ by electrolysis of $$100 \mathrm{~mL}$$ of $$0.1 \mathrm{~M} \mathrm{~NaOH}$$ (using current $$0.5 \mathrm{~A}$$ ) will be

For a cell reaction, $$A(s)+B^{2+}(a q) \longrightarrow A^{2+}(a q)+B(s)$$; the standard emf of the cell is $$0.295 \mathrm{~V}$$ at $$25^{\circ} \mathrm{C}$$. The equilibrium constant at $$25^{\circ} \mathrm{C}$$ will be

The Molar conductivity of $$0.05 \mathrm{M}$$ solution of $$\mathrm{MgCl}_2$$ is $$194.5 \mathrm{~ohm}^{-1} \mathrm{~cm}^2$$ per mole at room temperature. A Conductivity cell with electrodes having $$3.0 \mathrm{~cm}^2$$ surface area and $$1.0 \mathrm{~cm}$$ apart is filled with the solution of $$\mathrm{MgCl}_2$$. What would be the resistance offered by the conductivity cell?