$$0.1 \mathrm{M}$$ solution of $$\mathrm{AgNO}_3$$ is taken in a Conductivity cell and a potential difference of $$40 \mathrm{~V}$$ is applied across the ends of a column of this solution whose diameter is $$4.0 \mathrm{~cm}$$ and length of the column is $$12 \mathrm{~cm}$$. The current used is $$0.4 \mathrm{~A}$$. The Molar conductivity of the solution is _________.
Match the chemical reactions taking place at the Anode of the cell with the correct cell in which the reaction occurs.
| No. | Anodic reaction | No. | Cell |
|---|---|---|---|
| A | $$ \mathrm{Zn}(\mathrm{Hg})+2 \mathrm{OH}^{-} \rightarrow \mathrm{ZnO}_{(\mathrm{S})}+\mathrm{H}_2 \mathrm{O}+2 e $$ |
P | Fuel cell |
| B | $$ 2 \mathrm{H}_2(g)+4 \mathrm{OH}^{-} \rightarrow 4 \mathrm{H}_2 \mathrm{O}(l)+4 e $$ |
Q | Leclanche cell |
| C | $$ Z n_{(S)} \rightarrow Z n^{2+}+2 e $$ |
R | Nickel Cadmium storage cell |
| D | $$ \mathrm{Cd}_{(S)}+2 \mathrm{OH}^{-} \rightarrow \mathrm{CdO}_{(S)}+\mathrm{H}_2 \mathrm{O}(l)+2 e $$ |
S | Mercury cell |
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