$\; \text{ (a) Given : }\; A = 0.625\ \mathrm{cm}^2, l = 50\ \mathrm{cm}$ $R = 5 \times 10^3\ \mathrm{ohm}, \rho = ?$ $m = 0.05\ \mathrm{m}, K = ?$ $\Lambda_m = ?$ $\; \text{ Cell constant }\; = \frac{\ell}{A} = \frac{50\ \mathrm{cm}}{0.625\ \mathrm{cm}^2} = 80\ \mathrm{cm}^{-1}$ $\; \text{ Resistivity }\; = \frac{R}{\text{ cell constant }} \; \text{ or }\; \frac{R \times A}{l}$ $\Rightarrow \frac{5 \times 10^3 \times 0.625}{50}$ $\Rightarrow \frac{5 \times 10^3\ \text{ ohm }}{80\ \mathrm{cm}^{-1}} \Rightarrow 62.5\ \text{ ohm }\ \mathrm{cm}$ $\Rightarrow 62.5\ \text{ ohm cm. }$ $\; \text{ Conductivity }\; = \frac{1}{\text{ Resistivity }} \times \frac{l}{A}$ $\frac{1}{5 \times 10^3} \times \frac{50}{0.625}$ $= \frac{50}{5 \times 10^3 \times 625 \times 10^{-3}}$$= \frac{10}{625} = 0.016\ \mathrm{S\,cm}^{-1}$ Molar conductivity $(\Lambda_m) = \frac{10^3 K}{M}$ $\Lambda_m = \frac{K}{M} \times 1000 = \frac{10 \times 1000}{625 \times 0.05} = 320\ \mathrm{S\,m^2\,mol}$ (b) Given: $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} +0.34\ \mathrm{V}$ $\; E^\circ_{\left(\frac{1}{2}\mathrm{Cl}_2/\mathrm{Cl}^-\right)} = +1.36\ \mathrm{V}$ $\; E^\circ_{\mathrm{H}^+/\mathrm{H}_2(\mathrm{g})'} \mathrm{Pt} = 0.00\ \mathrm{V}, \ E^\circ_{\left(\frac{1}{2}\mathrm{O}_2/\mathrm{H}_2\mathrm{O}\right)} = +1.23\ \mathrm{V}$ At cathode: $\; \mathrm{Cu}_{(\mathrm{aq})}^{2+} + 2e^- \rightarrow \mathrm{Cu}(\mathrm{s}) ;\ E^\circ = 0.34$ $\; \mathrm{H}_{(\mathrm{aq})}^{+} + e^- \rightarrow \frac{1}{2}\mathrm{H}_2(\mathrm{g}) ;\ E^\circ = 0.000\ \mathrm{V}$ The reaction with a higher value of $E^\circ$ takes place at the cathode, so deposition of copper will take place at the cathode. At anode: The oxidation reactions are possible at the anode. $\ \mathrm{Cl}_{(\mathrm{aq})}^{-} \rightarrow \frac{1}{2}\mathrm{Cl}_2(\mathrm{g}) + e^- ;\ E^\circ = 1.36\ \mathrm{V}$ $2\mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow \mathrm{O}_2(\mathrm{g}) + 4\mathrm{H}^{+}(\mathrm{aq}) + 4e^- ;$ $E^\circ = +1.23\ \mathrm{V}$ At the anode the reaction with a lower value of $E^\circ$ is preferred. But due to the over potential of oxygen, $\mathrm{Cl}^{-}$gets oxidised at anode to produce $\mathrm{Cl}_2$ gas. OR (a)$\; \mathrm{Zn}(\mathrm{s}) / \mathrm{Zn}^{2+}(0.1\ \mathrm{M}) \parallel$ $\ (0.01\ \mathrm{M})\ \mathrm{Ag}^{+} / \mathrm{Ag}(\mathrm{s})$$\; E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76\ \mathrm{V}$$\; E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} = +0.80\ \mathrm{V} \; \; \mathrm{emf}=?$$\; E_{\;\text{cell }} = E^\circ_{\;\text{cell }} - \frac{0.0591}{n} \log \frac{\;\text{ [Anode }\;]}{[\;\text{ Cathode }\;]}$$\; E^\circ_{\;\text{cell }} = E^\circ_{\;\text{cathode }} - E^\circ_{\;\text{anode }}$$\; = E^\circ_{\mathrm{Ag}/\mathrm{Ag}} - E^\circ_{\mathrm{Zn}/\mathrm{Zn}}$$\; = 0.80 - (-0.76) = 1.56\ \mathrm{V}$$\; E_{\;\text{cell }} = 1.56 - \frac{0.0591}{2} \log \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Ag}^{+}]^2}$ $\; = 1.56 - \frac{0.0591}{2} \log \frac{[0.1]}{[0.01]^2}$$\; = 1.56 - 0.0295 \log 1000$$\; = 1.56 - 3(0.0295)$$\; = 1.56 - 0.09 = 1.4715$ (b) $\Upsilon$ is a weak electrolyte as $n$ dilution complete dissociation of weak electrolyte takes place and thus a sharp increase in molar conductivity while in strong electrolyte it has already dissociated completely. So on dilution molar conductivity does not rises much.