$\mathrm{R}=5\times {10}^3\mathrm{ohm},\mathit{\rho }=?$
$\mathit{m}=0.05\mathrm{m},\mathrm{K}=?$
${\wedge }_{\mathit{m}}=?$


$\Rightarrow \frac{5\times {10}^3\times 0.625}{50}$



$\frac{1}{5\times {10}^3}\times \frac{50}{0.625}$ $=\frac{50}{5\times {10}^3\times 625\times {10}^{-3}}$
$=\frac{10}{625}=0.016{\mathrm{}\mathit{s}\mathit{c}\mathit{m}}^{-1}$
Molar conductivity $\left({\mathrm{\Lambda }}_{\mathit{m}}\right)=\frac{{10}^3\mathit{K}}{\mathit{M}}$
${\mathrm{\Lambda }}_{\mathit{m}}=\frac{\mathrm{K}}{\mathrm{M}}\times 1000=\frac{10\times 1000}{625\times 0.05}=320{\mathrm{}\mathit{s}\mathit{m}}^2\mathrm{mol}$
(b) Given: $\mathrm{E}{°}_{{\mathrm{Cu}}^{2+}∕\mathrm{Cu}}+0.34\mathrm{V}$
$\mathrm{E}{°}_{\left(1∕2{\mathrm{Cl}}_2∕{\mathrm{Cl}}^{-}\right)}=+1.36\mathrm{V}$
$\mathrm{E}{°}_{{\mathrm{H}}^{+}∕{\mathrm{H}}_2{\left(\mathrm{g}\right)}^{\prime }}\mathrm{Pt}=0.00\mathrm{V},\mathrm{E}{°}_{\left(1∕2{\mathrm{O}}_2∕{\mathrm{H}}_2\mathrm{O}\right)}=+1.23\mathrm{V}$
At cathode:
${\mathrm{Cu}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathit{e}}^{-}\to \mathrm{Cu}\left(\mathrm{}\mathit{s}\right);{\mathrm{E}}^{\circ }=0.34$
${\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}+{\mathit{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right);{\mathrm{E}}^{\circ }=0.000\mathrm{V}$
The reaction with a higher value of ${\mathrm{E}}^{\circ }$ 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.
$\mathrm{}{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}\to \frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathit{e}}^{-};{\mathrm{E}}^{\circ }=1.36\mathrm{V}$
$2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+4{\mathit{e}}^{-};$ ${\mathrm{E}}^{\circ }=+1.23\mathrm{V}$
At the anode the reaction with a lower value of ${\mathrm{E}}^{\circ }$ is preferred. But due to the over potential of oxygen, ${\mathrm{Cl}}^{-}$gets oxidised at anode to produce ${\mathrm{Cl}}_2$ gas.
OR
(a)
$\mathrm{Zn}\left(\mathrm{}\mathit{s}\right)∕{\mathrm{Zn}}^{2+}\left(0.1\mathrm{M}\right)\parallel$ $\left(0.01\mathrm{M}\right){\mathrm{Ag}}^{+}∕\mathrm{Ag}\left(\mathrm{}\mathit{s}\right)$
$\mathrm{E}{°}_{{\mathrm{Zn}}^{2+}∕\mathrm{Zn}}=-0.76\mathrm{V}$
$\mathrm{E}{°}_{{\mathrm{Ag}}^{+}∕\mathrm{Ag}}=+0.80\mathrm{V}\mathrm{emf}=?$


$=\mathrm{E}{°}_{\mathrm{Ag}∕\mathrm{Ag}}-\mathrm{E}{°}_{\mathrm{Zn}}∕\mathrm{Zn}$
$=0.80-\left(-0.76\right)=1.56\mathrm{V}$

$=1.56-\frac{0.0591}{2}\log \frac{\left[0.1\right]}{{\left[0.01\right]}^2}$
$=1.56-0.0295\log 1000$
$=1.56-3\left(0.0295\right)$
$=1.56-0.09=1.4715$
(b) $\mathit{\Upsilon }$ is a weak electrolyte as $\mathit{n}$ 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.