(a) (i) When the concentration of the electrolyte approaches zero, the molar conductivity is termed as limiting molar conductivity. It is represented by ${\mathrm{\Lambda }}_{\mathit{m}}$.
(ii) Fuel cell: Fuel cells are the galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
Given,
For $0.1\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{KCl}$ solution
Resistance $\left(\mathrm{R}\right)=100\mathit{\Omega }$
Conductivity $\left(\mathit{k}\right)=1.29\times {10}^{-2}{\mathit{\Omega }}^{-1}{\mathrm{cm}}^{-1}$
Cell constant $\left(\mathrm{G}*\right)=\mathrm{k}\times \mathrm{R}$
$=1.29\times {10}^{-2}{\mathit{\Omega }}^{-1}{\mathrm{cm}}^{-1}\times 100\mathit{\Omega }$
$=1.29{\mathrm{cm}}^{-1}$
For $0.02\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{KCl}$ solution
Resistance $\left(\mathit{R}\right)=520\mathit{\Omega }$

$\mathrm{K}=\frac{1.29{\mathrm{cm}}^{-1}}{520\mathit{\Omega }}$
$\mathrm{K}=2.48\times {10}^{-3}{\mathit{\Omega }}^{-1}{\mathrm{cm}}^{-1}$
$\left(\mathrm{C}\right)=0.02\mathrm{mol}{\mathrm{L}}^{-1}$
$=0.02\times {10}^{-3}\mathrm{mol}{\mathrm{cm}}^{-3}$
Concentration $\left(\mathrm{C}\right)=0.02\mathrm{mol}{\mathrm{L}}^{-1}$
$=0.02\times {10}^{-3}\mathrm{}\mathrm{mol}\mathrm{}{\mathrm{cm}}^{-3}$
Molar conductivity $\left({\mathit{\Lambda }}_{\mathit{m}}\right)$
$=\frac{2.48\times {10}^{-3}{\mathit{\Omega }}^{-1}{\mathrm{cm}}^{-1}}{0.02\times {10}^{-3}\mathrm{mol}{\mathrm{cm}}^{-3}}$
${\mathrm{\Lambda }}_{\mathit{m}}=124{\mathit{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$
OR
(a) Faraday's first law of electrolysis states that "The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed."
$\mathit{m}=\mathrm{Z}\times \mathrm{Q}$
where, $\mathit{m}=$ mass of a substance deposited or liberated at an electrode.
$\mathrm{Q}=$ amount of charge passed through it and
$\mathrm{Z}=$ electrochemical equivalent
The reduction of one $\mathrm{mol}$ of ${\mathrm{Cu}}^{2+}$ to $\mathrm{Cu}$ can be represented as:
${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$
Since, $2\mathrm{mol}$ of electrons are involved in the reduction, so the amount of charge required is $2\mathrm{F}$.
(b) The given cell reaction can be represented as
$\mathrm{Mg}\left(\mathrm{}\mathit{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right)$ $+\mathrm{Cu}\left(\mathrm{}\mathit{s}\right)$

$\log \frac{0.1}{0.01}$