When a current of $X$ Faraday is passed through a solution of aluminium nitrate,$1 \ mol$ of $Al$ is deposited. Then,$X =$ . . . . . . ?

Aluminium oxide may be electrolysed at $1000 \, ^\circ C$ to furnish aluminium metal (At. Mass $= 27 \, amu$; $1 \, F = 96,500 \, C$). The cathode reaction is $Al^{3+} + 3e^- \to Al^0$. To prepare $5.12 \, kg$ of aluminium metal by this method would require:

In a copper voltameter experiment,current is decreased to one-fourth of the initial value but it is passed for four times the earlier duration. The amount of copper deposited will be:

For the oxidation of $0.2 \text{ M} \text{ FeSO}_4$ solution,$0.965 \text{ A}$ current is passed through it for $1 \text{ hour}$. The volume of the solution that is oxidised in $\text{mL}$ is:

Potassium chlorate is prepared by the electrolysis of $KCl$ in basic solution:
$6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$
If only $60\%$ of the current is utilized in the reaction,the time (rounded to the nearest hour) required to produce $10 \ g$ of $KClO_{3}$ using a current of $2 \ A$ is:
(Given: $F = 96,500 \ C \ mol^{-1}$,molar mass of $KClO_{3} = 122 \ g \ mol^{-1}$)

The electrochemical equivalent of a metal is $3.3 \times 10^{-7} \ kg/C$. The mass of the metal liberated at the cathode when a $3 \ A$ current is passed for $2 \ s$ will be: