Three Faraday's of electricity are passed through molten ${\mathrm{Al}}_2{\mathrm{O}}_3$, aq. solution of ${\mathrm{CuSO}}_4$ and molten NaCl taken in three different electrolytic cells. The amounts of Al, Cu and Na deposited at the electrodes will be

$$\begin{gathered} \mathrm{Al}, {\mathrm{Al}}_2{\mathrm{O}}_3 \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 3+. \\ {\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al} \\ \mathrm{Therefore}, 3 \mathrm{Faraday} \mathrm{are} \mathrm{needed} \mathrm{to} \mathrm{produce} 1 \mathrm{mole}. \\ \mathrm{This} \mathrm{means} \mathrm{that} 3\mathrm{F} \mathrm{will} \mathrm{produce} 1 \mathrm{mole} \mathrm{of} {\mathrm{Al}}_2{\mathrm{O}}_3. \\ \mathrm{Cu}, {\mathrm{CuSO}}_4 \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 2+. \\ {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu} \\ \mathrm{Therefore}, 2 \mathrm{Faraday} \mathrm{are} \mathrm{needed} \mathrm{to} \mathrm{power} 1 \mathrm{mole}. \\ \mathrm{Therefore}, 3\mathrm{F} \mathrm{will} \mathrm{result} \mathrm{in} 1.5 \mathrm{moles} \mathrm{of} {\mathrm{CuSO}}_4. \\ \mathrm{Na}, \mathrm{NaCl} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 1+ \\ {\mathrm{Na}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Na} \\ \mathrm{As} \mathrm{a} \mathrm{result}, 1 \mathrm{Faraday} \mathrm{is} \mathrm{needed} \mathrm{to} \mathrm{produce} 1 \mathrm{mole}, \mathrm{and} 3 \\ \mathrm{Faradays} \mathrm{will} \mathrm{produce} 3 \mathrm{moles} \mathrm{of} \mathrm{NaCl}, \mathrm{giving} \mathrm{us} \mathrm{the} \\ \mathrm{ratio} \mathrm{of} 1 \mathrm{mole} \mathrm{to} 1.5 \mathrm{moles} \mathrm{to} 3 \mathrm{moles}. \end{gathered}$$

Therefore, the correct option is B