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Aluminium oxide may be electrolysed at $1000 ° C$ to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is ${Al}^{3 +} + 3 e^{-} ⟶ {Al}^{O}$
To prepare 5.12 kg of aluminium metal by this method would require

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$5.49 \times 10^{7} C of electricity$

$1.83 \times 10^{7} C of electricity$

$5.49 \times 10^{4} C of electricity$

$5.49 \times 10^{1} C of electricity$

answer is A.

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Detailed Solution

$Q = \frac{m F Z}{M} = \frac{5.12 \times 10^{5} \times 96500 \times 3}{27}$

$= 5.49 \times 10^{7} C$

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