${Al}_{2} O_{3}$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^{4}$ amperes of current is passed through molten ${Al}_{2} O_{3}$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. Atomic mass of Al = 27 g ${mol}^{- 1}$ )

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$2.4 \times 10^{5} g$

$9.0 \times 10^{3} g$

$1.3 \times 10^{4} g$

$8.05 \times 10^{4} g$

answer is C.

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Detailed Solution

Eq. of Al formed $= \frac{m}{F} = \frac{it}{96500}$

$\begin{array}{l} ∴ m_{Al} = \frac{4 \times 10^{4} \times 6 \times 60 \times 60 \times 27}{96500 \times 3} (E = \frac{27}{3}) \\ ∴ m_{Al} = 8 .05 \times 10^{4} g \end{array}$

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