In the Hall process,aluminium is produced by | vedclass

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(B) Total mass of $Al$ required = $24 	imes 5.0 \ g = 120 \ g$. Using Faraday's law: $w = \frac{M}{n 	imes F} 	imes I 	imes t 	imes 	ext{efficie

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$148.14$

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(B) Total mass of $Al$ required = $24 	imes 5.0 \ g = 120 \ g$. Using Faraday's law: $w = \frac{M}{n 	imes F} 	imes I 	imes t 	imes 	ext{efficiency}$. Here,$w = 120 \ g$,$M = 27 \ g/mol$,$n = 3$ (for $Al^{3+} + 3e^- ightarrow Al$),$F = 96500 \ C/mol$,$I = 9650 \ A$,and efficiency = $0.90$. $120 = \frac{27}{3 	imes 96500} 	imes 9650 	imes t 	imes 0.90$. $120 = \frac{27}{30} 	imes t 	imes 0.90$. $120 = 0.9 	imes 0.9 	imes t$. $120 = 0.81 	imes t$. $t = \frac{120}{0.81} \approx 148.14 \ s$.

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