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(D) According to Faraday's law of electrolysis,the mass deposited $m$ is given by $m = \frac{M 	imes I 	imes t}{n 	imes F}$,where $n$ is the valenc

Vedclass · Accepted Answer

$66.67$

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(D) According to Faraday's law of electrolysis,the mass deposited $m$ is given by $m = \frac{M 	imes I 	imes t}{n 	imes F}$,where $n$ is the valency factor (change in oxidation state).For $Mg^{2+} + 2e^- ightarrow Mg$,the valency factor $n = 2$.For $Al^{3+} + 3e^- ightarrow Al$,the valency factor $n = 3$.The charge required to deposit $1 \ g$ of $Mg$ is proportional to $\frac{n}{M} = \frac{2}{24} = \frac{1}{12} 	ext{ mol of electrons}$.The charge required to deposit $10 \ g$ of $Al$ is proportional to $10 	imes \frac{n}{M} = 10 	imes \frac{3}{27} = 10 	imes \frac{1}{9} = \frac{10}{9} 	ext{ mol of electrons}$.Since the cost is proportional to the number of moles of electrons required:Cost for $1 \ g$ of $Mg = 5.00$ rupees for $\frac{1}{12} 	ext{ mol}$.Cost per mole of electrons $= 5.00 	imes 12 = 60$ rupees.Cost for $10 \ g$ of $Al = \frac{10}{9} 	imes 60 = \frac{600}{9} = 66.67$ rupees.

The cost of energy required to deposit $1 \ g$ of $Mg$ is $5.00$ rupees. What is the cost to deposit $10 \ g$ of $Al$? (Atomic mass of $Al = 27, Mg = 24$)

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