The density of Cu is 8.94 , g , cm^{-3}. The | vedclass

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(B) First,calculate the volume of the copper deposit: $V = 	ext{Area} 	imes 	ext{thickness} = (10 \, cm 	imes 10 \, cm) 	imes 10^{-2} \, cm = 1 \

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$27172$

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(B) First,calculate the volume of the copper deposit: $V = 	ext{Area} 	imes 	ext{thickness} = (10 \, cm 	imes 10 \, cm) 	imes 10^{-2} \, cm = 1 \, cm^3$. Next,calculate the mass $(W)$ of $Cu$ deposited: $W = 	ext{density} 	imes V = 8.94 \, g \, cm^{-3} 	imes 1 \, cm^3 = 8.94 \, g$. Using Faraday's law of electrolysis: $W = \frac{M}{n 	imes F} 	imes Q$,where $M = 63.5 \, g \, mol^{-1}$,$n = 2$ (for $Cu^{2+} + 2e^- ightarrow Cu$),and $F = 96500 \, C \, mol^{-1}$. Substituting the values: $8.94 = \frac{63.5}{2 	imes 96500} 	imes Q$. Solving for $Q$: $Q = \frac{8.94 	imes 2 	imes 96500}{63.5} \approx 27172 \, C$.

The density of $Cu$ is $8.94 \, g \, cm^{-3}$. The quantity of electricity needed to plate an area $10 \, cm \times 10 \, cm$ to a thickness of $10^{-2} \, cm$ using $CuSO_4$ solution would be,if atomic mass of $Cu$ is $63.5$ .................. $C$

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