(A) The volume of $Cu$ deposited is $V = \text{Area} \times \text{thickness} = (20 \times 5) \ cm^2 \times (3.6 \times 10^{-4} \ cm) = 0.036 \ cm^3$.

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(A) The volume of $Cu$ deposited is $V = \text{Area} \times \text{thickness} = (20 \times 5) \ cm^2 \times (3.6 \times 10^{-4} \ cm) = 0.036 \ cm^3$. The mass of $Cu$ deposited is $W = V \times d = 0.036 \ cm^3 \times 8.9 \ g/cm^3 = 0.3204 \ g$. Using Faraday's law,$W = \frac{M \times Q}{n \times F}$,where $M = 63.5 \ g/mol$,$n = 2$,and $F = 96500 \ C/mol$. $Q = \frac{W \times n \times F}{M} = \frac{0.3204 \times 2 \times 96500}{63.5} \approx 973.81 \ C \approx 974 \ C$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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Copper is to be electrodeposited on a nickel block of $(20 \times 5) \ cm^2$ area by using $CuSO_4$ as electrolyte. How much quantity of electricity is needed to deposit a $3.6 \ \mu m$ layer of copper (in $C$)? [Atomic weight of $Cu = 63.5 \ g \ mol^{-1}$,Density of $Cu = 8.9 \ g/cm^3$]

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A
$974$
B
$580$
C
$1080$
D
$365$

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Copper is to be electrodeposited on a nickel block of $(20 \times 5) \ cm^2$ area by using $CuSO_4$ as electrolyte. How much quantity of electricity is needed to deposit a $3.6 \ \mu m$ layer of copper (in $C$)? [Atomic weight of $Cu = 63.5 \ g \ mol^{-1}$,Density of $Cu = 8.9 \ g/cm^3$]

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One litre solution of $MgCl_2$ is electrolyzed completely by passing a current of $1 \ A$ for $16 \ min \ 5 \ sec$. The original concentration of $MgCl_2$ solution was (Atomic mass of $Mg = 24$)

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