(C) According to Faraday's law of electrolysis,the mass of substance deposited is given by $m = \frac{Q \times M}{n \times F}$.Here,$Q = 19296 \ C$,$M

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(C) According to Faraday's law of electrolysis,the mass of substance deposited is given by $m = \frac{Q \times M}{n \times F}$.Here,$Q = 19296 \ C$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$ (for $Cu^{2+} + 2e^- \rightarrow Cu$),and $F = 96500 \ C \ mol^{-1}$.Substituting the values: $m = \frac{19296 \times 63.5}{2 \times 96500}$.$m = \frac{1225296}{193000} = 6.35 \ g$.Therefore,the mass of copper deposited is $6.35 \ g$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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$A$ current of $19296 \ C$ is passed through an aqueous solution of copper sulphate using copper electrodes. What is the mass (in $g$) of copper deposited at the cathode? (Molar mass of $Cu = 63.5 \ g \ mol^{-1}$)

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A
$3.17$
B
$1.58$
C
$6.35$
D
$0.79$

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$A$ current of $19296 \ C$ is passed through an aqueous solution of copper sulphate using copper electrodes. What is the mass (in $g$) of copper deposited at the cathode? (Molar mass of $Cu = 63.5 \ g \ mol^{-1}$)

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