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Two Faraday of electricity is passed through a solution of $CuSO_4$ . The mass of copper deposited at the cathode is : (at . mass of Cu=63.5 amu)

$\begin{array}{1 1} 2\;g \\ 127\;g \\ 0\;g \\ 63.5\;g \end{array} $

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Answer :
63.5 g
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