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Q.
Two Faraday of electricity is passed through a solution of $CuSO_4$. The mass of copper deposited at the cathode is
(At. mass of $Cu\, = \,63.5\, u$)
Equivalent mass of copper $=\frac{\text { Atomic mass }}{ V \text { alency }}$
$=\frac{63.5}{2}=31.75$
Amount of copper deposited by $2$ faraday $=2 \times 31.75$
$=63.5 \,g$.