Time required for a current of 10 A to deposit 0.635 g of Cu from `CuSO_4` solution is about (Given `1 F = 96500 C mol^(-1) , M(Cu) = 63.5 g mol^(-1)`)

Options

A. 181 s
B. 193 s
C. 220 s
D. 249 s

Answer

B

Explanation

`W = (M_w xx i xx t)/(96500 xx n)`
`0.635 = (63.5 xx 10 xx t)/(96500 xx 2)`
`122555 = 635 t`
`t = (122555)/(635) = 193 sec`