A current of 5.00 A is passed through $CuSO X_{4}$ solution for 3860 s with a current efficiency of 80%. The mass of copper deposited (in g, two decimals) is. (Cu = 63.5)

Numerical answer type, answer in g, tolerance 0.02.

Answer: 5.08 g

Useful charge

= 5.00 \times 3860 \times 0.80 = 15440 C = 0.160 F

Cu X^{2 +} + 2 e X^{-} Cu

: moles Cu

= 0.080

, mass

= 0.080 \times 63.5 = 5.08 g