The mass of copper deposited from a solution of `CuSO_4` by passage of `45 A` current for `965` second is
(Mol . Wt. pf Copper `= 63.5` ).

Options

A. ` 15. 875 g`
B. ` 1. 585 g`
C. ` 4825 g`
D. ` 96 500 g`

Answer

B

Explanation

Current `I = 5 A` and time ` T = 965 sec` We know that equivalent weight of copper
` = ("Molecular weight")/("Valancy") = (63.5)/2`
and quantitu of electricity passed in coulomb
= current xx time ` =5 xxx 965 = 4825 C`. Since `96500` coulombs will deposit `(63.5)/2 g` of copper , therefore ` 4825` coulombs will deposit
` = (63. 5 xx 4825)/(96 500 xx 2) = 1. 5875 g`.