The mass of copper deposited from a solution of `CuSO_4` by passage of `5 A` current for `965` second is
(Mol . Wt. pf Copper `= 63.5` ).

Options

A. 15.875 g
B. 1.5875 g
C. 4825 g
D. 96500 g

Answer

B

Explanation

Current (I) = 5A and time (t) = 965 sec. We know that equivalent weight of copper
`= ("Molecular weight")/("Valancy")= (63.5)/(2)` and quantity of electricity passed in coulomb = current `xx` time `=5 xx 965`= 4825C. Since 96500 coulombs will deposit `(63.5)/(2)g` of copper therefore 4825 coulombs will deposit `=(63.5 xx 4825)/(96500 xx 2)= 1.5875g`