The mass of copper deposited from a solution of ${\mathrm{CuSO}}_4$ by passage of 5 A current for 965 second is

(Mol. wt. of Copper = 63.5)

Current (I) = 5 A and time (t) = 965 sec

We know that equivalent weight of copper 

$=\frac{\text{Molecular weight}}{\text{Valency}}=\frac{63.5}{2}$

and quantity of electricity passed in coulomb $=\mathrm{current}\times \mathrm{time}=5\times \mathrm{965=4825 }\mathrm{C}$. Since 96500 coulombs will deposit $\frac{63.5}{2}\mathrm{g}$ of copper, therefore 4825 coulombs will deposit 

$=\frac{63.5\times 4825}{96500\times 2}=1.5875\mathrm{g}$