When a current of 0.75 A is passed through a CuSO₄ solution for 25 min, .36 g of copper is deposited at the cathode. Calculate the atomic mass of copper.

Time  = 25 min  = 25 x 60 sec
Current     = 0.75 A
${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-} \to \mathrm{Cu}$
Electricity passed  = 25 x 60 x 0.75  = 1125 C of electricity deposit copper  = 0.369 g. 2 x 96500 C
of electricity will deposit copper.
M= ZxIxT

$= \frac{2 \times 96500 \times 0.369 }{1125}$
Atomic mass of copper = 63.3 u.