when current of 0.75 is passed through CuSO4 solution for 25 minutes ,0.36 9 g of copper is deposited at cathode. calculate the atomic mass of copper
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Using Faraday’s 1st Law of Electrolysis
W = ZIt
0.369 = Z*0.75*(25*60)
Z = eq mass/96500
eq mass = 31.65
Atomic mass= (eq mass)x(charge on ion)
= 31.65 x 2 = 63.3 amu
W = ZIt
0.369 = Z*0.75*(25*60)
Z = eq mass/96500
eq mass = 31.65
Atomic mass= (eq mass)x(charge on ion)
= 31.65 x 2 = 63.3 amu
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