What time will be required for a current of 0.1 A to deposit all copper from 50 cm3 of 0.1 N solution of CuSO4 .
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● Answer- 26.8 hrs
● Explanation-
Weight of copper in 50 cm^3 of 0.1 N-
W = N × V/100 × M
W = 0.1 × 50/100 × 63
W = 3.15 g
Quantity of charge passed is
Q = W/(M/2) × 96500
Q = 3.15/(63/2) × 96500
Q = 9650 C
Q = I × t
t = Q/I
t = 9650/0.1
t = 96500 s
t = 26.8 hrs
Hope this is useful...
● Answer- 26.8 hrs
● Explanation-
Weight of copper in 50 cm^3 of 0.1 N-
W = N × V/100 × M
W = 0.1 × 50/100 × 63
W = 3.15 g
Quantity of charge passed is
Q = W/(M/2) × 96500
Q = 3.15/(63/2) × 96500
Q = 9650 C
Q = I × t
t = Q/I
t = 9650/0.1
t = 96500 s
t = 26.8 hrs
Hope this is useful...
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