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A current of strength 2.5 A was passed ...

A current of strength ` 2.5 A` was passed through ` CuSO_4` solution for 6 minute `265` seconds. The amount of copper deposited is (At. Of ` Cu = 63. 5, 1 F = 96500 C`) .

A

0,3175 g

B

0.0031 g

C

6.35 g

D

3.175 g

Text Solution

Verified by Experts

The correct Answer is:
A
`W = (I t E)/(96500) = (2.6 xx380 xx 31.75)/(96500) ~~ 0.32 g`
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