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A current of strength 2.5 A was passed through CuSO4 solution for 6 minute 265 seconds. The amount of copper deposited is (At. wt. of Cu = 63.5; 1 F = 96500 C)

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answer is 3.

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Detailed Solution

W=ZIt=E96500×I×t=t=6 minute + 265 sec=360+265=625 secI=2.5 AE=63.52W=63.52×96500×2.5×625=0.514 g

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