6.40g
0.67g
0.64g
0.45g
Correct answer is B
Cu2++2e−→Cu(s)
2×96500C=193000Cdeposits64gofCu
M∝It
Q=It=0.45A×1hr15mins
= 0.45×75×60 [converting the time to seconds]
Quantity of electricity passed = 2025C
193000C→64gCu
1C→64193000
2025C→64193000×2025
=0.6715g≊
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