Q.
What will be the weight of silver deposited, if 96.5 A of current is passed into aqueous solution of AgNO3 for 100 s?
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a
1.08 g
b
10.8 g
c
108 g
d
1080 g
answer is B.
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Detailed Solution
Q=it =96.5 ×100=9650 C 96500 C charge deposited weight of Ag = 108 g9650 C deposited weight of Ag =108×965096500=10.8 g
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