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A current 0.5 ampere when passed through `AgNO_(3)` solution for 193 sec. Deposited 0.108 g of Ag. Find the equivalent weight of Ag.A. 108B. 54C. 10.8D. 5.4 |
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Answer» Correct Answer - A `W = ( I xx t(s) xx E)/(F)` `therefore E = (W xx F)/(I xx t) = (0.108 xx 96500)/(0.5 xx 193) = 108` |
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