1.

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

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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