A) 125.1 seconds
B) 12.5 seconds
C) 155.2 seconds
D) 200 seconds
Correct Answer: A
Solution :
[a] \[Volume=Area\times thickness\] \[Mass=Volume\times density\] \[\therefore \] Mass of Ag to be deposited \[=\frac{80\times 0.005}{10}\times 10.5=0.42\text{ }g\] \[Amount\text{ }deposited=\frac{i\times t\times E.wt}{96500}\] \[\therefore 0.42=\frac{108\times 3\times t}{98500}\therefore t=125.1\,seconds\]You need to login to perform this action.
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