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When 9.65 coulomb of electricity is pass...

When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g `mol^(-1)`, the amount of silver deposited is :

A

`10.8` mg

B

`5.4` mg

C

`16.2` mg

D

`21.2` mg

Text Solution

Verified by Experts

The correct Answer is:
A
`w = (E xxi xxt)/(96500) = (108 xx 9.65)/(96500) = 0.0108` g = 10.8 mg .
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