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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 gm

B

5.4 gm

C

16.2 gm

D

21.2 gm

Text Solution

Verified by Experts

The correct Answer is:
A
`W_(Ag)=(E_(Ag)xxQ)/(96500)=(108xx9.65)/(96500)`
`=1.08xx10^(-2)gm=10.8mg`.
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