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 gmB. 5.4 gmC. 16.2 gmD. 21.2 gm

When 9.65 coulomb of electricity is passed through a solution of silve

1.	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 gmB. 5.4 gmC. 16.2 gmD. 21.2 gm
Answer» Correct Answer - A `W_(Ag)=(E_(Ag)xxQ)/(96500)=(108xx9.65)/(96500)` `=1.08xx10^(-2)gm=10.8mg`.

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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 gmB. 5.4 gmC. 16.2 gmD. 21.2 gm

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