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` mgB. `5.4` mgC. `16.2` mgD. `21.2` mg

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` mgB. `5.4` mgC. `16.2` mgD. `21.2` mg
Answer» Correct Answer - A `w = (E xxi xxt)/(96500) = (108 xx 9.65)/(96500) = 0.0108` g = 10.8 mg .

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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` mgB. `5.4` mgC. `16.2` mgD. `21.2` mg

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