When $96.5$ coulomb of electricity is passed through a solution of $A g N O$ $_{3}$ .The amount of $A g$ deposited as :

108 m g

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5.4 m g

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16.2 m g

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21.2 m g

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Solution

The correct option is A $108 m g$
The quantity of electricity passed = 96.5 coulombs.
Moles of electrons passed $= \frac{Q (C)}{96500 \times \frac{C}{m o l e^{-}}}$
$= \frac{96.5 C}{96500 \frac{C}{m o l e^{-}}}$

$= \frac{96.5}{96500} m o l e^{-} .$
The mass of silver produced $= m o l a r m a s s o f A g \times m o l e r a t i o \times m o l e s o f e l e c t r o n s p a s s e d$

$= 108 \frac{g}{m o l} \times 1 \frac{m o l A g}{m o l e^{-}} \times \frac{96.5}{96500} m o l e^{-}$

$= 0.108 g = 108 m g .$

Option A is correct.

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[EAMCET 1992; KCET 2000]

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When 96.5 coulomb of electricity is passed through a solution of AgNO3.The amount of Ag deposited as :

When $96.5$ coulomb of electricity is passed through a solution of $A g N O$ $_{3}$ .The amount of $A g$ deposited as :