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Diagonal Relationship

The weight of...

Question

The weight of Ag deposited by passing 241.25 coulombs of current through AgNO $_{3}$ solution is :

0.27g

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2.7g

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27g

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0.54g

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Solution

The correct option is A 0.27g
The quantity of electricity passed = 241.25 coulombs.
Moles of electrons passed $= \frac{Q (C)}{96500 \frac{C}{m o l e^{-}}} = \frac{241.25 C}{96500 \frac{C}{m o l e^{-}}} = \frac{241.25}{96500} m o l e^{-}$
The mass of silver produced $= m o l a r m a s s o f A g \times m o l e r a t i o \times m o l e s o f e l e c t r o n s p a s s e d = 108 \frac{g}{m o l} \times 1 \frac{m o l A g}{m o l e^{-}} \times \frac{241.25}{96500} m o l e^{-} = 0.27 g .$

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Similar questions

When

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The weight of Ag deposited by passing 241.25 coulombs of current through AgNO3 solution is :

The correct option is A 0.27gThe quantity of electricity passed = 241.25 coulombs.Moles of electrons passed =Q(C)96500Cmole−=241.25C96500Cmole−=241.2596500mole−The mass of silver produced =molarmassofAg×moleratio×molesofelectronspassed=108gmol×1molAgmole−×241.2596500mole−=0.27g.

The weight of Ag deposited by passing 241.25 coulombs of current through AgNO $_{3}$ solution is :