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Q.

The amount of silver deposited by passing 241.25 C of current through silver nitrate solution is 

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a

0.27 g

b

0.54 g

c

2.7 g

d

2.7 mg

answer is D.

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Detailed Solution

                       Given, q = 241.25 C 

           96500 C current will deposit 108 g of Ag

∴      241.25 C  current will deposit = ?

                                               W = 0.27 g silver. 

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