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The amount of silver deposited by passin...

The amount of silver deposited by passing `241. 25 C` of current through silver nitrated solution is .

A

2.7g

B

2.7mg

C

0.27g

D

0.54g

Text Solution

Verified by Experts

The correct Answer is:
C

Given, current=241.25 coloumb.
1 coulomb current will depsite`=1.118xx10^(-3)gm` Ag.
`therefore241.25` current will deposite`=1.118xx10^(-3)xx241.25`
`=0.27gm` silver.
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