What is the time ( in sec ) required for depositing all the silver present in `125 mL` of `1M AgNO_(3)` solution by passing a current of `241.25 A ? ( 1F=96500 C)`

Options

A. 10
B. 50
C. 100
D. 1000

Answer

B

Explanation

Given, 125 ml of 1 M `AgNO_(3)`
which is equal to 13.5 g of Ag )
`because ` 1000 ml of `AgNO_(3)` = 108 g of Ag
`:. ` 125 ml of `AgNO_(3) = (108 xx 125)/(1000) = 13.5 ` g
Also `because ` 108 g of Ag is deposited by = 96500 C
`:. ` 13.5 g of Ag is deposited by `= (96500 xx13.5)/(108)` = 12062.5 C
`because ` Q = it
`:.` t (time ) = Q /i =` (12062.50)/(241.25) ,t=50` sec