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```
Calculate the time required to deposit 56g of silver from a silver nitrate solution using a current of 4.5A.

```
9 years ago

```
mass Ag deposited = 56g
MM = 107.9 g mol-1 (from Periodic Table)
n (Ag) 56 ÷ 107.9 = 0.519 mol = n(e)
n(e) = 0.519 mol F = 96,500 C mol-1Q = 0.519 x 96,500 = 50,083.5 CQ = 50,083.5 C   I = 4.5 A

t = Q/It = 50,083.5 ÷ 4.5 = 11,129.67 seconds t = 11,129.67 ÷ 60 = 185.5 minutes t = 185.5 ÷ 60 = 3.1 hours
```
9 years ago
```							WIEGHT DEPOSITED=EQUIVALENT WIEGHT*CURRENT(i)*TIME TAKEN FOR THE DEPOSIT(t)/96500
AND THUS 45=108*4.5*t/96500
```
9 years ago
```							Mass of Ag= 56 gmMolar mass of Ag= 107.9 gm/mol           Number of moles of Ag= mass÷molar mass                                                     = 56÷107.9                                                     = 0.518999                         Ag+ + e-= Ag       To deposite 1 mole Ag, 1 mole e- is passed         So,  to deposite 0.518999 mole,    Number of e- passed=0.518999×1=0.518999 moles                •1 mole consist of 96500C            So, 0.518999 mole charge contain is,      Amount of charge passed= 96500×0.518999                                                   =50083.4035C          Time taken = charge÷current                               = 50083.4035÷4.5                     time = 11129.645 second                      time= 1129.645÷60=185.494 minutes                      time = 185.494÷60=3.092 hour
```
one year ago
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