A) 10
B) 50
C) 1000
D) 100
Correct Answer: B
Solution :
Given, 125 mL of \[\text{1 M AgN}{{\text{O}}_{\text{3}}}\] solution. It means that \[\because \] 1000 mL of \[\text{AgN}{{\text{O}}_{\text{3}}}\]solution contains \[\text{= 108}\,\text{g}\,\text{Ag}\] \[\therefore \] 125 mL of \[\text{AgN}{{\text{O}}_{\text{3}}}\]solution contains \[=\frac{108\times 125}{1000}\text{g}\,\text{Ag}\] 108 g of Ag is deposited by = 96500 C \[\therefore \] 13.5 g of Ag is deposited by \[=\frac{96500}{108}\times 13.5\] \[=12062.5\,\text{C}\] \[Q=it\] or \[t=\frac{Q}{i}=\frac{12062.5}{241.25}=50\]
You need to login to perform this action.
You will be redirected in
3 sec
