A) 10
B) 50
C) 1000
D) 100
Correct Answer: B
Solution :
Given 125 mL of \[\text{1}\,\text{M}\,\text{AgN}{{\text{O}}_{\text{3}}}\]solution. It means that \[\because \] \[\,\,1000\,\text{ML}\] of \[\text{AgN}{{\text{O}}_{\text{3}}}\]solution contains \[=108\,g\,Ag\] \[\therefore \] \[\text{125}\,\text{mL}\]of \[\text{AgNO}{{ & }_{\text{3}}}\]solution contains \[=\frac{108\times 125}{1000}g\,Ag=13.5\,Ag\] \[\because \] \[108\,g\]of Ag is deposited by 96500 C \[\therefore \] 13.5 g of Ag is deposited by \[=\frac{96500}{108}\times 13.5\] = 12062.5 C \[\because \] \[Q=it\] \[\therefore \] \[t=\frac{Q}{i}=\frac{12062.5}{241.25}=50\]
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