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 mL of \[\text{AgN}{{\text{O}}_{\text{3}}}\]solution contains \[=108\,g\,Ag\] \[\therefore \] 125 mL of\[\text{AgN}{{\text{O}}_{\text{3}}}\] solution contains \[\text{=}\frac{108\times 125}{1000}g\,Ag\] \[=13.5\,g\,Ag\] \[\because \]\[108\,g\,\]of \[Ag\]deposited by \[=96500\,C\] \[\therefore \]13.5 g of Ag is deposited by \[=\frac{96500}{108}\times 13.5\] \[=12062.5\,C\] \[Q=it\] or \[t=\frac{Q}{i}=\frac{12062.5}{241.25}=50\]
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