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