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