A) 1.1 hr
B) 46 hr
C) 53.6 hr
D) 24.00 hr
Correct Answer: C
Solution :
Moles of \[Zn=\frac{100}{65}\]= 1.53 mol 1 L of 1M \[CuS{{O}_{4}}\] solution contains 1 mol of \[CuS{{O}_{4}}\]. \[\therefore \] \[Zn+CuS{{O}_{4}}\xrightarrow{\,}\,ZnS{{O}_{4}}+Cu\]\[CuS{{O}_{4}}\] is the limiting reagent. \[\therefore \] To deposit completely 1 mole of Cu. Electricity required \[=2\times 96500\,C\]\[t=\frac{Q}{I}=\frac{2\times 96500}{1}s\] \[=\frac{2\times 96500}{3600}hr\]\[=53.6\,hr\]
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