A) 63.60 g
B) 31.80 g
C) 15.90 g
D) 7.95 g
E) 4.00 g
Correct Answer: D
Solution :
According to the second law of Faradays electrolysis. \[\frac{{{m}_{1}}}{{{E}_{1}}}=\frac{{{m}_{2}}}{{{E}_{2}}}\] ?.. (i) where,\[{{m}_{1}}\]and\[{{m}_{2}}=\]masses of elements \[{{E}_{1}}\]and\[{{E}_{2}}=\]equivalent masses of elements \[E=\frac{atomic\text{ }mass}{valency}\] \[{{E}_{(Ag)}}=\frac{108}{1}=108\] \[{{E}_{(Cu)}}=\frac{63.6}{2}=31.8\] Putting the values in Eq. (i) \[\frac{27.0}{108}=\frac{{{m}_{2}}}{31.8}\] \[{{m}_{2}}=\frac{31.8\times 27.0}{108}=7.95g\]
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