A) 2.8 g
B) 3.2 g
C) 4.6 g
D) 6.8 g
Correct Answer: C
Solution :
Key Idea According to Faraday second law of electrolysis \[W\propto E\] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\] Equivalent weight \[=\frac{atomic\,\,weight}{valency}\] \[{{W}_{Al}}=1.8\,g\] \[{{E}_{Na}}=23\] \[{{E}_{Al}}=\frac{27}{3}=9\] \[\because \] \[\frac{{{W}_{Na}}}{{{W}_{Al}}}=\frac{{{E}_{Na}}}{{{E}_{Al}}}\] \[\therefore \] \[\frac{{{W}_{Na}}}{1.8}=\frac{23}{9}\] \[{{W}_{Na}}=4.6\,\,g\]You need to login to perform this action.
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