A) 0.9 gm
B) 0.3 gm
C) 0.27 gm
D) 2.7 gm
Correct Answer: A
Solution :
At cathode; \[A{{l}^{3+}}+3{{e}^{-}}\to Al\] \[{{E}_{Al}}=\frac{27}{3}=9\] \[{{W}_{Al}}={{E}_{Al}}\times \text{No}\text{.}\,\text{of}\,\text{faradays}\]\[=9\times 0.1=0.9\,gm\].
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