A) 27 gm
B) 36 gm
C) 45 gm
D) 39 gm
Correct Answer: C
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\to Al\] \[{{E}_{Al}}=\frac{27}{3}=9\] \[{{W}_{Al}}={{E}_{Al}}\times \text{No}\text{.}\,\text{of }\text{faradays}\]\[=9\times 5=45\,\,gm\].You need to login to perform this action.
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