A) 27 g
B) 36 g
C) 45 g
D) 39 g
Correct Answer: C
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] Equivalent weight of Al \[=\frac{atomic\text{ }weight\text{ }of\text{ }Al}{3}=\frac{27}{3}=9\] Weight of Al = equivalent weight of Al \[\times \] number of Faradays \[=9\times 5=45\,g\]You need to login to perform this action.
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