A) 32166.3 C
B) 96500 C
C) 3216.33 C
D) 9650 C
E) \[8.685\times {{10}^{5}}C\]
Correct Answer: B
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] 1 mol of\[Al\]requires 3 mol of electron or \[3\times 96500C\] \[1\text{ }mol\,of\,Al=27g\] 27 g ofAl require\[=3\times 96500\text{ }C\] 9 g ofAl require\[=\frac{3\times 96500}{27}\times 9\] \[=96500C\]You need to login to perform this action.
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