A) \[536111\,\,C\]
B) \[536.111\,\,C\]
C) \[96500\,\,C\]
D) \[38600\,\,C\]
Correct Answer: A
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] 1 mole of \[Al\] requires 3 moles of electrons or\[3\times 96500\,\,C\] 1 mole of\[Al=27\,\,g\] \[27\,\,g\]of\[Al\] require\[=3\times 96500\,\,C\] \[50\,\,g\] of \[Al\] require\[=\frac{3\times 96500\times 50}{27}\] \[=536111\,\,C\]
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