A) \[9.65\times {{10}^{4}}C\]
B) \[4.34\times {{10}^{5}}C\]
C) \[9.65\times {{10}^{3}}C\]
D) \[1.93\times {{10}^{4}}C\]
Correct Answer: C
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\to Al\] Atomic mass of \[Al=27\] \[\therefore \] 27 g of Al is deposited by \[=3\times 96500\,C\] \[\therefore \] 0.9 g of \[Al\] is deposited by \[=\frac{3\times 96500\times 0.9}{27}\] = 9650 C \[=9.65\times {{10}^{3}}C\]
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