A) \[5.49\times {{10}^{1}}C\]
B) \[5.49\times {{10}^{4}}C\]
C) \[1.83\times {{10}^{7}}C\]
D) \[5.49\times {{10}^{7}}C\]
Correct Answer: D
Solution :
[d] 1 mole of \[{{e}^{-}}=1F=96500\text{ }C\] 27g of Al is deposited by \[3\times 96500C\] 5120 g of Al will be deposited by \[=\frac{3\times 96500\times 5120}{27}=5.49\times {{10}^{7}}C\]You need to login to perform this action.
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