A) 0.6 g
B) 0.09 g
C) 5.4 g
D) 10.8 g
Correct Answer: C
Solution :
Total charge flowing through electrolyte Q = it \[=50\times 20\times 60\] \[=6\times {{10}^{4}}\,C\] 105 C release = 9 g of Al \[\therefore 6\times {{10}^{4}}\,C\,\]would release \[=\frac{9\times 6\times {{10}^{4}}}{{{10}^{5}}}g\] \[=5.4\,g\,of\,Al\]You need to login to perform this action.
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