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