A) \[0.5\,\,F\]
B) \[1.0\,\,F\]
C) \[1.5\,\,F\]
D) \[2.0\,\,F\]
Correct Answer: C
Solution :
Chemical equivalent of\[Al=\frac{27}{3}=9\] So, the amount of electricity required to deposit the\[9gm\]of\[Al\](\[1gm.\]equivalent) is\[1F\]. The given amount \[13\cdot 5gm\] is \[9\times 1\cdot 5\] \[i.e.,\] the given amount contains \[1.5\,\,gm\] equivalent of\[Al\], so it will require \[1\cdot 5F\] of electricity.You need to login to perform this action.
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