A) \[1.6\times {{10}^{20}}\]
B) \[8\times {{10}^{19}}\]
C) \[3.1\times {{10}^{19}}\]
D) \[6.2\times {{10}^{19}}\]
Correct Answer: C
Solution :
Number of Cu ions liberated is \[n=\frac{I.\,t}{e\,\text{(valence}\,\text{electron)}}\] \[n=\frac{1\times 10}{1.6\times {{10}^{-19}}\times 2}\] \[n=3.125\times {{10}^{19}}\] \[n\approx 3.1\times {{10}^{19}}\]You need to login to perform this action.
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