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 \[n=\frac{1}{e(valence\,\,electron)}\] \[n=\frac{1\times 10}{1.6\times {{10}^{-19}}\times 2}\] \[n=3.125\times {{10}^{19}}\] \[\therefore \] \[n\approx 3.1\times {{10}^{19}}\]You need to login to perform this action.
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