A) \[4.68\times {{10}^{18}}\]
B) \[4.68\times {{10}^{15}}\]
C) \[4.68\times {{10}^{12}}\]
D) \[4.68\times {{10}^{9}}\]
Correct Answer: A
Solution :
Quantity of electricity passed = \[\frac{25}{1000}\times 60=1.5\] \[2F=2\times 96500\,C\] deposit \[Ca=1mole\] \ 1.5 C will deposit \[Ca=\frac{1}{2\times 96500}\times 1.5\,mole\] \[=\frac{1}{2\times 96500}\times 1.5\times 6.023\times {{10}^{23}}\] atom \[=4.68\times {{10}^{18}}\].
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