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 :
Charge \[q=i\times t=\frac{25}{1000}\times 60=1.5C\] \[\underset{2\times 96500C}{\mathop{C{{a}^{2+}}}}\,+2{{e}^{-}}\xrightarrow{{}}\underset{6.02\times {{10}^{23}}}{\mathop{Ca}}\,\] \[\because \] At \[2\times 96500\text{ }C,\] the deposited \[Ca\] atoms \[=6.02\times {{10}^{23}}\] \[\therefore \] At\[1.5C\], the deposite \[Ca\] atoms \[=\frac{6.02\times {{10}^{23}}\times 1.5}{2\times 96500}\] \[=4.68\times {{10}^{18}}\]
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