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 :
Given, Current (i) = 25 mA = 0.025 A Time (t) = 60 sec Q = i t\[=60\times 0.025=1.5\]coulombs No. of electrons \[=\frac{1.5\times 6.023\times {{10}^{23}}}{96500}\] \[{{e}^{-}}=9.36\times {{10}^{18}}\] \[Ca\to C{{a}^{2+}}+2{{e}^{-}}\] \[2{{e}^{-}}\] are required to deposite one Ca atom \[9.36\times {{10}^{18}}\ {{e}^{-}}\] will be used to deposite \[=\frac{9.36\times {{10}^{18}}}{2}\] \[=4.68\times {{10}^{18}}\].
You need to login to perform this action.
You will be redirected in
3 sec
