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A current of $1A $ is passing through $ CuSO_4 $ solution. How many $Cu^{+2}ions$ ions will be deposited at the cathode in 5 seconds?

$\begin {array} {1 1} (A)\;0.3 \times 10^{-19} & \quad (B)\;4 \times 10^{-19} \\ (C)\;15.625 \times 10^{-19} & \quad (D)\;1.5625 \times 10^{-19} \end {array}$

 

1 Answer

Numbers of ions are equal to the total charge passed through the
solution per unit charge on the ions
Ans : (D)
answered Feb 10, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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