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In an electrolysis of a metallic chloride 3.283 g of the metal (molar mass $197gmol^{-1}$) was deposited on the cathode by the passage of 4825C of electric charge.The charge number of metal ion is

$\begin{array}{1 1}0.5\\1\\2\\3\end{array} $

1 Answer

Answer : 3
Using the expression $m=(Q/F)(m/|v_c|)$,we get
$|v_c|=\large\frac{Q}{F}\frac{M}{m}=\big(\large\frac{4825C}{96500Cmol^{-1}}\big)\big(\frac{197gmol^{-1}}{3.283g}\big)$$=3$
answered Jul 23, 2014 by sreemathi.v
 

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