logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...