During electrolysis, 25 g of a metal is deposited on the cathode of a metal nitrate solution by a current of 5 A passing for 4 hours. If the atomic weight of the metal is 100 amu, find the valency of the metal in the metal nitrate:
A
1
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B
2
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C
3
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D
4
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Solution
The correct option is C 3 Let the valency of the metal be n
Using Faraday's law of electrolysis, w=E.i.tF
Given,
w = 25 gm ; i = 5 A ; t = 4 hrs = 4×3600 seconds
E = atomicmassvalency=100n
put up the values,
25 =100n×5×4×360096500
solving for n we get n= 3