1.

Mass of copper deposited by the passage of 2 A of current for 965 seconds through a 2M solution of CuSO_(4) is (At.mass of Cu = 63.5)

Answer»

`0.325` g
0.635 g
1 g
1.2 g

Solution :W = ? , Z = 2 A , t = 965 SEC
`W = Z xx ixx t ` , But `Z = (E)/(96500)`
[E = EQ. MASS of `Cu = (63.5)/(2)` in `CuSO_(4)`, Cu is in + 2 state]
`= (63.5)/(2) xx (2 xx 965)/(96500) = 0.635` g


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