If 0.3605 g of a metal is deposited on the electrode by passing 1.2A current for 15 min through its salt, what will be its valence ? (Atomic weight of the metal is 96)

Answer

3

Explanation

We know that q= w/E and `q= I xxt`. So, `(w)/( E)= (I xx t)/(96500) rArr (0.3605)/(96//n) = (1.2 xx 15 xx 60)/(96500)`
`0.3605 xx n xx 96500= 1.2 xx 15 xx 60 x 96, n= (1.2 xx 15 xx 60 xx 96)/(0.3605 xx 96500)= 2.9803 ~=3`