# a current of 0.965 ampere is passed through 500ml of 0.2M solution of ZnSO4 for 10 min . the molarity of Zn+2 after deposition of zinc is1) 0.1M2) 0.5M3) 0.194M

The reaction taking place would be as follows:
​Zn2+ + 2e- $\to$Zn
So, 2 moles of electrons are used to reduce 1 mole of Zn2+. Or 2 x 96500 = 193000 C is used to reduce 1 mole of Zn2+
By given data we will calculate the amount of charge used.
Q = I x t
​   = 0.965 A x 10 min.
= 0.965 A x 10 x 60 sec.
= 579 C
Since 193000 C is used to reduce 1 mole of Zn2+
Therefore 579 C will reduce = 579/193000 = 0.003 mol.
Number of moles of Zn2+ in initial solution = C x V
= 0.2 M x 500ml/1000 ml L-1
= 0.1 mol
​Number of moles of Zn2+ left after the electrolysis = 0.1-0.003 = 0.097
Molarity of ​Zn2+ after deposition = moles of ​Zn2+ left/volume of solution per litre
= 0.097 $÷$ 500ml/1000 ml L-1
= 0.194 M

So, the correct option is 3.

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