A current of `3.6A` is a passed for 6 hrs between Pt electrodes in `0.5L` of `2M` solution of `Ni(NO_(3))_(2)`. What will be the molarity of solution at the end of electrolysis?

Answer

`1.172M`

Explanation

`Ni^(2+) +2e^(-) rarr Ni`
`"ne"^(-) = (3.7 xx 6.3600)/(96500)= 0.828`
`n_(Ni^(2+)) = 0.414`
mole of `Ni^(2+)` left in
Solution `= 1 - 0.414`
`=0.5859`
`M = (0.5899)/(500) xx 1000 = 1.172M`