During the electrolysis of an aqueoius nitric acid solution using pt electrodes

Answer

`0.154M`

Explanation

`I=(1.70xx90)/(100)A`
`:.` Equivalent of `Zn^(2+)` lost `=(It)/(96500)`
`=(1.70xx90xx230)/(100xx96500)`
`=3.646xx10^(-13)`
Milliequivalent of `Zn^(2+)` lost `=3.646xx10^(-3)`
Initially milliequivalent of `Zn^(2+)=300xx0.160xx2=96`
`( :. Mxx2=N fo r Zn^(2+)impliesmEq=NxxV_((mL)))`
`:.` Milliequivalent of `Zn^(2+)` left in solution `=96-3.646`
`=92.354`
`:. [ZnSO_(4)]=(92.254)/(2xx300)=0.154M`