What weight of `KmNO_4` is required to react completely with `500mL` of 0.4 M Cus in acidic medium?

Explanation

`Cu^(2+)` is not oxidised but `S^(2-)` ion is oxidised to `SO_2`. (n-factor`=6`) n-factor for `KMnO_4=5`
(Mw of `KMnO_4=158gmol^(-1)`
`CuS-=KMnO_4`
`1mEq-=1mEq`
`N_1V_1(mL)-=("weight")/("equivalent weight")xx10^(3)` mEq
`0.4xx6` (n-factor) `xx500=("Weight"xx10^(3))/((158)/(5))`
`therefore` Weight of `KMnO_4=(0.4xx6xx500xx158)/(5xx1000)=37.92g`