`20 mL 0.2M MnSO_(4)` are completely oxidised by `16 mL` of `KMnO_(4)` of unknown normality each forming `Mn^(4+)` oxidation state. The normality and molarity of `KMnO_(4)` are respectively:

Explanation

`Mn^(2+) to Mn^(4+) +2e ("oxidation")`
`3e^(-)+Mn^(7+) to Mn^(4+) ("reduction")`
Meq of `KMnO_(4)="Meq. of "MnSO_(4)`
`N xx 16 =20 xx 0.2 xx 2, therefore N=0.5`
M=0.5/3=0.167
by mole concept method
`overset(+2)(Mn) to overset(+4)(Mn)+2e^(-) xx3`
`overset(+7)(Mn)+3e^(-) to overset(+4)(Mn xx 4)`
`3 overset(+2)(Mn)+2 overset(+7)(Mn) to overset(+4)(5Mn)`
3 mole 2 mole
No. of moles of `MnSO_(4)` is 0.2 M, 20 ml. solution =`(0.2 xx 20)/(1000)" moles"`
Now, 3 mole `MnSO_(4)` required to 2 mole `. MnO_(4)^(-)`
i.e, 1 mole would required `to 2/3 " mole of "MnO_(4)^(-)" i.e. "(0.2 xx 20)/(1000) "would required"to 2/3 xx (0.2 xx 20)/(1000)" mole of "MnO_(4)^(-)`
Now, molarity of `KMnO_(4)`
`=2/3 xx (0.2 xx 20)/(1000 xx (16//1000)=0.167M`
i.e. `N=M xx n-"factor"`
`N=0.167 xx 3=N=0.5N`