`KMnO_(4)` solution is to be standardised by titration against `As_(2)O_(3) (s)`. A 0.1097 g sample of `As_(2)O_(3)` requires 26.10 mL of the `KMnO_(4)` solution for its titration. What are the molarity and normality of the `KMnO_(4)` solution (Mol. Wt. of As=75)

Explanation

`Mn^(7+)+5e to Mn^(2+) ("Reduction")`
`As_(2)^(3+) to 2As^(5+) +4e ("reduction")`
Meq. Of `As_(2)O_(3)="Meq. Of KMnO_(4)`
`[0.1097//(198//4)] xx 1000=26.10 xx N`
`[E_(A) S_(2)O_(3))=M//4]`
`N_(KMnO_(4))=0.085, M_(KMnO_(4)=0.085//5=0.017` by mole concept method
`As_(2)overset(+3)(O_(3)) to 2 overset(+5)(As)`
n-factor =2|5-3|=+4
`[overset(+3)(As_(2)) to 2 overset(+5)(As)+4e^(-)] xx 5`
`[overset(+7)(Mn)+5e^(-) to overset(+2) (Mn)]xx 4`
`5 overset(+3)(As_(2))+4 overset(+7)(Mn) to 10 overset(+5)(As)+4 overset(+2)(Mn)`
5 mole 4 mole
5 mole of `As_(2)^(+3) "require to "4 mole of "KMnO_(4)`
thus, 1 mole require to `4/5" mole of KMnO"_(4)`
thus, `(0.1097)/(198) "mole require "to " mole "KMnO_(4)`
Thus, molarity of `KMnO_(4)= 4/5 xx (0.1097)/(198 xx (26.10//1000))=0.017M`
i,e, `N=M xx n-"factor"`
`N=0.017 xx 5=0.085N`