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)

KMnO_(4) solution is to be standardised by titration against As_(2)O_(

1.	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)
Answer» Solution :`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`