`0.1 M KMnO_(4)` is used for the following titration. What volume of the solution in mL will be required to react with 0.158 g of `Na_(2)S_(2)O_(3)` ?
`underset(("not balanced"))(S_(2)O_(3)^(2-))+MnO_(4)^(-)+H_(2)O to MnO_(2)(s)+SO_(4)^(2-)+OH^(-)`

Explanation

Change in oxidation number of sulphur per molecule of `S_(2)O_(3)^(2-)=2xx(6-2)=8`
Change in oxidation number of Mn per molecule of `MnO_(4)^(-)=7-4=3`
No. of moles in 0.158 g of `Na_(2)S_(2)O_(3)=(0.158)/(158)=1xx10^(-3)`
No. of equivalents `=8xx10^(-3)`
Normality of 0.1 M `KMnO_(4)` solution `=0.1xx3=0.3`
Let V mL of volume of `KMnO_(4)` be required, then
`(V)/(1000)xx0.3=8xx10^(-3)`
or `V=(8)/(0.3)xx10^(-3)xx10^(3)=26.7 mL`