A sample of `MnSO_(4). 4H_(2)O` is strongly heated in air. The residue `( Mn_(3)O_(4))` left was dissolved in `100 mL` of `0.1NFeSO_94)` containing dil. `H_(2)SO_(4)`. This solution was completely reacted with `50 mL` of `KMnO_(4)` solution. `25 mL` of this `KMnO_(4)` solution was completely reduced by `30 mL` of `0.1N FeSO_(4)` solution. Calculate the amount of `MnSO_(4).4H_(2)O` in sample.

Explanation

`MnSO_(4). 4H_(2)O overset("Heat") to Mn_(3)O_(4)`
`Mn_(3)O_(4)` is dissolved in ferrous sulphate solution and is reduced from `Mn^((8//3)+) " to " Mn^(2+)` . The excess of `FeSO_(4)` is estimated by doing titration with `KMnO_(4)` solution. The normality of `KMnO_(4)` solution is determined by another ferrous sulphate solution.
For normality of `KMnO_(4)` solution :
`25xxN=30xx0.1`
`N=(30xx0.1)/(25)=(3)/(25)`
Let the volume of unreacted `FeSO_(4)` solution beV mL
V mL of `0.1 N FeSO_(4)=50 mL " of" (3)/(25)N KMnO_(4)`
or `" " V=(50xx3)/(0.1xx25)=60 mL`
`therefore` Volume of `FeSO_(4)` used for `Mn_(3)O_(4)=(100-60) mL`
=40 mL
`40 mL " of " 0.1 N FeSO_(4)-=40 mL " of " 0.1N Mn_(3)O_(4)`
`-=40 mL " of " 0.1 N MnSO_(4). 4H_(2)O`
Mass of `MnSO_(4). 4H_(2)O=(Exx0.1xx40)/(1000)=(E )/(250)g`x
Equivalent mass of
`MnSO_(4). 4H_(2)O=(M)/((8)/(3)-2)=(3M)/(2)=(3xx223)/(2)`
`therefore` Mass of `MnSO_(4). 4H_(2)O=(3xx223)/(2xx250)=1.338 g`