`25 grams` of a sample of ferrous sulphate is dissolved in dilute sulphuric acd. By adding water, its volume is made up to 1 litre.` 25 mL` of this solution requries `20 mL` of `N//10 KMnO_(4)` solution for oxidation. Calculate the percentage of `FeSO_(4) . 7H_(2)O` in the sample.
Strategy: Percentage of `FeSO_(4).7H_(2)O`
`=("mass"_("ferrous sulphate"))/("mass of sample")xx100%`
To get the mass of ferrous suphate, we need to calculate equivalents of ferrous sulphate. By law of equivalrnce, the milli equivalents of `KMnO_(4)` must be equal to the miliequivalents of ferrous sulphate. We can find the milliequivalents of `KMnO_(4)` by taking the product of millilitres of solution and its normality.

Explanation

Millieqivalents of `KMnO_(4)= V_(mL)xxN`
=`(20)(1//10)`
`=2`
Thus, meq. Of `FeSO_(4).7H_(2)O` in `25 mL` of solution =2. Consequently, meq. Of `FeSO_(4) 7H_(2)O` present in 1 litre of solution will be
`=2/25xx1000=80`
Equivalents (eq.) of `FeSO_(4).7H_(2)O=(meq.)/(1000)=80/1000`
Mass of `FeSO_(4).7H_(2)O=(Eq.)("gram equivalent mass")`
`=(80/1000 eq.)(278 g eq^(-1))`
`=22.24 g`
Note that equivalent mass of `FeSO_(4). 7H_(2)O` is equal to its formula mass because change in oxidation number of Fe is by just one unit: (from +2 to +3).
`MnO_(4)^(-)(aq.)+8H^(+)(aq.)+5Fe^(2+)(aq.) rarr 5Fe^(3+)(aq.)+Mn^(2+)(aq.)+4H_(2)O(l)`
Thus, the percentage of `FeSO_(4). 7H_(2)O` in the given sample
`=(22.24 g)/(25 g)xx100%`
`=88.96 %`