A sample of hydrazine sulphate `(N_(2)H_(6)SO_(4))` was dissolved in `100 mL` water. `10 mL` of this solution was reacted with excess of `FeCl_(3)` solution and warmed to complete the reaction. Ferrous ions formed were estimated and it required `20 mL` of `M//50 KMnO_(4)` solutions. Estimate the amount of hudrazine sulphate in one litre of solution.
Given `4Fe^(3+)+N_(2)H_(4)rarrN_(2)+4Fe^(2+)+4H^(+)`
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O`

Explanation

`20 mL (M)/(50) KMnO_(4)=20 mL (N)/(10) KMnO_(4)`
`["Equivalent mass of " KMnO_(4)=("Molecular mass")/(5)]`
`20 mL (N)/(10)KMnO_(4)-=20 mL(N)/(10)` Ferrous ion
`-=20 mL(N)/(10)FeCl_(3)`
`-=20 mL(N)/(10)N_(2)H_(6)SO_(4)`
Eq. mass `N_(2)H_(6)SO_(4)=("Mol. mass")/(4)=(130)/(4)=32.5`
[Since, change in O.N.`(N_(2)H_(4)to N_(2))` per molecule =4]
Amount of hydrazine sulphate in 10 mL of solution
`=(1)/(10)xx(32.5)/(1000)xx20=0.065 g`
Amount of hydrazine sulphate in one litre of solution
`(0.065)/(10)xx1000=6.50 g`