Grams of a sample of ferrus sulphate was dissolved in dilute sulphuric acid and water and its volume was made up to 1 litre . 25 mL of this solution required 20 mL of `N/10 KmnO_(4)` solution for complete oxidation . Calculate the percentage of `FeSO_(4). 7H_(2)O` in the sample .

Explanation

m.e of `KMnO_(4)" solution " = 1/10 xx 20 = 2 `
` :. ` m.e of 25 mL of `FeSO_(4) . 7H_(2)O " solution " = 2 " " …. (Eqn . 1)`
` :. ` m.e of 100m L of `FeSO_(4). 7H_(2)O" solution " = 2/25 xx 1000 = 80`
Equivalent of `FeSO_(4) . 7H_(2)O = 80/1000 = 80 `
Equivalent of `FeSO_(4) . 7H_(2)O` solution = `2/25 xx 1000 = 80`
Equivalent of `FeSO_(4) . 7H_(2)O = 80/1000 " "...(Eqn . 3)`
` :. ` weight of `FeSO_(4) .7H_(2)O` = equivalent `xx` eq.wt
` = 80/100 xx278 = 22.4 g `
`{ "As " Fe^(2+) to Fe^(3) ," eq . wt of " FeSO_(4) . 7H_(2)O = (mol . w.t )/(" change in ON ") = 278 / 1 } `
thus the percentage of `FeSO_(4). 7H_(2)O ` in the sample ` = (22.24)/25 xx 100`
` = 88.96 %`