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25 g of a sample of FeSO(4) was dissolve...

25 g of a sample of `FeSO_(4)` was dissolved in water containing `dil. H_(2)SO_(4)` and the volume made upto 1 litre. 25 mL of this solution required 20mL of `N/ 10 KMnO_(4)` for complete oxidation. Calculate % of `FeSO_(4).7H_(2)O` in given sample.

Text Solution

Verified by Experts

The correct Answer is:
0.8896
The redox changes are
`Fe^(2+) rarr Fe^(3+) + e`
`Mn^(7+) + 5e rarr Mn^(2+)`
`rArr` meq of `FeSO_(4).7H_(2)O` in 25 mL sample = meq of `KMnO_(4) = 20 xx (1/10)`
`:.`meq of `FeSO_(4).7H_(2)O` in 1 litre sample `= (20/10) xx (1000/25) = 80`
`(W/E) xx 1000 = 80`
Or `[w/(278/1)] xx 1000 = 80 ` or w 22.24 g
Molar mass of ` FeSO_(4).7H_(2)O ` = 278 `:. E = 278/1`
`:.`% purity of `FeSO_(4).7H_(2)O` sample `= 22.24 xx 100 /25 = 88.96 %`
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