3 " mol of "a mixture of `FeSO_(4)` and `Fe_(2)(SO_(4))_(3)` requried 100 " mL of " 2 M `KMnO_(4)` solution in acidic medium. Hence, mole fraction of `FeSO_(4)` in the mixture is

Answer

1

Explanation

Let mole of `FeSO_(4)=x`
Now, `KMnO_(4)` oxidises only `FeSO_(4)`
equivalent of `FeSO_(4)` = equivalent of `KMnO_(4)`
`x xx1 = (100)/(1000)xx2xx5 rArrx=1`