0.222 g of iron ore was brought into solution, `Fe^(3+)` is reduced to `Fe^(2+)` with `SnCl_(2)`. The reduced solution required 20 mL of `0.1 N KMnO_(4)` solution. The percentage of iron present in the ore is (equivalent weight of iron is 55.5 ) :

Options

A. `55.5 %`
B. `45.0%`
C. `50.0 %`
D. `40.0 %`

Answer

C

Explanation

Number of equivalent of iron
=Number of equivalents of `KMnO_(4)`
`(NV)/(1000)=(0.1xx20)/(1000)=0.002`
Mass of iron (pure)`=0.002xx55.5=0.111 g`
% purity `=("Mass of pure iron")/("Mass of iron ore")xx100=(0.111)/(0.222)xx100=50%`