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 ) :

0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to F

1.	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 ) :
Answer» `55.5 %` `45.0%` `50.0 %` `40.0 %` Solution :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%`