Sol 1.
As Reaction :
MnO4- + 8H+ + 5Fe2+ <---> Mn2+ + 5Fe3+ + 4H2O
Moles of MnO4- ( KMnO4) = Conc. of KMnO4 × Volume of KMnO4 / 1000
= 0.120 × 44.82 / 1000 = 0.0053784 mol
From reaction , 1 mole of MnO4- combines with 5 moles of Fe2+
So , 0.0053784 moles of MnO4- combines with
= 0.0053784 × 5 = 0.026892 moles of Fe2+
As Molar Mass of Fe2+ = 55.845 g/mol
So , Mass of iron (Fe2+ ) = 0.026892 × 55.845 = 1.5017 g
As Mass of steel sample = 2.264 g
So , Concentration of iron in the steel sample
= Mass of iron / Mass of steel sample
= 1.5017 / 2.264
= 0.6633 g Fe / g steel
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