Using atomic masses $Fe = 56$ and $O = 16$ , $23.2 g$ of $Fe X_{3} O X_{4}$ is dissolved in dilute $H X_{2} SO X_{4}$ (no oxidation takes place during dissolution). The volume of $0.100 M$ $KMnO X_{4}$ (acidic) needed to oxidise the iron(II) in this solution is

Answer: C

Fe X_{3} O X_{4}

FeO \cdot Fe X_{2} O X_{3}

, so

Fe X_{3} O X_{4} + 8 H X^{+} \to Fe X^{2 +} + 2 Fe X^{3 +} + 4 H X_{2} O

. Moles of

Fe X_{3} O X_{4} = 23.2/232 = 0.100

, giving

0.100 mol

Fe X^{2 +}

. Each

MnO X_{4} X^{-}

oxidises 5

Fe X^{2 +}

, so

0.0200 mol

is needed, i.e.

0.0200/0.100 = 0.200 L = 200 mL