0.800 g of pure ferrous oxalate, FeC2O4 (Fe = 56, C = 12, O = 16), is dissolved in dilute H2SO4. Both Fe^2+ and the oxalate are oxidised by 0.0200 M KMnO4 in acid (to Fe^3+ and CO2, with MnO4^- -> Mn^2+). Calculate the volume of KMnO4 solution required, in mL, to the nearest integer.

JEE Advanced levelNumerical answerDifficulty 3 of 3df104

$0.800 g$ of pure ferrous oxalate, $FeC X_{2} O X_{4}$ ( $Fe = 56$ , $C = 12$ , $O = 16$ ), is dissolved in dilute $H X_{2} SO X_{4}$ . Both $Fe X^{2 +}$ and the oxalate are oxidised by $0.0200 M$ $KMnO X_{4}$ in acid (to $Fe X^{3 +}$ and $CO X_{2}$ , with $MnO X_{4} X^{-} \to Mn X^{2 +}$ ). Calculate the volume of $KMnO X_{4}$ solution required, in $mL$ , to the nearest integer.

Numerical answer type, answer in mL, tolerance 0.5.

Molar mass of

FeC X_{2} O X_{4} = 56 + 24 + 64 = 144

, so moles

= 0.800/144 = 0.00556

. Electrons lost per formula unit:

Fe X^{2 +} \to Fe X^{3 +}

gives 1 and

C X_{2} O X_{4} X^{2 -} \to 2 CO X_{2}

gives 2, a total of 3 (n-factor 3). Equivalents

= 3 \times 0.00556 = 0.01667

0.0200 M

KMnO X_{4}

0.100 N

(n-factor 5), so

V = 0.01667/0.100 = 166.67 mL

, i.e. 167 mL.