An acidified solution contains 3.00x10^-3 mol FeSO4 and 1.20x10^-3 mol Fe2(C2O4)3. It is titrated with 0.0250 M KMnO4 (acidic medium). Only Fe^2+ and oxalate are oxidised; Fe^3+ is untouched. Volume of KMnO4 needed, in mL (one decimal)?

JEE Advanced levelNumerical answerDifficulty 3 of 3eq034

An acidified solution contains $3.00 \times 1 0^{- 3}$ mol $FeSO X_{4}$ and $1.20 \times 1 0^{- 3}$ mol $Fe X_{2} (C X_{2} O X_{4}) X_{3}$ . It is titrated with $0.0250 M$ $KMnO X_{4}$ (acidic medium). Only $Fe X^{2 +}$ and oxalate are oxidised; $Fe X^{3 +}$ is untouched. Volume of $KMnO X_{4}$ needed, in $mL$ (one decimal)?

Numerical answer type, answer in mL, tolerance 0.1.

FeSO X_{4}

gives

3.00 \times 1 0^{- 3}

eq (

n = 1

). Each

Fe X_{2} (C X_{2} O X_{4}) X_{3}

contains

3

oxalate ions at

2

eq each

= 6

eq, so

1.20 \times 1 0^{- 3} \times 6 = 7.20 \times 1 0^{- 3}

eq. Total

= 0.01020

eq. For

KMnO X_{4}

N = 0.0250 \times 5 = 0.125

, so volume

= 81.6 mL