JEE Advanced 2021 Paper 2 · Q11 · Mole Concept & Stoichiometry

Correct answer:1.875

Solution

Equivalents of Fe$^{2+}$ = Equivalents of KMnO$_4$ (MnO$_4^-$/Mn$^{2+}$, n=5).

Eq of KMnO$_4$ used = $0.03 \text{ M} \times 5 \times 12.5 \times 10^{-3}$ L = $1.875 \times 10^{-3}$ eq.

Moles of Fe$^{2+}$ in 25.0 mL aliquot = $1.875 \times 10^{-3}$ mol (n = 1 for Fe$^{2+}$ $\rightarrow$ Fe$^{3+}$).

Moles of Fe$^{2+}$ in 250 mL = $10 \times 1.875 \times 10^{-3} = 1.875 \times 10^{-2}$ mol $= x \times 10^{-2}$.

Hence $x = 1.875$.

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