$\mathrm{~A} ~0.081 \mathrm{~g}$ sample of pyrolusite ore (impure $\mathrm{MnO}_2$ ) is treated with $1.651 \mathrm{~g}$ of oxalic acid $\left(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{O}\right)$ in an acidic medium. Following this reaction the excess oxalic acid is titrated with $30.6 \mathrm{~mL}$ of $0.1 \mathrm{~M} \mathrm{~KMnO}_4$, the percentage of $\mathrm{~MnO}_2$ in the ore will be $(\mathrm{Mn}=55)$ $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{MnO}_2+2 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{CO}_2$ $5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2$

NEET UG CHEMISTRY MCQ: d & f- BLOCK ELEMENTS

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Application +4 | -1

\mathrm{~A} ~0.081 \mathrm{~g}

sample of pyrolusite ore (impure

\mathrm{MnO}_2

) is treated with

1.651 \mathrm{~g}

of oxalic acid

\left(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{O}\right)

in an acidic medium. Following this reaction the excess oxalic acid is titrated with

30.6 \mathrm{~mL}

0.1 \mathrm{~M} \mathrm{~KMnO}_4

, the percentage of

\mathrm{~MnO}_2

in the ore will be

(\mathrm{Mn}=55)

\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{MnO}_2+2 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{CO}_2

5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2

View answer and explanation

Correct option: D

82 \%

Option D (82%) is correct because it is the calculated percentage of MnO2 in the pyrolusite ore based on the given reactions and titration data.