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Question

3.2 g of pyrolusite (MnO2) was treated with 50 mL of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 mL in a flask. 25 mL of this solution when treated with 0.02 M KMnO4 required 32 mL of the solution. Find the percentage of MnO2 in the sample.
Given: Atomic mass of Mn is 55 g mol1

A
4.5 %
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B
24.4 %
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C
54.0 %
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D
11.5 %
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Solution

The correct option is B 24.4 %
n-factor=|Change in O.S.|×Number of atoms

Redox changes are
+3C2O242+4CO2 (n-factor = 2)
+7MnO4+2Mn2+ (n-factor = 5)
+4MnO2+2Mn2+ (n-factor = 2)

Milliequivalents of oxalic acid left (in 25 mL)=Milliequivalents of KMnO4=32×0.02×5=3.2
Milliequivalents of oxalic acid left (in 250 mL)=3.2×25025=32

Milliequivalents of oxalic acid taken=50×0.5×2=50
Milliequivalents of MnO2=Milliequivalents of (oxalic acid taken - oxalic acid left (in 250 mL))
Milliequivalents of MnO2=5032=18
WMnO2MMnO2×2×1000=18
WMnO255+2(16)×2×1000=18
WMnO2=0.783 g
% of MnO2=0.7833.2×100=24.4 %

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