The correct option is B 5.43%
MnO2(1.6 g)+H2C2O450 mL/0.1 N↓excess25 mL≡0.01 N/30 mLofKMnO4→Mn2+
mEq of KMnO4≡ mEq of H2C2O4]excessin25 mL
⇒ Excess mEq of H2C2O4 in 25 mL
=0.01×30=0.30
Excess mEq of H2C2O4 in 250 mL=3.0
mEq of H2C2O4 used =5−3=2
mEq of MnO2=2
Weight87/2×1000=2⇒Weight=0.087 g
% MnO2=0.0871.6×100=5.43%