The volume of 0.05  M  KMnO4  solution required to oxidize completely 2.70g of oxalic acid (H2C2O4)  in acidic medium will be

The balanced equation is 2 KMnO4+3H2SO4+5H2C2O4→                      K2SO4+2MnSO4+8H2O+   10  CO2 5 moles of oxalic acid react with KMnO4=2 moles2.70g  oxalic acid =2.7090 mole∴2.7090  mole of oxalic acid will react with                 KMnO4=25×2.7090=0.012 mole0.05  mole of KMnO4  sol an present in 1000  cm3 ∴0.012  mole of KMnO4  will be present in                   10000.05×0.012 cm3=240  cm3