Potassium permanganate is prepared from the mineralpyrolusite, `MnO_(2)`. Its crystals have deep purple colour. It acts as an oxidizing agentin the neutral, alkaline as well as acidic medium. In acidic medium , it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of `H_(2)SO_(4)`. However ,before using itas a titrant, it is first standized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration , 13.4 g of dry pure sodium oxalate ( molar mass `=134 g mol^(-1)` ) was dissolved in 100mLof distilled water and then 100 mL of 2M `H_(2)SO_(4)` were added. The solution, was cooled to `25.30^(@) C`. Now to this solution , 0.1 M `KMnO_(4)` solution was added till a very faint pink colour persisted. Th evoluem of `KMnO_(4)` solution that must have been added to obtain the faint pink colour at the end point must beA. 100 mLB. 200 mLC. 300 mLD. 400mL

Potassium permanganate is prepared from the mineralpyrolusite, `MnO_(2

1.	Potassium permanganate is prepared from the mineralpyrolusite, `MnO_(2)`. Its crystals have deep purple colour. It acts as an oxidizing agentin the neutral, alkaline as well as acidic medium. In acidic medium , it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of `H_(2)SO_(4)`. However ,before using itas a titrant, it is first standized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration , 13.4 g of dry pure sodium oxalate ( molar mass `=134 g mol^(-1)` ) was dissolved in 100mLof distilled water and then 100 mL of 2M `H_(2)SO_(4)` were added. The solution, was cooled to `25.30^(@) C`. Now to this solution , 0.1 M `KMnO_(4)` solution was added till a very faint pink colour persisted. Th evoluem of `KMnO_(4)` solution that must have been added to obtain the faint pink colour at the end point must beA. 100 mLB. 200 mLC. 300 mLD. 400mL
Answer» Molesof `( COONa)_(2) = 13.4 // 134 = 0.1` . The balanced equation for the reaction involved is `2MnO_(4)^(-) + 16 H^(+) + 5CrO_(4)^(2-) rarr 2Mn^(2+) + 8 H_(2)O + 10 CO_(2)` `:. ` Moles of `KMnO_(4)` used `= ( 2)/( 5) xx 0.1 = 0.04 ` mol As `KMnO_(4)` solution used is `0.1 M` , i.e., 0.1 mol are present in 1000 mL, therefore , 0.04 mol will be present in 400 mL.