Question

Calculate the weight of kmno4 contained in 2 litres of 0.15 m solution

Answer 1

N (normality) describes a solution that contains 1 gram equivalent weight (gEW) per litre solution. An equivalent weight is equal to the molecular weight divided by the valence (here it gets a little tricky, for acids ands bases it refers to the number of H+ or OH-, in salts it must be expressed which ion is meant unless the ratio is 1:1).

In the case of KMnO4, equivalent wt is reaction specific. When KMnO4 is used in acid medium as oxidiser, 5 electrons are gained by Mn atom. So equivalent wt of KMnO4 in acid medium = Molecular wt/no.of electrons gained in redox reaction = 158/5 =31.6.  

So for 0.1N KMnO4 solution, you have to dissolve 3.16g KMnO4 in 1L water. (Usually a little bit excess is taken, say 3.25g, since some crystals of KMnO4 will be remained undissolved that have to be removed by filtration. So even though u r preparing 0.1N KMnO4 solution by accurate weighing, it is not a primary standard and u have to standardise it against a primary std such as oxalic acid or sodium oxalate.

In alkaline or neutral medium, reaction of KMnO4 is different and Mn gains 3 electrons in redox reaction. So, for alkaline medium redox titrations, equivalent wt of KMnO4 will be 158/3 = 52.6.  

So for 0.1N KMnO4 solution in alkaline medium redox titration, dissolve 5.26g in 1L water.