In a quantitative determination of iron in an ore, an analyst converted `0.42` g of the ore into its ferrous form . This required `42.00` mL of `0.1` N solution of `KMnO_(4)` for titration.
How many moles of `KMnO_(4)` were used for titration ? (Fe =56)

Explanation

In this problem `Fe^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)`
Moles of `KMnO_(4) =("equivalents")/("factor relating mo.wt and eq.wt")" "...(Eqn.6ii)`
` = (0.0042)/5 = 0.00084` mole .
Note : Thus we see how the equations 1 to 7 have been put into use in the problems given above . However , application of these euations and the ability to use them come only from practice. The students are advised to apply the said rules in as many problems as possible ] .