Titrations!

My first chemistry answer !! ( I have referred to m.e. as milli-equivalents )

Starting from back ; the normality of $\ce{KMnO4} = 30*0.10/25 = 0.12 N$ .The m.e. of excess ferrous ions = m.e. of $\ce{KMnO4}$ is $50*0.12 = 6 m.e.$ (Law of Equivalence).

Originally there were $100*0.1 = 10 m.e.$ of ferrous ions.That means $4 m.e.$ reacted with $Mn_{3}O_{4}$ .Again from Law of Equivalence this is also the milli equivalents of $\ce{Mn3O4}$ .According to rxn : $\ce{Mn3O4} + Fe^{2+} ----> Mn^{2+} + Fe^{3+}$ .We can say that $Eq wt = Mol wt / \Delta O.N. = 229/2 = 114.5 g$ .Hence the mass of $Mn_{3}O_{4}$ is $4/1000*114.5 = 0.458 g$ .Now applying POAC for Mn atoms in rxn : $\ce{MnSO4.4H2O} ----> \ce{Mn3O4}$ mass is $1.338 g$ .This corresponds to $66.9 \approx 67$ %.

Hey! My solution for this awesome question....

Firstly we will be starting from the end and going in the beginning as the question is like that.

So as given 25 ml of $\ce{KMnO4}$ is completely reduced by 30 ml of 0.1 N $\ce{FeSO4}$ , We have

$\text{meq of } \ce{KMnO4} = \text{meq of } \ce{FeSO4}$

$\implies 25 \times N_{\ce{KMnO4}} = 30 \times 0.1$

$\implies N_{\ce{KMnO4}} = \dfrac{3}{25} \ce{N}$

So now as said we have the same stock solution means the normality will be same for the previous step which was done, i.e

The excess of ferrous ions left in solution were titrated with 50 ml of $\ce{KMnO4}$ solution in acidic medium.

$\implies V_{\text{excess}} \times 0.1 = 50 \times \dfrac{3}{25}$

$\implies V_{excess} = 60$ ml.

So the used up volume in the previous step is $100-60=40$ ml.

So now...

From the equation

$\ce{MnSO4.4H2O} \rightarrow \ce{Mn3O4}$ , the n-factor comes out to be = $\dfrac{8}{3} -2 =\dfrac{2}{3}$ .

So Again by law of Equivalence we have

$\dfrac{x}{\dfrac{223}{\dfrac{2}{3}}} \times 1000 = 40 \times 0.1$

$x=1.338 g$ .

So the % of purity will be = $\dfrac{1.338}{2} \times 1 00 \implies 67 \% \text{(approx)}$