When 10 mL of an aqueous solution of KMnO$$_4$$ was titrated in acidic medium, equal volume of 0.1M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO$$_4$$ in grams per litre is _________ $$\times 10^{-2}$$. (Nearest integer) [Atomic mass of K = 39, Mn = 55, O = 16]

We have a titration between acidic potassium permanganate and ferrous sulphate. In acidic medium the relevant half-reactions are first stated:

$$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$

For the electrons to cancel, we multiply the iron half-reaction by 5 and add:

$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$

Hence, $$1$$ mole of $$\text{KMnO}_4$$ reacts with $$5$$ moles of $$\text{FeSO}_4$$.

The volume of the ferrous sulphate solution used is $$10\ \text{mL}=0.010\ \text{L}$$ and its molarity is $$0.1\ \text{M}$$. By the definition of molarity, $$\text{Molarity} = \dfrac{\text{moles}}{\text{volume in L}}$$, so the moles of $$\text{FeSO}_4$$ taken are

$$n_{\text{FeSO}_4}=0.1\ \text{mol L}^{-1}\times0.010\ \text{L}=0.001\ \text{mol}$$

The stoichiometric ratio tells us that

$$n_{\text{KMnO}_4}=\dfrac{n_{\text{FeSO}_4}}{5}=\dfrac{0.001}{5}=0.0002\ \text{mol}$$

The volume of the potassium permanganate solution is also $$10\ \text{mL}=0.010\ \text{L}$$. Again using $$\text{Molarity} = \dfrac{\text{moles}}{\text{volume}}$$, we get

$$M_{\text{KMnO}_4}=\dfrac{0.0002\ \text{mol}}{0.010\ \text{L}}=0.02\ \text{M}$$

The strength (gram per litre) is obtained from the formula $$\text{Strength} = \text{Molarity}\times\text{Molar mass}$$.

The molar mass of $$\text{KMnO}_4$$ is calculated as $$39\ (\text{for K}) + 55\ (\text{for Mn}) + 4\times16\ (\text{for O}) = 158\ \text{g mol}^{-1}$$.

Substituting, we obtain

$$\text{Strength}=0.02\ \text{mol L}^{-1}\times158\ \text{g mol}^{-1}=3.16\ \text{g L}^{-1}$$

Finally, writing $$3.16\ \text{g L}^{-1}$$ in the form $$\times10^{-2}$$ gives $$3.16\ \text{g L}^{-1}=316\times10^{-2}\ \text{g L}^{-1}$$.

Taking the nearest integer as required, we have $$316$$.

So, the answer is $$316$$.