Write the reaction in acidic medium

In acidic medium, permanganate oxidizes ferrous ion to ferric ion:

$MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

So, $1\text{ mol } KMnO_4 \equiv 5\text{ mol } FeSO_4$

Given data

• Volume of $KMnO_4$ solution $= 10\,\text{mL}$
• Volume of $FeSO_4$ solution used $= 10\,\text{mL}$
• Molarity of $FeSO_4 = 0.1\,M$

Moles of ferrous sulphate used

$n(FeSO_4) = M \times V = 0.1 \times \frac{10}{1000} = 1.0 \times 10^{-3}\,\text{mol}$

Moles of $KMnO_4$ present in 10 mL

From stoichiometry, $n(KMnO_4) = \frac{1.0 \times 10^{-3}}{5} = 2.0 \times 10^{-4}\,\text{mol}$

Molarity of $KMnO_4$

This amount is present in $10\,\text{mL} = 0.01\,\text{L}$:

$M = \frac{2.0 \times 10^{-4}}{0.01} = 2.0 \times 10^{-2}\,M$

Molar mass of $KMnO_4$

$39 + 55 + 4(16) = 39 + 55 + 64 = 158\,\text{g mol}^{-1}$

Strength in g/L

$\text{Strength} = M \times \text{molar mass} = 2.0 \times 10^{-2} \times 158 = 3.16\,\text{g L}^{-1}$

Match with required form

Given: $\text{Strength} = \underline{\hspace{1cm}} \times 10^{-2}$

Since $3.16 = 316 \times 10^{-2}$

the required integer is:

$\boxed{316}$