The value of x is _________.

The purple permanganate ion $$\mathrm{MnO_4^-}$$ oxidises $$\mathrm{Fe^{2+}}$$ to $$\mathrm{Fe^{3+}}$$ in acidic medium according to

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

Thus, $$1$$ mole of $$\mathrm{KMnO_4}$$ reacts with $$5$$ moles of $$\mathrm{Fe^{2+}}$$.

Step 1 : Moles of $$\mathrm{KMnO_4}$$ used in the titration
Volume of permanganate solution used $$= 12.5\;\text{mL}=12.5\times10^{-3}\;\text{L}$$
Molarity $$=0.03\;\text{M}$$

$$\text{Moles of } KMnO_4 = 0.03 \times 12.5\times10^{-3}=3.75\times10^{-4}\;\text{mol}$$

Step 2 : Moles of $$\mathrm{Fe^{2+}}$$ present in the 25.0 mL aliquot

Using the $$1:5$$ stoichiometry,
$$\text{Moles of } Fe^{2+}=5\times3.75\times10^{-4}=1.875\times10^{-3}\;\text{mol}$$

Step 3 : Moles of $$\mathrm{Fe^{2+}}$$ in the entire 250 mL solution

The aliquot is one-tenth of the total solution volume (25.0 mL out of 250 mL). Therefore,
$$\text{Total moles of } Fe^{2+}=10\times1.875\times10^{-3}=1.875\times10^{-2}\;\text{mol}$$

This equals $$x\times10^{-2}$$ mol, so

$$x = 1.875$$

Rounded to two decimal places, $$x$$ lies in the range $$1.87-1.88$$.

Hence, the required value of $$x$$ is 1.87 - 1.88.