The volume, in mL, of $$0.02\,\text{M}\,\text{K}_2\text{Cr}_2\text{O}_7$$ solution required to react with $$0.288\,\text{g}$$ of ferrous oxalate in acidic medium is............ (Molar mass of Fe = $$56\,\text{g mol}^{-1}$$)

To determine the volume of $$0.02\text{ M }K_2Cr_2O_7$$ required for the titration, we use the concept of equivalent weights in redox reactions.

In an acidic medium, potassium dichromate acts as an oxidizing agent and undergoes the reduction

$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O.$$

Since two chromium atoms are reduced from an oxidation state of $$+6$$ to $$+3$$, one mole of $$Cr_2O_7^{2-}$$ gains a total of 6 electrons. Therefore, the $$n$$-factor of $$K_2Cr_2O_7$$ is $$6$$.

Ferrous oxalate ($$FeC_2O_4$$) acts as the reducing agent. In this process,

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

and

$$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-.$$

Hence, each mole of $$FeC_2O_4$$ loses a total of

$$1 + 2 = 3 \text{ electrons},$$

so its $$n$$-factor is $$3$$.

The molar mass of ferrous oxalate is

$$56 + 24 + 64 = 144\text{ g mol}^{-1}.$$

Therefore, the number of moles present in $$0.288\text{ g}$$ is

$$\frac{0.288}{144} = 0.002\text{ mol}.$$

The number of equivalents of ferrous oxalate is

$$0.002 \times 3 = 0.006\text{ equivalents}.$$

At the equivalence point, the equivalents of the oxidizing agent and reducing agent are equal. Thus,

$$0.02 \times V(\text{L}) \times 6 = 0.006.$$

Solving for the volume,

$$V(\text{L}) = \frac{0.006}{0.12} = 0.05\text{ L}.$$

Converting to millilitres,

$$V = 0.05 \times 1000 = 100\text{ mL}.$$

Hence, the required volume of $$0.02\text{ M }K_2Cr_2O_7$$ solution is

$$100\text{ mL}.$$