The volume of 0.1 M strong dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of $$OH^-$$ in aqueous solution is:

First we interpret the data. A strong dibasic acid dissociates completely and gives two hydrogen ions per mole. Hence one mole of this acid can neutralize two moles of hydroxide ions because the neutralization reaction is

$$H^+ + OH^- \rightarrow H_2O$$

Since the acid is dibasic, one mole supplies $$2H^+$$, so

$$1 \text{ mol acid} \longrightarrow 2 \text{ mol } H^+ \longrightarrow 2 \text{ mol } OH^- \text{ neutralized}.$$

Now, the base sample releases $$0.04$$ mole of $$OH^-$$. Let the number of moles of acid required be $$n_{\text{acid}}$$. Because each mole of acid neutralizes two moles of hydroxide, we set up the simple stoichiometric relation

$$2 \times n_{\text{acid}} = 0.04.$$

Solving for $$n_{\text{acid}}$$ gives

$$n_{\text{acid}} = \frac{0.04}{2} = 0.02 \text{ mol}.$$

Next we use the molarity relation. The molarity $$M$$ of a solution is defined by the formula

$$M = \frac{n}{V},$$

where $$n$$ is the number of moles of solute and $$V$$ is the volume of the solution in litres. Rearranging for volume, we have

$$V = \frac{n}{M}.$$

The molarity of the acid solution is given as $$0.1 \text{ M}$$, so substituting $$n = 0.02 \text{ mol}$$ and $$M = 0.1 \text{ mol L}^{-1}$$, we obtain

$$V = \frac{0.02}{0.1} = 0.2 \text{ L}.$$

To express this volume in millilitres, we recall that $$1 \text{ L} = 1000 \text{ mL}$$, therefore

$$0.2 \text{ L} = 0.2 \times 1000 \text{ mL} = 200 \text{ mL}.$$

Hence, the correct answer is Option C.