Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is $$x \times 10^{-1}$$. The value of $$x$$ is ______ (Nearest integer). Assume complete dissociation.

We need to find the millimoles of Ca(OH)$$_2$$ required to prepare 100 mL of an aqueous solution with pH = 12.

Find [OH$$^-$$] from pH.

$$\text{pH} = 12 \Rightarrow \text{pOH} = 14 - 12 = 2$$

$$[\text{OH}^-] = 10^{-2} = 0.01 \text{ M}$$

Relate [OH$$^-$$] to [Ca(OH)$$_2$$].

Ca(OH)$$_2$$ dissociates completely:

$$\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$$

So $$[\text{OH}^-] = 2 \times [\text{Ca(OH)}_2]$$:

$$[\text{Ca(OH)}_2] = \frac{0.01}{2} = 0.005 \text{ M}$$

Calculate millimoles in 100 mL.

$$\text{Moles} = 0.005 \times 0.1 = 5 \times 10^{-4} \text{ mol} = 0.5 \text{ mmol}$$

Given that the answer is $$x \times 10^{-1}$$ millimoles:

$$0.5 = x \times 10^{-1} \Rightarrow x = 5$$

The value of $$x$$ is $$\boxed{5}$$.