1. Given pH, find pOH

Since $\text{pH} + \text{pOH} = 14$ we get $\text{pOH} = 14 - 12 = 2$

2. Find hydroxide ion concentration

Using $[\text{OH}^-] = 10^{-\text{pOH}}$ so $[\text{OH}^-] = 10^{-2}\,\text{M}$

3. Find moles of } \text{OH}^- \text{ in 100 mL solution}

Volume of solution: $100\,\text{mL} = 0.1\,\text{L}$

Hence moles of $\text{OH}^-$ required are $n(\text{OH}^-) = [\text{OH}^-]\times V = 10^{-2}\times 0.1 = 10^{-3}\,\text{mol}$

Converting to millimoles: $10^{-3}\,\text{mol} = 1\,\text{mmol}$

4. Relate } \text{Ca(OH)}_2 \text{ to } \text{OH}^-

Assuming complete dissociation: $\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$

So, 1 mole of $\text{Ca(OH)}_2$ gives 2 moles of $\text{OH}^-$. Therefore required millimoles of $\text{Ca(OH)}_2$ are $\text{mmol of } \text{Ca(OH)}_2 = \frac{1}{2} = 0.5\,\text{mmol}$

5. Match with the form } x\times10^{-1}

We have $0.5\,\text{mmol} = 5\times10^{-1}\,\text{mmol}$

Thus, $x=5$

6. Comparison with stored answer

Derived answer is $5$, which matches the stored correct answer.