**Answer: 0.5**

*pH = 12*

*pOH = 14 - pH = 14 - 12 = 2*

*$\therefore [OH^-] = 10^{-2} M$*

*$\therefore [Ca(OH)_2] = 0.5 \times 10^{-2} M$*

*(Assuming complete dissociation at this concentration)*

*Mili moles of $Ca(OH)_2$ in solution = MV (in mL) = $0.5 \times 10^{-2} \times 100 = 0.5$*