$pH$ of $Ca(OH)_2$ solution is 12. Millimoles of ($Ca(OH)_2$) present in 100 mL of solution will be

(a) 1

(b) 0.5

(c) 0.05

(d) 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) = $0.5 \times 10^{-2} \times 100 = 0.5$