$\therefore\left[ H ^{+}\right]=10^{-12}\,M$
$\therefore\left[ OH ^{-}\right]=10^{-2}\,M$
$\therefore\left[ Ca ( OH )_2\right]=5 \times 10^{-3}\,M$
$5 \times 10^{-3}=\frac{\text { milli moles of } Ca ( OH )_2}{100\,mL }$
$\text { milli moles of } Ca ( OH )_2=5 \times 10^{-1}$
$\text { Ans. }=5$