Millimoles of calcium hydroxyide required to produce 100mL of the...

We know,
$pH = - log [ {H}^{+} ]$ , $pH = - log [ {OH}^{-} ] , pH + pOH = 14$
$∵ pH = 12$

$∴ \left[{H}^{+}\right] = {10}^{- 12}M$

$∴ \left[{OH}^{-}\right] = {10}^{- 2}M$
We also know,
$Ca ( OH{)}_{2} ⇌ 2{OH}^{-} + {Ca}^{2 +}$

$∴ \left[Ca ( OH{)}_{2}\right] = 5 \times {10}^{- 3}M$

$5 \times {10}^{- 3} = \dfrac{\text{milli moles of}Ca{\left(OH\right)}_{2}}{100mL}$

milli moles of $Ca{\left(OH\right)}_{2} = 5 \times {10}^{- 1}$

Ans. $= 5$