pH of Ca(OH)₂ is 12. Mlilliequivalents of Ca(OH)₂ present in 100 mL solution will be

pH = 12

$$\begin{gathered} \therefore \mathrm{pOH}=14-12=2 \\ \therefore \left[{\mathrm{OH}}^{-}\right]={10}^{-2}\mathrm{N}={10}^{-2}\text{ equivalent/L } \\ ={10}^{-2}\times {10}^3\text{ milliequivalent/L } \\ =\frac{{10}^{-2}\times {10}^3}{{10}^3}\times 100\text{ milliequivalent }/100\mathrm{mL} \end{gathered}$$
= 1