Answer
0.0250 M
Work Step by Step
We find:
$\text{Moles of solute}=0.185\,g\times\frac{1\,mol\,Ca(OH)_{2}}{74.093\,g}=0.002497\,mol$
$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Liters of solution}}=\frac{0.002497\,mol}{100\times10^{-3}\,L}=0.0250\,M$
