# If 35.9 grams of Ca(OH)_2 are dissolved in 535 mL of solution, what is the molarity of the solution?

$\frac{35.9 \cdot g}{74.10 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{0.535 \cdot L}$ $=$ $0.906 \cdot m o l \cdot {L}^{-} 1$
This value, $0.906 \cdot m o l \cdot {L}^{-} 1$ with respect to $\text{calcium hydroxide}$ makes no sense at all, as calcium hydroxide has fairly limited water solubility.
This site list a solubility product for calcium hydroxide, ${K}_{s p} = 5.5 \times {10}^{-} 6$.