Question

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

Answer 1

`(35.9*g)/(74.10*g*mol^-1)xx1/(0.535*L)` `=` `0.906*mol*L^-1`

Explanation:

This value, `0.906*mol*L^-1` with respect to `"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_(sp)=5.5xx10^-6`.