# How much water do I need to dissolve 95.2*g magnesium hydroxide?

Sep 10, 2017

Good luck with this endeavour ...... it is chemically UNREASONABLE.

#### Explanation:

Magnesium hydroxide is fairly insoluble in water, ${K}_{\text{sp}} = 5.61 \times {10}^{-} 12$..which corresponds to a solubility of approx. $10 \cdot m g \cdot {L}^{-} 1$.

On the other hand we could make a solution of magnesium chloride with a concentration of $1.25 \cdot m o l \cdot {L}^{-} 1$.

And so we take a mass of $1.25 \cdot m o l \times 95.2 \cdot g$ $M g C {l}_{2}$..and place this mass in a $1 \cdot L$ volumetric flask, and make it up to volume....

$\text{Concentration"xx"Volume"-="Moles}$

And $\text{mass"="number of moles"xx"molar mass}$

$\equiv \frac{1.25 \cdot m o l \times 95.2 \cdot g \cdot m o {l}^{-} 1}{1.0 \cdot L}$ $=$ $119 \cdot g$.....WITH RESPECT to MAGNESIUM CHLORIDE, i.e. I have altered the boundary conditions of your problem.