How much water do I need to dissolve $95.2 \cdot g$ $95.2*g$ magnesium hydroxide?

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How much water do I need to dissolve $95.2 \cdot g$ $95.2*g$ magnesium hydroxide?

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Answer 1 · 2017-09-10T14:35:13

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

Explanation:

Magnesium hydroxide is fairly insoluble in water, $K_{sp} = 5.61 \times 10^{- 12}$ $K_"sp"=5.61xx10^-12$ ..which corresponds to a solubility of approx. $10 \cdot m g \cdot L^{- 1}$ $10*mg*L^-1$ .

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

And so we take a mass of $1.25 \cdot m o l \times 95.2 \cdot g$ $1.25*molxx95.2*g$ $M g C l_{2}$ $MgCl_2$ ..and place this mass in a $1 \cdot L$ $1*L$ volumetric flask, and make it up to volume....

$Concentration \times Volume \equiv Moles$ $"Concentration"xx"Volume"-="Moles"$

And $mass = number of moles \times molar mass$ $"mass"="number of moles"xx"molar mass"$

$\equiv \frac{1.25 \cdot m o l \times 95.2 \cdot g \cdot m o l^{- 1}}{1.0 \cdot L}$ $-=(1.25*molxx95.2*g*mol^-1)/(1.0*L)$ $=$ $=$ $119 \cdot g$ $119*g$ .....WITH RESPECT to MAGNESIUM CHLORIDE, i.e. I have altered the boundary conditions of your problem.

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How much water do I need to dissolve $95.2 \cdot g$ $95.2*g$ magnesium hydroxide?

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How much water do I need to dissolve 95.2⋅g95.2*g magnesium hydroxide?

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How much water do I need to dissolve $95.2 \cdot g$ $95.2*g$ magnesium hydroxide?