Question

How many grams of Mg(OH)₂ are required to obtain a 0.01M hydroxide solution.

Answer 1

If we wish to prepare 1L of 0.01M of solution then

The molecular weight of Mg(OH)2 = atomic weight of Mg + 2 X atomic weight of oxygen + 2X atomic weight of hydrogen

Molecular weight of Mg(OH)2 = 24 + 32 + 2 = 58g / mole

Molarity = Mass of solute / Molecular weight of solute X volume of solution in litres

molarity = 0.01 M

Mass = ?

Mol wt = 58 g / mole

volume =1L

0.01 = Mass / 58 X 1

mass = 58 X 0.01 grams = 0.58grams

so we will dissolve 0.58 grams of Magnesium hydroxide in 1L of solution

NOTe: it will not give a true solution as solubility of magnesium hydroxide is very low in water (6.4 X 10^-3 g / L)