# Grams of solution of magnesium chloride given molarity of chloride ion [closed]

To make 100. mL of a solution that is 0.25 M in chloride ion, how many grams of magnesium chloride would you need to dissolve?

I had a few different approaches to this, but my most immediate thought was to think of each element (monoatomic or polyatomic) as a part of a larger compound. Therefore, molarity of the chloride ion is 0.25, which is 0.66 moles of the entirety of the compound; the rest being magnesium.

This approach led me to 3.52 grams, since I have 0.037 moles (molarity is 0.037 moles/liter). However, the answer is 1.2 grams.

What am I doing wrong?

• If you use algebra, you spot any mistake or error much more easily. I am very curious, how you get 0.66 mol. How many moles of chloride and magnesium contains 1 mol of magnesium chloride ? Mar 28 at 17:47
• @Poutnik Yes, I am doing a crash course in chemistry, so I constantly make silly mistakes :D But that’s the fun part. Chloride is just Cl2, so magnesium chloride should contain 2/3 moles of Cl2 and 1/3 mole of magnesium, right?
– Alex
Mar 28 at 17:49
• @Alex No, no, and no. Chloride is not $\ce{Cl2}$. Magnesium chloride doesn't contain any $\ce{Cl2}$. And why do you assume that magnesium chloride contains 1/3 mole magnesium? Do you mean 1 mol $\ce{MgCl2}$ contains 1/3 mol $\ce{Mg}$? No, that would be wrong, too. Mar 28 at 18:35

In your problem, the concentration of the $$\ce{Cl-}$$ ion is $$\pu{0.25 M}$$. This means that the concentration of Magnesium is half of it, so that $$[\ce{MgCl2}]$$ = $$\pu{0.125 M}$$. As $$\ce{MgCl2}$$ molar weight is $$\pu{95.3 g/mol}$$, it means that $$1$$ liter of $$\ce{MgCl2 0.125 M}$$ contains $$\pu{0.125 · 95.3 g = 11.9 g \ce{MgCl2}}$$.
As a consequence, $$\pu {100 mL}$$ of this solution contains $$\ce{1.19 g MgCl2}$$
• The ion $\ce{Cl-}$ is called CHLORIDE, and not chloRINE. The ion CHLORINE does not exist, or it may be the unstable $\ce{Cl+}$. Mar 28 at 19:27