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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?

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  • $\begingroup$ 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 ? $\endgroup$
    – Poutnik
    Mar 28 at 17:47
  • $\begingroup$ @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? $\endgroup$
    – Alex
    Mar 28 at 17:49
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    $\begingroup$ @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. $\endgroup$
    – Loong
    Mar 28 at 18:35
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It does not make sense to suppose that a solution is a compound. In a solution, the proportion of one compound can change continuously. In a compound the proportion of an atom is fixed and cannot change continuously.

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 $\ce{1.19 g MgCl2}$

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  • $\begingroup$ This was very helpful. I think I got confused and read chloride ion as chlorine, which would be Cl2, when we are really referring to the chlorine ION, i.e. Cl-. $\endgroup$
    – Alex
    Mar 28 at 19:01
  • $\begingroup$ The ion $\ce{Cl-}$ is called CHLORIDE, and not chloRINE. The ion CHLORINE does not exist, or it may be the unstable $\ce{Cl+}$. $\endgroup$
    – Maurice
    Mar 28 at 19:27

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