Question

What is the molarity of the chloride ion in 250 mL of a solution containing 1.90 g of `MgCl_2`?

Answer 1

`"0.16 mol L"^(-1)`

Explanation:

The trick here is to realize that one formula unit of magnesium chloride, `"MgCl"_2`, contains one magnesium cation, `"Mg"^(2+)`, and two chloride anions, `"Cl"^(-)`.

Since magnesium chloride is a soluble ionic compound, this implies that every mole of the salt that dissolves in water will produce `1` mole of magnesium cations and `color(red)(2)` moles of chloride anions.

`"MgCl"_ (color(red)(2)(aq)) -> "Mg"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)`

The first thing to do here is figure out how many moles of magnesium chloride were needed to make this solution. To do that, use the compound's molar mass, which essentially tells you the mass of one mole of magnesium chloride

`1.90 color(red)(cancel(color(black)("g"))) * "1 mole MgCl"_2/(95.211color(red)(cancel(color(black)("g")))) = "0.01996 moles MgCl"_2`

Now, each mole of magnesium chloride will produce `color(red)(2)` moles of chloride anions in solution

`0.01996 color(red)(cancel(color(black)("moles MgCl"_2))) * (color(red)(2)color(white)(a)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = "0.03392 moles Cl"^(-)`

At this point, all you have to do is figure out how many moles of chloride anions you get in `"1 L"` of this solutions by suing the fact that `"250 mL"`, the equivalent of `1/4"th"` of a liter, contains `0.03992` moles

`1 color(red)(cancel(color(black)("L solution"))) * "0.03992 moles Cl"^(-)/(1/4color(red)(cancel(color(black)("L solution")))) = "0.16 moles Cl"^(-)`

Since molarity tells yo uthe number of moles of solute you get per liter of solution, you can say that the molarity of the chloride anions will be

`"molarity Cl"^(-) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.16 mol L"^(-))color(white)(a/a)|)))`

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution, i.e. `"250 mL"`.