# Question 3f934

Aug 28, 2016

${\text{1.2 g MgCl}}_{2}$

#### Explanation:

The most important thing to keep in mind when dealing with magnesium chloride, ${\text{MgCl}}_{2}$, is that it is soluble in water, meaning that it dissociates completely to form magnesium cations, ${\text{Mg}}^{2 +}$, and chloride anions, ${\text{Cl}}^{-}$

${\text{MgCl"_ (color(red)(2)(aq)) -> "Mg"_ ((aq))^(2+) + color(red)(2)"Cl}}_{\left(a q\right)}^{-}$

Notice that every mole of magnesium chloride that dissociates produces $\textcolor{red}{2}$ moles of chloride anions. This means that in order to have

100 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.25 moles Cl"^(-)/(1color(red)(cancel(color(black)("L")))) = "0.025 moles Cl"^(-)

you need to dissolve

0.025 color(red)(cancel(color(black)("moles Cl"^(-)))) * "1 mole MgCl"_2/(color(red)(2)color(red)(cancel(color(black)("moles Cl"^(-))))) = "0.0125 moles MgCl"_2

Now all you have to do is convert the moles of magnesium chloride to grams by using the salt's molar mass

0.0125 color(red)(cancel(color(black)("moles MgCl"_2))) * "95.2 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.2 g")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have one sig fig for the volume of the solution.