# Question 0bd93

Oct 14, 2017

${\text{1200 g MgCl}}_{2}$

#### Explanation:

For starters, you know that magnesium chloride is soluble in water, which implies that it dissociates completely in aqueous solution to produce magnesium cations and chloride anions.

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

This tells you that the solution will contain twice as many moles of chloride anions than the number of moles of magnesium chloride dissolved to make the solution.

In your case, the solution is said to have a volume of $\text{100. L}$ and a molarity of $\text{0.25 M}$ for the chloride anions. This implies that the solution must contain

100. color(red)(cancel(color(black)("L solution"))) * "0.25 moles Cl"^(-)/(1color(red)(cancel(color(black)("L solution")))) = "25 moles Cl"^(-)

This implies that the solution was made by dissolving

25 color(red)(cancel(color(black)("moles Cl"^(-)))) * "1 mole MgCl"_2/(2color(red)(cancel(color(black)("moles Cl"^(-))))) = "12.5 moles MgCl"_2#

in enough water to make the volume of the solution equal to $\text{100. L}$.

To convert the number of moles of magnesium chloride to moles, you can use the molar mass of the compound.

$12.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles MgCl"_2))) * "95.211 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = color(darkgreen)(ul(color(black)("1200 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the chloride anions.