# Calculate the concentration in "g dm"^-3 of chloride ions in a "0.25-M" potassium chloride solution?

Feb 24, 2018

${\text{8.9 g dm}}^{- 3}$

#### Explanation:

Your goal here is to convert the concentration of chloride anions from moles per cubic decimeter to grams per cubic decimeter.

Now, you know that potassium chloride is soluble in water, so it will dissociate completely in aqueous solution to produce potassium cations and chloride anions.

${\text{KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Now, every mole of potassium chloride that you dissolve in this solution will produce $1$ mole of chloride anions, so you can say that

["Cl"^(-)] = ["KCl"] = "0.25 M"

The molarity of the chloride anions tells the number of moles of chloride anions present for every ${\text{1 L = 1dm}}^{3}$ of this solution.

In this case, you have $0.25$ moles of chloride anions for every ${\text{1 dm}}^{3}$ of the solution. To convert the number of moles to grams, use the molar mass of atomic chlorine, $\text{Cl}$--keep in mind that the mass of the chloride anion is approximately equal to the mass of the neutral atom!

0.25 color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1 color(red)(cancel(color(black)("mole Cl"^(-))))) = "8.86 g"

Since this represents the mass of chloride anions present in ${\text{1 dm}}^{3}$ of this solution, you can say that the concentration of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{concentration Cl"^(-) = "8.9 g dm}}^{- 3}}}}$

The answer must be rounded to two sig figs, the number of sig figs you have for the molarity of the solution.