# Question 9c19a

Mar 30, 2016

${\text{0.667 mol L}}^{- 1}$

#### Explanation:

Every time you're looking for a solution's molarity, you must determine how many moles of solute you get in one liter of solution.

That is what molarity essentially tells you - how many moles of solute you'd get if you had exactly one liter of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molarity" = "moles of solute"/"one liter of solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Notice that the problem provides you with the volume of the solution and with the mass of the solute, which in your case is potassium fluoride, $\text{KF}$.

Your first goal here will be to use potassium fluoride's molar mass to determine how many moles you get in that sample

116.2color(red)(cancel(color(black)("g"))) * "1 mole KF"/(58.097color(red)(cancel(color(black)("g")))) ~~"2.00 moles KF"

So, you know that this solution contains $2.00$ moles of potassium fluoride in a volume of $\text{3.00 L}$. In order to find the solution's molarity, you must determine how many moles you get in $\text{1.00 L}$ of solution.

This means that you can use the number of moles of solute present in this sample as a conversion factor to help you find the number of moles of solute in $\text{1.00 L}$

1.00color(red)(cancel(color(black)("L solution"))) * "2.00 moles KF"/(3.00color(red)(cancel(color(black)("L solution")))) = "0.667 moles KF"

The solution will thus have a molarity of ${\text{0.667 mol L}}^{- 1}$.

In other words, every liter of this solution will contain $0.667$ moles of potassium fluoride.

Notice that you can find the solution's molarity by dividing the number of moles of solute by the total volume of the solution

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this case, you would have

c = "2.00 moles"/"3.00 L" = color(green)(|bar(ul(color(white)(a/a)"0.667 mol L"^(-1)color(white)(a/a)|)))#

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.