# Question #9c19a

##### 1 Answer

#### Answer:

#### 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.

#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"one liter of solution"color(white)(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*,

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 **moles** of potassium fluoride in a volume of

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

#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

In other words, **every liter** of this solution will contain

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

#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(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.