Question

Question #9c19a

Answer 1

`"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.

`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, `"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 `"3.00 L"`. In order to find the solution's molarity, you must determine how many moles you get in `"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 `"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 `"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

`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.