Question

How many grams of water must be added to 455 grams of potassium sulfate in order to make a 1.50 M solution?

Answer 1

Here's what I got.

Explanation:

A solution's molarity essentially tells you how many moles of solute you get in one liter of solution.

This means that before you can say for certain how many grams of water are needed to make this solution, you must find

• the number of moles of potassium sulfate
• the volume of water that is needed for this solution

Now, to get the number of moles of potassium sulfate, `"K"_2"SO"_4`, present in that `"455-g"` sample, use potassium sulfate's molar mass, which tells you the mass of one mole of potassium sulfate.

`455color(red)(cancel(color(black)("g"))) * overbrace(("1 mole K"_2"SO"_4)/(174.26color(red)(cancel(color(black)("g")))))^(color(purple)("moles mass of K"_2"SO"_4)) = "2.611 moles K"_2"SO"_4`

Now, a `"1.50 mol L"^(-1)` solution implies that you get `1.50` moles of solute for every liter of solution. In this case, `2.611` moles would require a total volume of solution of

`2.611color(red)(cancel(color(black)("moles K"_2"SO"_4))) * "1.0 L solution"/(1.50color(red)(cancel(color(black)("moles K"_2"SO"_4)))) = "1.741 L solution"`

So, you know that this solution must have a total volume of `"1.741 L"`. In order to determine the mass of water needed, you must make a couple of assumptions

• the volume of the solvent will eventually be equal to the volume of the solution
• water has a density of `"1.00 g mL"^(-1)`

Since the problem didn't provide you with the density of this potassium sulfate solution, you cannot know for sure how much water will be needed to make the total volume of the solution equal to `"1.741 L"`.

You can say that you will need less than `"1.741 L"`, since the fact that adding that much potassium sulfate to `"1.741 L"` of water will definitely move the volume of the solution past that mark.

However, for the sake of simplicity, you can assume that you need `"1.741 L"` of water to make `"1.741 L"` of `"1.50 mol L"^(-1)` potassium sulfate solution.

Now, if you take water's density to be equal to `"1.00 g mL"^(-1)`, you can use the fact that

`color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(a/a)|)))`

to say that you have

`1.741 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "1741 g"`

Rounded to three sig figs, the answer would be

`"mass of water" = color(green)(|bar(ul(color(white)(a/a)"1740 g"color(white)(a/a)|)))`

Mind you, this is not a practical result, but given the lack of information provided with the problem, you can say that, from a conceptual point of view at least, this is the answer to the question.