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

##### 1 Answer

#### Answer:

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

thenumber of molesof potassium sulfatethevolumeof water that is needed for this solution

Now, to get the number of moles of potassium sulfate, **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 **moles** of solute **for every liter** of solution. In this case,

#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 **mass** of water needed, you must make a couple of *assumptions*

the volume of thesolventwill eventually be equal to the volume of thesolutionwater 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

You can say that you will need **less** than **definitely** move the volume of the solution past that mark.

However, for the sake of simplicity, you can *assume* that you need

Now, if you take water's density to be equal to

#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.