Question

Question #ba9ef

Answer 1

`0.0357`

Explanation:

The idea here is that you need to use the density of the solution to find the mass of a sample of known volume.

As you know, the molarity of a solution tells you the number of moles of solute present for every `"1 L" = 10^3` `"mL"` of this solution.

In your case, a `"0.220-mol L"^(-1)` potassium oxalate solution will contain `0.220` moles of potassium oxalate for every `10^3` `"mL"` of solution.

To make the calculations easier, let's pick a sample of this solution that has a volume of `10^3` `"mL"`. By definition, this sample will contain `0.220` moles of potassium oxalate.

Use the molar mass of potassium oxalate to convert the number of moles to grams

`0.220 color(red)(cancel(color(black)("moles K"_2"C"_2"O"_4))) * "166.22 g"/(1color(red)(cancel(color(black)("mole K"_2"C"_2"O"_4)))) = "36.57 g"`

Now, you know that this solution has a density of `'1.0235 g mL"^(-1)`. This tells you that every `"1 mL"` has a mass of `"1.0235 g"`.

Consequently, this particular sample will have a mass of

`10^3 color(red)(cancel(color(black)("mL solution"))) * "1.0235 g"/(1color(red)(cancel(color(black)("mL solution")))) = "1023.5 g"`

You know no the total mass of the solution and the mass of potassium oxalate it contains, so you can say that the mass fraction of the solute will be

`(36.57 color(red)(cancel(color(black)("g"))))/(1023.5color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(0.0357)))`

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