# Question ba9ef

Oct 8, 2017

$0.0357$

#### 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 $\text{1 L} = {10}^{3}$ $\text{mL}$ of this solution.

In your case, a ${\text{0.220-mol L}}^{- 1}$ potassium oxalate solution will contain $0.220$ moles of potassium oxalate for every ${10}^{3}$ $\text{mL}$ of solution.

To make the calculations easier, let's pick a sample of this solution that has a volume of ${10}^{3}$ $\text{mL}$. 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 $\text{1 mL}$ has a mass of $\text{1.0235 g}$.

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

$\left(36.57 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(1023.5color(red)(cancel(color(black)("g}}}}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.0357}}}$

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