# Oxalic acid, H_2C_2O_4, occurs as the potassium or calcium salt in many plants, including rhubarb and spinach. An aqueous solution of oxalic acid is 0.585 molal H_2C_2O_4. The density of the solution is 1.022 g/mL. What is the molar concentration?

Jan 15, 2018

${\text{0.568 mol L}}^{- 1}$

#### Explanation:

The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.

To make the calculations easier, let's pick a sample that contains exactly $\text{1 kg}$ of water, the solvent. By definition, this sample will contain $0.585$ moles of oxalic acid because its molality is equal to $\text{0.585 molal}$, or ${\text{0.585 mol kg}}^{- 1}$.

Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.

0.585 color(red)(cancel(color(black)("moles H"_2"C"_2"O"_4))) * "90.03 g"/(1color(red)(cancel(color(black)("mole H"_2"C"_2"O"_4)))) = "52.68 g"

This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to $\text{1 kg" = 10^3 quad "g}$, will be

${10}^{3} \quad \text{g" + "52.68 g" = "1052.68 g}$

Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly $\text{1 L" = 10^3 quad "mL}$ of this solution.

Use the density of the solution to find the total volume of the sample

1052.68 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.022color(red)(cancel(color(black)("g")))) = "1030.0 mL"

Since you know that this sample contains $0.585$ moles of oxalic acid, you can say that ${10}^{3} \quad \text{mL}$ of this solution will contain

10^3 color(red)(cancel(color(black)("mL solution"))) * ("0.585 moles H"_2"C"_2"O"_4)/(1030.0color(red)(cancel(color(black)("mL solution")))) = "0.568 moles H"_2"C"_2"O"_4

This means that the solution has a molarity of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity" = "0.568 mol L}}^{- 1}}}}$

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