# 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?

##### 1 Answer

#### Answer:

#### 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** **moles** of oxalic acid because its *molality* is equal to

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

#10^3 quad "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

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 **moles** of oxalic acid, you can say that

#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

#color(darkgreen)(ul(color(black)("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.