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?

Md Ariful Islam

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Md Ariful Islam

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

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

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 $"1 kg" = 10^3 quad "g"$, will be

$10^3 quad "g" + "52.68 g" = "1052.68 g"$

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

Use the 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 $10^3 quad "mL"$ 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

$color(darkgreen)(ul(color(black)("molarity" = "0.568 mol L"^(-1))))$

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

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