What is the molality of a 21.6% (by mass) aqueous solution of phosphoric acid (H_3PO_4)?

Jan 8, 2016

$\text{0.241 molal}$

Explanation:

As you know, molality is defined as moles of solute, which in your case is phosphoric acid, ${\text{H"_3"PO}}_{4}$, divided by kilograms of solvent, which is of course water.

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

This means that you need to use the solution's percent concentration by mass to determine how much solute and how much solvent you get in a random sample of solution.

Since you're dealing with percentages, you can make the calculations easier by picking a $\text{100.0-g}$ sample of this solution. According to the given percent concentration, this solution will contain $\text{21.6 g}$ of phosphoric acid.

Remember, percent concentration by mass is calculated using the mass of solute and the mass of the solution, which includes both the solute and the solvent.

This means that the sample will also contain

${m}_{\text{sample" = m_"solute" = m_"solvent}}$

${m}_{\text{water" = "100.0 g" - "21.6 g" = "78.4 g}}$

Do not forget that you need to convert the mass of the solvent from grams to kilograms by using the conversion factor

$\text{1 kg" = 10^3"g}$

78.4 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 78.4 * 10^(-3)"g"

Now, to find the number of moles of phosphoric acid present in this sample, use the compound's molar mass.

21.6 color(red)(cancel(color(black)("g"))) * ("1 mole H"_3"PO"_4)/(97.995color(red)(cancel(color(black)("g")))) = "0.2204 moles H"_3"PO"_4

Therefore, the solution's molality will be

b = "0.2204 moles"/(78.4 * 10^(-3)"kg") = color(green)("2.81 molal")

The answer is rounded to three sig figs.

SIDE NOTE The molality of the solution must be the same regardless of what sample you pick. I highly recommend redoing the calculations using a different starting sample $\to$ the molality will once again be $\text{0.241 molal}$.