Question

A solution of phosphoric acid was made by dissolving 10.0 g `H_3PO_4` in 100.0 mL water. The resulting volume was 104 mL. What is the density, mole fraction, molarity, and molality of the solution?

Assume water has a density of 1.00 `g` `/` `cm^3`.

Answer 1

Here's what I got.

Explanation:

The first thing to focus on here is finding the density of the solution.

As you know, density is defined as mass per unit of volume. You know that your solution has a volume of `"104 mL"`, which means that you only have to figure out its mass in order to be able to calculate its density.

Notice that you are given the volume of water and its density. Use this information to find the mass of water used to make the solution - keep in mind that you have `"1 cm"^3 = "1 mL"`.

`100.0 color(red)(cancel(color(black)("mL"))) * overbrace("1.00 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density of water")) = "100.0 g"`

So, the total mass of the solution will be equal to

`m_"sol" = m_(H_3PO_4) + m_"water"`

`m_"sol" = "10.0 g" + "100.0 g" = "110.0 g"`

The density of the solution will thus be

`rho = "110.0 g"/"104 mL" = color(green)("1.06 g mL"^(-1))`

To get the mole fraction of phosphoric acid in this solution, you need to know

• the number of moles of phosphoric acid
• the number of moles of water

Since you have the masses of the two compounds, you can use their molar masses to determine how many moles of each you have present in this solution.

`10.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_3"PO"_4)/(97.995color(red)(cancel(color(black)("g")))) = "0.10205 moles H"_3"PO"_4`

and

`100.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"`

The total number of moles in this solution will be

`n_"total" = "0.1025 moles" + "5.551 moles" = "5.6531 moles"`

The mole fraction of phosphoric acid will be

`chi_(H_3PO_4) = (0.10205 color(red)(cancel(color(black)("moles"))))/(5.6531 color(red)(cancel(color(black)("moles")))) = color(green)(0.0181)`

The molarity of the solution is defined as the number of moles of solute divided by the volume of the solution - expressed in liters.

`color(blue)(c = n_"solute"/V_"solution")`

In your case, you will have

`c = "0.10205 moles"/(104 * 10^(-3)"L") = color(green)("0.981 mol L"^(-1))`

Finally, the molality of the solution is defined as the number of moles of solute divided by the mass of the solvent - expressed in kilograms.

`color(blue)(b = n_"solute"/m_"solvent")`

Plug in your values to get

`b = "0.10205 moles"/(100.0 * 10^(-3)"kg") = color(green)("1.02 mol kg"^(-1))`