Question

Question #535af

Answer 1

Here's what I got.

Explanation:

The problem tells you that you're dealing with a `"1.37-M"` aqueous solution of citric acid,. `"C"_5"H"_8"O"_7`, that has a density of `"1.10 g mL"^(-1)`.

Your strategy when it comes to this type of problems will be to pick a convenient sample of this solution as a starting point for your calculations.

Since molarity is defined as moles of solute, which in your case is citric acid, per liters of solution, you can make the calculations easier by picking a `"1.0-L"` sample of this solution.

To find the solution's percent concentration by mass, you need to know

• the mass of the solute in this sample
• the total mass of the sample

When you're working with a `"1.0-L"` sample, the solution's molarity and the number of moles of solute are interchangeable. This means that you sample will contain

`n_"citric" = "1.37 moles C"_5"H"_8"O"_7`

Use citric acid's molar mass to find how many grams would contain this many moles

`1.37 color(red)(cancel(color(black)("moles C"_5"H"_8"O"_7))) * "192.12 g"/(1color(red)(cancel(color(black)("mole C"_5"H"_8"O"_7)))) = "263.2 g"`

Next, use the density of the solution to find its mass - do not forget to convert the volume from liters to milliliters!

`1.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.10 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("given density")) = "1100 g"`

The solution's percent concentration by mass will thus be

`color(blue)("% w/w" = "mass of solute"/"mass of solution" xx 100)`

In your case, you will have

`"% w/w" = (263.2 color(red)(cancel(color(black)("g"))))/(1100color(red)(cancel(color(black)("L")))) xx 100 = color(green)("23.9%")`

To find the solution's molality, you need to know

• the number of moles of solute `->` you already have this!
• the mass of the solvent expressed in kilograms

To find the mass of the solvent, subtract the mass of the solute from the total mass of the solution

`m_"solution" = m_"solvent" + m_"solute"`

`m_"solvent" = "1100 g" - "263.2 g" = "836.8 g"`

Convert this to kilograms

`836.8 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 836.8 * 10^(-3)"kg"`

The molality of the solution will be

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

In your case, you will have

`b = "1.37 moles"/(836.8 * 10^(-3)"kg") = color(green)("1.64 mol kg"^(-1))`

Finally, to get the mole fraction of the acid, you need to know

• the number of moles of solute `->` you already have this!
• the total number of moles present in the sample

Find the number of moles of water by using its molar mass

`838.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "46.45 moles H"_2"O"`

The total number of moles present in solution will be

`n_"total" = n_"citric" = n_"water"`

`n_"total" = "1.37 moles" + "46.45 moles" = "47.82 moles"`

The mole fraction of citric acid will be

`color(blue)(chi_"citric" = n_"citric"/n_"total")`

Plug in your values to get

`chi_"citric" = (1.37color(red)(cancel(color(black)("moles"))))/(47.82color(red)(cancel(color(black)("moles")))) = color(green)(0.0286)`

The given answers are rounded to three sig figs.

As practice, you can try solving a very similar problem

http://socratic.org/questions/a-14-78-m-solution-of-nh3-in-water-has-a-density-of-0-899-g-ml-at-20oc-what-is-t

Also, you must keep in mind that the results do not depend on the sample used! To test this, I recommend redoing the calculations using another sample of solution.