# Question #535af

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The problem tells you that you're dealing with a *citric acid*,.

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

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

themass of the solutein this samplethetotal mass of the sample

When you're working with a **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

thenumber of molesof solute#-># you already have this!themass of the solventexpressed inkilograms

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

thenumber of molesof solute#-># you already have this!thetotal number of molespresent 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

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.