# Question #7ea5f

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

So, you know that you're dealing with a

Your strategy here will be to pick a *sample* of this solution. Since you're dealing with molarity, which as you know is defined as *moles of solute* per **liter** of solution, you can make the calculations easier by picking a

This sample will contain

#color(blue)(c = n/V implies n = c * V)#

#n_(NH_3) = "14.78 M" * "1.00 L" = "14.78 moles NH"_3#

Now, in order to find the **mole fraction** of ammonia in this sample, you need to figure out how many moles of water you have present.

To do that, use the given density to find the **mass** of the sample - keep in mind that density is given in *grams per milliliter*, and your sample is measured in *liters*

#1.00 color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.899 g"/(1 color(red)(cancel(color(black)("mL")))) = "899 g"#

This sample will have a mass of **molar mass** to help you find how many *grams* of ammonia you get in this sample

#14.78 color(red)(cancel(color(black)("moles NH"_3))) * "17.03 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "251.7 g NH"_3#

This means that the **mass of water** in this sample will be equal to

#m_"sample" = m_"water" = m_(NH_3)#

#m_"water" = "899 g" - "251.7 g" = "647.3 g H"_2"O"#

Use water's **molar mass** to find you many *moles* of water would be present in this many grams

#647.3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "35.93 moles H"_2"O"#

The **mole fraction** of ammonia will be equal to the number of moles of ammonia divided by the **total number of moles** present in solution.

In your case, you will have

#chi_(NH_3) = (14.78color(red)(cancel(color(black)("moles"))) )/((14.78 + 35.93)color(red)(cancel(color(black)("moles")))) = color(green)(0.291)#

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")#

In your case, you will have

#b = "14.78 moles"/(647.3 * 10^(-3)"kg") = color(green)("22.8 molal")#

The solution's percent concentration by mass is defined as the mass of solute divided by the **mass of the solution** and multiplied by

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

In your case, the

#"%w/w" = (251.7 color(red)(cancel(color(black)("g"))))/(899color(red)(cancel(color(black)("g")))) xx 100 = color(green)(28.0%)#

Finally, to get the concentration of the solution in parts per million, simply divide the mass of the solute *in grams* by the mass of the solvent *in grams* and multiply the result by

#color(blue)("ppm" = m_"solute"/m_"solvent" xx 10^6)#

In your case, you will have

#"ppm" = (251.7 color(red)(cancel(color(black)("g"))))/(647.3color(red)(cancel(color(black)("g")))) xx 10^6 = color(green)(3.89 * 10^5"ppm")#

The answers are rounded to three sig figs, the number of sig figs you have for the density of the solution.

**SIDE NOTE** *I highly recommend redoing the calculations using a different sample of the solution. The important thing to realize is that the values must come out exactly the same regardless of what volume you pick.*