Question

Question #7ea5f

Answer 1

Here's what I got.

Explanation:

So, you know that you're dealing with a `"14.78-M"` ammonia solution. Moreover, this solution is said to have a density of `"0.899 g/mL"`.

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 `"1.00-L"` sample of this solution.

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 `"899 g"`. You know by looking at its molarity that it contains `14.78` moles of ammonia. This means that you can use ammonia's 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 `100`.

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

In your case, the `"1.00-L"` sample has a mass of `"899 g"`, which means that the solution's percent concentration by mass is equal to

`"%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 `10^6`.

`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.