Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# An aqueous solution is 12.00% ammonium chloride, NH4Cl, by mass. The density of the solution is 1.036 g/mL. What are the molality, mole fraction, and molarity of NH4Cl in the solution?

Here's what I got.

In order to be able to calculate the , mole fraction, and of the solution, you first need to pick a volume sample.

Since is defined as moles of per **liters** of solution, a ##"1.00-L"## sample will make the calculations easier.

So, let's say that we have a ##"1.00-L"## sample of this solution. You can use its given to determine the **mass** of this sample

##1.00 color(red)(cancel(color(black)("L"))) * (1000 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.036 g"/(1color(red)(cancel(color(black)("mL")))) = "1036 g"##

Now, you know that this solution is ##12.00%## ammonium chloride **by mass**. This means that every ##"100 g"## of solution will contain ##"12.0 g"## of ammonium chloride, ##"NH"_4"Cl"##.

In your case, the sample will contain

##1036 color(red)(cancel(color(black)("g solution"))) * ("12.00 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g solution")))) = "124.32 g NH"_4"Cl"##

Next, use ammonium chloride's **molar mass** to determine how many moles of the compound will be present in this many grams

##124.32 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"Cl")/(53.49color(red)(cancel(color(black)("g")))) = "2.3242 moles NH"_4"Cl"##

This means that the molarity of the solution will be

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

##c = "2.3242 moles"/"1.00 L" = color(green)("2.324 M")##

To get the **mole fraction** of ammonium chloride, you need to know the **total number of moles** present in the solution sample. More specifically, you need to figure out how many moles of water, the , are present in this sample.

To do that, use the mass of the solution and the mass of ammonium chloride

##m_"water" = m_"solution" - m_"ammonium chloride"##

##m_"water" = "1036 g" - "124.32 g" = "911.68 g H"_2"O"##

Use water's molar mass to figure out how many moles can be found in this many grams

##911.68 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015 color(red)(cancel(color(black)("g")))) = "50.607 moles H"_2"O"##

The total number of moles present in solution will thus be

##n_"total" = 2.3242 + 50.607 = "52.931 moles"##

fraction of ammonium chloride, which is equal to the number of moles of ammonium chloride divided by the total number of moles in solution, will be

##chi_(NH_4Cl) = (2.3242 color(red)(cancel(color(black)("moles"))))/(52.931color(red)(cancel(color(black)("moles")))) = color(green)(0.04391)##

Finally, is defined as moles of solute divided by **kilograms** of solvent. Since this sample contain ##"911.68 g"## of water, its molality will be

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

##b = "2.3242 moles"/(911.68 * 10^(-3)"kg") = color(green)("2.549 molal")##

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