# Question 74985

Dec 16, 2015

#### Answer:

Here's what I got.

#### Explanation:

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

Since molarity is defined as moles of solute per liters of solution, a $\text{1.00-L}$ sample will make the calculations easier.

So, let's say that we have a $\text{1.00-L}$ sample of this solution. You can use its given density 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 $\text{100 g}$ of solution will contain $\text{12.0 g}$ of ammonium chloride, $\text{NH"_4"Cl}$.

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

$\textcolor{b l u e}{c = \frac{n}{V}}$

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 solvent, are present in this sample.

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

${m}_{\text{water" = m_"solution" - m_"ammonium chloride}}$

${m}_{\text{water" = "1036 g" - "124.32 g" = "911.68 g H"_2"O}}$

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

${n}_{\text{total" = 2.3242 + 50.607 = "52.931 moles}}$

The mole 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

${\chi}_{N {H}_{4} C l} = \left(2.3242 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(52.931color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{0.04391}$

Finally, molality is defined as moles of solute divided by kilograms of solvent. Since this sample contain $\text{911.68 g}$ of water, its molality will be

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

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

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