# Question ff5e9

Sep 24, 2015

$\text{8.9665 molal}$

#### Explanation:

The idea here is that you need to take a sample of your solution and calculate its mass by using the given density.

Once you know the mass of the solution, you can find the mass of acetic acid it contains by using the solution's molarity. This will get you the mass of water, and thus the solution's molality.

So, let's say that we take a $\text{1.0000-L}$ sample of this solution.

Since one liter is equal to one cubic decimeter, you can use the solution's density to find its mass

1.0000color(red)(cancel(color(black)("dm"^3))) * "1.0438 g"/(1color(red)(cancel(color(black)("cm"^3)))) * ("1000 cm"""^3)/(1color(red)(cancel(color(black)("dm"^3)))) = "1043.8 g"

You know that this solution has a molarity of $\text{6.0835 M}$, which means that it contains

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{acetic acid" = 6.0835"moles"/color(red)(cancel(color(black)("L"))) * 1.00color(red)(cancel(color(black)("L"))) = "6.0835 moles HAc}}$

Molality is defined as moles of solute, in your case acetic acid, per kilogram of solvent.

Use acetic acid's molar mass to determine how many grams would contain this many moles

6.0835color(red)(cancel(color(black)("moles HAc"))) * "60.052 g HAc"/(1color(red)(cancel(color(black)("mole HAc")))) = "365.33 g HAc"

This means that the solution must contain

${m}_{\text{sol" = m_"HAc" + m_"water}}$

${m}_{\text{water" = m_"sol" - m_"HAc}}$

${m}_{\text{water" = "1043.8 g" - "365.33 g" = "678.47 g water}}$

The molality of the solution will thus be

$b = {n}_{\text{HAc"/m_"water in kg}}$

b = "6.0835 moles"/(678.47* 10^(-3)"kg") = color(green)("8.9665 molal")#

I'll leave the answer rounded to five sig figs, the number of sig figs you gave for the density and molarity of the solution.