# Question #269cf

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to pick a sample of this solution and use its *molarity* to find how many **grams** of solute, which in this case is acetic acid,

Once you know the mass of the solute, you can use the *density* of the solution to find the mass of solvent, which is water.

To keep the calculations simple, pick a **moles** of acetic acid.

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

#2color(red)(cancel(color(black)("moles CH"_3"COOH"))) * "60.05 g"/(1color(red)(cancel(color(black)("mole CH"_3"COOH")))) = "120.1 g"#

Now, a density of **every**

#1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.2 g"/(1color(red)(cancel(color(black)("mL")))) = "1200 g"#

The mass of the solution is equal to

#m_"solution" = m_"solute" + m_"solvent"#

This sample will thus contain

#m_"solvent" = "1200 g" - "120.1 g" = "1079.9 g water"#

Now, **molality** is defined as moles of solute **per kilogram of solvent**. Convert the mass of water from *grams* to *kilograms* and plug in your values to find

#b = "2 moles"/(1079.9 * 10^(-3)"kg") = color(green)(|bar(ul(color(white)(a/a)"1.9 mol kg"^(-1)color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**.

**SIDE NOTE** *The molaity of the solution, much like its molarity, must be the same regardless of the sample you pick. I recommend starting with different volumes of the solution to see that this is indeed the case.*