Question

Question #269cf

Answer 1

`"1.9 mol kg"^(-1)`

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, `"CH"_3"COOH"`, it contains.

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 `"1-L"` sample of the solution. Since the solution is said to have a molarity of `"2 mol L"^(-1)`, it follows that this sample will contain `2` 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 `"1.2 g mL"^(-1)` tells you that you that every `"mL"` of solution has a mass of `"1.2 g"`. This means that the sample you picked will have a mass of

`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.