Question

Question #9b21a

Answer 1

`"17.47 M"`

Explanation:

Molarity is defined as moles of solute per liters of solution.

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

Glacial acetic acid is actually anhydrous acetic acid, which implies that the acetic acid is not actually dissolved in a solvent and therefore is not ctually a solute.

However, you can still calculate its molarity based on the number of moles of you get per liter of glacial acetic acid.

To do that, start with a sample o `"1.000 L"` of glacial acetic acid. You know that at `25^@"C"`, glacial acetic acid has a density of `"1.049 g/mL"`, which tells you that every mililiter of glacial acetic acid has a mass of `"1.049 g"`.

This means that the mass of the sample will be

`1.000color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1.000color(red)(cancel(color(black)("L")))) * "1.049 g"/(1color(red)(cancel(color(black)("mL")))) = "1049 g"`

To find how many moles you get in the sample, use the given molar mass, which tells you how many grams of acetic acid you get per mole

`1049color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) = "17.469 moles CH"_3"COOH"`

Now that you know the volume of the sample and how many moles it contains, you can say that

`c = "17.469 moles"/"1.000 L" = color(green)("17.47 M")`

I'll keep the number of sig figs given for the density of glacial acetic acid.