# What is the molarity of glacial acetic acid (CH3COOH, Mr = 60.05g/mol) at 25 degrees C given that the density of acetic acid at that temperature is 1.049g/mL?

##”17.47 M”##

is defined as moles of 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 and therefore is not ctually a solute.

However, you can still calculate its 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 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”https://studydaddy.com/”1.000 L” = color(green)(“17.47 M”)##

I’ll keep the number of given for the density of glacial acetic acid.

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