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QUESTION

# 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"/"1.000 L" = color(green)("17.47 M")

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