##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_3COOH)/(60.05color(red)(cancel(color(black)(g)))) = 17.469 moles CH_3COOH##
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.