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Report (any type)/Brief report
22/09/2021
##”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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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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