Calculating Molarity of Ethyl Alcohol in Whiskey

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SUMMARY

The molarity of ethyl alcohol (CH3CH2OH) in eighty-six proof whiskey, which is 43% ethyl alcohol by volume, is calculated to be 7.37 M. The calculation incorporates the density of ethyl alcohol at 0.79 kg/L and the molar mass of ethyl alcohol at 46.07 g/mol. The initial attempt of 17.15 M was incorrect due to the omission of the 43% concentration factor. The correct formula applied is 0.79 kg/L * 0.43 L / 1 L * 1000 g / 1 kg * 1 mol / 46.07 g.

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Homework Statement


Eighty-six proof whiskey is 43 percent ethyl alcohol, CH3CH2OH, by volume. If the density of ethyl alcohol is 0.79 kg/L, what is the molarity in whiskey.


Homework Equations


Molar Mass of CH3CH2OH = 46.07 g


The Attempt at a Solution



0.79 kg/L * 1000 g/kg * 1 mole / 46.07 g = 17.15 M

I was just looking for a check on this particular problem since I haven't taken a chemistry course since my freshman year ha!
 
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Not bad - you are on the right track - but wrong. You have not used 43% in your calculations and this is an important information.
 
Last edited by a moderator:
I thought you would have to use that 43% in there somewhere, but I'm not sure how to use it. It would be nice if I still owned my chem book.

UPDATE:

Ok I tried re-working the problem again, here's what I came up with:

0.79 kg/L * 0.43 L / 1 L * 1000 g / 1 kg * 1 mol / 46.07 g = 7.37 M

how does that look?
 
Last edited:
Much better.
 

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