Calculating Molarity of Ethyl Alcohol in Whiskey

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Discussion Overview

The discussion revolves around calculating the molarity of ethyl alcohol in whiskey, specifically focusing on an 86 proof whiskey that is 43 percent ethyl alcohol by volume. The participants explore the application of density and percentage concentration in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant initially calculates the molarity of ethyl alcohol without considering the 43% concentration, arriving at a value of 17.15 M.
  • Another participant points out the oversight regarding the 43% concentration, indicating its importance in the calculations.
  • A later reply expresses uncertainty about how to incorporate the 43% concentration, reflecting a lack of recent chemistry knowledge.
  • The participant reworks the problem, incorporating the 43% concentration and arrives at a new molarity of 7.37 M, seeking feedback on this revised calculation.
  • Another participant acknowledges the revised calculation as an improvement.

Areas of Agreement / Disagreement

Participants generally agree that the 43% concentration is crucial for the calculation, but there is no consensus on the initial calculation method or the correctness of the final value presented.

Contextual Notes

The discussion highlights the dependence on the correct application of percentage concentration and density in molarity calculations. There are unresolved aspects regarding the initial calculation method and the assumptions made in the reworked solution.

Jim4592
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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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