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Question: Volume occupied by 10% alcohol, 20%, 30%, 40%...

 
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Feb21-04, 04:32 PM   #1
 

Question: Volume occupied by 10% alcohol, 20%, 30%, 40%...


I have looked for months in handbooks, online, in encyclopaedias and so on, tried piecing things together fro mvarious sources, and I still cannot find this. What I basically wish for is a list that tells me what volume a certain mass of a solution of alcohol occupies at 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90%, 100%. The mass can be a kilogramme or a pound or a gramme or whatever. I know that the relative density of 100% alcohol is 0.79, but do not know of the rest.

Please could somebody help me, as this information is relatively important to me. Thank you and cheers for any help offered :)

-Bryan
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Feb21-04, 10:37 PM   #2
 
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International Critical Tables (old, but no less accurate than today's measurements); failing that, hit whatever abstract/citation indices are available in the library for "systems: water-_____ol."
Feb22-04, 05:16 PM   #3
 
Mentor
With the info you have, you can build yourself that table in Excel in about 30 seconds. Don't use units, but assuming you mean alcohol percentage mass vs volume, the (unitless) equation is (someone check me):

volume = .79*(mass fraction of alcohol) + (1-mass fraction of alcohol)
Feb22-04, 05:24 PM   #4
 

Question: Volume occupied by 10% alcohol, 20%, 30%, 40%...


Originally posted by russ_watters
With the info you have, you can build yourself that table in Excel in about 30 seconds. Don't use units, but assuming you mean alcohol percentage mass vs volume, the (unitless) equation is (someone check me):

volume = .79*(mass fraction of alcohol) + (1-mass fraction of alcohol)
I am not quite sure I follow what you are saying. Could you give this to me referring to measuring units so I can get this point. Thank you :)
Feb23-04, 06:29 AM   #5
 
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Failing Excel, you could try doing the arithmetic yourself!

You have already told us that you know that alcohol has a relative density of 0.79. Okay, a 30% solution of alcohol consists of 0.3 alcohol and 0.7 water. (Although you didn't say if that was 50% by weight or by volume-those are two completely different things. I'll assume "by volume" which is more common.)

You are seeking the volume of this solution that will have a mass of 1 (kg, say). Let X be the volume of Water and Y the volume of alchol. This is a 30% solution so Y/(X+Y)= 0.30. Since the density of alcohol is 0.79 times the density of water, the mass of the solution is (0.79Y+ X)times the density of water= 1. If I recall correctly the density of water is 1 gram/cm3 so 1 kg/liter. Using that: 0.79Y+ X= 1.

Now solve the two equation: Y/(X+Y)= 0.30 (which is the same as
Y= 0.3X+ 0.3Y or 0.7Y= 0.3X) and 0.79Y+ X= 1.

For a general percentage of alcohol, P, the same analysis gives the two equations as (1-P)Y= PX and 0.79Y+ X= 1. Solve those two equations for X and Y.
Feb23-04, 06:56 AM   #6
 
Originally posted by HallsofIvy
Failing Excel, you could try doing the arithmetic yourself!

You have already told us that you know that alcohol has a relative density of 0.79. Okay, a 30% solution of alcohol consists of 0.3 alcohol and 0.7 water. (Although you didn't say if that was 50% by weight or by volume-those are two completely different things. I'll assume "by volume" which is more common.)

You are seeking the volume of this solution that will have a mass of 1 (kg, say). Let X be the volume of Water and Y the volume of alchol. This is a 30% solution so Y/(X+Y)= 0.30. Since the density of alcohol is 0.79 times the density of water, the mass of the solution is (0.79Y+ X)times the density of water= 1. If I recall correctly the density of water is 1 gram/cm3 so 1 kg/liter. Using that: 0.79Y+ X= 1.

Now solve the two equation: Y/(X+Y)= 0.30 (which is the same as
Y= 0.3X+ 0.3Y or 0.7Y= 0.3X) and 0.79Y+ X= 1.

For a general percentage of alcohol, P, the same analysis gives the two equations as (1-P)Y= PX and 0.79Y+ X= 1. Solve those two equations for X and Y.
Oh my God. [8)] [:D] A very interesting and informative response. Physics, maths etc are not really my strong point- I am an English student. Regardless I shall attempt this as I need the info...

Are you saying to me that the volume occupied by 1 kilogramme of a solution that is 30% alcohol is equal to 1080 millilitres? That is my figure, anyway.
Erm, that is, pure alcohol would be 1/0.79 * 1000ml = 1266ml. This 266ml difference then * 0.3 = 80ml (plus the original 1000ml)

Is this right?


So a 75% alcohol solution, for instance, which weighs a kilogramme, will have a volume of 1200ml.
So a 40proof solution which weighs a pound will have a volume of 29.15cuin.


[?] [?] [?]
Feb23-04, 11:49 AM   #7
 
Mentor
Originally posted by Bryan Parry
I am not quite sure I follow what you are saying. Could you give this to me referring to measuring units so I can get this point. Thank you :)
Halls (the math wiz) gave a great explanation of the derivation, but for the units thing, apply whatever units you want using the appropriate conversion. Metric is easiest, since 1cc of water is 1g.

And yeah - looks like you understand the equations.
Feb23-04, 12:02 PM   #8
 
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Are you guys sure of the assumption that volume is conserved when alcohol and water are mixed?
Feb23-04, 12:16 PM   #9
 
Originally posted by NateTG
Are you guys sure of the assumption that volume is conserved when alcohol and water are mixed?
Oh dear... so what is the problem now? :D
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