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Topic: Atmospheric Pressure Correction  (Read 14352 times)

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Offline mspike6

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Atmospheric Pressure Correction
« on: September 23, 2010, 07:33:44 PM »
Hey !

I got this question for a lab report... it looks so easy, and i got an answer, but for some reason am doubting this answer 1 so if someone can confirm it to me or give me some Hints, that would be great .

"5.  Using the Boiling point correction formula, determine the boiling point of Isopropyle Acohol at 722 mm HG. ( Boiling point at 1 atm is 83 C )"

So here is what i did :

The formula is :
BP = T - (P- 760)0.037
Where BP is Boiling Point at standard atmospheric pressure ( 760 mmHG/1atm)
T = Observed Boiling point
P = Observed Atmospheric pressure (in mmHG)


So i know that BP = 83, and we need T when P=722
So  83= T-(722 - 760) * 0.037
solve for T and T = 81.59 C

i am doubting this because the difference is so small, so... i kinda dont know if i calculated it wrong , or if i got the wrong idea ?

i also have an other question that i dont understand at all...i literally dont understand what he is asking for

4.  Using the Information from the website (i will Copy past all the info in a sec), what is the Mass % of the Unkown Liquid /Water Mixture ? (Note if your Density indicates a mass % lower than 0%, answer as 0%. If your density indicates a mass % that is greater than 100% answer as 100% )

Here are the "information" in the website

Observed Atmospheric Pressure:

Sept 15th: 711.7 mmHg

 
Identification of Unknown:

The chemical in your unknown may be any of the following. Identify the chemical by the boiling point. (Note that water in your sample will affect the boiling point.)

 

Substance                              Boiling Point ∘C

Acetone                                  56

Isopropyl alcohol                    83

Methyl alcohol                       65

Water                                      100
--------------------------------------------------------------------------

My "Unkown Substance Turned out to be Isopropyle Alcohol, with Density 0.977 g/mL3






Thanks guys, any help is really really appreciated



Edit:  This Isopropyle Smelled like a really really good whiskey ...i am surprised that i didn't take a sip here or there :D


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