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Forum Rules: Read This Before Posting Topic: Ideal gases: calculating Mol ratio  (Read 3534 times)

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• Mole Snacks: +0/-0 Ideal gases: calculating Mol ratio
« on: May 01, 2006, 08:03:12 PM »
Greetings,

I am having difficulty with calculating the mol ratio of argon (MW 39.95) to nitrogen (28.01) if the calculated MW is 28.15.  This problem is taken from Zumdahl's Chemical Principals 2nd Ed. (Chapter 5 #29) and the answer is given at the back of the book as 1.18 X 10^-2.  Can someone nudge me in the right direction on how to arrive at this response?  Thanks! Borek Re: Ideal gases: calculating Mol ratio
« Reply #1 on: May 01, 2006, 08:32:13 PM »
Try with molar fractions.

Assuming there is x of one and (1-x) of another, calculated molar mass is 39.95x + (1-x)*28.01.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info tennis freak Re: Ideal gases: calculating Mol ratio
« Reply #2 on: May 09, 2006, 09:43:07 PM »
to do this wouldnt you need to know how many moles of each there were in the reaction? and where does that mass come from anyway?
AP Chemistry Squad Member 

Don't be afraid to try something new, remember amateurs built the ark and professionals built the titanic! Borek Re: Ideal gases: calculating Mol ratio
« Reply #3 on: May 10, 2006, 03:57:48 AM »
There is no reaction here, there is a mixture of two gases that behaves like gas of given molar mass.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info Donaldson Tan Re: Ideal gases: calculating Mol ratio
« Reply #4 on: May 11, 2006, 08:51:13 AM »
let mol% of Argon be x

=> mole% of Nitrogen = 1 - x

Average Molar Mass = x*(molar mass of Argon) + (1-x)*(molar mass of N2)

Solve for x
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