Solve CCl3F & CCl2F2 Molar Concentration in Tropo/Stratosphere

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SUMMARY

The discussion focuses on calculating the molar concentration of CCl3F and CCl2F2, two chlorofluorocarbons linked to ozone depletion, under specific atmospheric conditions. The gases were found in quantities of 261 ppt and 509 ppt, respectively, as of 1994. The ideal gas law (pV=nRT) is utilized to determine the number of moles of air in a liter, allowing for the calculation of the moles of CFCs based on their ppt ratio. The key takeaway is that 1 part per trillion by volume translates directly to a ratio of molecules in a given volume of air.

PREREQUISITES
  • Understanding of the ideal gas law (pV=nRT)
  • Knowledge of molar mass calculations
  • Familiarity with parts per trillion (ppt) as a concentration unit
  • Basic principles of atmospheric chemistry
NEXT STEPS
  • Research the impact of chlorofluorocarbons on ozone depletion
  • Learn about the calculation of molar concentrations in different atmospheric conditions
  • Explore the significance of parts per trillion in environmental science
  • Study the properties and uses of CCl3F and CCl2F2 in industrial applications
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Environmental scientists, atmospheric chemists, students studying chemistry or environmental science, and anyone interested in the effects of CFCs on the ozone layer.

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Thermo Question!

Homework Statement



CCl3F and CCl2F2 have been linked to ozone depletion in Antarctica. As of 1994, these gases were found in quantities of 261 and 509 parts per trillion (1012) by volume (World Resources Institute, World resources 1996–97). Compute the molar concentration of these gases under conditions typical of (a) the mid-latitude troposphere (10°C and
1.0 atm) and (b) the Antarctic stratosphere (200 K and 0.050 atm).

Homework Equations


pV=nRT


The Attempt at a Solution


Okay ppt is hurting my head. Basically ppt = 1ng/L

CCl3F MW = 137.37 g/mol

So I did (137.37 g/mol)/(251g/1x10^9L )to get mol/L

But I end up getting 0.526 nL/mol..

So when I go to plug in n = pV/RT to solve for my molecular weight concentration I don't know how the units are going to cross out... because its 0.526 nL/mol..
Can I just multiply 0.526(1x10^-9nL= to get L? Or is that against multiplication rules. I've never really dealt with ppt.
 
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Hello Workout,

Just stick to the definition of ppt given in your problem statement. 1 part per trillion is exactly what is says. It's just a RATIO. So, for every trillion molecules of air*, there is 1 molecule of the chloro-fluoro-carbon, or whatever that is (I'm not a chemist).

So what you have to do is use the ideal gas equation to compute how many moles of air you have in a volume of 1 litre under each set of atmospheric conditions. Based on that number of moles of air, you can figure out the number of moles of the CFC in question (because you know the ratio in ppt). This will give you, directly, the number of mol/L of the CFC.

*note: it was 1 part per trillion by volume, not by mass. Since, in a gas at a given temperature and pressure, the volume occupied is proportional to the number of particles, this means that it's one litre for every trillion litres, or equivalently, one particle for every trillion particles. In contrast, if it had been one ppt BY MASS, then it would not be one particle for every trillion particles, it would have been one gram for every trillion grams. If the substance is heavier than air, this would mean *fewer* than one particle for every trillion particles.
 

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