Lindeburg: Toxicology

[tucents]

All,

In the Practice Problems book, Chapter 45: Toxicology Problem # 2, there are 2 workers exposed to trichloroethylene. He gives a concentration of 0.6 ppm in the problem statement, and then it materializes in the solution as 3.22 mg/cu. m.

My question is: :smileyballs: HOW? Anybody have any insight on this? He does state the MW 131.4 g/mol, but I dont see this being relevant at this point.

Thanks in advance!

Henry

 

[mcap8]

All,
In the Practice Problems book, Chapter 45: Toxicology Problem # 2, there are 2 workers exposed to trichloroethylene. He gives a concentration of 0.6 ppm in the problem statement, and then it materializes in the solution as 3.22 mg/cu. m.

My question is: :smileyballs: HOW? Anybody have any insight on this? He does state the MW 131.4 g/mol, but I dont see this being relevant at this point.

Thanks in advance!

Henry

I don't have the problem in front of me, but from the information you provided and assuming standard conditions, a form of the ideal gas law will get you to 3.22 mg/m^3. Cmass = (Cppm*MW)/24.45.

 

[Wyatt72]

Yep that appears to be correct he just does the ppm to mass/volume conversion. Lindeburg can be a pain with problems like that when they dont fully explain the answers.

 

[tucents]

Gents,

You're both scholars! I appreciate your prompt response and feedback.

Thanks and enjoy the rest of your weekend.

Henry

 

[Guest]

That answer is true as long as you assume (know) that pressure = 760 mg Hg and Temp = 25C

If you are not at those STP conditions, you have to incorporate the Ideal Gas Law to correct for non-STP conditions; therefore,

Cmass = {P (mm Hg) x MW x Cppm } / {62.4 x (273.2 + T (deg C))}

JR

 

[RIP - VTEnviro]

Questions like that are all just manipulations of the various forms of the ideal gas law. I bet I got most of those wrong on test day.