Posted by manny on .
Hi,
I need help with this problem I have no idea how to go about it
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An average adult breathes about 8.5 x 10^3 L of air per day. The concentration of lead in higly polluted urban air is 7.0 x 10^6 g of lead per one m3 of air. Assume that 75% of the lead is present as particles less than 1.0 x 10^6 m in diameter, and that 50% of the particles below that size are retained in the lungs. Calculate the mass of lead absorbed in this manner in one year by an average adult living in this environment.
***Note answer has two sigfigs
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Thanks for all the help you guys can provide,
manny

chem 
DrBob222,
How about this hint.
(7.0 x 10^6 g/m^{3}) x 0.75 x 1/2 = amount of consumable lead. Convert 8.5 x 10^{3} L to cubic meters (or the other way aroundit doesn't matter which one you convert) and calculate the amount consumed per day. Then multiply by 365 to convert to amount per year. 
chem 
manny,
ok! thanks... Dr. Bob your the best!!!
I assumed you meant (7.0 x 10^6 g/m3)
I got 8.1 x 10^3 g Pb / year
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