Section 1.7 showed that in 1997 los angeles county air had carbon monoxide (co) levels of 15.0 ppm. an average human inhales about 0.50 l of air per breath and takes about 20 breaths per minute. part a how many milligrams of carbon monoxide does the average person inhale in an 6 hour period for this level of carbon monoxide pollution? assume that the carbon monoxide has a density of 1.2 g/l. (hint: 15.0 ppm co means 15.0 lco per 106l air.)

First, let's determine the amount of air a human intakes per minute:

0.5 L/breath * (20 breaths/min) = 10 L/min

So, for 6 hours, the amount of volume would be:

6 hrs = 360 min
10 L/min ( 360 min = 3,600 L of total air inhaled

Next, we convert the concentration to g/L. The solution is as follows:
15 L CO/10⁶ L air * 1.2 g CO/L = 18 g CO/L air
So, we multiply to this the volume of air inhaled.

Amount of CO inhaled = (18 g CO/L air)(3,600 L of total air)(1,000 mg/1 g)
Amount of CO inhaled = 64.8×10⁶ mg of CO inhaled

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