Answer
\[\underline{9.0\,\times \,{{10}^{1}}\,mg}\]
Work Step by Step
(1) Calculate the volume of air inhaled by an average person per minute:
\[\begin{align}
& \text{Inhaled air per minute= Volume of CO inhaled per breath}\,\,\times \,\text{number}\,\text{of}\,\text{breaths}\,\text{per}\,\text{minute} \\
& \text{ }\left( \text{0}\text{.50 L/breath} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{20 breaths/min} \right) \\
& \text{= 10 L/min }
\end{align}\]
(2) Convert the volume of air inhaled by an average person per minute into per hour:
Since, \[60\ \ \min \,\,=\,\,1\,\,hour\] the volume of air inhaled by an average person per hour is : \[10\ \frac{L}{\min }\,\,\times \,\frac{60\,\min }{1\,\,hour}\,\,=\,600\,L/hour\]
(3) Calculate the volume of air inhaled by an average person in \[8.0\,hour\]
The volume of air inhaled by an average person in \[8.0\,hour\]:\[\begin{align}
& =\,600\,\frac{L}{hour}\,\times \,8.0\,hour \\
& =\,\,4,800\,L \\
\end{align}\]
(4) Calculate the volume of \[\text{CO}\] inhaled by an average person in \[8.0\,hour\] :
\[\begin{align}
& \because \text{1}{{\text{0}}^{6}}\text{ L air}\,\text{inhaled contains = 15}\text{.0 L CO} \\
& \therefore \text{4800 L air inhaled contains}=\text{ 15}\text{.0 L }\left( \frac{\text{4800 L}}{\text{1}{{\text{0}}^{6}}\text{ L}} \right)\text{CO}\,\,\text{=}\,\text{ 0}\text{.072 L CO}
\end{align}\]
(5) Density (d) is defined as mass per unit volume.
\[d=m/V\]
Here, m is the mass of \[\text{CO}\]and V is the volume of \[\text{CO}\].
Rearrange this equation to calculate the mass of \[\text{CO}\].
\[m=V\times d\] (1)
Calculate the mass of \[\text{CO}\] by substituting the values in equation (1) as follows:
\[\begin{align}
& m=V\times d \\
& =0.072\text{ }L\times 1.2\text{ g/L} \\
& =0.0864\,\,g
\end{align}\]
Convert this mass of \[\text{CO}\]into milligrams by the use of the following relation:
\[1\ g\,\,=\,\,1000\,mg\]
\[\begin{align}
& Thus,in\text{ }milligrams=0.0864\,\,g\,\,\times \,\frac{1000\ mg}{1\,\,g} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,86.4\,mg \\
\end{align}\]
Round of the answer to \[9.0\,\times \,{{10}^{1}}\,mg\].
Amount of \[\text{CO}\] inhaled by an average person in \[\text{8 h}\] is \[\underline{9.0\,\times \,{{10}^{1}}\,mg}\].
