## Chemistry: Molecular Approach (4th Edition)

$\underline{9.0\,\times \,{{10}^{1}}\,mg}$
(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 $\underline{9.0\,\times \,{{10}^{1}}\,mg}$.