The dimensions of room are [tex](11.5\times 12\times 20.5) ft.[/tex]. Thus, volume of room will be:
[tex]V=(11.5\times 12\times 20.5) ft^{3}=2829 ft^{3}[/tex]
Converting [tex]ft^{3}[/tex] to [tex]m^{3}[/tex]
Since, [tex]1 ft^{3}=0.02831 m^{3}[/tex]
Thus, [tex]2829 ft^{3}=(0.02831\times 2829)m^{3}=80.08 m^{3}[/tex]
Now, concentration of carbon monoxide in the urban apartment is [tex]48 \mu g/m^{3}[/tex], thus, mass of carbon monoxide can be calculated as follows:
[tex]C=\frac{m}{V}[/tex]
Rearranging, [tex]m=C\times V[/tex]
Putting the values,
[tex]m=(48\mu g/m^{3})\times (80.08 m^{3})=3.84\times 10^{3}\mu g[/tex]
converting [tex]\mu g\rightarrow g[/tex]
Since, [tex]1\mu g=10^{-6}g[/tex] thus,
[tex]3.84\times 10^{3}\mu g=3.84\times 10^{3}\times 10^{-6}=3.84\times 10^{-3}g[/tex]
Therefore, mass of carbon monoxide present in the room is [tex]3.84\times 10^{-3}g[/tex].
