Question

The concentration of carbon monoxide in an urban apartment is 48 μg/m3. what mass of carbon monoxide in grams is present in a room measuring 11.5 × 12.0 × 20.5 ft?

Answer 1

The dimensions of room are
`(11.5* 12* 20.5) ft.`. Thus, volume of room will be:

`V=(11.5* 12* 20.5) ft^(3)=2829 ft^(3)`

Converting
`ft^(3)` to
`m^(3)`

Since,
`1 ft^(3)=0.02831 m^(3)`

Thus,
`2829 ft^(3)=(0.02831* 2829)m^(3)=80.08 m^(3)`

Now, concentration of carbon monoxide in the urban apartment is
`48 \mu g/m^(3)`, thus, mass of carbon monoxide can be calculated as follows:

`C=(m)/(V)`

Rearranging,
`m=C* V`

Putting the values,

`m=(48\mu g/m^(3))* (80.08 m^(3))=3.84* 10^(3)\mu g`

converting
`\mu g\rightarrow g`

Since,
`1\mu g=10^(-6)g` thus,

`3.84* 10^(3)\mu g=3.84* 10^(3)* 10^(-6)=3.84* 10^(-3)g`

Therefore, mass of carbon monoxide present in the room is
`3.84* 10^(-3)g`.