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?

Question

asked Jun 3, 2019 207k views

1 Answer

Melix · Answer 1 · 2019-06-07T02:49:19+0000

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 .