Question

Carbon monoxide (CO) detectors sound an alarm when peak levels of carbon monoxide reach 100 parts per million (ppm). This level roughly corresponds to a composition of air that contains 400,000 /(/mu/)g carbon monoxide per cubic meter of air (400,000 /(/mu/)g/m3). Assuming the dimensions of a room are 18 ft x 12 ft x 8 ft, estimate the mass (in grams) of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector. (1 meter = 3.28 feet)

Answer 1

Answer:

19.61kg

Explanation:

The first thing to do is divide the amount of feet into 3.28:

$18ft*12ft*8ft=5.49m*3.66m*2.44m$

With this information, we calculate the volume:

$V=5.49m*3.66m*2.44m=49.03m^{3}$

As we have the concentration and the volume, we can calculate the mass of CO:

$m=400,000 \mu g/m^{3}*49.03m^{3}=19611158.4 \mu g$

As this is a very big number, we can express it in kg by dividing in 10^6:

$19611158.4 \mu g/10^{6}=19.61 kg$