Suppose on an average day air contains 6.5 mg of carbon monoxide per 1000 m 3 3 of air. how many milligrams of carbon monoxide are in a room measuring 10 ft x 10 ft x 10 ft? (1m = 3.28 ft). do not use scientific notation.

Answer is: there is 0,184 mg of carbon monoxide in a room.
10 ft = 3,048 m.
V(room) = 3,048 m · 3,048 m · 3,048 m.
V(room) = 28,33 m³.
d (carbon monoxide) = 6,5 mg/1000 m³ = 0,0065 mg/m³.
m(CO) = d(CO) · V(room)
m(CO) = 0,0065 mg/m³ · 28,33 m³.
m(CO) = 0,184 mg.

The answer to your question would be c. mineral oil  

Anyone know the answer?