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The concentration of carbon monoxide in an urban apartment is 48μg/m3. What mass of carbon monoxide in grams is present in a room measuring 10.6ft×14.8ft×20.5ft?

1 Answer

5 votes

Answer:

m =
4.37 * 10^(-3) g

Step-by-step explanation:

Volume = 10.6 × 14.8 × 20.5 ft = 3216.04 ft³

1 feet³ = 0.0283168 meter ³

Now convert feet³ to meter³

Volume = 3216.04 ft³ × (3.048/1 m³/ft³) = 91.06811131 m³

This question gives the density as 48μg/m3

μ =
10^(-6)

The formula for density is
p = (m)/(V)


48*10^(-6) = (m)/(91.06811131)

Solve for m

m =
4.37 * 10^(-3) g

User Antonio Espinosa
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