Question

the concentration of suspended particulates in the atmosphere can be measured by forcing air through a semi - porous paper that filters the particulates from the air. determine the suspended particulate concentration (in micrograms/m³) from the measurements presented as follows: airflow rate = 0.71 cubic meters per minute air sample time = 24 hours weight of filter paper before sampling = 4.5722 grams weight of filter paper after sampling = 4.7424 grams.

Explanation:

Step1: Convert sampling time to minutes

Since 1 hour = 60 minutes, 24 hours = 24×60 = 1440 minutes.

Step2: Calculate the total volume of air sampled

The airflow rate is 0.71 cubic - meters per minute. The total volume of air sampled $V$ (in cubic meters) is the product of the airflow rate and the sampling time in minutes. So $V=0.71\times1440 = 1022.4$ cubic meters.

Step3: Calculate the mass of suspended particulates

The mass of suspended particulates $m$ is the difference between the weight of the filter - paper after sampling and before sampling. So $m = 4.7424 - 4.5722=0.1702$ grams. Since 1 gram = 1000000 micrograms, $m = 0.1702\times1000000 = 170200$ micrograms.

Step4: Calculate the suspended particulate concentration

The suspended particulate concentration $C$ (in micrograms per cubic meter) is given by the formula $C=\frac{m}{V}$. Substituting the values of $m$ and $V$ we get $C=\frac{170200}{1022.4}\approx166.5$ micrograms/$m^{3}$.

Answer:

166.5 micrograms/$m^{3}$