Answer:
887.89 μg/m³
Explanation:
Given:
Duration of mass flow = 24 hours = 24 × 60 = 1440 minutes
Flowrate = 1.4 m³/min
starting weight of the filter paper before sampling = 2.05 grams
Filter's weight = 3.84 grams
Now,
Mass of the particular matter collected
= Mass of filter after 24 hours - Initial weight of the filter
= 3.84 - 2.05
= 1.79
Now,
The total volume passing through filter = Flowerate × Time
= 1.4 × 1440
= 2016 m³
Particulate matter concentration = ![\frac{\textup{Mass of particulare matter collected}}{\textup{Volume of the sample}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7BMass%20of%20particulare%20matter%20collected%7D%7D%7B%5Ctextup%7BVolume%20of%20the%20sample%7D%7D)
or
Particulate matter concentration = ![\frac{\textup{1.79}}{\textup{2016}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7B1.79%7D%7D%7B%5Ctextup%7B2016%7D%7D)
or
Particulate matter concentration = 0.0008878 g/m³
also,
1 gram = 10⁶ micrograms
thus,
Particulate matter concentration = 0.0008878 × 10⁶ μg/m³
= 887.89 μg/m³