Question

An air sampling procedure for benzene (MW = 78.1) vapor in air is conducted using activated charcoal and calibrated vacuum pump. The air flow rate was 0.001 mº/min and the total sampling time was 24 hours. GC-MS analysis of the activated charcoal indicated that a total of 315 pg of benzene was collected on the sorbent. a) Calculate the average air concentration in pg/m?) of benzene. b) Assuming normal temperature and pressure (25°C and 1 atm), convert the air concentration in pg/ m to ppb.

Please Help!!

Answer 1

a) Air flow Rate = 0.001m³/min

Total time = 24hours= 24*60mins = 1440mins

Thus, Total air volume= air flow rate* total time = 0.001*1440 = 1.44m

Now, total benzene collected = 315ug

Thus average air concentration = mass/volume = 315/1.44 = 218.75 ug/m³(Ans)

b) The general equation for conversion is μg/m3 = (ppb)*(12.187)*(M_w) / (273.15 + C) , where M_w is the molecular weight

Thus, at 25C, 1 ppb of benzene = 3.19 μg/m³ , Molecular weight of Benzene = 78.11

Thus, 3.19μg/m³ = 1ppb

Thus, 218.75 μg/m³ = 218.75/3.19 = 68.57ppb (Ans)