Question

3,4-benzopyene (BP, C20H12) is a toxic aromatic hydrocarbon frequently found in polluted air. 5.0 liters of air (at standard temperature and pressure) is bubbled through 10 ml of dilute sulfuric acid, capturing the BP. 3.0 ml of this solution is diluted to 10 ml, and this final solution gives a fluorescence intensity of 205 (arbitrary units). A blank gave an intensity of 35 (same units). A solution containing 0.75 mg of BP per ml is prepared and 1.5 ml of the this solution is added to 3.0 ml of the initial BP solution, prior to dilution to 10 ml. This solution gives a fluorescence intensity of 294 (arbitrary units).

a) Calculate the mass of BP per liter of air.

b) Express this answer in terms of ppm and ppmv.

Answer 1

a) If the mass of BP in 5L of air is xmg that amount will get extracted in to 10mL of diluted sulphuric acid.

From that 3mL is taken and diluted into 10mL

Therefore the concentration of BP in final diluted solution =(3x/10)/10 mg/ml

In fluorescence the concentration is dirctly proportional to the intesity of the fluorescence obtained

the intensity of the sample=205-35=170

   equation 1

the amount of BP in the next mixture = the mass of BP from the standard solution + the mass of BP from 3mL of the sample

                                                      =0.75*1.5+3x/10=1.125+3x/10

the concentration of BP in this solution=(1.125+3x/10)/4.5mL

the intensity of fluoresence of this solution=294-35=259

equation 2

equation1/equation 2

/

3x*4.5/(1.125+3x/10)=170/259

259*4.5*3x=1.125*170+3x/10*170

3496.5x=191.25+51x

3445.5x=191.25

x=18mg

therefore the mass of BP in 5L of air = 18mg

mass of BP per L=18/5=3.6mg/L

b)in ppm=3.6ppm

in ppmv

ppmv=mg/m³ *0.08205*T/M

ppmv=parts per million in volume

mg/m³ =mg of pollutant per m3 at sea level in atmospheric pressure and T

T=ambient temperature in K

0.08205=universal gas constant in at,.m3/(kmol.K)

M=molecular mass of the air pollutant

M=252g/mol

concentration of BP in mg/m³   =3600mg/m³

ppmv=3600mg/m³ .0.08205.298K/252g/mol

       =349.3ppmv