Question

A particular air sample at 1 atm pressure has a density of ~ 1.3 g/L and contains SO2 at a concentration of 25 ug/m3. What is the concentration in terms of the following units: a. Parts per million (mass SO2 per 106 units mass of air)? b. Moles SO2 per 100 mole of air?

Answer 1

Answer:

a. 0.0192 ppm, and b. 8.70×10^{-7}

Explanation:

a. Calculation of parts per million (PPM) of SO2

The concentration of SO2 according to the question is 25 micrograms (ug) per 1 cubic meter (m3), so we need to convert both values to the same mass unit in order to calculate the parts per million. In this case, we will convert both values to grams:

25 ug = 25×10^{-6} g ......(i)

1 m3 = 1,000 L, but the density of the air is 1.3 g/L, so 1 L weighs 1.3 g. Using a rule of three:

1 L → 1.3 g

1,000 L → ?

? = 1,300 g

So, 1 m3 (cubic meter) of air weighs 1,300 g ......(ii)

Now we can calculate the parts per million by dividing the number of grams of SO2 (i) times the number of grams of air (ii) times 10^{6}:

PPM = (25×10^{-6} g)/(1,300 g)×10^{6} = 0.0192 ppm

b. Calculation of moles of SO2 per 100 mol of air

As we do not have any temperature mentioned in the question, we can not use the ideal gas equation to calculate the number of moles of air. So we will use the average weight of a mol of air, that is 28.97 g/mol

From (ii) we know that 1 m3 weighs 1,300 g, so we can apply a rule of three to get the number of moles we have in 1 m3:

28.97 g → 1 mol

1,300 g → ? moles of air

? = 1,300/28.97 = 44.874 moles of air

Also, from (i) we know that in 1 m3 of air we have 25×10^{-6} g of SO2. This amount can be expressed in moles by using the molar mass of SO2 that is 64 g/mol (S=32, O=16 and 32+2×16 = 64), using a rule of three:

64 g → 1 mol

25×10^{-6} g → ? moles of SO2

? = 3.91×10^{-7} moles of SO2

Now, we can calculate the moles of SO2 per 100 moles of air by doing the following:

moles of SO2 per 100 moles of air = (3.91×10^{-7})/(44.874)×100 = 8.70×10^{-7}