Question

At a pressure of 1 atmosphere and temperature of 25\(^{\circ}\)C, 365 µg m\(^{-3}\) of a pollutant corresponds to a mixing ratio of 139 parts per billion (ppb). The atomic weights: C – 12, H – 1, O – 16, N – 14 and S – 32. Which one of the following options most closely represents the pollutant?

• A. SO\(_2\)
• B. NO\(_2\)
• C. O\(_3\)
• D. CO

Hint:

To identify gaseous pollutants using µg/m\(^3\) and ppb values, apply: \[ \mu g/m^3 = \frac{{ppb} \times {Molecular Weight}}{24.45} \] This formula assumes standard temperature (25°C) and pressure (1 atm). Always cross-check the molecular weight.

Answer 1

The Correct Option is A

Step 1: Use the formula to relate ppb and µg/m³.
At 25°C and 1 atm, the concentration in µg/m³ is given by:
\[ \text{Concentration} = \frac{{\text{ppb}} \times {\text{Molecular Weight}}}{24.45} \] Substitute the known values:
\[ 365 = \frac{139 \times M}{24.45} \Rightarrow M = \frac{365 \times 24.45}{139} \approx \frac{8924.25}{139} \approx 64.20 \] Step 2: Match the computed molecular weight with known pollutants.
SO₂: \(32 + 2 \times 16 = 64\)
NO₂: \(14 + 2 \times 16 = 46\)
O₃: \(3 \times 16 = 48\)
CO: \(12 + 16 = 28\)

Conclusion:
The calculated molecular weight is approximately 64, which matches with SO₂. Hence, the pollutant is SO₂.