Answer: 110 mg/L.
- A 110 mg/L
- B 5.5 mg/L
- C 55 mg/L
- D 275 mg/L
Correct answer: A. 110 mg/L
Explanation: DO consumed by the diluted sample = 9.0 − 3.5 = 5.5 mg/L. Since this is the demand of the 1/20 diluted sample, the original water's BOD = 5.5 × 20 = 110 mg/L. This severely exceeds the 17 ppm threshold for very polluted water, indicating raw sewage or highly concentrated industrial effluent.
Classical smog is a reducing mixture of SO₂ and fog, while photochemical smog is an oxidising mixture generated by UV-driven reactions of NOx and hydrocarbons.
Concept context
Understand how human activities alter the atmosphere, water, and soil. Covers air pollutants, smog types, ozone depletion, acid rain, greenhouse effect, water and soil pollution, and the principles of green chemistry.