If 10 mL of raw sewage is diluted to 250 mL and the dissolved oxygen decreases from 9 mg/L to 5 mg/L after 5 days at 20°C, the BOD5 of raw sewage is
- A4 mg/L
- B10 mg/L
- C40 mg/L
- D100 mg/LCorrect
Solution
Concept
BOD5 dilution calculation
Formula
BOD5 = (DOi - DOf) × (Vdiluted / Vsample)
Step-by-step solution
1. Dissolved oxygen depletion Delta DO = DOinitial - DOfinal = 9 mg/L - 5 mg/L = 4 mg/L. 2. Dilution factor DF = Diluted mixture volume / Sample volume = 250 mL / 10 mL = 25. 3. BOD5 = Δ DO × DF = 4 mg/L × 25 = 100 mg/L. 4. Hence, the BOD5 of the raw sewage is 100 mg/L. Answer: D
Explanation
Biochemical Oxygen Demand (BOD5) measures the dissolved oxygen consumed by microorganisms during aerobic stabilization of organic matter over 5 days at 20°C. With DO dropping from 9 mg/L to 5 mg/L (depletion = 4 mg/L) and a dilution ratio of 250 mL / 10 mL = 25, BOD5 = 4 mg/L × 25 = 100 mg/L.
Why the other options are wrong
A) 4 mg/L is merely the unadjusted DO loss in the diluted incubation bottle. B) 10 mg/L is the raw sample volume. C) 40 mg/L assumes an incorrect dilution factor of 10.
Shortcut
BOD5 = (Initial DO - Final DO) × (250 / 10) = 4 × 25 = 100 mg/L.
Common mistake
Omitting the dilution factor and reporting raw DO depletion (4 mg/L) as the BOD.
Correct answer: D) 100 mg/L