Biochemical Oxygen Demand (BOD₅) calculation: A 2.5 ml sewage sample is diluted to 250 ml and incubated for 5 days at 20°C. If the dissolved oxygen depletion measured is 2.5 mg/litre, what is the BOD of the sewage (mg/l)?

Introduction / Context:
BOD₅ is a standard test for quantifying biodegradable organic load in wastewater. It measures the oxygen consumed by microorganisms over five days at 20°C in a diluted sample. Correctly applying the dilution factor is critical to obtain the actual BOD of the original sewage.

Given Data / Assumptions:

• Sewage aliquot volume Vs = 2.5 ml.
• Total diluted volume Vt = 250 ml.
• Observed dissolved oxygen depletion in the diluted bottle = 2.5 mg/l.
• Incubation: 5 days at 20°C.

Concept / Approach:
The BOD of the undiluted sample equals the observed depletion multiplied by the dilution factor. Dilution factor DF = Vt / Vs. The BOD is then BOD = DO depletion * DF, provided seed corrections are not required and initial and final DO are within the acceptable range.

Step-by-Step Solution:
Compute dilution factor: DF = 250 / 2.5 = 100.Compute BOD: BOD = 2.5 mg/l * 100 = 250 mg/l.Select the matching option.

Verification / Alternative check:
If a 1% aliquot (2.5 ml of 250 ml) causes 2.5 mg/l depletion in the diluted bottle, the undiluted would demand roughly 100 times more oxygen, giving 250 mg/l, consistent with medium-strength municipal sewage ranges.

Why Other Options Are Wrong:
50, 100, 150, 200 mg/l: each ignores or under-applies the dilution factor.

Common Pitfalls:

• Forgetting to account for dilution volume precisely.
• Confusing mg/l depletion in the bottle with BOD of the original sewage without scaling.
• Neglecting seeding corrections when seed is used; here none is indicated.

Final Answer:
250 mg/l