A 2.5% dilution of a sewage sample shows an oxygen depletion of 5 mg/L after 5 days at 21°C. What is the 5-day biochemical oxygen demand (BOD₅) of the undiluted sewage (in mg/L)?

Introduction / Context:
BOD testing often uses dilutions to keep the dissolved oxygen (DO) depletion within measurable limits. The observed depletion must be scaled by the dilution factor to estimate the original BOD.

Given Data / Assumptions:

• Dilution = 2.5% = 0.025 (by volume).
• Observed 5-day DO depletion at 21°C = 5 mg/L.
• No seed correction or blank depletion specified (assume negligible).

Concept / Approach:
BOD of original sample = observed depletion / dilution fraction, provided no additional corrections are needed.

Step-by-Step Solution:
Step 1: Determine dilution fraction f = 0.025.Step 2: Compute BOD₅ = 5 mg/L / 0.025.Step 3: BOD₅ = 200 mg/L.

Verification / Alternative check:
Check units and logic: a 40× dilution would multiply the depletion by 40 to recover the undiluted demand; 5 × 40 = 200 mg/L matches.

Why Other Options Are Wrong:

• 50, 100, 150 mg/L: Underestimate because they ignore the dilution factor magnitude.
• 250 mg/L: Overestimates beyond the simple dilution scaling.

Common Pitfalls:
Forgetting to subtract seed/control depletion when present, or misreading the dilution percentage.

Final Answer:
200 mg/L