A 2% dilution of a sewage sample is incubated for 5 days at 20°C, and the measured dissolved oxygen depletion is 8 mg/L. Compute the 5-day BOD (BOD₅) of the undiluted sewage.

Introduction / Context:The biochemical oxygen demand (BOD) test quantifies biodegradable organic load. Because raw sewage typically depletes oxygen rapidly, dilutions are used so that the final dissolved oxygen (DO) remains above zero after five days. The undiluted BOD is then inferred from the observed depletion and the dilution fraction.

Given Data / Assumptions:

• Dilution fraction f = 2% = 0.02.
• Observed 5-day DO depletion = 8 mg/L.
• Incubation temperature: 20°C (standard for BOD₅).
• No seed correction or blank depletion specified; assume negligible.

Concept / Approach:For a simple BOD test without seed corrections, BOD₅ (undiluted) = observed DO depletion / dilution fraction. This rescales the consumption measured in the diluted bottle back to the original strength of the sewage.

Step-by-Step Solution:Compute BOD₅ = 8 mg/L / 0.02.BOD₅ = 400 mg/L.Select the matching option.

Verification / Alternative check:A 2% dilution is a 50:1 factor. Multiplying the depletion by 50 yields the same result: 8 * 50 = 400 mg/L.

Why Other Options Are Wrong:

• 100–300 mg/L: Underestimate; they do not account for the full 50× dilution.
• 500 mg/L: Overestimate relative to the measured depletion.

Common Pitfalls:Forgetting to adjust for seed oxygen demand (if seed is used), or misreading the dilution percentage as 0.2 instead of 0.02; both lead to substantial errors.

Final Answer:400 mg/L