BOD₅ calculation for a diluted sample: 3.0 mL of raw sewage is diluted to 300 mL. The initial dissolved oxygen (DO) of the diluted sample is 8 mg/L, and after 5 days at 20°C the DO is 5 mg/L. What is the BOD₅ of the raw sewage (in mg/L)?

Introduction / Context:
Biochemical Oxygen Demand over 5 days at 20°C (BOD₅) is a standard test in environmental engineering to estimate biodegradable organic pollution. When samples are diluted, the oxygen depletion measured on the diluted bottle is scaled back by the dilution factor to obtain the BOD of the original sewage.

Given Data / Assumptions:

• Raw sewage volume used = 3.0 mL.
• Total diluted volume = 300 mL.
• Initial DO (diluted) = 8 mg/L.
• Final DO after 5 days (diluted) = 5 mg/L.
• Incubation temperature = 20°C; nitrification not considered.

Concept / Approach:

BOD₅ of the original sample = (DO initial − DO final) * dilution factor. The dilution factor equals total volume / sample volume when the seed correction is negligible and blank is acceptable.

Step-by-Step Solution:

Verification / Alternative check:

If a seed/blank correction were required, it would be subtracted before multiplying by the dilution factor. Here none is indicated, so 300 mg/L stands.

Why Other Options Are Wrong:

100 mg/L and 200 mg/L understate the effect of the 100× dilution. 400 mg/L overstates depletion beyond the observed 3 mg/L.

Common Pitfalls:

Forgetting to apply the dilution factor; mixing up mL and L; neglecting seed correction when present; using temperature other than 20°C for BOD₅.

Final Answer:

300 mg/L