Biochemical Oxygen Demand (BOD5) from dilution bottle test A wastewater sample is diluted 100 times with aeration water. Initial DO = 7.0 mg/L; after 5 days at 20°C, DO = 0.0 mg/L. Determine the BOD of the original wastewater.

Introduction / Context:
BOD5 quantifies the oxygen consumed by microorganisms to oxidize biodegradable organics over 5 days at 20°C. In practice, samples are diluted so that residual DO remains measurable; the depletion multiplied by the dilution factor gives the BOD of the undiluted wastewater.

Given Data / Assumptions:

• Dilution factor = 100 (i.e., bottle contains 1 part wastewater + 99 parts dilution water).
• Initial DO = 7.0 mg/L.
• Final DO after 5 days = 0.0 mg/L.
• Nitrification is assumed suppressed or negligible, and seed correction is ignored (typical for basic problems).

Concept / Approach:
BOD5 (original) = (DO_initial − DO_final) * dilution factor. Here, the entire initial DO is consumed in the diluted bottle, indicating a strong wastewater.

Step-by-Step Solution:
1) DO depletion in diluted bottle = 7.0 − 0.0 = 7.0 mg/L.2) Multiply by dilution factor 100.3) BOD5 (original) = 7.0 * 100 = 700 mg/L.4) Select 700 mg/L.

Verification / Alternative check:
If a seed correction were present, it would be subtracted before multiplying; none is given, so direct multiplication applies.

Why Other Options Are Wrong:
100 mg/L ignores the full depletion and factor; 7 mg/L is just the diluted bottle depletion; “cannot be determined” is incorrect since sufficient data are provided.

Common Pitfalls:
Forgetting dilution factor; misreading units; not ensuring final DO is non-negative (bottle should not go anoxic in real tests).

Final Answer:
700 mg/L