Question

The following data pertain to a sewage sample: Initial DO = 10 mg/L, Final DO = 2.5 mg/L, Dilution = 2%. The BOD of given sewage sample is:

• A. 7.5 mg/L
• B. 15 mg/L
• C. 93.75 mg/L
• D. 375 mg/L

Hint:

Always convert dilution percentage into fraction before calculating BOD to avoid numerical errors.

Answer 1

The Correct Option is D

Step 1: Understanding BOD.
Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen required by microorganisms to decompose organic matter present in sewage. It is calculated based on the depletion of dissolved oxygen after incubation and corrected for dilution.
Step 2: Calculating oxygen depletion.
Initial DO = 10 mg/L
Final DO = 2.5 mg/L
Oxygen consumed = 10 − 2.5 = 7.5 mg/L
Step 3: Applying dilution factor.
Dilution = 2% = 0.02
BOD is calculated using the formula:
\[ \text{BOD} = \frac{\text{DO depletion}}{\text{Fraction of sample}} \] \[ \text{BOD} = \frac{7.5}{0.02} = 375 \, \text{mg/L} \]
Step 4: Conclusion.
The Biochemical Oxygen Demand of the given sewage sample is 375 mg/L.

Answer 2

BOD accounts for the oxygen consumed by microorganisms breaking down organic matter, scaled up from the diluted test sample back to the full-strength sewage. Let's check each candidate:

1. 7.5 mg/L: This is simply the raw oxygen depletion in the diluted sample (\(10 - 2.5 = 7.5\)), but it hasn't been scaled back up to account for the fact that the sample was diluted to just 2% of its original strength, so it understates the true BOD of the undiluted sewage by a large factor.
2. 15 mg/L: This would only be double the raw depletion, but the dilution factor here is 2%, meaning the true sewage is 50 times stronger than the diluted sample, not just twice as strong, so doubling isn't the right correction.
3. 93.75 mg/L: This would correspond to dividing the depletion by 8% rather than 2% (\(7.5 / 0.08 = 93.75\)), an inconsistent dilution value that doesn't match the 2% given in the problem.
4. 375 mg/L: Dividing the observed oxygen depletion by the actual dilution fraction, \(\frac{7.5}{0.02} = 375\), correctly scales the small, diluted-sample depletion back up to represent the full-strength sewage sample.

Only dividing by the correct dilution fraction of 2% (0.02) reproduces a consistent, correctly scaled BOD value.

Therefore, the correct answer is 375 mg/L.