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.