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: Oxygen consumed in the diluted sample \(= 10 - 2.5 = 7.5\) mg/L.
Step 2: Since the sewage was diluted to only 2% of its original strength, the undiluted sample would show 50 times ( \(\frac{1}{0.02} = 50\) ) as much oxygen depletion.
Step 3: BOD \(= 7.5 \times 50 = 375\) mg/L.
\[ \boxed{375 \text{ mg/L}} \]

Answer 2

We can also verify this by checking what dilution fraction each option implicitly assumes, working backward from \(\text{BOD} = \frac{\text{DO depletion}}{\text{dilution fraction}}\):

1. 7.5 mg/L: Implies a dilution fraction of \(\frac{7.5}{7.5} = 1\), i.e. no dilution at all, which contradicts the given 2% dilution.
2. 15 mg/L: Implies a dilution fraction of \(\frac{7.5}{15} = 0.5\), i.e. 50% dilution, far from the stated 2%.
3. 93.75 mg/L: Implies a dilution fraction of \(\frac{7.5}{93.75} = 0.08\), i.e. 8% dilution, again not matching the given value.
4. 375 mg/L: Implies a dilution fraction of \(\frac{7.5}{375} = 0.02\), i.e. exactly 2%, matching the dilution stated in the problem.

Only one option, when reverse-checked against the depletion value, reproduces the actual 2% dilution given.

Therefore, the correct answer is 375 mg/L.