Question

The 2.4 ml of raw sewage is diluted to 240 ml. The Dissolved Oxygen (DO) of the diluted sample at the beginning of Biochemical Oxygen Demand (BOD) test was 8 mg/l and it was 6 mg/l after 5-day incubation at 20°C.
The BOD₅ (in mg/l) of the raw sewage is

• A. 200
• B. 100
• C. 250
• D. 150

Hint:

BOD of raw sewage = (drop in DO of the diluted sample) times (dilution factor = diluted volume divided by raw sewage volume).

Answer 1

The Correct Option is A

Step 1: Understanding the Question.
Raw sewage is far too concentrated to test directly, because the bacteria in it would use up all the dissolved oxygen in the sample bottle long before 5 days pass. So a small, known volume of raw sewage is diluted with clean, oxygen-rich water before running the BOD test, and the oxygen drop measured in this diluted sample is then scaled back up to find the BOD of the original, undiluted sewage.

Step 2: Key Formula or Approach.
The BOD of the original (raw) sample is found from:
\[ \text{BOD}_5(\text{raw}) = (DO_i - DO_f) \times \text{Dilution Factor} \]
where \(DO_i\) and \(DO_f\) are the initial and final dissolved oxygen readings of the DILUTED sample, and the Dilution Factor tells us how many times the sewage was diluted:
\[ \text{Dilution Factor} = \frac{\text{Volume of diluted sample}}{\text{Volume of raw sewage used}} \]

Step 3: Detailed Explanation.
The oxygen used up in the diluted sample over 5 days is:
\[ DO_i - DO_f = 8 - 6 = 2\text{ mg/l} \]
The dilution factor, from the given volumes (2.4 ml of raw sewage made up to 240 ml total):
\[ \text{Dilution Factor} = \frac{240}{2.4} = 100 \]
This means every litre of the diluted sample represents only 1/100th of a litre of undiluted sewage, so the oxygen drop measured must be multiplied by 100 to get back to the concentration in the raw sewage.
\[ \text{BOD}_5(\text{raw}) = 2 \times 100 = 200\text{ mg/l} \]

Step 4: Why the other options are wrong.
Option (B), 100, would come from forgetting to multiply by the oxygen drop and using only the dilution factor on its own. Option (C), 250, and option (D), 150, do not come from either the correct dilution factor (100) or the correct oxygen drop (2 mg/l) multiplied together, so they do not match this direct calculation.

Final Answer:
\[ \boxed{\text{BOD}_5(\text{raw sewage}) = 200\text{ mg/l}} \]