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

Let's work this out by tracking exactly how much oxygen was used per unit of RAW sewage, instead of applying the dilution-factor formula directly.

The diluted sample used up $8-6=2$ mg of oxygen per litre over the 5 days. This 2 mg/l of oxygen use came from only the raw sewage that is present in that litre of diluted sample, since the dilution water itself has no biological oxygen demand of its own.

Now let's find how much raw sewage sits inside each litre of the diluted mixture. The dilution was made by taking 2.4 ml of raw sewage and making it up to a total of 240 ml with clean water. So the fraction of the diluted sample that is actually raw sewage is:

\[ \frac{2.4}{240} = \frac{1}{100} \]

This means each litre of diluted sample contains only $1/100$ of a litre worth of raw sewage. The 2 mg/l of oxygen used, therefore, was used up by just $1/100$ of a litre of raw sewage, not by a whole litre of it.

To find the oxygen demand of a FULL litre of raw sewage, scale this up by the same factor of 100:

\[ \text{BOD}_5(\text{raw sewage}) = 2\ \text{mg/l} \times 100 = 200\ \text{mg/l} \]

Let's summarize:

• Only 1/100 of a litre of raw sewage was present in each litre of the diluted test sample.
• The 2 mg/l oxygen drop measured must be scaled up by that same factor of 100 to represent the full-strength raw sewage.

So the BOD5 of the raw sewage works out to 200 mg/l, matching option (A).