3.0 ml of raw sewage is diluted to 300 ml. The D.O. concentration of the diluted sample at the beginning of the test was 8 mg/l. After 5 day-incubation at 20°C, the D.O. concentration was 5 mg/l. The BOD of raw sewerage is A. 100 mg/l B. 200 mg/l C. 300 mg/l D. 400 mg/l

The correct answer is B. 200 mg/l.

BOD stands for Biochemical Oxygen Demand. It is a measure of the amount of oxygen that is used up by microorganisms in a given sample of water over a period of time. The higher the BOD, the more organic matter is present in the water.

In this case, 3.0 ml of raw sewage is diluted to 300 ml. The D.O. concentration of the diluted sample at the beginning of the test was 8 mg/l. After 5 day-incubation at 20°C, the D.O. concentration was 5 mg/l. This means that 3 mg/l of oxygen was used up by the microorganisms in the sample during the 5-day period.

The BOD of the raw sewage can be calculated using the following formula:

BOD = (D.O. at the beginning – D.O. at the end) x dilution factor

In this case, the BOD is:

BOD = (8 mg/l – 5 mg/l) x 300/3 = 200 mg/l

Therefore, the correct answer is B. 200 mg/l.

The other options are incorrect because they do not take into account the dilution factor. Option A is incorrect because it does not take into account the fact that the D.O. concentration decreased over time. Option C is incorrect because it does not take into account the fact that the sample was diluted. Option D is incorrect because it is twice the actual BOD.