In a BOD test, 1.0 ml of raw sewage was diluted to 100 ml and the dissolved oxygen concentration of diluted sample at the beginning was 6 ppm and it was 4 ppm at the end of 5 day incubation at 20°C. The BOD of raw sewage will be A. 100 ppm B. 200 ppm C. 300 ppm D. 400 ppm

The correct answer is A. 100 ppm.

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

In this question, 1.0 ml of raw sewage was diluted to 100 ml. This means that the final concentration of the raw sewage in the sample was 1%. The dissolved oxygen concentration of the diluted sample at the beginning was 6 ppm and it was 4 ppm at the end of 5 day incubation at 20°C. This means that the microorganisms in the sample used up 2 ppm of oxygen during the 5 day incubation period.

The BOD of the raw sewage is calculated by multiplying the initial dissolved oxygen concentration by the dilution factor and the percent decrease in dissolved oxygen. In this case, the BOD is calculated as follows:

BOD = (6 ppm)(100/1) x (1 – 4/6) = 100 ppm

Therefore, the correct answer is A. 100 ppm.

Option B is incorrect because it is the BOD of the diluted sample, not the BOD of the raw sewage.

Option C is incorrect because it is the BOD of the raw sewage after 5 days of incubation, not the BOD of the raw sewage at the beginning of the incubation period.

Option D is incorrect because it is the BOD of the raw sewage after 1 day of incubation, not the BOD of the raw sewage at the beginning of the incubation period.