Question

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

Answer 1

The BOD of the raw sewage is calculated by multiplying the change in dissolved oxygen concentration (2 ppm) by the dilution factor (100), resulting in a BOD of 200 ppm.

Step-by-step explanation:

The student has performed a biochemical oxygen demand (BOD) test to measure the amount of dissolved oxygen consumed by bacteria in the presence of organic matter in a sample of raw sewage. In this case, a 1.0 ml sample was diluted to 100 ml. The dissolved oxygen concentration was 6 ppm at the start of the test and decreased to 4 ppm after 5 days of incubation at 20°C.

To calculate the BOD of the raw sewage, we consider the change in dissolved oxygen concentration, which is 2 ppm (6 ppm - 4 ppm) for the diluted sample. Because the sample was diluted 100 times, we multiply this change in concentration by the dilution factor to obtain the BOD for the raw sewage. Therefore, the BOD is 2 ppm × 100 = 200 ppm.