3 A 10 mL sample is pipetted directly into a 300 mL incubati

3. A 10 mL sample is pipetted directly into a 300 mL incubation bottle. The initial DO of the diluted sample is 9.0 mg/L and its final DO is 2.0 mg/L. The dilution water is incubated in a 200 mL bottle, and the initial and final DO are, respectively, 9.0 and 8.0 mg/L. If the sample and the dilution water are incubated at 20°C for 5 days, what is the BODs of the sample at this temperature?

Solution

Its solution is simple:

since it is asked to find the BOD of the sample,so we will ony pick the data of related to the incubation of sample only,therefore,the BOD of the sample will become:

BOD= (initial D.O - Final D.O) X (Volume of bottle/Volume of sample used)

so, BOD= (9-2)*(300/10)=210mg/L