Consider a sample of water from a stream found to have an initial do of 10.0 mg/l. it is placed directly into a bod bottle and incubated for 5 days at 20c. after 5 days, the do is found to be 5.0 mg/l. what is the bod5 for this sample (mg/l)?

The BOD5 for this sample is 5 mg/L

Explanation:

The BOD5 takes into consideration the difference between the initial DO and the final DO divided by the fraction of the sample tested in a 300 mL BOD flask. The problem states that the sample was directly placed in a BOD bottle so we assume that the volume used was 300 mL of the water sample. The fraction of water is 1 (300/300).

BOD5 = (10-5)/1

BOD 5 = 5 mg/L