Answer:
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