If the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C, B.O.D. of the sewage is
A. 50 mg/l
B. 100 mg/l
C. 150 mg/l
D. 250 mg/l
Answer: Option D
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BOD of sewage = DO×dilution factor
Df= 250/2.5=100
Bod= 2.5×100 =250mg/l
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