If a 2% solution of sewage sample is incubated for 5 days at 20°C and the dissolved oxygen depletion was found to be 8 mg/l. The BOD of the sewage is
A. 100 mg/l
B. 200 mg/l
C. 300 mg/l
D. 400 mg/l
Answer: Option D
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Thank you @Aziz Panna
B.O.D of sewage= D.O × Dilution factor
e.g., if Sewage solution is 5% then Dilution factor will be= 100/5. (Not 5/100)
So, finally B.O.D = 8 × 100/2 = 8 × 50 = 400 mg/l