In a BOD test, 1.0 ml of raw sewage was diluted to 100 ml and the dissolved oxygen concentration of diluted sample at the beginning was 6 ppm and it was 4 ppm at the end of 5 day incubation at 20°C. The BOD of raw sewage will be
A. 100 ppm
B. 200 ppm
C. 300 ppm
D. 400 ppm
Answer: Option B
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BOD=(DO1-DO2)D.F.
=(6-4)100/1=200ppm
Thanks@sonali
{(DO)initial -(DO)final }* 100/%dilution
(6)-(4)}* 100/1= 200ppm
B.O.D.=(6-4)/(1/100)=200ppm