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This question belongs to Civil Engineering Environmental Engineering

Environmental Engineering

?

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 & Solution

Correct Answer: Option B

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Question posted by Examveda

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5 Comments

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BHIMRAO SHINDE 2 years ago

DF = A+B/A =1+100/1=101
BOD=(6-4) *101=202PPM

Kirdar Husain 7 years ago

BOD=(DO1-DO2)D.F.
=(6-4)100/1=200ppm

Muhammad Saqib 7 years ago

Thanks@sonali

Sonali Gawai 7 years ago

{(DO)initial -(DO)final }* 100/%dilution
(6)-(4)}* 100/1= 200ppm

Sanjay Chandrawal 8 years ago

B.O.D.=(6-4)/(1/100)=200ppm

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