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The initial DO of a river water sample is 8.5 mg/L After incubation at $20^{\circ} \mathrm{C}$ for 5 days, the DO was reduced to 5.2 mg/L. What is the BOD of the sample?
Option: 1
3.3 mg/L

Option: 2
1.6 mg/L

Option: 3
6.9 mg/L

Option: 4
2.8 mg/L

Answers (1)

best_answer

BOD = Initial DO - Final DO

BOD = 8.5 - 5.2

BOD = 3.3 mg/L

Posted by

Rakesh

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