The dissolved oxygen in water changes form 9 mg/1 to 3 mg/1 after 5 days. If the dilution factor is 3 ∶ 50 then will be the BOD₅(in mg/l) of the solution?

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Question

This question was previously asked in

JKSSB JE Civil Jal Shakti 6 Dec 2022 Official Paper (Shift 2)

Answer 1

Concept:

BOD:

BOD is defined as the amount of oxygen demanded by microorganisms to decompose biodegradable organic matter present in wastewater under anaerobic conditions.
BOD is the measure of the strength of wastewater.
Within 5 days 60-70% of organic content is oxidized.
Therefore 5 days BOD is taken as the standard reference
BOD = (Initial BOD Final BOD) x Dilution Factor

Calculation:
Given:

Initial DO =9 mg/l

Final DO =3 mg/l

Dilution factor = 50/3

BOD₅ = (9-3) × 50 /3

BOD₅ = 100 mg/l

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