Question
Download Solution PDFIf 2% diluted wastewater sample incubated at 20 degrees centigrade resulted in DO depletion of 8 mg/l, BOD of the sample is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFDilution factor \(= \frac{{100}}{{{\rm{\% \;solution}}}} = \frac{{100}}{2} = 50\)
Depletion of oxygen = 8 ppm or mg/l and we know BOD5 = Depletion of oxygen × Dilution factor
BOD5 = 8 × 50 = 400 ppm
∴ BOD of the sewage is 400 ppm.Last updated on Jun 4, 2025
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