Question
Download Solution PDFlf 2% solution of a sewage sample is incubated for 5 days at 20 °C and depletion of oxygen was found to be 5 ppm, BOD of the sewage is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Biological Oxygen Demand(BOD):
Biological Oxygen Demand or Biochemical oxygen demand (BOD) is the amount of dissolved oxygen used by microorganisms to break down organic matter in water.
A low BOD is an indicator of good quality water, while a high BOD indicates polluted water.
BOD = ( DOi - DFf ) × Dilution factor
Where,
BOD = Biochemical oxygen demand in ppm or mg/lit
DOi = Initial dissolved oxygen in mg/lit.
DOf =Final dissolved oxygen in mg/lit.
Dilution factor \(= \frac{{Volume\ of\ the\ diluted\ sample}}{{{\rm{Volume \ of\ the\ undiluted\ sewage\ sample}}}} \)
Calculation:
Given:
( DOi - DFf ) = 5
Volume of diluted sample = 100
Volume of the undiluted sewage sample = 2
Dilution factor = \(\frac{100}{2}\) = 50
BOD5 = 5 × 50 = 250 ppm
∴ BOD of the sewage is 250 ppm.Last updated on Jul 12, 2025
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