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BOD of a waste water sample is estimated to be 180 mg/l. Assuming 4 mg/l BOD can be consumed in the BOD bottle, the volume of undiluted sample to be added to a 300 ml bottle is nearly
1. 6.7 ml
2. 5.6 ml
3. 4.4 ml
4. 3.3 ml

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Correct Answer - Option 1 : 6.7 ml

Concept:

BOD = BOD consumed × dilution factor

Dilution factor = volume of diluted sample/volume of undiluted sample

Calculation:

Given that, BOD of wastewater sample = 180 mg/l

BOD consumed = 4 mg/l

Volume of diluted sample = 300 ml

Let the volume of undiluted sample is X ml.

180 = 4 × (300/X)

⇒ X = 6.67 ml

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