A wastewater is expected to have BOD₃²⁷ of about 300 mg/L. The initial DO of dilution water is 8·5 mg/L. The minimum DO that should remain is 1·5 mg/L. The dilution requirement for BOD determination is :
- A100 times dilution of the sample
- B50 times dilution of the sampleCorrect
- C200 times dilution of the sample
- D150 times dilution of the sample
Solution
Concept
Dilution factor for a BOD test
Formula
Dilution factor ≥ BOD / (DOinitial − DOremaining)
Step-by-step solution
1. Usable oxygen = 8.5 − 1.5 = 7 mg/L. 2. Dilution ≥ 300 / 7 = 42.9, i.e. about 50 times.
Explanation
The diluted sample should consume no more oxygen than the DO available. For a BOD of 300 mg/L and 7 mg/L usable DO, the sample must be diluted by at least 43 times; 50 times is the nearest convenient dilution.
Why the other options are wrong
100, 150 and 200 times dilutions are too high; they would give very small DO depletion, which makes the test inaccurate.
Shortcut
Dilution ≈ BOD ÷ 7.
Common mistake
Using the initial DO 8.5 without subtracting the residual DO.
Correct answer: B) 50 times dilution of the sample