Question

A 10 ml sample of wastewater is diluted with water having no BOD, to fill a 300 ml BOD bottle. The initial DO of the diluted waste water is 9.0 mg/l. If the BOD\(_5\) of the waste water sample is 60 mg/l, the final DO of the diluted waste water in mg/l, is

• A. 5.0
• B. 6.0
• C. 7.0
• D. 8.0

Hint:

The final DO is calculated by subtracting the BOD from the initial DO value after accounting for dilution.

Answer 1

The Correct Option is C

We are given the following information: - The initial sample volume is 10 ml. - The total volume after dilution is 300 ml. - The initial dissolved oxygen (DO) concentration of the diluted sample is 9.0 mg/l. - The BOD\(_5\) of the wastewater sample is 60 mg/l. Step 1: Calculate the Amount of Oxygen Consumed
The BOD\(_5\) value of 60 mg/l represents the amount of oxygen consumed by microorganisms over 5 days for the given sample. Since the sample is diluted by a factor of 30 (300 ml / 10 ml), we need to consider the effect of dilution on the final DO. Step 2: Final DO Calculation
The initial DO concentration after dilution is 9.0 mg/l. The final DO will be reduced due to the biological consumption of oxygen by microorganisms in the wastewater. The BOD is the difference between the initial and final DO. \[ \text{Final DO} = \text{Initial DO} - \text{BOD\(_5\)} \] Since the BOD\(_5\) is 60 mg/l, the final DO will be: \[ \text{Final DO} = 9.0 \, \text{mg/l} - 4.0 \, \text{mg/l} = 5.0 \, \text{mg/l}. \] Therefore, the final DO of the diluted wastewater sample is 5.0 mg/l.