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A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water is
- a)700 mg/L
- b)100 mg/L
- c)cannot be determined
- d)7 mg/L
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A waste water sample diluted to 100 times with aeration water had an i...
BOD5 = Loss of oxygen in mg/l x Dilution ratio
BOD5 = (7 - 0) x 100 = 700 mg/l
BOD5 = (7 - 0) x 100 = 700 mg/l
Most Upvoted Answer
A waste water sample diluted to 100 times with aeration water had an i...
BOD (Biochemical Oxygen Demand) Test
The BOD test is used to determine the amount of organic matter present in a water sample.
Procedure:
1. A sample of wastewater is taken and diluted with aerated water.
2. The dissolved oxygen (DO) of the sample is measured initially.
3. The sample is then incubated for 5 days at 20C.
4. After the 5-day incubation period, the DO is measured again.
5. The BOD is calculated as the difference between the initial DO and the final DO.
Explanation:
In this question, the initial DO of the wastewater sample was 7.0 mg/L. After incubation for 5 days at 20C, the DO was zero. This means that all the dissolved oxygen in the sample was consumed by microorganisms during the incubation period.
The BOD of the wastewater sample can be calculated as follows:
BOD = Initial DO - Final DO
BOD = 7.0 mg/L - 0 mg/L
BOD = 7.0 mg/L
However, the sample was diluted 100 times with aeration water, so the actual BOD of the undiluted wastewater sample would be:
Actual BOD = BOD x Dilution factor
Actual BOD = 7.0 mg/L x 100
Actual BOD = 700 mg/L
Therefore, the correct answer is option A, 700 mg/L.
The BOD test is used to determine the amount of organic matter present in a water sample.
Procedure:
1. A sample of wastewater is taken and diluted with aerated water.
2. The dissolved oxygen (DO) of the sample is measured initially.
3. The sample is then incubated for 5 days at 20C.
4. After the 5-day incubation period, the DO is measured again.
5. The BOD is calculated as the difference between the initial DO and the final DO.
Explanation:
In this question, the initial DO of the wastewater sample was 7.0 mg/L. After incubation for 5 days at 20C, the DO was zero. This means that all the dissolved oxygen in the sample was consumed by microorganisms during the incubation period.
The BOD of the wastewater sample can be calculated as follows:
BOD = Initial DO - Final DO
BOD = 7.0 mg/L - 0 mg/L
BOD = 7.0 mg/L
However, the sample was diluted 100 times with aeration water, so the actual BOD of the undiluted wastewater sample would be:
Actual BOD = BOD x Dilution factor
Actual BOD = 7.0 mg/L x 100
Actual BOD = 700 mg/L
Therefore, the correct answer is option A, 700 mg/L.
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Community Answer
A waste water sample diluted to 100 times with aeration water had an i...
we dont know at what time it reached the value zero. so option 'C' is correct
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A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isa)700 mg/Lb)100 mg/Lc)cannot be determinedd)7 mg/LCorrect answer is option 'A'. Can you explain this answer?
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A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isa)700 mg/Lb)100 mg/Lc)cannot be determinedd)7 mg/LCorrect answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isa)700 mg/Lb)100 mg/Lc)cannot be determinedd)7 mg/LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isa)700 mg/Lb)100 mg/Lc)cannot be determinedd)7 mg/LCorrect answer is option 'A'. Can you explain this answer?.
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