250+ TOP MCQs on Determination of Chemical Characteristics and Answers

Waste Water Engineering Multiple Choice Questions on “Determination of Chemical Characteristics”.

1. _______ are found in agricultural wastes.
a) Pesticides
b) Phenols
c) Proteins
d) Grease
Answer: a
Clarification: There are different chemical constituents found in wastewater. The wastewater resulting from agriculture consists of pesticides. The industrial wastewater contains phenols.

2. _________ is determined by measuring the dissolved oxygen used by microorganisms during the biochemical oxidation of organic matter in 5 days at 20˚C.
a) BOD5
b) COD
c) TOC
d) ThOD
Answer: a
Clarification: BOD5 is the oxygen equivalent of organic matter. It is determined by measuring the dissolved oxygen used by microorganisms during the biochemical oxidation of organic matter in 5 days at 20˚C.

3. How are many forms of nitrogen present in wastewater?
a) 3
b) 4
c) 2
d) 5
Answer: b
Clarification: There are different types of nitrogen present in wastewater. Basically, our types of nitrogen are present. They are organic nitrogen, ammonia nitrogen, nitrate nitrogen and nitrite nitrogen.

4. After how many days will nitrogen will be oxidized to nitrite and nitrate?
a) 9
b) 10
c) 10-12
d) 14-15
Answer: c
Clarification: After 10-12 days, nitrogen will be oxidized to nitrite and then nitrate. Nitrogen in the presence of oxygen gets converted to nitrite and then gets converted to nitrate.

5. ______ is determined by measuring the dissolved oxygen used during the chemical oxidation of organic matter in 3 hours.
a) COD
b) BOD
c) ThOD
d) TOC
Answer: a
Clarification: It is the oxygen equivalent of organic matter. Chemical Oxygen Demand is determined by measuring the dissolved oxygen used during the chemical oxidation of organic matter in 3 hours.

6. Which of the following is used for a small concentration of organic matter?
a) COD
b) TOC
c) BOD
d) ThOD
Answer: b
Clarification: The organic carbon existing in the wastewater is injected to a sample of the wastewater in a special device in which the carbon is oxidized to carbon dioxide then carbon dioxide is measured and used to quantify the amount of organic matter in the wastewater. This method of TOC measurement is only used for small concentration of organic matter.

7. Which of the following indicates that the water body has been used for waste disposal?
a) Chlorides
b) Nitrates
c) Phosphates
d) Ammonia
Answer: a
Clarification: High concentrations of chlorides indicate that the water body has been used for waste disposal. Chlorides affect the biological process when present in high concentrations.

8. What is the maximum concentration of total solids present in wastewater?
a) 350 mg/L
b) 720 mg/L
c) 1200 mg/ L
d) 850 mg/L
Answer: c
Clarification: The concentration of total solids present in weak wastewater is 350 mg/ L. The concentration of total solids present in strong wastewater is 1200 mg/ L. The medium range consists of 720 mg/ L.

9. What is the TSS effluent concentration for 7 days?
a) 30 mg/L
b) 45 mg/L
c) 25 mg/L
d) 40 mg/L
Answer: b
Clarification: TSS stands for total suspended solids. Effluent TSS concentration for average of 30 days is 30 mg/L. The average 7 day concentration for effluent containing TSS is 45 mg/L.

10. The BOD test is carried out for how many days?
a) 1 day
b) 2 days
c) 5 days
d) 6 days
Answer: c
Clarification: The BOD test is carried out for 5 days. BOD is a biological oxygen demand. It is the amount of Dissolved oxygen utilized by bacteria to carry out biochemical reactions.

11. At what temperature the bottles for the BOD test are incubated?
a) 25 degree Celsius
b) 35 degree Celsius
c) 20 degree Celsius
d) 30 degree Celsius
Answer: c
Clarification: The bottles to be tested in order to calculate the amount of BOD present is kept at 20 degree Celsius. This is incubated for 5 days. This is done to estimate the ultimate BOD.

12. What is the mathematical expression of BOD?
a) BOD = [(D1-D2)-(B1-B2)f]/P
b) BOD = [(D1-D2)-(B1-B2)f].
c) BOD = [(D1-D2) f]/P
d) BOD = [(D1-D2)-(B1-B2)]/P
Answer: a
Clarification: BOD = [(D1-D2)-(B1-B2) f]/P. Where D1 is the dissolved oxygen of the diluted sample right after the preparation. D2 is the Dissolved Oxygen of the sample after incubation.B1 is the Dissolved oxygen of the control before incubation. B2 is the dissolved oxygen of the control after incubation. F is the fraction of diluted water in the sample to the dilution of water in the control. P is the fraction of waste water sample volume to the total combined volume.

13. In terms of percentage how much BOD is oxidised in 5 days?
a) 90%
b) 70-90%
c) 60-70%
d) 50%
Answer: c
Clarification: In 5 days 60-70% BOD is oxidised. It takes around 20 days to oxidise the carbonaceous organic matter to 90-95%. The rate of oxidation is proportional to the amount of organic matter present in the water.

14. How is COD calculated?
a) Waste water is oxidised chemically using sodium in acid solutions
b) Waste water is oxidised chemically using dichromate in acid solutions
c) Waste water is oxidised chemically using bromine in acid solutions
d) Waste water is oxidised chemically using strontium in acid solutions
Answer: b
Clarification: Waste water is oxidised chemically using dichromate in acid solutions. High COD value indicates that the presence of inorganic compounds is high. Inorganic compounds get chemically oxidised and this results in the increase of organic compounds in the sample.

15. What is the ratio of BOD/COD in untreated waste?
a) 1-3
b) 0.3-0.8
c) 0.1-0.2
d) 3-5
Answer: b
Clarification: The ratio of BOD/COD is 0.3-0.8 in untreated waste. The ratio of BOD/TOC in untreated waste is 1.2-2. The ratio of BOD/COD after primary settling is 0.4-0.6.

16. What is the ratio of BOD/COD in the final effluent?
a) 0 8-1.2
b) 0.2-0.5
c) 0.1-0.3
d) 0.4-0.6
Answer: c
Clarification: The ratio of BOD/COD is 0.1-0.3 in the final effluent. The ratio of BOD/TOC in the final effluent is 0.2-0.5. The ratio of BOD/TOC from the primary settling tank is 0.8-1.2.

17. How is TSS calculated?
a) MPN
b) HPLC
c) Filtration
d) Mass spectrometer
Answer: c
Clarification: TSS is calculated by the filtration method. A filter paper is used to filter the water sample and the filtrate is then weighed. The amount of filtrate obtained is considered as the amount of TSS present.

18. Which of these is the used as the indicator when the titration is carried out to determine the amount of COD present in a sample.
a) Methyl Orange
b) Methyl blue
c) Ferroin
d) Phenolphthalein
Answer: c
Clarification: Ferroin is the indicator used while titration is carried out to determine the COD present in a water sample. The sample is titrated against ferrous ammonium sulphate. Titration is carried out until the solution turns to reddish brown.

19. Which of these is not a method to determine the number of colonies of bacteria present in a sample?
a) Multiple Tube fermentation
b) Pour and spread plate method
c) Membrane filter technique
d) Toxicity test
Answer: d
Clarification: Toxicity test is used to assess the environmental conditions of aquatic life. This is used to assess the effects of wastewater treatment methods. This is not used to determine the number of colonies of bacteria present in a water sample.