Category: Engineering Chemistry Objective Questions

Engineering Chemistry Questions & Answers for Exams focuses on “Synthetic Gasoline from Non-Petroleum Sources – 2”.

1. Which type of coal is used in Bergius process?
a) Anthracite coal
b) Steam coal
c) Pulverised coal
d) Gas coal
Answer: d
Clarification: Bergius process is used to produce synthetic gasoline form low ash pulverised coal. This coal is finely divided into powder and is converted into a paste.

2. Which substance is mixed with the pulverised coal in the Bergius process?
a) Sulphuric acid
b) Heavy oil and catalyst
c) Heavy oil
d) Catalyst
Answer: b
Clarification: The paste of pulverised coal is mixed with a heavy oil and a catalyst. Nickel oleate is used as a catalyst in this process. This mixture is then sent to a converter.

3. At what temperature does the Bergius process is carried out?
a) 200-300 ⁰C
b) 700-800 ⁰C
c) 350-500 ⁰C
d) 0-150 ⁰C
Answer: c
Clarification: The temperature of converter is set to about 350-500 ⁰C and a pressure of about 200-250 atmospheres where the paste of coal meets hydrogen. The Fischer-Tropsch process is carried out at 200-300 ⁰C.

4. Which product is obtained from the converter in the Bergius process?
a) Hydrocarbons
b) Olefins
c) Fuel oil
d) Gasoline
Answer: a
Clarification: The combination of hydrogen with the carbon framework of the coal yields various hydrocarbons from wax to gases. The high molecular weight hydrocarbons are decomposed further at the high temperature prevailing in the converter giving lower hydrocarbons.

5. Same catalyst is used for production of oil in the Bergius process as used in the production of hydrocarbons.
a) True
b) False
Answer: b
Clarification: For production of gasoline, oils produced in the first stage of the process are subjected to further hydrogenation in the presence of different catalyst. Nickel oleate is not used for this process.

6. Which compounds are rich in gasoline obtained from the Bergius process?
a) Tetraethyl lead
b) Ethanol
c) Olefins
d) Aromatic compounds
Answer: d
Clarification: The gasoline obtained from the Bergius process is rich in aromatic and branched-chain hydrocarbons. Tetraethyl lead is an anti-knocking agent which is used for increasing the octane number.

7. What happens to the middle oil fraction in the Bergius process?
a) The oil is converted into olefins
b) The oil is converter into paraffin
c) The oil is converted into gasoline
d) The oil is converted into hydrocarbons
Answer: c
Clarification: The middle oil fraction is subjected to hydrogen in the presence of a catalyst to produce gasoline. Same process is carried out in the top fraction of the reactor. The heavy oil fraction is recycled to make a paste with a fresh batch of coal powder.

8. Which undesirable compound is present in gasoline?
a) Sulphur compounds
b) Phosphorous compounds
c) Chloride compounds
d) Magnesium compounds
Answer: a
Clarification: The gasoline obtained by any process contains some sulphur compounds and unsaturated hydrocarbons, which get oxidized and polymerised to gums and sludge’s. Hence it has to be refined by chemical processes.

9. How much percentile of sulphuric acid is treated with gasoline for the removal of its impurities?
a) 99 %
b) 60 %
c) 80 %
d) 40 %
Answer: c
Clarification: The impure gasoline is treated with cold 80 % sulphuric acid. This does not react with paraffin, naphthenes and aromatic compounds. It reacts with unsaturated hydrocarbons to produce alcohols and ethers.

10. Removal of gaseous impurities like methane, ethane, propane and butane from gasoline is called as _____________________
a) Stabilization of gasoline
b) Sweetening of gasoline
c) Doping of gasoline
d) Blending of gasoline
Answer: a
Clarification: The stabilization of gasoline is achieved by passing the impure gasoline into a fractionating tower having 49 plates. The lower plates are heated by steam, while the gasoline is introduced near the upper plates.

exam questions on Engineering Chemistry,

Engineering Chemistry Quiz focuses on “Calgon Conditioning”.

1. What is Calgon?
a) Potassium hexa meta sulphate
b) Magnesium hexa meta phosphate
c) Sodium hexa meta phosphate
d) Calcium hexa meta sulphate
Answer: c
Clarification: Calgon is sodium hexa meta phosphate. Calgon prevents limescale build-up in washing machines all over the world to protect the heating element, pipes and drum of the machines.

2. Calgon is added to boiler water to prevent ___________
a) Foaming
b) Sludge and scale formation
c) Priming
d) Corrosion
Answer: b
Clarification: Calgon is added to boiler water to prevent sludge and scale formation. Calgon prevents limescale build-up in washing machines all over the world to protect the heating element, pipes and drum of the machines.

3. Calgon converts the scale forming impurity like CaSO₄ into ___________
a) Soluble complex compound
b) Insoluble complex compound
c) Soluble acids
d) Insoluble acids
Answer: a
Clarification: Calgon converts the scale forming impurity like CaSO₄ into soluble complex compound which are harmless to boiler.

4. The quantity of calgon to be added to prevent scale and sludge formation is ___________
a) 10-20 ppm
b) 30-50 ppm
c) 0.005-0.1 ppm
d) 0.5-5 ppm
Answer: d
Clarification: The quantity of calgon to be added to prevent scale and sludge formation is 0.5-5 ppm. It is used in small quantity.

5. At high temperature and pressure, calgon is converted into sodium ortho-phosphate which reacts with ____________ salts.
a) Magnesium
b) Calcium
c) Sodium
d) Potassium
Answer: b
Clarification: At high temperature and pressure, calgon is converted into sodium orthophosphate which reacts with calcium salts to form calcium ortho-phosphate.

6. Calcium ortho-phosphate appears as loose sludge and thus can be removed by blow-down operation.
a) True
b) False
Answer: a
Clarification: Calcium orthophosphate appears as loose sludge and thus can be removed by blow-down operation. It is formed by the reaction of sodium ortho-phosphate with calcium salts.

7. Calgon conditioning is not applicable for the prevention of ___________
a) Iron oxide only
b) Copper depositions only
c) Sulphurdioxide only
d) Iron oxide and copper depositions
Answer: d
Clarification: Calgon conditioning is not applicable to the prevention of iron oxide and copper depositions. Calgon prevents limescale build-up in washing machines all over the world to protect the heating element, pipes and drum of the machines.

8. Iron oxide and copper depositions can be prevented by adding ___________ to the boiler water.
a) Calgon
b) Carbonate
c) EDTA
d) EBT
Answer: c
Clarification: Iron oxide and copper depositions can be prevented by adding EDTA to the boiler water. Ethylenediaminetetraacetic acid (EDTA), is an amino poly-carboxylic acid and a colorless and a water-soluble solid.

9. Carbonate conditioning is not done in high pressure boilers because it may lead to caustic embrittlement.
a) True
b) False
Answer: a
Clarification: Carbonate conditioning is not done in high pressure boilers because it may lead to caustic embrittlement. Caustic embrittlement is the phenomenon in which the material of a boiler becomes brittle due to the accumulation of caustic substances.

10. Which of the following is not used in colloidal conditioning?
a) Kerosene
b) Tannin
c) Agar-agar
d) Vinegar
Answer: d
Clarification: Kerosene, tannin and agar-agar are used in colloidal conditioning. Vinegar is not used for this purpose.

areas of Engineering Chemistry for Quizzes,

Engineering Chemistry Questions and Answers for Experienced people focuses on “Oils and Fats – 2”.

1. The catalyst used in the addition of iodine is ____________
a) Ni/ Pt
b) Lewis acids
c) CH₃MgCl
d) HgCl₂
Answer: d
Clarification: The catalyst used in the addition of iodine is HgCl₂. Other catalyst that are mentioned cannot be used for this purpose.

2. Drying oils are used as the medium of paints and varnishes.
a) True
b) False
Answer: a
Clarification: Drying oils are used as the medium of paints and varnishes. When paints and varnishes are applied in thin layers, linseed oil quickly dries up to form a protective layer.

3. Which of the following is responsible for rancidity?
a) Alkalies
b) Ketones
c) Aldehydes
d) Alcohols
Answer: c
Clarification: Volatile acids and aldehydes are responsible for rancidity due to their offensive odour.

4. The number of milligrams of KOH required for the saponification of one gram of oil or fat is called ____________
a) Acid number
b) Iodine number
c) Richert-Meissl number
d) Saponification number
Answer: d
Clarification: The number of milligrams of KOH required for the saponification of one gram of oil or fat is called saponification number. The long chain fatty acids found in fats have a low saponification value because they have a relatively fewer number of carboxylic functional groups per unit mass of the fat as compared to short chain fatty acids.

5. Iodine number is defined as number of grams of iodine needed for the iodination of ________ gram/grams of oil or fat.
a) 1
b) 5
c) 100
d) 1000
Answer: c
Clarification: Iodine number is defined as number of grams of iodine needed for the iodination of 100 grams of oil or fat. Iodine numbers are often used to determine the amount of unsaturation in fatty acids. This unsaturation is in the form of double bonds, which react with iodine compounds. The higher the iodine number, the more C=C bonds are present in the fat.

6. Richert-Meissl number is defined as the volume of 0.1M KOH solution required for the neutralisation of _______ gram/grams of fat or oil.
a) 1
b) 5
c) 100
d) 1000
Answer: b
Clarification: Richert-Meissl number is defined as the volume of 0.1M KOH solution required for the neutralisation of 5grams of fat or oil. It is an indicator of how much volatile fatty acid can be extracted from fat through saponification.

7. Which of the following tells the amount of free fatty acids present in fat or oil?
a) Acid number
b) Iodine number
c) Saponification number
d) Richert-Meissl number
Answer: a
Clarification: Acid number tells the amount of free fatty acids present in fat or oil. It is the mass of potassium hydroxide (KOH) in milligrams that is required to neutralize one gram of chemical substance.

8. Which of the following helps in the classification of oils into drying, semi-drying and non-drying categories?
a) Acid number
b) Iodine number
c) Saponification number
d) Richert-Meissl number
Answer: b
Clarification: Iodine number helps in the classification of oils into drying, semi-drying and non-drying categories. Iodine numbers are often used to determine the amount of unsaturation in fatty acids. This unsaturation is in the form of double bonds, which react with iodine compounds. The higher the iodine number, the more C=C bonds are present in the fat.

9. Which of the following is of special value in testing the purity of butter and desi-ghee?
a) Acid number
b) Iodine number
c) Saponification number
d) Richert-Meissl number
Answer: d
Clarification: Richert-Meissl number is of special value in testing the purity of butter and desi-ghee. It is an indicator of how much volatile fatty acid can be extracted from fat through saponification.

10. The smaller is the saponification value, the higher will be the molecular weight of oil or fat.
a) True
b) False
Answer: a
Clarification: The smaller is the saponification value, the higher will be the molecular weight of oil or fat. The long chain fatty acids found in fats have a low saponification value because they have a relatively fewer number of carboxylic functional groups per unit mass of the fat as compared to short chain fatty acids.

areas of Engineering Chemistry for Experienced people,

Engineering Chemistry Multiple Choice Questions on “Molecular Aggregation”.

1. The main cause of molecular aggregation is ___________
a) Conduction
b) Induction
c) Inter-molecular interactions
d) Resonance
Answer: c
Clarification: The main cause of molecular aggregation is inter-molecular interactions. Inter-molecular interactions stem from the electric properties of atoms. Being the cause of molecular aggregation, inter-molecular forces are at the roots of chemistry and are the fabric of the world.

2. From where does the inter-molecular interaction originate?
a) From the electric properties of atoms
b) From the cohesive properties of atoms
c) From the magnetic properties of atoms
d) None of the mentioned
Answer: a
Clarification: Inter-molecular interactions originate from the electric properties of atoms. Inter-molecular interactions stem from the electric properties of atoms. Being the cause of molecular aggregation, inter-molecular forces are at the roots of chemistry and are the fabric of the world.

3. Which of the following is not a state of aggregation of matter?
a) Solid
b) Liquid and Gas
c) Plasma
d) None of the mentioned
Answer: d
Clarification: Solid, liquid, gas and plasma are the four states of aggregation of matter.

4. The process by which individual molecules form the defined aggregate is called ___________
a) Self-aggregation
b) Self-assembly
c) Aggregation
d) Assembly
Answer: b
Clarification: The process by which individual molecules form the defined aggregate is called self-assembly. Molecular self-assembly is the process by which molecules adopt a defined arrangement without guidance or management from an outside source.

5. In water, non-polar molecules tend to aggregate because they are forced to come into close proximity with each other due to ___________
a) Hydrophobic interactions
b) Hydrophilic interactions
c) Vander Waals interactions
d) Electrovalent interactions
Answer: a
Clarification: In water, non-polar molecules tend to aggregate because they are forced to come into close proximity with each other due to hydrophobic interactions.

6. Which type of molecules forms aggregate through a self-assembly process that is driven by the hydrophobic effect?
a) Electrophilic
b) Nucleophilic
c) Ambiphilic
d) None of the mentioned
Answer: c
Clarification: Ambiphilic molecules form aggregate through a self-assembly process that is driven by the hydrophobic effect. Molecular self-assembly is the process by which molecules adopt a defined arrangement without guidance or management from an outside source.

7. Why metallic bonds are non-directional?
a) Because there is a uniform distribution of mobile valence electrons around the kernels
b) Because there is a non-uniform distribution of mobile valence electrons around the kernels
c) There is no change in inter-kernel distance
d) There is a change in inter-kernel distance
Answer: a
Clarification: Metallic bonds are non-directional because there is a uniform distribution of mobile valence electrons around the kernels, provided there is no change in inter-kernel distance.

8. The number of molecules in a conduction band ___________
a) Increases with fall in temperature
b) Decreases with a rise in temperature
c) Increases with a rise in temperature
d) Does not depend upon the temperature
Answer: c
Clarification: The number of molecules in the conduction band increases with the rise in temperature. At high temperatures, randomness increases and more molecules move towards the conduction band.

Engineering Chemistry Multiple Choice Questions on “Conductivity”.

1. Conductivity is defined as the ability to carry ____________
a) Voltage
b) Resistance
c) Current
d) All of the mentioned
Answer: c
Clarification: Conductivity is defined as the ability to carry current. It is measured by the flow of electrons and charges through a conductor.

2. The reciprocal of conductivity is ____________
a) Viscosity
b) Resistivity
c) Turbidity
d) None of the mentioned
Answer: b
Clarification: The reciprocal of conductivity is resistivity. It is the measure of resistance provided in the path of electrons.

3. Which of the following is a specific conductivity reagent?
a) KCl
b) HCl
c) NaCl
d) MgCl₂
Answer: a
Clarification: KCl is a specific conductivity reagent. Specific conductance is a measure of the electric current in the water sampled carried by the ionized substances.

4. The internationally recommended unit for conductance is ____________
a) Poise
b) Dyne
c) Ohm
d) Siemens
Answer: d
Clarification: The internationally recommended unit for conductance is Siemens (S). 1 siemen = 1 ohm^-1

5. The cell constant is defined as the ratio of ____________
a) Area of either electrodes to the length between the electrodes
b) Length between the electrodes to the area of either electrodes
c) Length between the electrodes to the volume of either electrode
d) Resistivity to conductivity
Answer: b
Clarification: The cell constant is defined as the ratio of length between the electrodes to the area of either electrodes. It only depends upon the physical dimension of the cell.

6. Choose the correct order of molar ionic conductivities of the following ions.
a) Li⁺ < Na⁺ < K⁺ < Rb⁺
b) Li⁺ < K⁺ < Rb⁺ < Na⁺
c) Li⁺ < Na⁺ < Rb⁺ < K⁺
d) Li⁺ < Rb⁺ < Na⁺ < K⁺
Answer: a
Clarification: the correct order of molar ionic conductivity is- Li⁺ < Na⁺ < K⁺ < Rb⁺.

7. On which factor does the conductance of electrolytic solutions depend?
a) Temperature and pressure
b) Number of charge carriers
c) Dielectric constant of the solvent
d) All of the mentioned
Answer: d
Clarification: The factors on which conductance of electrolytic solution depends are- Temperature, pressure, number of charge carriers and dielectric constant of the solvent.

8. On dilution, the specific conductance ____________
a) Increases
b) Remains same
c) Decreases
d) None of the mentioned
Answer: c
Clarification: On dilution, the specific conductance decreases because dilution decreases the concentration of the solution.

9. The equivalent conductance of 0.1 H₂SO₄offering a resistance of 50ohms when placed in a conductivity cell whose electrodes are 1cm apart with a cross-sectional area of 2cm² at 25⁰C is?
a) 100 S cm² eq^-1
b) 1000 S cm² eq^-1
c) 10 S cm² eq^-1
d) 1 S cm² eq^-1
Answer: a
Clarification: The equivalent molar conductance is 100 S cm² eq^-1.
Conductivity = Conductance * cell constant
Cell constant = Length / Area

10. Dilution means an increase in the amount of ____________
a) Solute
b) Solvent
c) Electrolyte
d) All of the mentioned
Answer: b
Clarification: Dilution means an increase in the amount of solvent and hence it decreases the concentration of solute particles.

Engineering Chemistry Interview Questions and Answers for Experienced people focuses on “Electrophillic Substitution Reaction – 2”.

1. An activating substituent group activates ___________
a) Ortho position
b) Para position
c) Both ortho and para positions
d) Meta position
Answer: b
Clarification: An activating substituent group activates the ortho and para positions. It does not activate the meta position.

2. A deactivating substituent group directs ___________
a) Ortho position
b) Para position
c) Both ortho and para positions
d) Meta position
Answer: d
Clarification: A deactivating substituent group directs only meta position. It does not activate ortho or para positions.

3. Which of the following is ortho-para directing group?
a) –NHCOCH₃
b) –NO₂
c) –CN
d) –CHO
Answer: a
Clarification: NHCOCH₃ is ortho-para directing group. NO, CN and CHO are meta-directing groups.

4. Which of the following is a meta directing group?
a) –NHCOCH₃
b) –COOH
c) –OH
d) –OCH₃
Answer: b
Clarification: -COOH is a meta directing group. All the other options are ortho-para directing groups.

5. The most stable carbonium ion is ___________
a) Methyl carbonium ion
b) 2⁰ carbonium ion
c) 1⁰ carbonium ion
d) 3⁰ carbonium ion
Answer: d
Clarification: The most stable crbonium ion is 3⁰ carbonium ion. The least stable is primary or 1⁰ carbonium ion.

6. Which of the following has the highest activation of the benzene ring?
a) – NHCOCH₃
b) –OH
c) –NH₂
d) –C₆H₅
Answer: c
Clarification: –NH₂ among the following has the highest activation of the benzene ring. In cases where the substituents are esters or amides, they are less activating because they form resonance structure that pull the electron density away from the ring.

7. HBr reacts fastest with ___________
a) 2-methylpropan-2-ol
b) Propan-1-ol
c) Propan-2-ol
d) 2-methylpropan-1-ol
Answer: a
Clarification: HBr reacts fastest with 2-methylpropan-2-ol. Hydrogen bromide is the diatomic molecule and a colorless compound.

8. The compound that can be most readily sulphonated is ___________
a) Benzene
b) Nitrobenzene
c) Toluene
d) Chlorobenzene
Answer: c
Clarification: The compound that can be most readily sulphonated is toluene. Toulene contains CH₃ group.

9. In Cannizaro reaction, two molecules of aldehydes are reacted to produce ___________
a) Alcohol only
b) Carboxylic acid only
c) Alcohol and carboxylic acid
d) Alcohol, carboxylic acid and ketone
Answer: c
Clarification: In Cannizaro reaction, two molecules of aldehydes are reacted to produce alcohol and carboxylic acid using a hydroxide base.

areas of Engineering Chemistry for Interviews,