Category: Engineering Chemistry Objective Questions

Engineering Chemistry Multiple Choice Questions on “Engineering Materials”.

1. Based on the important category, concrete and fibre glass are the examples of ___________
a) Ceramics
b) Polymers
c) Composites
d) Semi-conductors
Answer: c
Clarification: Example of ceramic are Al₂O₃ and ZrO₂, Polymers are Thermoplastic and Thermosets, Composites are Fibre glass and Concrete and Semi-conductors are Silicon and Germanium.

2. Which of the following is not an inorganic functional material?
a) Ferroelectric
b) Reverse micelles
c) Magnetic field sensor
d) Light detectors
Answer: b
Clarification: Reverse micelles is an organic functional material.

3. Which of the following is not an aerospace material?
a) Plastics
b) Silica
c) Aluminium alloys
d) Polymers
Answer: d
Clarification: Polymers are not categorized under aerospace materials.

4. Select the incorrect statement from the following option.
a) Metals are extremely good conductors of heat and electricity
b) The properties of metal degrade rapidly with temperature
c) Metals have poor corrosion resistant
d) A polished metal surface has a dull appearance
Answer: d
Clarification: A polished metal surface has a lustrous appearance. All the other options are correct. Metals are extremely good conductors of heat and electricity, its properties degrade rapidly with temperature and have poor corrosion resistance.

5. Which one of the following is the best heat and corrosion resistant material?
a) Metals
b) Ceramics
c) Polymers
d) Semi-conductors
Answer: b
Clarification: Ceramics materials generally consist of oxides, nitrides, carbides, silicates or borides of various metals. Their heat and corrosion resistance is best compared to metals and polymers.

6. Polymers are used in the chemical industry because of their ___________
a) Inert nature
b) Light weight
c) Low cost
d) Easiness in fabricability
Answer: a
Clarification: Polymers are used in the chemical industry because of their inert nature.

7. Select the incorrect statement from the following option.
a) Composites are optically opaque materials
b) Carbon fiber reinforced composite materials are not used in space vehicles
c) Re-cyclability of composite material is poor
d) Processing of composite material is difficult
Answer: b
Clarification: Carbon fiber reinforced composite materials are used in space vehicles. All the other options are correct. Composites are optically opaque materials, re-cyclability of composite material is poor and its processing is difficult.

8. Which type of material expands and contract in response to an applied electric field?
a) Advanced material
b) Smart material
c) Biomaterial
d) Nanomaterial
Answer: b
Clarification: Smart materials(Piezoelectric) expand and contract in response to an applied electric field(or voltage).

9. Which one of the following is non-linear material?
a) Zirconium oxide
b) Magnetite
c) Maghemite
d) Lithium niobate
Answer: d
Clarification: Lithium niobate is a non-linear material used for telecommunication.

10. Which of the following is not an application of nanomaterials?
a) TV and computer monitors
b) Cardiology
c) Magnetic Resonance Imaging (MRI)
d) Sunscreens and fuel cells
Answer: b
Clarification: Cardiology is not an application of nanomaterials. TV and computer monitors, Magnetic Resonance Imaging (MRI) and sunscreens and fuel cells are some of the applications of nanomaterials.

Engineering Chemistry Multiple Choice Questions on “Carbocations”.

1. Stability of free radicals can be explained on the basis of __________
a) Inductive effect
b) Electromeric effect
c) Hyperconjugation
d) Mesomeric effect
Answer: c
Clarification: Stability of free radicals can be explained on the basis of hyperconjugation effect and ease of formation.

2. The hybridisation of carbocation is __________
a) Sp
b) Sp²
c) Sp³
d) Sp³d
Answer: b
Clarification: The hybridisation of carbocation is sp². A carbocation is molecule in which a carbon atom bears three bonds and a positive charge.

3. Arrange the following carbocations in the order of increasing stability.
a) Benzyl > 3⁰ > 2⁰ > 1⁰
b) Benzyl > 1⁰ > 2⁰ > 3⁰
c) 3⁰ > 2⁰ > 1⁰ > Benzyl
d) 1⁰ > 2⁰ > 3⁰ > Benzyl
Answer: a
Clarification: The correct stability order of carbocation is- Benzyl > 3⁰ > 2⁰ > 1⁰. Benzyl carbocation is the most stable and 1⁰ carbocation is least stable.

4. The shape of carbocation is __________
a) Pyramidal
b) Bent
c) Linear
d) Trigonal planar
Answer: d
Clarification: The shape of the carbocation is a trigonal planar. It is sp² hybridised. A carbocation is molecule in which a carbon atom bears three bonds and a positive charge.

5. Carbonium ions are the intermediates in which the positive charge is carried by the carbon atom with ___________ electrons in the valence shell.
a) 6
b) 5
c) 4
d) 3
Answer: a
Clarification: Carbonium ions are the intermediates in which the positive charge is carried by the carbon atom with six electrons in the valence shell. It is an organic cation in which the positive charge is located on a carbon atom.

6. Positive charge of carbocations can be dispersed by __________
a) (+I) effect of alkyl group
b) Resonance in allyl or benzyl carbocation
c) Hyperconjugation in 1⁰, 2⁰ and 3⁰ carbocations
d) All of the mentioned
Answer: d
Clarification: Positive charge of carbocations can be dispersed by (+I) effect of alkyl group or by resonance in allyl or benzyl carbocation or by hyperconjugation in 1⁰, 2⁰ and 3⁰ carbocations.

7. Alkyl substitution at the carbon bearing positive charge stabilizes carbocations.
a) True
b) False
Answer: a
Clarification: Alkyl substitution at the carbon bearing positive charge stabilizes carbocations. Carbocation is a molecule in which a carbon atom bears three bonds and a positive charge.

8. The formal charge at the carbocation is equal to __________
a) -1
b) 0
c) +1
d) +2
Answer: c
Clarification: The formal charge at the carbocation is equal to +1. A carbocation is molecule in which a carbon atom bears three bonds and a positive charge.

9. The homolytic bond dissociation energy is inversely proportional to the __________
a) Bond length
b) Ease of formation
c) Dipole moment
d) All of the mentioned
Answer: b
Clarification: The homolytic bond dissociation energy is inversely proportional to the ease of formation of free radicals. Free radicals do not carry any charge.

10. Which of the following free radical has the maximum ease of formation?
a) 1⁰
b) 2⁰
c) 3⁰
d) CH₃
Answer: c
Clarification: 3⁰ free radical has the maximum ease of formation. Free radicals do not carry any charge.

Engineering Chemistry Multiple Choice Questions on “Two Component System”.

1. What is the definition of a lower critical temperature?
a) The minimum temperature at which equilibrium is achieved
b) The lowest temperature at which two components will attain vapor state
c) The lowest temperature at which two components sublimates
d) The lowest temperature at which two components form a mixture
Answer: a
Clarification: The lower critical solution temperature is the lowest temperature at which two components are insoluble and attains equilibrium. Conversely, the upper critical solution temperature is the highest temperature at which two components are soluble as single phase.

2. With reference to a two component system, a vapor line indicates ________
a) A region where the temperature and pressure remains stable
b) A region where the solubility remains constant
c) An area below which components do not mingle
d) An area above which both the components mingle to form a single mixture
Answer: a
Clarification: A vapor line is another name for an isotherm. It represents constant temperature and constant pressure relation at which the equilibrium is obtained. A vapor line helps in finding out equilibrium concentrations.

3. With reference to a two component system, an isobar indicates _______
a) A region where the temperature remains constant
b) An area below phase end products remains
c) An area above which only the liquid vapors remains
d) A region where the composition lies in equilibrium
Answer: d
Clarification: An isobar is a line that indicates a region where the composition remains constant and in the equilibrium state. In a two component solid-liquid system, it is drawn vertically covering a wide range of temperatures.

4. Which of the following statements is not true for a system that has reached the eutectic temperature?
a) The system is a open system
b) The system is a miscible fluid
c) Components lie in liquid state
d) A region between sublimation and eutectic curve lies stable
Answer: a
Clarification: The eutectic temperature is achieved by heating a liquid in the point where the density of the liquid state is equal to the density of the vapor state. Here, the interface between the liquid and vapor vanishes resulting in a miscible fluid. This can only occur in a closed system (otherwise the vapor would escape into the surroundings).

5. What does the term “ metastable” indicate?
a) A place where the composition of the system remains constant
b) A place where the pressure remains constant with low value
c) An area below which a vapor- liquid mixture is obtained
d) A region where the temperature remains constant
Answer: d
Clarification: A metastable is a line that indicates a place where the temperature remains constant. In a two component system it is drawn horizontally covering a wide range of compositions.

6. Which of the following statements are correct about the equilibrium point on a two component system?
a) All the compounds are liquid
b) The boiling point of the mixture is less than the boiling points of the individual compounds
c) The degree of freedom is 0
d) Invariant reaction takes place
Answer: b
Clarification: The Eutectic point on a phase diagram is a point at which the mixture boils at a temperature lower than that of the pure substances. The substance attains equilibrium state.

7. During metastable state, the size of the particle_________
a) Increases
b) Decreases
c) Won’t change
d) Not related
Answer: a
Clarification: Since for a system to be in equilibrium condition, its size automatically increases by Mond’s effect which is why the particle’s size increases during metastable state due to changes in molecular size.

8. Which of the following is not responsible for phase deposition?
a) Container wall
b) Grain size
c) Stacking effect
d) Disjoints
Answer: a
Clarification: Under equilibrium conditions, the molecules adhere to the center of the container and under non-equilibrium conditions, the vapors are let out through the walls of the container hence the container wall has no role to play in the reaction.

9. Where does the particle growth occur?
a) Movement of grains
b) Movement of equilibrium liquid
c) Equilibrium mixture
d) Non equilibrium mixture
Answer: a
Clarification: During diffusion process in the transformation of solid to vapor state, the molecules move from regions of higher concentration to regions of lower concentrations and hence growth takes place n a controlled way.

10. Overall transformation rate changes with temperature as follows under what condition?
a) Decreases
b) Increases then decreases
c) Follows a linear path
d) Increases with temperature
Answer: b
Clarification: According to the phase graph, at the solidus region, the temperature is high. As the melting point increases, the temperature decreases and vice versa (applicable to all component systems).

Engineering Chemistry Multiple Choice Questions on “Elastomers – 1”.

1. Which of the following is not the essential structural requirement of an elastomer?
a) Long flexible chains
b) Weak intermolecular forces
c) Rigidity in structure
d) Occasional cross-linking
Answer: c
Clarification: Elastomers are a long flexible chain; they are not rigid in nature.

2. Natural rubber is ____________
a) Poly isoprene
b) Ethylene glycol
c) Butadiene
d) Acrylonitrile
Answer: a
Clarification: Natural rubber is poly (cis) isoprene. Malaysia is one of the leading producers of rubber.

3. The structural formula of isoprene is ____________
a) 2-benzyl-1,3-butadiene
b) 2-methyl-1,3-butadiene
c) 3-benzyl-1,2-butadiene
d) 3-methyl-1,2-butadiene
Answer: b
Clarification: The structural formula of isoprene is 2-methyl-1,3-butadiene. It is a common organic compound with the formula CH₂=C(CH₃)-CH=CH₂. In its pure form, it is a colorless volatile liquid. Isoprene is produced by many plants.

4. Select the incorrect statement from the following option.
a) Raw rubber is weak and have low tensile strength
b) Raw rubber is attacked by oxidizing agent
c) In organic solvents, it undergoes swelling and disintegration
d) Raw rubber is durable
Answer: d
Clarification: Raw rubber is not durable due to its oxidation in air. All the other options are correct.

5. The temperature at which raw rubber is heated for the vulcanisation process is ____________
a) 0-10 ⁰C
b) 10-50 ⁰C
c) 50-100 ⁰C
d) 100-140 ⁰C
Answer: d
Clarification: The temperature at which raw rubber is heated for the vulcanisation process is 100-140 ⁰C. A typical vulcanization temperature for a passenger tire is 10 minutes at 177 °C.

6. In the vulcanisation process, the raw rubber is heated with ____________
a) Oxygen
b) Sulphur
c) Carbon
d) Calcium
Answer: b
Clarification: In the vulcanisation process, the raw rubber is heated with sulphur. It is a chemical process for converting natural rubber or related polymers into more durable materials via the addition of sulphur or other equivalent curatives or accelerators. These additives modify the polymer by forming cross-links (bridges) between individual polymer chains.

7. The amount of sulphur added determines the extent of stiffness of vulcanised rubber.
a) True
b) False
Answer: a
Clarification: The amount of sulphur added determines the extent of the stiffness of vulcanised rubber. This sulphur modify the polymer by forming cross-links (bridges) between individual polymer chains.

8. Select the incorrect statement from the following option.
a) Vulcanised rubber has excellent resilience
b) Vulcanised rubber has only slight tackiness
c) Vulcanised rubber has high elasticity
d) Vulcanised rubber has tensile strength 10 times more than raw rubber
Answer: c
Clarification: Vulcanised rubber has low elasticity and decreases with the extent of vulcanisation. All the other options are correct.

9. The tensile strength(kg/cm ²) of vulcanised rubber is ____________
a) 200
b) 2000
c) 500
d) 5000
Answer: b
Clarification: The tensile strength of vulcanised rubber is 2000 kg/cm ². Cross-linking introduced by vulcanization prevents the polymer chains from moving independently.

10. The percentage of butadiene and styrene in Buna-S is ____________
a) 50% each
b) 60% and 40% respectively
c) 80% and 20% respectively
d) 75% and 25% respectively
Answer: d
Clarification: The percentage of butadiene and styrene in Buna-S is 75% and 25% respectively. These materials have good abrasion resistance and good aging stability when protected by additives.

Tough Engineering Chemistry Questions and Answers focuses on “Determination of Calorific Value of Gaseous and Volatile Liquid Fuels”.

1. Apart from Boy’s calorimeter, which process can be used for finding the calorific value of gases?
a) Bomb calorimeter
b) Heat balance calorimeter
c) Junker’s calorimeter
d) Heat flow calorimeter
Answer: c
Clarification: In this calorimeter, a known volume of gas is burned and imparting heat with maximum efficiency so as to find out the rise in temperature of the water.

2. How many burners are used in Boy’s calorimeter for finding the calorific value for gases and volatile liquid?
a) 4
b) 3
c) 1
d) 2
Answer: d
Clarification: The two flat flame burners are situated in chimney in the calorimeter. Two are used to impart efficient heat to find the rise in temperature.

3. The water condensed from the products of combustion is collected and settles in the Boy’s bomb itself.
a) True
b) False
Answer: b
Clarification: The water condensed from the products of combustion is removed through the side tube, so that its volume can be measured.

4. What is the use of insulating baffles in Boy’s thermometer?
a) To deflect the gas upwards so that it can pass through the holes of the calorimeter
b) To make the gas stable in copper tube
c) To decrease the rate of heating
d) To decrease the temperature of cold water which is obtained from inlet
Answer: a
Clarification: The cold water is sent from the water inlet and the gas is in a chimney which is deflected downwards, for the mixing of gases with water the gases deflected upwards.

5. What is the basic formula for calculating GCV in Boy’s calorimeter?
a) GCV = w(t₂+t₁)/V
b) GCV = w(t₂-t₁)/V
c) GCV = V(t₂+t₁)/w
d) GCV = V(t₂-t₁)/w
Answer: b
Clarification: In this (t₂-t₁) is the rise in temperature of the water which is multiplied to its weight and divide by the volume of gas burned at standard conditions of temperature and pressure. Its unit is in cal/m³.

6. What is the main source of error in conducting the experiment of Boy’s calorimeter?
a) The temperature of gas is greater than the room temperature
b) Due to presence of moisture
c) The mass of the gas is comparatively low
d) Due to the rise in ambient temperature
Answer: d
Clarification: Ambient temperature is the temperature of the air in the room. If this temperature changes suddenly, it will affect the readings of the experiment.

7. What is the basic formula for calculating NCV in Boy’s Calorimeter?
a) NCV = GCV – (m/V)×587
b) NCV = GCV – (V/m)×587
c) NCV = GCV + (m/V)×587
d) NCV = GCV + (V/m)×587
Answer: a
Clarification: Since the calorific value of gases in expressed on volumetric bases, it is essential to define conditions of temperature and pressure. Its unit is same as GCV which is cal/m³.

8. Determine the gross calorific value (GCV) of a gas, having following results after using Boy’s calorimeter: Volume of the gaseous fuel burnt = 0.093 m³, weight of the water used for cooling = 30.5 kg, Weight of the steam condensed = 31 gm, Temperature of inlet water = 26.1 ⁰C and Temperature of outlet water = 36.5 ⁰C?
a) 3100 Kcal/m³
b) 3310 cal/gm
c) 3410 Kcal/m³
d) 3780 Kcal/gm
Answer: c
Clarification: By using relation GCV = w(t₂ – t₁)/ V,
We get GCV = 3410 Kcal/m³.

9. Determine net calorific value of a volatile liquid fuel, having following results after using Boy’s calorimeter: Volume of the liquid fuel burnt = 0.152 m3, weight of the water used for cooling = 48.5 kg, Weight of the steam condensed = 91 gm, Temperature of inlet water = 35.2 ⁰C and Temperature of outlet water = 42.6 ⁰C?
a) 2361.184 Kcal/m³
b) 2013.348 Kcal/m³
c) 2256.95 Kcal/m³
d) 2185.679 Kcal/m³
Answer: b
Clarification: By using relation GCV = w(t₂ – t₁)/ V,
We get GCV = 2361.184 Kcal/m³
NCV = GCV – (m/V)×587
NCV = 2013.348 Kcal/m³.

10. Junker’s calorimeter cannot be used for measuring the calorific value of volatile liquid fuel.
a) True
b) False
Answer: a
Clarification: Junker’s calorimeter can be used for measuring the calorific value of both gaseous and volatile liquid fuel. Also Boy’s calorimeter is used for measuring the calorific values of both gaseous and volatile liquid fuel. For solids and non-volatile liquid fuel Bomb calorimeter is used.

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