Category: Engineering Chemistry Objective Questions

Engineering Chemistry Multiple Choice Questions on “Detergents”.

1. Detergents are synthetic soaps like cleansing agents and are also known as ____________
a) Artifacts
b) Detritus
c) Syndets
d) Collagen
Answer: c
Clarification: Detergents are synthetic soaps like cleansing agents and are also known as syndets. They are a mixture of surfactants with “cleaning properties in dilute solutions.

2. Which of the following is also known as invert detergents?
a) Anionic detergents
b) Cationic detergents
c) Non-ionic detergents
d) All detergents comes in the category of invert detergents
Answer: b
Clarification: Cationic detergents are also known as invert detergents. In cationic detergent, the ammonium center is positively charged.

3. Which of the following is an example of non-ionic detergent?
a) Alkyltrimethyl ammonium chloride
b) Sodium salts of alkyl sulphates
c) Sodium salts of alkyl benzene sulphonic acids
d) Polyethers
Answer: d
Clarification: Polyethers derived from ethylene oxide is an example of non-ionic detergent.

4. Which of the following is an example of cationic detergent?
a) Alkyltrimethyl ammonium chloride
b) Sodium salts of alkyl sulphates
c) Sodium salts of alkyl benzene sulphonic acids
d) Polyethers
Answer: a
Clarification: Alkyltrimethyl ammonium chloride is an example of cationic detergent as its ammonium center is positively charged.

5. The detergents which bear negative charge at the soluble end of the chain are called anionic detergents.
a) True
b) False
Answer: a
Clarification: The detergents which bear negative charge at the soluble end of the chain are called anionic detergents. Typical anionic detergents are alkyl benzene sulfonates.

6. If the carbon chain is highly branched, the corresponding detergent will be ____________
a) Soft and biodegradable
b) Soft and non-biodegradable
c) Hard and biodegradable
d) Hard and non-biodegradable
Answer: d
Clarification: If the carbon chain is highly branched, the corresponding detergent will be hard and non-biodegradable.

7. If the carbon chain is linear, the corresponding detergent will be ____________
a) Soft and non-biodegradable
b) Soft and biodegradable
c) Hard and biodegradable
d) Hard and non-biodegradable
Answer: b
Clarification: If the carbon chain is linear, the corresponding detergent will be soft and biodegradable.

8. The % weight of detergent in washing powders is ____________
a) 5 – 10
b) 50 – 70
c) 15 – 30
d) 30 – 45
Answer: c
Clarification: The weight of detergent in washing powders is 15-30% which is quite high as compared to soaps.

9. Which of the following chemical is added in washing powder for keeping the dirt suspended in water?
a) Sodium silicate
b) Inorganic phosphate
c) Carboxy-methyl cellulose
d) Sodium perborate
Answer: c
Clarification: Carboxy-methyl cellulose is added in washing powder for keeping the dirt suspended in water while others are used for bleaching and stabilizing action.

10. Which of the following chemical is added in washing powder for keeping it dry?
a) Sodium silicate
b) Inorganic phosphate
c) Carboxy-methyl cellulose
d) Sodium perborate
Answer: a
Clarification: Sodium silicate is added in washing powder for keeping it dry while others are used for fragrance and cleansing action.

Engineering Chemistry Multiple Choice Questions on “Hybridisation”.

1. Hybridisation of Acetylene is _____________
a) sp
b) sp²
c) sp³
d) dsp²
Answer: a
Clarification: The Acetylene molecule is C₂H₂. It is sp hybridised.

2. Shape of PCl₅ molecule is _____________
a) Tigonal Planar
b) Linear
c) Trigonal bipyramidal
d) Tetrahedral
Answer: c
Clarification: PCl₅ is trigonal bipyramidal with its bond angle equal to 120⁰ and 90⁰.

3. Predict the shape of the H₂O compound based upon concepts of hybridisation.
a) Tetrahedral
b) Angular or bent structure
c) Trigonal Planar
d) Pyramidal
Answer: b
Clarification: The water molecule has angular or bent structure due to the presence of two lone pairs which repels strongly.

4. Which of the following is an example of sp₃d₂ hybridised molecule?
a) HCHO
b) ClO₄^–
c) SF₆
d) PF₅
Answer: c
Clarification: The SF₆ molecule is sp₃d₂ hybridised and the shape is regular octahedron.

5. Number of chlorine atoms which form equatorial bonds in PCl₅ molecule are ____________
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: In PCl₅ molecule, three chlorine atoms which lie at an angle of 120⁰ in the same plane forms equatorial bonds and the other two atoms forms axial bonds at an angle of 90⁰.

6. Due to unsymmetrical structure, the molecules having sp₃d hybridisation are ____________
a) more stable and more reactive
b) less stable and more reactive
c) more stable and less reactive
d) less stable and less reactive
Answer: b
Clarification: In sp₃d hybridisation, due to its unsymmetrical structure the molecules are in the shape of trigonal bipyramidal which is less stable and more reactive.

7. The bond angles in sp₃d₂ hybridisation is ____________
a) 90⁰
b) 120⁰
c) 109.5⁰
d) 180⁰
Answer: a
Clarification: All the bond angles are equal in sp₃d₂ hybridisation i.e., 90⁰.

8. The percentage p-character in sp₃ hybridisation is ____________
a) 25%
b) 50%
c) 75%
d) 66.67%
Answer: c
Clarification: In sp₃ hybridisation, s-character is 25% and p-character is 75%.

9. All the hybridised orbital are not equal in energy and shape.
a) True
b) False
Answer: b
Clarification: All the hybridised orbital are equal in energy and shape. They are similar to each other and possess equal energy levels.

10. The hybridisation of BeF_3- is _____________
a) sp₃
b) sp
c) sp₂
d) d₂sp₃
Answer: c
Clarification: The hybridisation of BeF₃– is sp₂ and it is trigonal planar.

Engineering Chemistry Multiple Choice Questions on “Crystallisation”.

1. Which of the following is not a common method used for purification?
a) Sublimation
b) Crystallisation
c) Electrolysis
d) Chromatography
Answer: c
Clarification: Sublimation, crystallisation, distillation, differential extraction and chromatography are some of the common methods used for purification. Electrolysis process is not used for this purpose.

2. Crystallisation is based on the ____________
a) Difference in melting point
b) Difference in boiling point
c) Difference in pressure
d) Difference in solubility
Answer: d
Clarification: Crystallisation is based on the difference in the solubility of the compound and the impurities in a suitable solvent.

3. Which of the following is the example of crystallisation process?
a) Purification of alum
b) Purification of sea water
c) Separation of gases from air
d) None of the mentioned
Answer: a
Clarification: An impure sample of alum or copper sulphate is purified by crystallisation process. Purification of sea water is example of distillation process.

4. At room temperature, the impure compound in crystallisation is ____________
a) Soluble
b) Sparingly soluble
c) Insoluble
d) None of the mentioned
Answer: b
Clarification: The impure compound is sparingly soluble in a solvent at room temperature but appreciably soluble at higher temperatures.

5. Which of the following is known as mother liquor?
a) Solvent
b) Solute
c) Solution
d) Filtrate
Answer: d
Clarification: The filtrate is also known as mother liquor. A mother liquor is the part of a solution that is left over after crystallization. It is encountered in chemical processes including sugar refining.

6. The solution of impure compound and solvent is concentrated to get ____________
a) Unsaturated solution
b) Undersaturaed solution
c) Saturated solution
d) Oversaturated solution
Answer: c
Clarification: The solution of impure compound and solvent is concentrated to get a saturated solution. The solution is filtered to remove insoluble impurities.

7. Insoluble impurities from solution during crystallization are removed by ____________
a) Drying
b) Filtration
c) Heating
d) Cooling
Answer: b
Clarification: Insoluble impurities from solution during crystallization are removed by filtration. The solution is filtered to remove insoluble impurities.

8. The solution which is obtained after filtration is ____________
a) Suspended solution
b) Clear solution
c) Colloidal solution
d) None of the mentioned
Answer: b
Clarification: The solution which is obtained after filtration is a clear solution. The solution is filtered to remove insoluble impurities.

9. Crystal phases can be inter-converted by varying ____________
a) Temperature
b) Pressure
c) Size
d) Viscosity
Answer: a
Clarification: Crystal phases can be inter-converted by varying factors such as temperature.

10. The nature of the crystallization process is governed by _____________
a) Thermodynamics
b) Kinetic factors
c) Thermodynamics and Kinetic factors
d) None of the mentioned
Answer: c
Clarification: The nature of the crystallization process is governed by both thermodynamic and kinetic factors.

Engineering Chemistry Multiple Choice Questions on “Molecular Spectroscopy”.

1. The different types of energies associated with a molecule are __________
a) Electronic energy
b) Vibrational energy
c) Rotational energy
d) All of the mentioned
Answer: d
Clarification: The different types of energies associated with a molecule are electronic energy, vibrational energy, rotational energy and translational energy.

2. During the motion, if the centre of gravity of molecule changes, the molecule possess __________
a) Electronic energy
b) Rotational energy
c) Translational energy
d) Vibrational energy
Answer: c
Clarification: During the motion, if the centre of gravity of molecule changes, the molecule possess translational energy. Translational refers to the movement in horizontal or vertical direction.

3. The correct order of different types of energies is __________
a) E_el >> E_vib >> E_rot >> E _tr
b) E_el >> E_rot >> E_vib >> E _tr
c) E_el >> E_vib >> E_tr >> E _rot
d) E_tr >> E_vib >> E_rot >> E _el
Answer: a
Clarification: The correct order is: E_el >> E_vib >> E_rot >> E _tr. Electronic enrgy is the highest whereas translational energy is the lowest.

4. The region of electromagnetic spectrum for nuclear magnetic resonance is __________
a) Microwave
b) Radio frequency
c) Infrared
d) UV-rays
Answer: b
Clarification: The region of the electromagnetic spectrum for nuclear magnetic resonance is radio frequency.

5. Which of the following is an application of molecular spectroscopy?
a) Structural investigation
b) Basis of understanding of colors
c) Study of energetically excited reaction products
d) All of the mentioned
Answer: d
Clarification: The various applications of molecular spectroscopy are- Structural investigation, basis of understanding of colors and study of energetically excited reaction products.

6. Select the correct statement from the following option.
a) Spectroscopic methods require less time and more amount of sample than classical methods
b) Spectroscopic methods require more time and more amount of sample than classical methods
c) Spectroscopic methods require less time and less amount of sample than classical methods
d) Spectroscopic methods require more time and less amount of sample than classical methods
Answer: c
Clarification: Spectroscopic methods require less time and less amount of sample than classical methods (1 mg).

7. The results obtained by spectroscopic methods are less reliable, less reproducible and incorrect than classical methods.
a) True
b) False
Answer: b
Clarification: The results obtained by spectroscopic methods are reliable, reproducible and correct than classical methods.

8. The transition zone for Raman spectra is __________
a) Between vibrational and rotational levels
b) Between electronic levels
c) Between magnetic levels of nuclei
d) Between magnetic levels of unpaired electrons
Answer: a
Clarification: The transition zone for Raman spectra is between vibrational and rotational levels. Raman spectroscopy is a spectroscopic technique used to observe vibrational, rotational, and other low-frequency modes in a system.

9. The criteria for electronic spin resonance is ____________
a) Periodic change in polarisability
b) Spin quantum number of nuclei > 0
c) Presence of unpaired electron in a molecule
d) Presence of chromophore in a molecule
Answer: c
Clarification: The criterion for electronic spin resonance is the presence of unpaired electron in a molecule. This spectroscopy is a method for studying materials with unpaired electrons.

10. Sample recovery is possible after spectroscopic analysis because the sample is not chemically affected.
a) True
b) False
Answer: a
Clarification: Sample recovery is possible after spectroscopic analysis because the sample is not chemically affected.

Engineering Chemistry Multiple Choice Questions on “Integrated Rate Equation and Half Lives”.

1. For a zero order reaction, the rate of reaction is independent of ___________
a) Temperature
b) Nature of reactants
c) Concentration of reactants
d) Effect of catalyst
Answer: c
Clarification: For a zero order reaction, the rate of reaction is independent of the concentration of reactants. The rate law for a zero order reaction is R = K.

2. The half-life period of zero order reaction is directly proportional to the ___________
a) Rate constant
b) Initial concentration of reactants
c) Final concentration of reactants
d) Concentration of products
Answer: b
Clarification: The half-life period of zero order reaction is directly proportional to the initial concentration of reactants. It is given by:
t_1/2 = [A₀] / 2K .

3. Which of the following is not an example of zero order reaction?
a) Photochemical combination of hydrogen and chlorine
b) Decomposition of ammonia over molybdenum surface
c) Thermal decomposition of HI on gold surface
d) Inversion of cane sugar in the presence of mineral acids
Answer: d
Clarification: Inversion of cane sugar in the presence of mineral acids is not an example of zero order reaction. All the other options are the example of zero order reaction.

4. For a first order reaction, the half life period is independent of the ___________
a) Initial concentration of the reactants
b) Final concentration of the reactants
c) Rate constant
d) Concentration of products
Answer: a
Clarification: For a first order reaction, the half life period is independent of the initial concentration of the reactants. t_1/2 = ln2 / K .

5. The thermal decomposition of nitrogen pentaoxide in the gaseous phase is the example of ___________
a) Zero order reaction
b) Half order reaction
c) First order reaction
d) Second order reaction
Answer: c
Clarification: The thermal decomposition of nitrogen pentaoxide in the gaseous phase is the example of first order reaction. A first order reaction depends on the concentration of only one reactant (a unimolecular reaction).

6. A compound decomposes according to the first order rate law with a half life period of 30 min. What will be the fraction of the remaining compound after 120 min?
a) 0.625
b) 0.0625
c) 0.25
d) 0.025
Answer: b
Clarification: k = 0.693/t_1/2
K = (2.303/t) log ([R₀]/[R_t]).

7. In a first order reaction, the time required for the completion of 99% is _______ for its 90% completion.
a) Same
b) Negligible
c) Thrice
d) Twice
Answer: d
Clarification: In a first order reaction, the time required for the completion of 99% is twice for its 90% completion. ln [A] = -kt + ln [A₀].

8. In a first order reaction, the time required for the completion of 99% is _______ times the time required for the completion of half of the reaction.
a) 5
b) 10
c) 20
d) 50
Answer: b
Clarification: In a first order reaction, the time required for the completion of 99% is ten times the time required for the completion of half of the reaction. ln [A] = -kt + ln [A₀].

9. The half life period of first order reaction is 15 min. Its rate constant will be equal to ___________
a) 0.0462 min^-1
b) 0.462 min^-1
c) 0.00462 min^-1
d) 0.562 min^-1
Answer: a
Clarification: k = 0.693/t_1/2

10. For a second order reaction, the rate is proportional to the square root of the concentration of the same reactant.
a) True
b) False
Answer: b
Clarification: For a second order reaction, the rate is proportional to the square of the concentration of the same reactant.

11. Which of the following is not an example of second order reaction?
a) Oxidation of hydrogen bromide
b) Saponification of ester
c) Acid catalyzed hydrolysis of ester
d) Gaseous decomposition of hydrogen iodide
Answer: c
Clarification: Acid catalyzed hydrolysis of an ester is not an example of second order reaction. All the other options are examples of second order reaction.

12. The half life period of n^th order reaction varies inversely to ____________ ^th power of the initial concentration of the reactants.
a) n
b) 2n
c) (n+1)
d) (n-1)
Answer: d
Clarification: The half life period of n^th order reaction varies inversely to (n-1) ^th power of the initial concentration of the reactants. The half-life of a reaction describes the time needed for half of the reactant to be depleted.

13. The order of the reaction, if the time of half-completion is changed from 50 sec to 25 sec, when the initial concentration is changed from 0.5 to 1M will be ___________
a) 0
b) 1
c) 2
d) 3
Answer: c
Clarification: T_b / T_a = (a/b)^n-1

Engineering Chemistry Multiple Choice Questions on “Polymer Classification”.

1. A polymer is any substance made up of many repeating units called ______________
a) Resins
b) Plastic
c) Mers
d) Blocks
Answer: c
Clarification: A polymer is any substance made up of many repeating units, building blocks called mers. Polymers are made up of many many molecules all strung together to form really long chains.

2. Small molecules which combine to form polymer are called ____________
a) Resins
b) Monomers
c) Plastic
d) Blocks
Answer: b
Clarification: Small molecules which combine with each other to form polymer are called monomers. Polymers are made up of many many molecules all strung together to form really long chains.

3. Select the incorrect statement from the following options.
a) When in form ready for further working, polymers are called resins
b) The chemical process leading to the formation of polymer is known as polymerisation
c) The number of monomeric units contained in the polymer is called the degree of polymerisation
d) Due to their small size, polymers are also called micro-molecules
Answer: d
Clarification: Due to their large size, polymers are also called macro-molecules. All the other options are correct. When in form ready for further working, polymers are called resins. The chemical process leading to the formation of polymer is known as polymerisation and the number of monomeric units contained in polymer is called degree of polymerisation.

4. Below threshold degree of polymerisation (DP), the polymer does not possess any strength and exist either as liquid resin or friable powder.
a) True
b) False
Answer: a
Clarification: Below threshold degree of polymerisation, the polymer does not possess any strength and exist either as liquid resin or friable powder. The number of monomeric units contained in the polymer is called degree of polymerisation.

5. The optimum DP value of cellulose is ____________
a) 150
b) 250
c) 400
d) 500
Answer: b
Clarification: The optimum DP value of cellulose is 250. The number of monomeric units contained in the polymer is called degree of polymerisation (DP).

6. The functionality of ethylene glycol is ____________
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: The functionality of ethylene glycol is 2. It is a colorless, odorless, viscous dihydroxy alcohol and has a sweet taste, but is poisonous if ingested.

7. Which of the following polymer is not classified under the category of configuration?
a) Syndiotactic
b) Atactic
c) Cross-linked
d) Isotactic
Answer: c
Clarification: Cross-linked polymer is not classified under the category of configuration. Syndiotactic, atactic and isotactic are classified under the category of configuration. Tacticity is the relative stereochemistry of adjacent chiral centers within a macromolecule.

8. Which of the following is not an example of semi-crystalline polymer?
a) HDPE
b) Nylon
c) Polyesters
d) LDPE
Answer: d
Clarification: LDPE is not an example of a semi-crystalline polymer. It is an example of a branched amorphous polymer. All the other options are examples of semi-crystalline polymer.

9. Styrene acrylonitrile is an example of ____________
a) Co-polymer
b) Homopolymer
c) Linear polymer
d) Amorphous polymer
Answer: a
Clarification: Styrene acrylonitrile resin is a copolymer plastic consisting of styrene and acrylonitrile. It is also known as SAN. It is widely used in place of polystyrene.

10. Which of the following polymer is not classified under the category of end use?
a) Fibers
b) Adhesives
c) Elastomers
d) Synthetic
Answer: d
Clarification: Synthetic polymer is not classified under the category of end use. It is classified under the category of origin. Fibers, adhesives and elastomers are classified under the category of end use.

11. Select the incorrect statement from the following option.
a) Thermosets are formed by condensation polymerisation reactions
b) Thermosets have 3-D, cross-linked network structure
c) Thermosets soften on heating and stiffen on cooling
d) Thermosets are generally insoluble in any solvent
Answer: c
Clarification: Thermosets do not soften on heating. All the other options are correct. Thermosets are formed by condensation polymerisation reactions, have 3-D, cross-linked network structure and are generally insoluble in any solvent.

12. Which one of the following is not an example of thermoplastic?
a) Polyvinyl chloride
b) Nylon
c) Polyesters
d) Epoxy
Answer: d
Clarification: Epoxy is not an example of thermoplastic. It is an example of a thermoset polymer. Polyvinyl chloride, nylon and polyesters are examples of thermoplastic.

13. The weight average molecular weight for PP given its degree of polymerisation as 10,000 will be ____________
a) 3,00,000 gm/mol
b) 4,20,000 gm/mol
c) 6,70,000 gm/mol
d) 8,40,000 gm/mol
Answer: b
Clarification: The weight average molecular weight for PP given its degree of polymerisation as 10,000 will be 4,20,000 gm/mol.

14. The cryoscopy refers to ____________
a) Osmotic pressure measurement
b) Elevation in boiling point measurement
c) Freezing point depression measurement
d) Increase in solubility measurement
Answer: c
Clarification: The cryoscopy refers to freezing point depression measurement. A technique for determining the molar concentration of a solution by measuring the freezing point.

15. Higher molecular weight polymers are tougher and more heat resistant.
a) True
b) False
Answer: a
Clarification: Higher molecular weight polymers are tougher and more heat resistant. Polymeric mixtures are far less miscible than mixtures of small molecule materials.