Category: Engineering Chemistry Objective Questions

Engineering Chemistry Multiple Choice Questions on “Optical Isomerism”.

1. Optical activity is the property in which a substance is unable to rotate the plane of polarisation.
a) True
b) False
Answer: b
Clarification: Optical activity is the property in which a substance is able to rotate the plane of polarisation of plane polarised light.

2. Select the incorrect statement from the following option.
a) Racemic modification is an equimolar mixture of dextrorotatory and levorotatory isomers
b) Meso compounds contain more than one chiral carbon centre
c) Meso compounds are externally compensated
d) Racemic mixture is designated as dl-pair
Answer: c
Clarification: Meso compounds are internally compensated form whereas racemic mixtures are externally compensated modification. All the other options are correct.

3. Select the correct statement from the following option.
a) Meso compound possess both plane of symmetry and centre of symmetry
b) Meso compound possess either plane or centre of symmetry
c) Meso compound does not possess either plane or centre of symmetry
d) Meso compounds are externally compensated form
Answer: b
Clarification: Meso compound possess either plane or centre of symmetry. They do not possess both plane of symmetry and centre of symmetry and also they are internally compensated form.

4. How many optical isomers are possible in a compound with one chiral carbon?
a) 5
b) 4
c) 2
d) 3
Answer: d
Clarification: A compound with one chiral carbon has three optical isomers (+), (-) and (±).

5. Which of the following compound would show optical isomerism?
a) CH₃ – CH(OH) COOH
b) H₂N CH(CH₃)₂
c) (CH₃)₂ CHCHO
d) H₂N CH₂ COOH
Answer: a
Clarification: CH₃ – CH(OH) COOH will show optical isomerism as it possess chiral carbon.

6. The number of configurational isomers of molecules having (n) different chiral carbons is ____________
a) 2n
b) 2ⁿ
c) 2^n-1
d) 2ⁿ⁺¹
Answer: b
Clarification: The number of configurational isomers of molecules having (n) different chiral carbons is 2ⁿ. It is calculated mathematically.

7. The number of racemic forms of molecules having (n) different chiral carbons is ____________
a) 2n
b) 2ⁿ
c) 2^n-1
d) 2ⁿ⁺¹
Answer: c
Clarification: The number of racemic forms of molecules having (n) different chiral carbons is 2^n-1 whereas the number of configurational isomers of molecules having (n) different chiral carbons is 2ⁿ.

8. For a molecule with two like chiral carbon atoms, the number of optically inactive form is?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: For a molecule with two like chiral carbon atoms, the number of optically inactive form is one. This is always used for the calculation of optically inactive forms.

9. For a molecule with two like chiral carbon atoms, the number of optically active form is?
a) 4
b) 3
c) 1
d) 2
Answer: d
Clarification: For a molecule with two like chiral carbon atoms, the number of optically active form is two. This is always used for the calculation of optically active forms.

10. Find the number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃.
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: The number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃ is four. This is calculated by the formula 2ⁿ⁺¹.

11. The necessary condition for showing optical activity is the chirality of a molecule as a whole.
a) True
b) False
Answer: a
Clarification: The necessary condition for showing optical activity is the chirality of a molecule as a whole. It is the most important condition for showing optical activity.

12. The sufficient condition for showing optical activity is _______________
a) Molecule should have measurable amount of optical activity only
b) Polarimeter should have capacity of recording low-degree optical activity only
c) Chirality of molecule as a whole only
d) Both, the molecule should have measurable amount of optical activity and polarimeter should have capacity of recording low-degree optical activity
Answer: d
Clarification: The molecule should have measurable amount of optical activity and polarimeter should have capacity of recording low-degree optical activity.

13. Which of the following is an example of optically active compounds without chirality?
a) Tartaric acid
b) Sulfhonium salt
c) Diphenic acid
d) Glyceraldehyde
Answer: b
Clarification: Sulphhonium salt is an example of optically active compounds without chirality. It is a special case. There are various such molecules which are optically active compounds without chirality.

14. Spiranes exhibit optical isomerism because of restricted rotation.
a) True
b) False
Answer: a
Clarification: Spiranes exhibit optical isomerism because of restricted rotation. They cannot rotate about any axis.

15. Which of the following is not optically active compound?
a) 1,7- Dicarboxylic spirocycloheptane
b) 1,3- Diphenyl propadiene
c) Meso-tartaric acid
d) Glyceraldehyde
Answer: c
Clarification: Meso-tartaric acid is an optically inactive molecule with a chiral carbon atom. It is a special case of optical activity.

Engineering Chemistry Multiple Choice Questions on “Column Chromatography”.

1. Column chromatography is based on the principle of _______________
a) Ion-exchange
b) Exclusion principle
c) Differential adsorption
d) Absorption
Answer: c
Clarification: Column chromatography is based on the principle of differential adsorption. It means different compounds have different tendencies to get adsorb on a particular surface.

2. Arrange the following compounds in order of their increasing adsorption tendencies.
a) Cellulose >> starch >> calcium carbonate >> alumina
b) Cellulose >> starch >> alumina >> charcoal
c) Charcoal >> cellulose >> alumina >> starch
d) Calcium carbonate >>; alumina >> starch >> cellulose
Answer: a
Clarification: Cellulose >> starch >> calcium carbonate >> charcoal >> alumina is the correct order of increasing adsorption tendencies.

3. What is the factor responsible for the separation in column chromatography?
a) Polarity differences between the solvent
b) Polarity differences between the solute
c) Polarity indifference between the solvent
d) Polarity indifference between the solute
Answer: b
Clarification: Polarity differences between the solute molecules are responsible for the separation in column chromatography.

4. Select the correct statement from the following options.
a) The lesser the polarity of solute, more strongly it will be adsorbed on a polar surface
b) The greater the polarity of solute, more weakly it will be adsorbed on a polar surface
c) The greater the polarity of solute, more strongly it will be adsorbed on a polar surface
d) All of the mentioned option
Answer: c
Clarification: The greater the polarity of solute, more strongly it will be adsorbed on a polar surface.

5. The correct order of increasing strength of adsorption is ____________
a) Alkanes >> Esters >> Aldehydes >> Phenols >> Ketones
b) Aldehydes >> Phenols >> Ketones >> Esters >> Alkanes
c) Aldehydes >> Ketones >> Esters >> Alkanes >> Phenols
d) Alkanes >> Esters >> Ketones >> Aldehydes >> Phenols
Answer: d
Clarification: the correct order increasing strength of adsorption is: Alkanes >> Esters >> Ketones >> Aldehydes >> Phenols.

6. The components of the mixture in column chromatography are eluted in order of ____________
a) Increasing polarity and decreasing distribution ratio
b) Increasing polarity and increasing distribution ratio
c) Decreasing polarity and increasing distribution ratio
d) Decreasing polarity and decreasing distribution ratio
Answer: b
Clarification: The components of the mixture in column chromatography are eluted in order of increasing polarity and increasing distribution ratio.

7. The elution power of a solvent is determined by ____________
a) Its overall polarity
b) The polarity of the stationary phase
c) The nature of the sample components
d) All of the mentioned
Answer: d
Clarification: The elution power of a solvent is determined by its overall polarity, the polarity of the stationary phase and the nature of the sample components.

8. Which of the following is separated through column chromatography?
a) Chlorophyll and carotenoids
b) Inorganic cations or complexes
c) Sugar derivatives
d) Amino acids formed by hydrolysis of a protein molecule
Answer: a
Clarification: The main application of column chromatography is to separate plant pigment (Chlorophyll and carotenoids).

9. The mixture of petroleum ether and benzene is used in the elution ratio of ____________
a) 1 : 2
b) 1 : 5
c) 1 : 9
d) 1 : 12
Answer: c
Clarification: The mixture of petroleum ether and benzene is used in the elution ratio of 1:9 (V/V).

10. Chloroform fraction is eluted from the column by passing chloroform through the column which acts as ____________
a) Eluter
b) Eluant
c) Elution
d) None of the mentioned option
Answer: c
Clarification: Chloroform fraction is eluted from the column by passing chloroform through the column which acts as eluant. The eluate is the mobile phase leaving the column. The eluent is the solvent that carries the analyte.

Engineering Chemistry online quiz focuses on “Mesomeric Effect”.

1. In mesomeric effect, the electrons are transferred from __________
a) A multiple bonds to an atom
b) A multiple bonds to a single covalent bond
c) An atom with a lone pair to the adjacent single covalent bond
d) All of the mentioned
Answer: d
Clarification: In the mesomeric effect, the electrons are transferred from multiple bonds to an atom, a multiple bond to a single covalent bond and an atom with the lone pair to the adjacent single covalent bond.

2. Which of the following is a resonance effect?
a) Inductive effect
b) Electromeric effect
c) Mesomeric effect
d) Inductomeric effect
Answer: c
Clarification: Mesomeric effect is also known as the resonance effect. The mesomeric effect is a permanent effect and operates in compounds containing at least one double bond.

3. The phenomenon in which 2 or more structures, involving identical position of atoms can be written for a particular molecule, is called __________
a) Conjugation
b) Resonance
c) Hyper conjugation
d) Vibration
Answer: b
Clarification: The phenomenon in which 2 or more structures, involving the identical position of atoms can be written for a particular molecule, is called resonance. The mesomeric effect is a permanent effect and operates in compounds containing at least one double bond.

4. Select the incorrect option from the following option.
a) Resonating structures have a real existence
b) The actual structure lies between various possible resonating structures
c) Resonating structures are useful as they allow us to describe molecules
d) None of the mentioned
Answer: a
Clarification: Resonating structures have no real existence. All the other options are correct. The actual structure lies between various possible resonating structures, resonating structures are useful as they allow us to describe molecules.

5. The resonance energy is defined as a difference in energy between __________
a) Two consecutive resonating structures
b) Resonance hybrid and most unstable resonating structure
c) Resonance hybrid and most stable resonating structure
d) First and last resonating structures
Answer: c
Clarification: The resonance energy is defined as a difference in energy between resonance hybrid and most stable resonating structure. The resonance energy of a compound is a measure of the extra stability of the conjugated system compared to the corresponding number of isolated double bonds.

6. Which of the following is an application of the mesomeric effect?
a) Dipole moment
b) Strength of acids and bases
c) Bond length
d) All of the mentioned
Answer: d
Clarification: Dipole moment, strength of acids and bases and bond length are some of the applications of the mesomeric effect.

7. Dipole moment of CH₃-CH₂-Cl > CH₂=CH-Cl.
a) True
b) False
Answer: a
Clarification: Dipole moment of CH₃-CH₂-Cl > CH₂=CH-Cl. This is due to resonance in vinyl chloride.

8. Select the correct statement from the following option.
a) Benzene ring has two different types of bond length for single and double bonds
b) All the bond length in benzene ring is equal due to hyperconjugation
c) All the bond length in benzene ring is equal due to resonance
d) All of the mentioned
Answer: c
Clarification: All the bond length in the benzene ring is equal due to resonance. All the bond lengths are equal in benzene ring i.e. double bond as well as single bond due to resonance.

9. Greater the number of resonating structures for a given intermediate __________
a) Less will be its stability
b) More will be its stability
c) It will not affect its stability
d) None of the mentioned
Answer: b
Clarification: Greater the number of resonating structures for a given intermediate, more will be its stability. More resonating structure increases stability.

10. Phenyl group show __________
a) (+M) effect
b) (+E) effect
c) (+I) effect
d) (-M) effect
Answer: d
Clarification: Phenyl group show (-M) effect. The phenyl group or phenyl ring is a cyclic group of atoms with the formula C₆H₅.

To practice all areas of Engineering Chemistry for online Quizzes,

Engineering Chemistry Multiple Choice Questions on “Phases and Components – 1”.

1. What is the number of phases and components present in the following reaction?
MgCO₃ ⟶ MgO +CO₂ (Gas)
a) 3, 2
b) 1, 3
c) 2, 3
d) 1, 2
Answer: a
Clarification: Since we have two solids and one gaseous component in the above reaction, we have 3 phases. Also since under equilibrium, the gaseous component may evaporate, we have only two components.

2. What is the number of phases and components in the following reaction?
NH₄Cl ⟶ NH₃ (Gas) + HCl (Gas)
a) 1, 2
b) 2, 1
c) 3, 1
d) 3, 2
Answer: b
Clarification: Since we have one solid component and two gaseous components (n gaseous components contribute to one phase and n solid components contribute to n phases), we have 2 phases. Since the gaseous components may evaporate, we have only one component.

3. What is the number of phases and components in an aqueous solution of glucose?
a) 2, 1
b) 3, 2
c) 1, 3
d) 1, 2
Answer: d
Clarification: Since we have a solid mixed in a liquid, we have two components. Since the above solution is homogeneous, we have only one phase.

4. What is the number of phases and components in the following reaction?
Fe + H₂O (Gas) ⟶ FeO + H₂ (Gas)
a) 3, 3
b) 2, 3
c) 1, 3
d) 2, 2
Answer: a
Clarification: Here we have two solids and two gases, so the number of phases is two. Also the number of species is 4 (i.e N=4) and the number of reactions is one (i.e E=1). Therefore, the number of components is C = N-E = 4-1 = 3.

5. What is the number of components in a nonreactive system of KCl-NaCl-H₂O?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: Since it is a nonreactive system, we should also take the ions into consideration (sodium ions, potassium ions, chlorine ions) and also the individual components (NaCl, KCl, H₂0). Hence N=6. Also E=3. Therefore the number of components are C = N-E = 6-3 = 3.

6. What is the number of components present in a reactive system of NaCl-KBr-H₂0?
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: Since it is a reactive system, we have N=9 (taking all the ions into consideration along with individual species) and also we have E=5 (considering all the neutral reactions). Therefore C = N – E = 9-5 = 4.

7. What is the degree of freedom for a water system?
a) 1
b) 2
c) 4
d) 0
Answer: d
Clarification: We know that the number of phases and components in a water system is 3 and 1 respectively. Since it is a one component system, F = C-P+2 = 1-3+2 = 0.

8. Calculate the degree of freedom for the following reaction.
CaCO₃ ⟶ CaO + CO₂ (Gas)
a) 1
b) 2
c) 3
d) 0
Answer: a
Clarification: We know that the number of phases and components in the above reaction is 3 and 2 respectively. Since it is a one component system, F = C-P+2 = 2-3+2 = 1.

9. Calculate the degree of freedom for the following reaction.
NaCl (Solid) ⟶ NaCl (Liquid) + H₂0 (Gas)
a) 2
b) 3
c) 1
d) 0
Answer: c
Clarification: We know that the number of phases and components in the above reaction is 3 and 2 respectively. Since it is a one component system, F = C-P+2 = 2-3+2 = 1.

10. Calculate the degree of freedom for an aqueous solution of glucose.
a) 1
b) 2
c) 4
d) 3
Answer: d
Clarification: We know that the number of phases and components in the above reaction is 1 and 2 respectively. Since it is a one component system, F = C-P+2 = 2-1+2 = 3.

11. Calculate the degree of freedom for the following reaction.
N₂O₄ (Gas) ⟶ 2NO₂ (Gas)
a) 1
b) 2
c) 3
d) 0
Answer: b
Clarification: We know that the number of phases and components in the above reaction is 1 and 1 respectively. Since it is a one component system, F = C-P+2 = 1-1+2 = 2.

12. Calculate the degree of freedom for a saturated solution of sodium chloride.
a) 0
b) 1
c) 3
d) 2
Answer: d
Clarification: We know that the number of phases and components in the above reaction is 2 and 2 respectively. Since it is a one-component system, F = C-P+2 = 2-2+2 = 2.

Engineering Chemistry Multiple Choice Questions on “Thermoplastic Polymers – 1”.

1. The thermoplastic is ____________
a) Cross-linked
b) Insoluble
c) Amorphous
d) Held by a covalent bond
Answer: c
Clarification: Thermoplastics are linear or branched, soluble in suitable solvents, amorphous or semi-crystalline and held together by weak vander waal’s force or hydrogen bonds.

2. Which of the following is not an example of thermoplastic?
a) Nylon
b) Polyester
c) PVC
d) Vulcanised rubber
Answer: d
Clarification: The examples of thermoplastic are nylon, polyester, PVC,PE, PP, PVA etc.

3. Polythene is prepared by the process of __________________ polymerisation of ethylene.
a) Addition
b) Condensation
c) Living
d) Free-radical
Answer: a
Clarification: Polythene is prepared by the process of addition polymerisation of ethylene. Polythene is of low strength, hardness and rigidity, but has a high ductility and impact strength as well as low friction.

4. LDPE is prepared by polymerising ethylene at a pressure of ____________
a) 100-200 atmospheres
b) 1000-5000 atmospheres
c) 10-100 atmospheres
d) 2-10 atmospheres
Answer: b
Clarification: None.

5. The crystallinity of LDPE is ____________
a) 10%
b) 30%
c) 55%
d) 80-90%
Answer: c
Clarification: The crystallinity of LDPE is low (55%). LDPE is defined by a density range of 0.910–0.940 g/cm³.

6. Select the incorrect statement from the following option.
a) LDPE is chemically inert and has excellent chemical resistance
b) LDPE is tough and flexible
c) LDPE is used for making films in general packaging, carrier bags etc
d) LDPE has high rigidity and is suitable for load bearing applications
Answer: d
Clarification: LDPE has low rigidity and is not suitable for load bearing applications. All other options are correct.

7. LDPE is suitable for the manufacture of pipes for distribution of gas.
a) True
b) False
Answer: b
Clarification: LDPE is not suitable for the manufacture of pipes for distribution of gas because it is permeable to gas molecules. LDPE is widely used for manufacturing various containers, dispensing bottles, wash bottles etc.

8. The ethylene is polymerised under in the presence of ____________
a) Zeigler Natta catalyst
b) Supported metal oxide catalyst
c) Lewis acids catalyst
d) Friedel crafts catalyst
Answer: a
Clarification: The ethylene is polymerised under 6-7 atmospheric pressure at 60-70⁰C in the presence of Zeigler Natta catalyst.

9. The softening temperature of HDPE is ____________
a) 20⁰C
b) 85⁰C
c) 135⁰C
d) 150⁰C
Answer: c
Clarification: The softening temperature of HDPE is 135⁰C. HDPE is known for its large strength-to-density ratio.The density of HDPE can range from 0.93 to 0.97 g/cm³ or 970 kg/m³.

10. Select the incorrect statement from the following option.
a) HDPE has excellent electrical insulation properties
b) HDPE is free from odour and toxicity
c) HDPE can be used for domestic water and gas piping
d) HDPE possess lower tensile strength compared to LDPE
Answer: d
Clarification: HDPE possess greater tensile strength compared to LDPE. All the other options are correct.

Engineering Chemistry Multiple Choice Questions on “Calorific Value – 2”.

1. Which of the following factor does not affect the gross calorific value?
a) CO₂ emissions from the fuel
b) Latent heat produced in the fuel
c) Size of the fuel
d) Moisture content of the fuel
Answer: c
Clarification: In calculating gross calorific value, the products of combustion are cooled down which does not depends upon the size of fuel.

2. A coal sample contains C=75.5%, H=11.25%, O=11.25, N=1%, S=1%. Caluclate gross calorific value?
a) 9568 cal/gm
b) 10000 cal/gm
c) 9799.010 cal/gm
d) 9518.893 cal/gm
Answer: d
Clarification: By using Dulong’s formulae –
GCV = [8080 %C + 34500(%H/1 – %O/8) + 2240 %S]/100,
Placing the values we get
GCV = 9518.893 cal/gm.

3. Calculate the net calorific value of the fuel if it’s ultimate analysis comes out to be containing C = 64%, O = 14.56%, H = 10.44%, N = 6.52%, S = 4.48%?
a) 8245.452 cal/gm
b) 7865.550 cal/gm
c) 7693.907 cal/gm
d) 8890.654 cal/gm
Answer: c
Clarification: NCV = [GCV – 0.09%H×587]cal/gm
For calculating GCV use Dulong’s formulae,
GCV = [8080 %C + 34500(%H/1 – %O/8) + 2240 %S]/100,
GCV = 8245.452 cal/gm
NCV = 7693.907 cal/gm.

4. How can we calculate the calorific value of a fuel at constant pressure?
a) QC.V. = QC.P. – (Δn)×R×T
b) QC.V. = QC.P. + (Δn)×R×T
c) QC.P. = QC.V. – (Δn)×R×T
d) QC.V. = QC.P. – (Δn)×R×T×S
Answer: c
Clarification: If there is a decrease in the number of gaseous molecules formed after reaction, then Δn will have a negative value and consequently will be higher than QC.V.
QC.P. = QC.V. – (Δn)×R×T
where QC.P. is calorific value at constant pressure, QC.V. is calorific value at constant volume, R is gas constant, T is temperature and S is entropy.

5. When does a fuel burn with high calorific intensity?
a) When solid fuel burns in high local fuel-bed temperature
b) When the fuel burns with an appreciable flame
c) When the combustion of fuel is not proper
d) When the fuel is of large size
Answer: a
Clarification: This is because when fuel burns without appreciable flame, the whole heat liberated is concentrated over a relatively smaller area.

6. Flaming fuels burn with lower calorific intensities then the flameless fuels.
a) True
b) False
Answer: a
Clarification: Since the total heat produced by the fuel is liberated over the entire area of the burning mass and the larger the area of flame the lesser is the concentration of heat.

7. On what factors does the temperature of flame depends?
a) Calorific value of gas, total gaseous products formed and pressure
b) Calorific value of gas, total gaseous products formed and specific heat
c) Calorific value of gas, total gaseous products formed and area
d) Calorific value of gas, total gaseous products formed and intensity
Answer: b
Clarification: The rate of heating of an object increases as the difference in temperature between the flame and the object increases which is depended on the following factors.

8. Why does the flame temperatures calculated solely are invariably higher?
a) Since the combustion takes place instantly and completely
b) Since some heat is lost as latent heat of steam produced
c) Since the dissociating gas molecules does not absorb a little heat
d) Since the specific heat of gases decreases with temperature.
Answer: b
Clarification: When we calculate flame temperature solely we cannot get an exact maximum temperature due to which there is a loss in latent heat of the seam produced.

9. How can fuel have good flexibility?
a) Should have high calorific value
b) Should have high moisture content
c) Should be easy to handle and control
d) Should have high cost
Answer: c
Clarification: Flexibility is defined as the rate of response in heat liberation with the variation in operating conditions.