Category: Engineering Chemistry Objective Questions

250+ TOP MCQs on Two Component System and Answers

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).

250+ TOP MCQs on Elastomers – 1 and Answers

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.

250+ TOP MCQs on Determination of Calorific Value of Gaseous and Volatile Liquid Fuels and Answers

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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250+ TOP MCQs on Cracking – 1 and Answers

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

1. Which of the following is a catalytic process?
a) Hydrogenation process
b) Coking
c) Pyrolysis
d) Thermal Cracking
Answer: a
Clarification: The processes in which catalysts are used for the generation of hydrocarbons are called catalytic process. Hydrogenation process is used for hydrofining of various fractions in the presence of elevated temperatures and pressures.

2. Which of the following is called as straight run petrol?
a) Intermediate oils
b) Residue fuel oils
c) Gasoline oils
d) Lubricants
Answer: c
Clarification: Gasoline fuels has the highest demand due to the automobile industries. The quality of straight-run petrol produced is not so good, so it is only used after suitable blending.

3. By which process we can get higher yields of gasoline in cracking?
a) Electrolytic decomposition
b) Thermal decomposition
c) Nuclear energy
d) Destructive distillation
Answer: b
Clarification: The process of cracking involves the thermal or catalytic decomposition of surplus heavier fractions, so as to obtain a higher yield of improved gasoline. It was noticed that cracked gasoline gives better engine performance.

4. Which of the following solid product is formed after thermal cracking of gasoline?
a) Charcoal
b) Sodium chloride
c) Butane
d) Coke
Answer: d
Clarification: The heavier petroleum fractions are decomposed at a temperature of 420 0C to form low molecular weight hydrocarbons. After this process coke, liquid and gaseous products are also formed.

5. In thermal cracking, by which method does the cracked products are separated?
a) Fractional distillation
b) Destructive distillation
c) Oil distillation
d) Steam distillation
Answer: a
Clarification: After the process of thermal decomposition, some of the products undergo polymerisation and the other crack products are separated using fractional distillation method. The yields are from 7-30 %.

6. Liquid-phase cracking can crack any type of oil.
a) True
b) False
Answer: a
Clarification: All types of oils, i.e., residue, fuel oil and gasoline oils can be cracked using liquid-phase cracking. The oils which can be vaporized at low temperatures can be cracked using vapour-phase cracking.

7. In which conditions does hydro-cracking are conducted?
a) In the presence of water
b) In the presence of electricity
c) In the presence of hydrogen
d) In the presence of oxygen
Answer: c
Clarification: The cracking which is brought about in the presence of a catalyst under a hydrogen atmosphere at slightly reduced temperature and high pressure is called hydro-cracking. It is a type of catalytic cracking.

8. Which type of catalyst is used in catalytic cracking?
a) Catalyst containing nitrogen
b) Catalyst containing alumina and silica
c) Catalyst containing noble elements
d) Catalyst with high electronegativity
Answer: b
Clarification: Earlier acid clays were used as a catalyst for catalytic cracking. Now silica and alumina with minor amounts of oxides of Ca, Mg, Na and rare earth are used as a catalyst.

9. Which compounds are formed by cracking of heavy hydrocarbons?
a) Saturated hydrocarbons
b) Unsaturated hydrocarbons
c) Saturated and unsaturated hydrocarbons
d) Inorganic matter
Answer: c
Clarification: In actual practice, during the cracking of heavy hydrocarbons both saturated and unsaturated compound are formed. These compounds are then separated by the process of fractional distillation.

10. Activity of a catalyst determines __________________
a) Yield of end-product
b) Ability to accelerate the reaction
c) Ageing
d) The nature of reactants
Answer: a
Clarification: Activity is a characteristic of alumina-silicate catalyst which determines the yield of the end-product, relative to the starting material used. Selectivity is the measure of the ability of the catalyst to accelerate the desired reaction.

250+ TOP MCQs on Boiler Feed Water and Answers

Engineering Chemistry Multiple Choice Questions on “Boiler Feed Water”.

1. Water is mainly used in boilers for the generation of ________________
a) Power
b) Electricity
c) Steam
d) Current
Answer: c
Clarification: Water is mainly used in boilers for the generation of steam. This steam is further used in the power plants for the generation of electricity.

2. Which of the following should not be a composition of boiler-feed water?
a) Hardness should be below 0.2ppm
b) Its caustic alkalinity should lie between 0.15 to 0.45 ppm
c) Its soda alkalinity should be 0.45-1 ppm
d) Its caustic alkalinity should be 1.5-2 ppm
Answer: d
Clarification: Its caustic alkalinity should be 0.15-0.45 ppm. All the other options are correct. Hardness should be below 0.2ppm, caustic alkalinity should lie between 0.15 to 0.45 ppm and soda alkalinity should be 0.45-1 ppm.

3. Which of the following is not a result of the excess of impurity in boiler-feed?
a) Scale and sludge formation
b) Decomposition
c) Corrosion, priming and foaming
d) Caustic embrittlement
Answer: b
Clarification: Excess of impurities in boiler-feed results in scale and sludge formation, corrosion, priming, foaming and caustic embrittlement. It does not contribute to the decomposition process.

4. If the precipitate formed is soft, loose and slimy, these are __________ and if the precipitate is hard and adhering on the inner wall, it is called _____________
a) Sludges, scale
b) Scale, sludges
c) Sludges, rodent
d) Scale, rodent
Answer: a
Clarification: If the precipitate formed is soft, loose and slimy, these are sludges and if the precipitate is hard and adhering on the inner wall, it is called scale. A rodent is a small furry mammal whose teeth never stop growing.

5. Which of the following option is incorrect about the sludges?
a) Sludges are soft, loose and slimy precipitate
b) They are non-adherent deposits and can be easily removed
c) Formed generally at heated portions of the boiler
d) Can be removed by blow down operation
Answer: c
Clarification: Sludges are formed generally at colder portions of the boiler. All the other options are correct. Sludges are soft, loose and slimy precipitate, non-adherent deposits and can be easily removed by blow down operation.

6. The scales decrease the efficiency of boiler and chances of explosions are also there.
a) True
b) False
Answer: a
Clarification: The scales decrease the efficiency of boiler and chances of explosions are also there. Scale is the precipitate that is hard and adhering on the inner wall.

7. The propulsion of water into steam drum by extremely rapid, almost explosive boiling of water at the heating surface is called ___________
a) Foaming
b) Priming
c) Corrosion
d) Caustic embrittlement
Answer: b
Clarification: The propulsion of water into steam drum by extremely rapid, almost explosive boiling of water at the heating surface is called priming. Priming is necessary as it removes all the air voids from the passage of water.

8. The phenomenon during which the boiler material becomes brittle due to accumulation of caustic substances is known as ___________
a) Foaming
b) Priming
c) Corrosion
d) Caustic embrittlement
Answer: d
Clarification: The phenomenon during which the boiler material becomes brittle due to accumulation of caustic substances is known as caustic embrittlement. This can be prevented by using sodium phosphate instead of sodium carbonate as softening reagents.

9. Foaming is caused by the formation of ___________
a) Acids
b) Alcohols
c) Oils and alkalis
d) Ketones
Answer: c
Clarification: Foaming is caused by the formation of oils and alkalis. Oils and Alkalis becomes sticky and greasy on the surface and hence results in foaming.

10. Corrosion is the decay or disintegration of boiler body material either due to chemical or electrochemical reaction with environment.
a) True
b) False
Answer: a
Clarification: Corrosion is the decay or disintegration of boiler body material either due to chemical or electrochemical reaction with environment. It converts a refined metal to a more stable form, such as its oxide, hydroxide, or sulphide.

250+ TOP MCQs on Electroplating Baths – 2 and Answers

Engineering Chemistry Multiple Choice Questions on “Electroplating Baths – 2”.

1. Which of the following is used as an electrolyte for antimony Plating?
a) KNO3
b) Dilute H2SO4
c) Silver in potassium solution
d) HCl
Answer: c
Clarification: Usually in the antimony plating method, the pure antimony bar acts as the anode, the object is to be plated (impure metal) acts as the cathode and the silver in potassium solution acts as the electrolyte. When electricity is passed between the electrodes, the positive silver ions from the electrolyte move to the negative anode (the object), where they are neutralized by the electrons and adhere to the object as silver. On further transmission we get antimony deposition.

2. The electrolyte used in soft metal Plating is _____________
a) Sodium cyanide with caustic soda
b) Cyanide hexa Meta phosphate
c) Cyanide oxide
d) Cyanide peroxide
Answer: a
Clarification: Usually in the soft metal plating method, the pure metal bar acts as the anode, the object is to be plated (impure metal) acts as the cathode and the sodium cyanide along with caustic soda solution acts as the electrolyte. When electricity is passed between the electrodes, the positive metal ions from the electrolyte move to the negative anode (the object), where they are neutralized by the electrons and adhere to the object as soft metal.

3. Which of the following metals cannot be electro plated?
a) Tungsten
b) Nickel
c) Silver
d) Copper
Answer: a
Clarification: Since tungsten does not have discharge capacity, it cannot transfer electrons from one electrode to another easily which is why it can be electroplated. Tungsten is the least electroplated metal.

4. Which of the following is not a characteristic of electrolyte?
a) Enables transportation of electrons
b) Determines the strength of the metal
c) Determines solubility
d) Identifies discharge scale
Answer: c
Clarification: Since electrolyte is the main component in electroplating, it aids in flow of electrons from one electrode to another but it cannot identify the solubility of the material (it can only determine discharge capacity).

5. Identify the strong electrolyte from the following.
a) Caustic soda
b) Potassium chloride
c) Potassium sulphide
d) Barium chloride
Answer: b
Clarification: Since caustic soda dissociates completely in water, it gives equal proportions of hydrogen and hydroxide ions. That is why it is considered to be a strong solution. Where weak electrolytes are opposite of this.

6. Calculate the amount of electrolyte required for electroplating 50 grams of magnetite.
a) 60 ml
b) 70 ml
c) 80 ml
d) 100 ml
Answer: d
Clarification: We know that the amount of electrolyte required for electroplating metal is twice the amount of metal. Thus here since 50 grams of magnetite, the amount of electrolyte required is 50 x 2 = 100 ml.

7. Calculate the amount of metal used when 200 ml of NaCl?
a) 100 g
b) 0 g
c) 10 g
d) 5 g
Answer: a
Clarification: We know that the amount of electrolyte required for electroplating metal is twice the amount of metal. Thus here since 200 ml, the amount of metal required is 200/2=100 ml.

8. Which of the following can be used as an electrolyte for metal oxide deposition?
a) Sodium solution
b) NaCl solution
c) Water
d) Immiscible liquids
Answer: c
Clarification: Since for the deposition of metal oxide layer, charge carriers need to be absorbed by the materials and hence water can be used as an electrolyte to absorb charge carriers from metals.

9. Calculate the amount of water to be mixed (in ml) with the electrolyte from the give data.
Amount of metal oxide used: 100 g
Amount of electrolyte used: 10 ml
a) 10
b) 20
c) 30
d) 0
Answer: a
Clarification: We know that from Falton’s theory of electroplating, the amount of water used with metal is equal to metal oxide/electrolyte. So here it is 100/10=10.

10. Calculate the amount of water to be mixed (ml) with the electrolyte from the give data.
Amount of metal oxide used=12 g
Amount of electrolyte: 6 ml
a) 3
b) 2
c) 0
d) 1
Answer: b
Clarification: We know that from Falton’s theory of electroplating, the amount of water used with metal is equal to metal oxide/electrolyte. So here it is 12/6=2.