Category: Applied Chemistry Objective Questions

Applied Chemistry Multiple Choice Questions on “Cloud Point and Pour Point”.

1. The temperature expressed in multiples of 1^oC at which the haze of the wax crystal appears when the oil is cooled under prescribed conditions is called ________
a) Pour point
b) Cloud point
c) Oxidation stability
d) Acid number
Answer: b
Clarification: The temperature expressed in multiples of 1^oC at which the haze of the wax crystal appears when the oil is cooled under prescribed conditions is called cloud point. The haze of wax crystals is also called as cloud. It appears on the walls of the jar.

2. The cloud point is the index of the ___________ temperature.
a) Lowest
b) Highest
c) Reference
d) Infinity
Answer: a
Clarification: The cloud point is the index of the lowest temperature of its utility for certain applications. Every lubricating oil has cloud point in most of cases it is true.

3. The cloud point is used as ______ control parameter.
a) Quantity
b) Temperature
c) Speed
d) Quality
Answer: d
Clarification: The cloud point is used as the quality control parameter in chemical, petrochemical, personal care and other industries. Mainly it is measured in the oils.

4. Olive oil starts to solidify at __________
a) 1^oC
b) 2^oC
c) 3^oC
d) 4^oC
Answer: d
Clarification: Olive oil starts to solidify at 4^oC. In winters or at very low temperatures around 0^oC, then the olive oil begins to develop white waxy clumps of solidified oil that sink to the bottom of the container.

5. The lowest temperature expressed in multiple of 2^oC at which the oil is observed to flow after cooled at prescribed conditions is called pour point.
a) True
b) False
Answer: b
Clarification: The lowest temperature expressed in multiple of 3^oC at which the oil is observed to flow after cooled at prescribed conditions is called pour point. In cloud point, the oil hazes into crystal but in pour point it flows.

6. Pour point indicates the _______ of lubricating oil.
a) Dissolved wax concentration
b) Total wax concentration
c) Initial wax concentration
d) Weight of the wax
Answer: a
Clarification: Pour point indicates the dissolved wax concentration of lubricating oil. In pour point experiment two thermometers are present to measure the temperature off cooling mixture and lubricating oil.

7. In pour point experiment, the diameter of the cylindrical tube is of the diameter ____________
a) 3-3.35mm
b) 3-3.35cm
c) 3-3.35km
d) 3-35nm
Answer: b
Clarification: In pour point experiment, the diameter of the cylindrical tube is of the diameter is 3-3.35cm. The height of the cylinder is about 11.5-12.5cm. There is mark up to which the sample must be taken.

8. Cooling bath contains suitable ________
a) Ice
b) Water
c) Cooling mixture
d) Cool water
Answer: c
Clarification: Cooling bath contains suitable cooling mixture. It may be water sometimes but it is selected according to the lubricant. It is mainly used for cooling purposes.

9. The jacket is provided with a gasket that prevents the test jar from __________
a) Cooling
b) Heating
c) Breakage
d) Touching the jacket
Answer: d
Clarification: The jacket is provided with a gasket that prevents the test jar from touching the jacket. The jacket is fitted with a disk. The jacket is of 11.5cm depth.

10. The cloudiness of the lubricant is observed for every ___________
a) 1^oC
b) 2^oC
c) 3^oC
d) 4^oC
Answer: a
Clarification: The cloudiness of the lubricant is observed for every 1^oC. First the oil is taken into the jar up to the marked level. The cork is fitted and then heated and observed for cloudiness for every 1^oC raise in temperature.

Applied Chemistry Multiple Choice Questions on “Mining of Petroleum”.

1. Removal of water is done by ___________
a) Orsteds process
b) Kjeldahl’s process
c) Filtration
d) Cottrells process
Answer: d
Clarification: Cottrells process is used to remove the water from the petroleum. kjeldahl’s process is used to know the percentage of nitrogen in the coal sample.

2. Crude oil is in the form of ____________
a) emulsion of oil and brine
b) emulsion of oil and impurities
c) emulsion of brine and impurities
d) emulsion of impurities and moisture
Answer: a
Clarification: The crude oil is obtained from the earth’s crust which will be in the form of emulsion of oil and brine. It is very stable. When this emulsion is charged then water gets separated from the oil.

3. To remove sulphur compounds from petroleum, it is treated with __________
a) copper nitrate
b) copper oxide
c) magnesium chloride
d) sodium chloride
Answer: b
Clarification: The sulphur compounds can be removed by treating with copper oxide as the sulphur compounds are get converted into the insoluble copper sulphide.

4. The scales formed during the removal of harmful impurities can be removed by __________
a) electrolysis
b) electroplating
c) electrotyping
d) dehydration
Answer: d
Clarification: The corrosion is will be there if there is presence of magnesium chloride or sodium chloride salts in crude oil and forms scales. These scales can be removed by dehydration process.

5. During refining the petroleum, in fractional distillation, what is the temperature in still?
a) 800^oC
b) 400^oC
c) 300^oC
d) 100^oC
Answer: b
Clarification: The crude oil is distilled and later it is sent into the pre-heater or else it can be heated with steam, inside the still the temperature is about 400^oC.

6. Which of the following fraction obtained on distillation used in dry cleaning?
a) petroleum ether
b) heavy oil
c) gas oil
d) naphtha
Answer: d
Clarification: Naphtha is used for dry cleaning. It is also called as solvent spirit. Heavy oil is mainly used in producing gasoline by cracking method.

7. What is the catalyst used in catalytic polymerization?
a) Phosphoric acid
b) Al₂O₃
c) Al₂(siO₃)₃
d) Zirconium oxide
Answer: a
Clarification: Phosphoric acid is used as catalyst in catalytic polymerization and Al₂O₃ , Al₂(siO₃)₃ are used as catalysts in catalytic thermal cracking.

8. Which of the following reactions indicates the Fischer-tropsch method?
a) nC+H_(2n+2) → C_nH_(2n+2)
b) nC+H_2n → C_nH_2n
c) nCO+2nH₂ → C_nH₂n+nH₂O
d) nC+H_(2n-1) → C_nH_(2n-1)
Answer: c
Clarification: The equation nCO+2nH₂ → C_nH₂n+nH₂O represents the Fischer-tropsch method, here the catalyst used is Ni or CO. The other three reactions indicates the Bergius process or hydrogenation of coal.

9. What is the catalyst used in Bergius process?
a) nickel oleate
b) phosphoric acid
c) zirconium oxide
d) aluminum oxide
Answer: a
Clarification: In Bergius process the catalyst used is Nickel oleate. In this process the low ash coal is finely powdered and turned into a paste using heavy oil and the catalyst is taken into temperature.

10. What is the pressure applied during hydrogenation of coal?
a) 200-280atm
b) 200-250atm
c) 300-380atm
d) 300-350atm
Answer: b
Clarification: A pressure of 200-250atm is applied in Bergius process for one hour 30 minutes.In polymerization the pressure of 70-350kg/cm² is applied.

11. The boiling range of un-condensed gas is ___________
a) < 300C
b) < 400C
c) > 500C
d) < 200C
Answer: a
Clarification: The boiling range of un-condensed gas must be very low that is less than 30^oC. As the gases are un-condensed they are ready to condense and become steam. So, a small amount of heat is sufficient to boil them.

12. Heavy oil on refractionation produces _________
a) Asphalt
b) diesel oil
c) grease
d) kerosene oil
Answer: c
Clarification: Heavy oil on refractionation produces grease. It also produces lubricating oil, petroleum jelly, paraffin wax.

13. How much percentage of petrol is produced by the synthesis of coal?
a) 40%
b) 50%
c) 80%
d) 30%
Answer: d
Clarification: 30% of petrol is produced from the synthesis of coal. Coal is mainly utilized in producing electricity and petrol.50% of petrol is produced form the process called cracking.

14. Which of the following can be used as a preservative for wood?
a) solvent spirit
b) petroleum coke
c) tar
d) gasoline
Answer: c
Clarification: Tar is used for preserving wood and the solvent spirit is mainly used for dry cleaning and as motor fuel.

15. What is the boiling range of diesel oil?
a) 120-180^oC
b) 180-250^oC
c) 320-400^oC
d) 250-320^oC
Answer: d
Clarification: Diesel oil’s boiling point is 250-320^oC. So, in cracking the high temperatures are applied.

Applied Chemistry Assessment Questions and Answers focuses on “Reverse Osmosis, Ultrafilteration, COD and BOD”.

1. When two solutions of the different concentration are separated by the semi permeable membrane then the solvent flows from low to higher concentration is called osmosis.
a) True
b) False
Answer: a
Clarification: Always the energy flows from lower to higher concentration. When two solutions of the different concentration is separated by the semi permeable membrane then the solvent flows from low to higher concentration is called osmosis.

2. Which of the following method is not the method of desalination?
a) Reverse osmosis
b) Multi flash distillation
c) Electro dialysis
d) Smelting
Answer: d
Clarification: The reverse osmosis, multi flash distillation and electro dialysis are the methods of desalination and smelting is not the process of desalination.

3. The natural tendency of the water can be reversed by applying ________ to the salty water part.
a) Low pressure
b) High pressure
c) Low temperature
d) High temperature
Answer: b
Clarification: The natural tendency of the water can be reversed by applying the high pressure with a piston to the salty part of the water.

4. In reverse osmosis, the water flows from __________ concentration to ___________ concentration.
a) Low, high
b) High, low
c) High, moderate
d) Moderate, low
Answer: a
Clarification: In reverse osmosis, the water flows from the lower to higher concentration. In osmosis the water flows from the higher to lower concentration.

5. The semi permeable membranes used in the reverse osmosis is _________
a) Cellulose
b) Glucose
c) Cellulose acetate
d) Glucose acetate
Answer: c
Clarification: The semi permeable membrane used in reverse osmosis is cellulose acetate and cellulose butyrate etc.

6. By ultra filtration aldrin can be removed by _________
a) 100%
b) 99%
c) 98%
d) 97%
Answer: b
Clarification: By ultra filtration process, many micro organisms can be removed by reacting the industrial waste with activated charcoal. Aldrin can be removed by 99%.

7. __________ co polymer can remove the chlorinated pesticides.
a) Styrene di vinyl benzene
b) Styrene
c) Benzoyl peroxide
d) Phenol
Answer: a
Clarification: The styrene di vinyl benzene can remove the chlorinated pesticides by absorption at the surface.

8. In reverse osmosis, the membrane pores are smaller in size about ___________
a) 0.04 to 600nm
b) 0.08 to 500nm
c) 0.1 to 600nm
d) 0.2 to 500nm
Answer: a
Clarification: In reverse osmosis, the size of the membrane used is very smaller. It is about 0.04 to 600nm.

9. COD values are always _________ BOD.
a) Less than
b) Higher than
c) Equal
d) Nearly equal
Answer: b
Clarification: COD values are always greater than that of the BOD values because the organic and the biologically oxidisable materials are oxidised in COD.

10. COD can be determined in ___________
a) 1 Hour
b) 2 Hours
c) 3 Hours
d) 4 Hours
Answer: c
Clarification: COD can be determined in 3 hours. The organic matter of sample is oxidised to the carbondioxide, ammonia and water.

11. Increase in the BOD value in the water indicates ___________
a) Decrease in pollution
b) Increase in pollution
c) Pollution is independent of BOD
d) Slight decrease in the BOD
Answer: b
Clarification: Increase in the value of the BOD in water indicates the increase in the water pollution.

12. The organic matter have the limit in water that is about __________
a) 0.2 to 1
b) 0.5 to 1
c) 1 to 2
d) 2 to 4
Answer: a
Clarification: The organic matter has the desired limit in water about 0.2 to 1 ppm. Increase in the value indicates the pollution.

13. In sewage the waste is about ___________
a) 0.01
b) 0.02
c) 0.03
d) 0.05
Answer: d
Clarification: Sewage contains 99.95% of water and 0.05% of the organic and municipal wastes. Strength of sewage is expressed in terms of BOD.

Applied Chemistry Assessment Questions,

Applied Chemistry Multiple Choice Questions on “Compounding of Plastics – 1”.

1. Usually the polymer resin is mixed with four to ten ingredients during the fabrication to impart useful properties to the finished articles. This is called mix.
a) True
b) False
Answer: a
Clarification: Usually the polymer resin is mixed with four to ten ingredients during the fabrication to impart useful properties to the finished articles. This is called mix. They are must and should to get good plastic.

2. __________ acts as a binder in plastics.
a) Resins
b) Plasticizers
c) Lubricant
d) Accelerators
Answer: a
Clarification: Resins acts as the binders in plastic which holds the different constituents together. The product of polymerisation is called resins.

3. ___________ increases the flexibility of the polymer.
a) Resins
b) Catalysts
c) Lubricants
d) Plasticizers
Answer: d
Clarification: Plasticizers are used to increase the flexibility of the polymer. It is also used to the increase of plasticity of the polymer.

4. Plasticizers are used to get the ___________ between the polymeric molecules.
a) Attraction
b) Freedom of movement
c) Pi-bond
d) Sigma bond
Answer: b
Clarification: Plasticizers are used to get the greater freedom of movement between the polymeric molecules. Plasticizers neutralise the intermolecular forces of attraction between polymer chains.

5. Plasticizers ___________ the strength of the plastics.
a) Increase
b) Decrease
c) Do not affect
d) May increase above the room temperature
Answer: b
Clarification: Plasticizer decreases the strength of the plastics. They also decrease the chemical resistance of the plasticizers.

6. __________ makes the polymers impermeable to x-rays.
a) Strontium salts
b) Ammonium salts
c) Calcium salts
d) Barium salts
Answer: d
Clarification: The barium salts makes the polymers impermeable to x-rays and asbestos provides heat and corrosion resistance to polymers.

7. The percentage of the fillers is up to ___________ total moulding mixture.
a) 20%
b) 30%
c) 40%
d) 50%
Answer: d
Clarification: The percentage of the fillers is up to the 50% of the total moulding mixture of the plastics. They reduce the cost of the polymers.

8. Which of the following are used as the fillers?
a) Vegetable oils
b) Camphor
c) Ester of stearic acids
d) Wood flour
Answer: d
Clarification: Vegetable oils, camphor and ester of stearic acids are used as the plasticizers. The wood flour, ZnO, PbO are used as the fillers.

9. __________ prevent moulded article from sticking to the fabrication equipment.
a) Catalyst
b) Fillers
c) Lubricants
d) Stabilizers
Answer: c
Clarification: Lubricants prevent moulded article from sticking to the fabrication equipment. They impart flawless and glossy finish to the products.

10. Lubricants are used to prevent the moulded article from _________
a) Rusting
b) Sticking
c) Not sticking
d) Burning
Answer: b
Clarification: Lubricants are used to prevent the moulded article from sticking to the fabrication equipment. Lubricants make the moulding plastics easier.

11. Which of the following are not used lubricants?
a) Oils
b) Waxes
c) Oleates
d) Salts
Answer: d
Clarification: The most commonly used lubricants are oils, waxes, oleates, stereates and soaps. The salts are not used as lubricants.

12. ___________ is used to improve the thermal stability during polymerisation.
a) Accelerators
b) Colouring materials
c) Stabilizers
d) Lubricants
Answer: c
Clarification: Stabilizers are used to improve the thermal stability during polymerisation. Accelerators are used to increase the polymerisation of fusion resin during modulation operation into linked infusible form.

13. Which of the following is a stabilizer?
a) White lead
b) PbO
c) ZnO
d) Metallic oxides
Answer: a
Clarification: The PbO, ZnO and metallic oxides are used as the fillers. ZnO and metal oxides are also used as the accelerators. The stabilizers are white lead, lead chromate, red lead.

14. Compression moulding is applicable to ___________
a) Thermoplastic resins
b) Thermo setting resins
c) Both thermo plastic and thermo setting resins
d) Neither thermo plastics nor thermo setting resins
Answer: c
Clarification: Compression moulding is applied to both the thermo plastics and thermo setting resins. It is the one of the method of fabrication.

15. How many types of fabrication methods are there?
a) 4
b) 5
c) 6
d) 7
Answer: d
Clarification: There are seven types of fabrication methods are there. They are compression moulding, injection moulding, transfer moulding, extrusion moulding, bubble moulding, casting and thermoforming.

Applied Chemistry Quiz focuses on “Conversion of Solar Energy and Applications”.

1. Non conventional energy sources are those energy sources that are _________
a) Renewable
b) Non-renewable
c) Produced from electricity
d) Produced from heat
Answer: a
Clarification: The non conventional energy sources are the sources that are renewable and economically safe such as solar energy, wind energy, biomass energy etc.

2. The importance of increasing the use of non conventional energy sources in India is in _________
a) 1940
b) 1950
c) 1960
d) 1970
Answer: d
Clarification: The importance of increasing the use of non conventional energy sources in India is in 1970. Some of them are geothermal energy, nuclear energy.

3. Solar energy in India is utilised through ________
a) Photo voltaic route
b) Photometric route
c) Photo power
d) Photosynthesis
Answer: a
Clarification: In India, the solar energy is utilised through the photo voltaic route. This can also be utilised from the thermal route.

4. Power generating system is based on _______
a) Biomass gasification
b) Coal gasification
c) Solar energy
d) Wind energy
Answer: a
Clarification: Power generating system is based on the biomass gasification. It can also be based on the biomass combustion that was launched in different places in central India.

5. Wind generator is used for ________
a) Power generation
b) Heat generation
c) Power dissipation
d) Power consumption
Answer: a
Clarification: Wind generation is used to produce power. The wind energy is mostly used for this purpose. The wind flows from high region to the lower region.

6. Geothermal energy is mostly produced in ________ in India.
a) Mumbai
b) Kerala
c) Chennai
d) Himachal pradesh
Answer: d
Clarification: Geothermal energy is mostly produced in Himachal Pradesh in India. It is also seen in Jammu and Kashmir in India. The geothermal energy is generated from the hot springs.

7. In Timarpur, the power is generated from __________
a) Urban waste
b) Solar energy
c) Hydro power
d) Wind energy
Answer: a
Clarification: In Timarpur in Delhi, The power is generated from the urban wastes and industrial wastes. This process is also carried out in Tamil nadu in India.

8. The main application of solar cells is to generate electricity from __________
a) Water
b) Sunlight
c) Wind
d) Biomass
Answer: b
Clarification: The main application of solar cells is to generate electricity from sunlight. Even it is small scale or large scale electricity from sun light.

9. Solar photovoltaic cells are often connected electrically in _______
a) Series
b) Parallel
c) Randomly
d) Neither series nor parallel
Answer: a
Clarification: Solar photovoltaic cells are often connected electrically in series. They encapsulated as a module to create additive voltage and to yield high current.

10. Solar power is a good __________ renewable source.
a) Commercial
b) Economical
c) Commercial and economical
d) Neither commercial nor economical
Answer: b
Clarification: Solar power is a good economic renewable source of energy. It helps preserving the environment.

11. _________ can be used for recharging the portable devices.
a) Solar photovoltaic cells
b) Solar panels
c) Solar batteries
d) Solar dishes
Answer: b
Clarification: The solar panels can be used for recharging portable devices and making power. They are also used as the heating water etc.

12. Solar cells can be used in calculators.
a) True
b) False
Answer: a
Clarification: Solar cells can be used in calculators. They can also be used in components like solar flash light, street lightening, home lightening and lanterns.

13. _________ in sunlight hits the panel.
a) Dust
b) Photons
c) Electrons
d) Protons
Answer: b
Clarification: The photons n sunlight hits the panel. The photons are absorbed by the semi conducting material that is used to make the panel. Silicon is mostly used to make the panels.

14. The __________ present in the material are de-localised.
a) Electrons
b) Protons
c) Neutrons
d) Positrons
Answer: a
Clarification: The electrons present in the semi conducting material are de-localised and allowed to flow. The flow of electrons produces the current.

15. Due to the special composition of the solar cells the electrons are allowed to flow in _________
a) Opposite direction
b) Multiple directions
c) Single direction
d) Random direction
Answer: c
Clarification: Due to the special composition of the solar cells the electrons are allowed to flow in single direction only. An array of solar cells converts solar energy into the direct current.

Applied Chemistry for Quizzes,

Applied Chemistry Interview Questions and Answers for Experienced people focuses on “Saponification Value – 2”.

1. The saponification value is given by ______
a) Saponification value = (volume of KOH +NKOH+56)/wt of the oil taken
b) Saponification value = (volume of KOH * NKOH/56)/wt of the oil taken
c) Saponification value = (volume of KOH * NKOH*56)/wt of the oil taken
d) Saponification value = (volume of KOH – NKOH-56)/wt of the oil taken
Answer: c
Clarification: The saponification value is given by Saponification value = (volume of KOH * NKOH*56)/wt of the oil taken. So, to determine the saponification value first we need to know the volume of KOH utilised.

2. What is the acid value of the groundnut oil?
a) 0.2-8.0
b) 0.4-0.8
c) 10.0-35.2
d) 0.4-2.2
Answer: a
Clarification: The acid value of the groundnut oil is 0.2-8.0. Castor oil has the acid value of 0.4-0.8. Coconut oil is having a high acid value that is about 10.0-35.2.

3. What is the saponification oil of coconut oil?
a) 194-196
b) 201-203
c) 253-260
d) 194-197
Answer: c
Clarification: The saponification value of the coconut oil is very high because it has very high acid value. The saponification value of the coconut is 253-203. The more acid value is the high the saponification value.

4. Which of the following has no acid value?
a) Cotton seed oil
b) Rape seed oil
c) Whale oil
d) Hard oil
Answer: d
Clarification: Hard oil has no acid value. Cotton seed oil has the acid value of 194-195. Rape seed oil has low acid value that is about 1.4-4.0. The whale oil has the acidic value of 0.3-51.4.

5. The saponification value of the hard oil is _________
a) 192-198
b) 194-195
c) 194-196
d) 190-191
Answer: a
Clarification: The saponification value of the hard oil is 192-198. Though it has no acid value it has saponification value. 194-195 is the saponification value of the cotton seed oil. Groundnut oil and whale oil has the saponification values of 194-196 and 190-191 respectively.

6. The percentage of fatty oil when compounded in the oil is given by _________
a) C*F/100
b) (C/F)*100
c) (C+F)*100
d) (C-F)*100
Answer: b
Clarification: The percentage of fatty oil when compounded in the oil is given by (C/F)*100. Here, C = saponification value of the compounded or lubricating oil and F = saponification of the fatty oil.

7. Triglyceride reacts with glycerol to form _________
a) Sodium of fatty acid
b) Potassium of fatty acid
c) Magnesium of fatty acid
d) Chlorides of fatty acids
Answer: b
Clarification: Triglyceride reacts with glycerol to produce the potassium of fatty acids. The potassium of fatty acids are called as soaps. Mainly alkyl groups of triglycerides and glycerol are made to react.

8. In determination of saponification value of the oil reaction _______ is unreacted.
a) NaOH
b) Nacl
c) KOH
d) Kcl
Answer: c
Clarification: In determination of the saponification value of the oil reaction, KOH is unreacted. The KOH is used as the catalyst and it remains unreacted and then titrated with standard acid using phenolphthalein as an indicator.

9. Find the acid value of the vegetable oil whose 10ml required 4.0ml of 0.01N KOH during titration (d = 0.92).
a) 0.243mg
b) 2.43mg
c) 24.3mg
d) 243mg
Answer: a
Clarification: To find out the acid value of an oil we know the formula, acid value= (volume of KOH*N KOH*56)/wt of the oil in mg. Wt of the oil can be known by calculating the product of density and volume to get 9.2gm. Substitute the values in formula to get 0.243mg as the answer.

10. The saponification value also indicates _______
a) Number of carbons
b) Number of alkyl groups
c) Number of fatty acids
d) Length of the carbon chain
Answer: d
Clarification: The saponification value also indicates the length of the carbon chain in the particular oil or fat. Higher the saponification value greater the percentage of the short chain acids present in glycerides.

Applied Chemistry for Interviews,