Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

## Applied Chemistry MCQ Set 1

1. Calculate the net calorific value of a coal sample having the following composition:
C=80%, H=5%, O=4%, N=3%, S=3.5% and ash=5%
a) 7251.8cal/g
b) 7780.5cal/g
c) 7621.5cal/g
d) 7830.75cal/g

Answer: d [Reason:] Apply the dulong’s formula that is: HCV = 1/100[8080C + 34500(H-O/8) + 22400S], here the C, S, O , H are the percentages of carbon, sulphur, oxygen and hydrogen. So, substitute all the given values in the formula and calculate so that you will get HCV (or) GCV as 8094.9cal/g and then apply the formula NCV=(GCV-0.09H*587), her 587cal/g is the latent heat of steam then you will get NCV=7830.75cal/g.

2. A coal has the following composition by weight:
C=90%, O=4%, N=1%, S=0.5% and ash=5%.The NCV of the fuel was found to be 8480cal/g. Calculate the percentage of hydrogen and HCV of the fuel.
a) H=4.21%, HCV=8621.80cal/g
b) H=4.521%, HCV=8221.80cal/g
c) H=4.686%, HCV=8727.37cal/g
d) H=4.1%, HCV=8221.37cal/g

Answer: c [Reason:] Apply the dulong’s formula that is: HCV = 1/100[8080C + 34500(H-O/8) + 22400S], here the C, S, O ,H are the percentages of carbon, sulphur, oxygen and hydrogen. So, substitute all the given values in the formula and calculate so that you will get HCV. As, we don’t know the value of H, you will get HCV= [7110.7+345H]Cal/g. Let it be equation -1 and then we know that NCV= (GCV-0.09H*587), here 587cal/g is the latent heat of steam. NCV is given then you will get GCV= [8480+52.83H] Cal/g and let it be equation-2. So equate both the equations to get the value of H. you will get H=4.686 and substitute in equation-1 to get the value of HCV=8727.37cal/g.

3. C=70%, O=10%, N=1%, S=5% and ash=4%.The NCV of the fuel was found to be 9210cal/g. percentage of hydrogen be x and HCV of the fuel be y. Find out y/x.
a) 747.7
b) 768
c) 777
d) 676.9

Answer: a [Reason:] Apply the dulong’s formula that is: HCV = 1/100[8080C + 34500(H-O/8) + 22400S], here the C, S, O ,H are the percentages of carbon, sulphur, oxygen and hydrogen. So, substitute all the given values in the formula and calculate so that you will get HCV. As, we don’t know the value of H, you will get HCV=[5336.75+345H]Cal/g. Let it be eauatin-1 and then we know that NCV= (GCV-0.09H*587), here 587cal/g is the latent heat of steam. NCV is given then you will get GCV=[9210+52.83H]Cal/g and let it be equation-2. So equate both the equations to get the value of H. you will get H=13.25% and let it be x and substitute in equation-1 to get the value of HCV=9908cal/g. Now let be y and divide y with x to get 747.7.

4.A formula giving the gross heating value of coal in terms of the weight fractions of carbon, hydrogen, oxygen and sulphur from the _________
a) ultimate analysis
b) proximate analysis
c) distillation
d) filtration

Answer: a [Reason:] The percentages of the coal components are given by ultimate analysis and the proximate analysis is used to know the percentages of volatile matter, moisture etc.

5. The dulong’s formula is not applicable for_____
a) solid fuel
b) gaseous fuel
c) liquid fuel
d) any fuel

Answer: b [Reason:] For calculating the components of coal and the petroleum the dulong’s formula is applied. So, the dulong’s formula is applied for both solids and liquids and not for gaseous.

6. Modification of the dulong’s formula can be done by considering the______
a) latent heat
b) heat
c) fuel state
d) liquid fuels

Answer: a [Reason:] The modification of the formula is done by considering the latent heat. It may be of water or vapour or steam accordingly. It represents the heating value of the fuels.

7. Calculate the HCV of the coal from the given data:
Weight of the fuel burnt=0.92
Initial temperature=120C
Final temperature=19.20C
Weight of the water in calorimeter is 1458g
Water equivalent of calorimeter =14g
a) 11520k.cal/m3
b) 11560k.cal/m3
c) 11000k.cal/m3
d) 11590k.cal/m3

Answer: a [Reason:] Use the formula HCV= [(W + w)(t2-t1)]/x, where W= Weight of the water in calorimeter, w= Water equivalent of calorimeter, t2= Final temperature, t1= Initial temperature, x= Weight of the fuel burnt. By substituting, you will get 11520k.cal/m3 as the final answer.

8. While calculating HCV, if we need to apply the fuse wire, acid and cooling corrections then what is the formula for HCV?
a) HCV= [(W + w)(t2-t1+cooling correction)-(acid correction + fuse correction)]/weight of fuel
b) HCV= [(W + w)(t2+t1+cooling correction)-(acid correction + fuse correction)]/weight of fuel
c) HCV= [(W + w)(t2-t1+ acid correction)-( cooling correction + fuse correction)]/weight of fuel
d) CV= [(W + w)(t2+t1+ acid correction)-( cooling correction + fuse correction)]/weight of fuel

Answer: a [Reason:] The cooling correction must be added to the temperature difference because the heat loss is may be conduction, convection or radiation. The sum of acid and fuse corrections are subtracted because the heat produced by exothermic reactions is already included in the change in temperature. So, it is to be subtracted to get HCV.

## Applied Chemistry MCQ Set 2

1. Poly carbonates are the durable materials with _________
a) Low impact resistances
b) High scratch resistance
c) Low scratch resistance
d) Low transparency

Answer: c [Reason:] Poly carbonates are the durable materials with the high impact resistances, low scratch resistances and high transparency.

2. Poly carbonates possess glass transition temperature of __________
a) 144oC
b) 145oC
c) 146oC
d) 147oC

Answer: d [Reason:] Poly carbonates has the transition temperature of 147oC. Poly carbonates are transparent and good electrical insulators.

3. _________ are used in the data storage.
a) Bakelite
b) TEFLON
c) Poly carbonates
d) Polyethylene

Answer: c [Reason:] Poly carbonates are used in the data storage. They are used in the production of compact discs and bluray discs as well as DVD’s.

4. Poly carbonates are used as ________ in the high stability capacitors.
a) Accelerator
b) Inhibitor
c) Dielectric
d) Initiator

Answer: c [Reason:] Poly carbonates are used as the dielectric in the high stability capacitors. Poly carbonates are used in the construction materials.

5. Polyurethanes are in the class of compounds __________
a) Reaction polymers
b) Chain polymers
c) Branched polymers
d) Linear polymers

Answer: a [Reason:] Polyurethanes are in the class of compounds called reaction polymers which include in epoxies, polyesters and phenolics.

6. Polyurethanes possess __________
a) No impact resistance
b) Softness
c) Hardness
d) No tear resistance

Answer: c The polyurethanes possess the impact resistance, tensile strength, hardness, abrasion and tear resistance.

7. The chair, house and car are some of the applications of the __________
a) Bakelite
b) TEFLON
c) Polyurethane
d) Poly vinyl chloride

Answer: c [Reason:] The chair, house and car are some of the applications of the polyurethane. It has many applications in the daily life also.

8. Polyurethanes are _________
a) Good conductors
b) Good insulators
c) Having no tensile strength
d) Affected by heat

Answer: b [Reason:] Polyurethanes are having the good insulating properties. They are having tensile strength and they are resistant to environment conditions like heat, moisture etc.

9. Polyurethanes are used for purifying the ________
a) Water
b) Alcohol
c) Acids
d) Salt solutions

Answer: a [Reason:] Polyurethanes are used for purifying the water. They are used in refrigeration and freezers also.

10. Sun glasses, computer cases and printing substrate can be done by using _________
a) Polyurethanes
b) Poly carbonates
c) Poly vinyl cyanide
d) Polyethylene

Answer: b [Reason:] Sun glasses, computer cases and printing substrate can be done by using poly carbonates. Theft proplastic packaging which cannot be opened by hand are also made from poly carbonates.

## Applied Chemistry MCQ Set 3

1. Soaps can be defined as the soap consuming capacity of the water sample.
a) True
b) False

Answer: a [Reason:] Soaps are defined as the soap consuming capacity of the water. Soaps are sodium salts of fatty acids like oleic acid and stearic acid.

2. The soft water contains the hardness of about ____________
a) 0-45ppm
b) 0-55ppm
c) 0-65ppm
d) 0-75ppm

Answer: d [Reason:] The soft water contains the hardness of about 0-75ppm. The hardness of water is mainly due to the salts of calcium and magnesium.

3. The hardness of moderately hard water is about _____________
a) 75-150ppm
b) 75-120ppm
c) 75-130ppm
d) 75-100ppm

Answer: a [Reason:] The hardness of CaCO3 of moderately hard water is about 75-150ppm. The hardness of the water can be calculated from amount of calcium and magnesium ions present in water along with bicarbonates, sulphates.

4. The very hard water has the hardness of CaCO3 is given by ___________
a) 100-200ppm
b) 100-300ppm
c) 200-300ppm
d) Above 300ppm

Answer: c [Reason:] The very hard water has the hardness of CaCO3 is given about above 300ppm. The hard water is having the hardness of CaCO3 is given by 150-300ppm.

5. The PH value of the drinking water is about _________
a) 6.5-8.5
b) 5.5-6.5
c) 4.5-5.5
d) 3.5-4.5

Answer: a [Reason:] The PH value of the drinking water is about 6.5 to 8.5. The odour of the drinking water is unobjectionable but the drinking water is generally having no odour.

6. The drinking water can have the magnesium limit about ____________
a) 10-150ppm
b) 20-150ppm
c) 30-150ppm
d) 40-150ppm

Answer: c [Reason:] The drinking water can have the magnesium limit about 30-150ppm. The calcium in the drinking water can be about 75-200ppm.

7. The chloride in drinking water range can be about ___________
a) 200-600ppm
b) 300-600ppm
c) 400-600ppm
d) 500-600ppm

Answer: a [Reason:] The chloride in drinking water range can be about 200-600ppm. The nitrate is about the range of 45ppm in drinking water.

8. The iron is about the range in drinking water is _________
a) 1-1.5ppm
b) 0.01-0.1ppm
c) 1-1.1ppm
d) 0.1-1ppm

Answer: d [Reason:] The iron is about the range in drinking water is 0.1-1.0ppm. The magnesium is about the range of 30-150ppm in drinking water.

9. The phosphate is about the range of __________ in drinking water.
a) 5-10ppm
b) 10-15ppm
c) 15-20ppm
d) 20-25ppm

Answer: b [Reason:] The phosphate is about the range of 10-15ppm in drinking water. The sulphate in drinking water is about 200-400ppm.

10. The organic matter in drinking water must be about _________
a) 0.2-1.0ppm
b) 1.0-2.0ppm
c) 2.0-3.0ppm
d) 3.0-4.0ppm

Answer: a [Reason:] The organic matter in drinking water must be about 0.2-1.0ppm. The phosphate is also low that is about 10-15ppm.

11. In _____________ when the eater is heated then the soluble salts turns into insoluble ones and removed by filtration.
a) Temporary hardness
b) Permanent hardness
c) Non-carbonate
d) Non-alkaline

Answer: a [Reason:] The hardness in the temporary hard water can be removed by heating and filtrating the insoluble salts.

12. Which of the following does not cause the permanent hardness in water?
a) Nitrates
b) Sulphates
c) Chlorides
d) Bicarbonates

Answer: d [Reason:] The bicarbonates of the calcium and magnesium cause the temporary hardness and the sulphides, nitrates and the chlorides cause the permanent hardness.

13. The total hardness can be given by _________
a) Temporary + permanent hardness
b) Temporary – permanent hardness
c) Temporary * permanent hardness
d) Temporary/permanent hardness

Answer: a [Reason:] The sum of the temporary and permanent hardness of the water gives the total hardness of the water.

14. Which of the following process does not remove the permanent hardness of water?
a) Lime-soda
b) Ion exchange process
c) Zeolite process
d) Heating

Answer: c [Reason:] Heating of the water removes the temporary hardness and the permanent hardness is removed by the zeolite process, lime soda process and the ion exchange method.

15. All carbonate and bicarbonates are _________
a) Alkaline
b) Acidic
c) Highly acidic
d) Neutral

Answer: a [Reason:] All the carbonates and bicarbonates are the alkaline in nature. So, the hardness due to them is called as carbonate hardness or alkaline hardness.

## Applied Chemistry MCQ Set 4

1. Redwood viscometer no1 is used to find the viscosity of the _________ liquids.
a) High viscous
b) Low viscous
c) Moderate viscous
d) No viscous

Answer: b [Reason:] The redwood viscometer no1 is used to find out the viscosity of the low viscous liquids. They have an efflux of 2000seconds or less. Redwood viscometer no2 is used to find the viscosity of the high viscous liquids like fuel oils.

2. What is the efflux of the redwood viscometer no2?
a) 2000
b) 20000
c) 100
d) 200

Answer: d [Reason:] The efflux of the redwood viscometer no2 is 200 seconds or less. Its jet for the outflow of the oil is having larger diameter. We can find the relative viscosities of the liquids using the redwood viscometer.

3. Test thermometer is one of the parts of the redwood viscometer.
a) True
b) False

Answer: a [Reason:] The parts of the redwood viscometer are test thermometer, bath thermometer, water inlet, oil stirrer, water bath, heating coil, metal indicator, brass oil cup, water bath stirrer blade, valve rod, water outlet, agate jet, kohlrausch flask and levelling screw.

4. The oil cup is made up of _________
a) Brass
b) Copper
c) Silver
d) Chromium

Answer: a [Reason:] The oil cup is one of the parts of the redwood viscometer. It is made up of brass. It is open at upper end. Its base is fitted with an agate jet, with the bore of diameter 1.62mm and internal length 10mm.

5. The height and diameter of the oil cup is __________ respectively.
a) 250mm, 46.5mm
b) 290mm, 46.5mm
c) 290mm, 40.5mm
d) 250mm, 40.5mm

Answer: b [Reason:] The height of the oil cup is 290mm and the diameter is about 46.5mm. The level to which the oil cup is fixed with oil is indicated by the stout wire fixed in the side of the oil cup. The wire is turned upwards and it is tapered to sharp point to indicate level properly.

6. The jet is opened or closed by ______
a) Cup
b) Valve rod
c) Stout wire
d) Screw

Answer: b [Reason:] The jet of the redwood viscometer is opened or close by using the valve rod. The lid of the cup is provided with the spirit level for the vertical levelling of the jet.

7. The cylindrical vessel will be around the oil cup which serves as a water bath made up of _______
a) Brass
b) Chromium
c) Copper
d) Aluminium

Answer: c [Reason:] The oil cup is surrounded by a cylindrical vessel and it acts as a water bath that is made up of the copper. It is provided by a tap. The tap is used for emptying the vessel by draining the water present in it.

8. The long side tube is used to heat the water by means of _______
a) Burner
b) Spirit lamp
c) Bunsen burner
d) Candle

Answer: b [Reason:] The long side tube is projected outwards. This is used to heat the water by means of gas or spirit lamp. The copper vessel is provided with thermometer to measure the temperature of the water.

9. The water bath is provided with stirrer having ______ blades.
a) One
b) Two
c) Three
d) Four

Answer: d [Reason:] The water bath is provided with a stirrer having four blades. It is provided with four blades because it helps to maintain the uniform temperature in bath to facilitate the uniform heating of the oil.

10. The entire redwood viscometer apparatus rests on _______
a) Table
b) Stand
c) Tripod stand
d) Burner

Answer: c [Reason:] The entire redwood viscometer apparatus rests on the tripod stand. It is provided with screws at the bottom. It has four legs. The screws are used for adjusting the legs for increasing or decreasing the diameter.

a) 100ml
b) 150ml
c) 50ml
d) 25ml

12. The absolute viscosity of the oil can be given as_______
a) ŋ = k{(t1D2)/(t2D1)}
b) ŋ = k{(t1D2)+(t2D1)}
c) ŋ = k{(t1D2)(t2D1)}
d) ŋ = k{(t1D2)-(t2D1)}

Answer: a [Reason:] By performing the redwood viscometer experiment, we can know the absolute viscosity of the oils. The absolute viscosity of the oils can be given by ŋ=k{(t1D2)/(t2D1)} where, t1=time in seconds, taken for the flow of the 50 c.c of oil, t2=time in seconds, taken for the flow of 50 c.c standard oil or liquid. D1=specific gravity of the oil to be tested. D2=specific gravity of the standard liquid. K= arbitrary constant.

13. The value of the arbitrary constant K is _________ for water.
a) 1
b) 100
c) 2
d) 50

Answer: a [Reason:] The absolute viscosity of the oil after performing the experiment in redwood viscometer can be given by ŋ=k{(t1D2)/(t2D1)}. Here k is called as the arbitrary constant. The value of the arbitrary constant for water is 1. The value of the arbitrary constant for rape seed oil is 100.

14. The kinematic viscosity of the oil can be calculated by _______
a) V = ŋ+ρ
b) V = ŋ/ρ
c) V = ŋ – ρ
d) V = ŋ*ρ

Answer: b [Reason:] The kinematic viscosity means the viscosity that determines the kinematic energy of the liquid that means it determines the speed of the oil. So, the kinematic viscosity of the oil is given by V=ŋ/ρ, where ŋ= absolute viscosity of the oil and ρ= density of the oil.

## Applied Chemistry MCQ Set 5

1. The molecular structure of the organic lubricants consists of _______
a) pyramidal
b) tetrahedral
c) trigonal planar
d) long chain

Answer: d [Reason:] The molecular structure of the organic lubricants is in the form of long chains. The molecular chains are parallel. The bonding strength between the molecules is very weak and may slide on each other.

2. Soaps are metal salts of _________
a) fatty acids
b) carboxylic acids
c) amino acids
d) inorganic acids

Answer: a [Reason:] Soaps are the metal salts of the fatty acids. Lithium, sodium, potassium and calcium are the metal salts. Soap molecules will be attached to substrate surface of the soap lubricant.

3. Soaps are prepared by chemical treating of _____ by strong alkaline solutions.
a) Oils and salts
b) Oils and fats
c) Salts and fats
d) Fats and proteins

Answer: b [Reason:] Soaps are made up of the chemical treatment of the oils and fats by strong alkaline solution. The soap molecule is the long non-polar tail which is hydrophobic end and the other is salt polar end which is the hydrophilic end.

4. Mobility of the solid lubricants on the adsorbents promotes _____
a) Self-healing
b) Maintain the molecular structure
c) Ineffectiveness
d) Effectiveness

Answer: a [Reason:] Mobility of solid lubricants on the surface of adsorbates promotes the self healing. It also prolongs the endurance of the films. As long as the film remains intact it can be protected.

5. Insulators making rubbing contact require _______
a) Low conductivity
b) High conductivity
c) Low pressure
d) High pressure

Answer: a [Reason:] Insulators making rubber contact needed the low electrical conductivity. In some other applications of the solid lubricants, the high electrical conductivities. Sliding electric contact requires high electrical conductivity.

6. Only lamellar structure lubricants provide lubrication.
a) True
b) False

Answer: b [Reason:] The statement lamellar structure lubricants provide lubrication is wrong because the non-lamellar lubricants also provide good lubrication. The lamellar structure is more likely to produce high lubrication.

7. If melting point increases, then the atomic bonds will ______
a) Become strong
b) Become weak
c) Be destroyed
d) Be with moderate strength

Answer: c [Reason:] As the melting point increases, the atomic bonds will be destroyed. It makes the lubrication less effective.

8. The lubricant should have _____ vapor pressure.
a) High
b) Extreme
c) Moderate
d) Low

Answer: d [Reason:] The ideal lubricant must have low vapour pressure for any application because in most of the applications of the lubricants, the lubricants must have low pressure and high temperature resistant.

9. What is the range of coefficient of friction for solid lubricants?
a) 0.005-0.01
b) 0.001-0.01
c) 0.0001-0.001
d) 0.0005-0.001