Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Preparation of Amines – 2”.

1. Which of the following reagents cannot be used to convert ethanenitrile to ethylamine?
a) H₂/Ni
b) LiAlH₄
c) Na(Hg), ethanol
d) Sn, HCl
Answer: d
Clarification: Nitriles on reduction with lithium aluminium hydride or catalytic hydrogenation or with Na(Hg)-ethanol produce primary amines.

2. How many more carbon atoms are present in the amine formed from the reduction of a nitrile, than in the nitrile itself?
a) 0
b) 1
c) 2
d) 1 or 2
Answer: a
Clarification: The -CN group of the nitrile is reduced to -CH₂-NH₂ group. This means that the parent carbon chain of the amine has one more carbon than the parent chain of the nitrile. However, the total number of carbons in both compounds is same as one of the carbon in nitriles is present in the cyanide group.

3. The reduction of phenyl isocyanide with H₂ and Ni catalyst gives a/an _________
a) primary amine
b) secondary amine
c) tertiary amine
d) arylalkyl amine
Answer: b
Clarification: The reduction of isocyanide compounds (where the CN group is attached through N atom) with H₂/Ni gives secondary amines or N-alkyl amines. For example, phenyl isocyanide gives N-methylaniline.

4. What is the product of the following reaction?

a) Phenylmethanamine
b) 1-Phenylethan-1-amine
c) 2-Phenylethan-1-amine
d) Toluene
Answer: c
Clarification: Benzyl cyanide reacts with LiAlH₄ to undergo reduction of the CN group to a CH₂NH₂ group and form a primary aromatic amine. The formula of this product will be C₆H₅CH₂CH₂NH₂, which is 2-phenylethan-1-amine.

5. Which of the following amines cannot be formed from the reduction of amides with LiAlH₄?
a) Ethylamine
b) Benzenamine
c) Benzylamine
d) Ethylmethylamine
Answer: b
Clarification: The -CONH₂ group of amides is reduced to a -CH₂NH₂ group and results in a compound having the same number of carbon atoms. The simplest aromatic amide is benzamide, which is reduced to give benzylamine. Hence, aniline cannot be formed as there is no corresponding amide that can be reduced to it with LiAlH₄. However, it can be achieved by Hoffmann bromamide degradation.

6. Only primary amines can be obtained from the reduction of amides with LiAlH₄.
a) True
b) False
View Answer

Answer: b
Clarification: Primary, secondary or tertiary amines can be formed depending on whether the amide used is primary, secondary or tertiary. This is because irrespective of the number of alkyl groups, the CO group of amides is directly reduced to CH₂.

7. Ethyldimethylamine is obtained from the reduction of _______ with LiAlH₄.
a) acetamide
b) benzamide
c) N-methylacetamide
d) N,N-dimethylacetamide
Answer: d
Clarification: Ethyldimethylamine is a tertiary amine with two methyl substituents directly attached to N atom. It can be obtained only from the reduction of a corresponding tertiary amide that has two methyl substituents attached to N atom.

8. Which path gives propan-1-amine as the product in the reaction below?

a) Only A
b) Only B
c) Both A and B
d) Not A nor B
Answer: a
Clarification: Propanamine has three carbon atoms and so does the parent amide. So, the reduction of CO to CH₂ will give the required compound. Whereas in path B, the amide undergoes Hoffmann bromamide degradation to give ethanamine.

9. Which of the following is not a by product of Hoffmann bromamide degradation of acetamide with alcoholic KOH?
a) KBr
b) KCN
c) K₂CO₃
d) H₂O
Answer: b
Clarification: Acetamide (CH₃CONH₂) reacts with Br₂ and 4 molecules of KOH to form methylamine (main product) along with 2 molecules KBr, 2 molecules of H₂O and one molecule of K₂CO₃.

10. Hoffmann bromamide degradation reaction is used for preparing _______ amines.
a) primary
b) secondary
c) tertiary
d) mixed
Answer: a
Clarification: Mixed amines are a classification of secondary and tertiary amines. The Hoffmann bromamide degradation is only possible with 1° amides so as to produce primary amines.

11. The best reagent for converting 2-phenylpropanamide to 1-phenylethanamine is ________
a) H₂/Ni
b) Na(Hg)/C₂H₅OH
c) LiAlH₄
d) NaOH/Br₂
Answer: d
Clarification: 2-Phenylpropanamide has 9 carbon atoms while 1-phenylethamine has 8 carbon atoms. This depicts stepping down of carbon from amide to amine conversion and this takes place in Hoffmann degradation reaction.

12. Which of the following compounds undergoes Hoffmann bromamide degradation reaction?
a) C₆H₅NH₂
b) C₆H₅NO₂
c) C₆H₅CONH₂
d) C₆H₅CH₂NH₂
Answer: c
Clarification: Only primary amides undergo this reaction with bromine in NaOH to form primary amines. Benzamide undergoes this reaction to form benzenamine or aniline.

13. Which of the following amines can be prepared from Gabriel phthalimide synthesis?
a) Benzylamine
b) Aniline
c) o-Toluidine
d) N-Methylbenzenamine
Answer: a
Clarification: Since Gabriel phthalimide synthesis is a nucleophilic substitution reaction, aryl halides cannot be prepared form this. This is because of the sp² hybridised aryl carbon and stable benzene ring.

14. Gabriel phthalimide synthesis can be used for preparing phenylmethanamine.
a) True
b) False
Answer: a
Clarification: Potassium phthalimide can undergo nucleophilic substitution with a benzyl halide to form N-phenylphthalimide, which on hydrolysis with NaOH gives benzylamine or phenylmethanamine.

15. Which of the following steps is not present in Gabriel phthalimide synthesis?
a) Treating phthalimide with alcoholic KOH
b) Heating potassium phthalimide with alkyl halide
c) Alkaline hydrolysis of N-alkylphthalimide
d) Heating phthalic acid with NaOH
Answer: d
Clarification: The chemical compounds involved in the Gabriel phthalimide synthesis are phthalimide, potassium phthalimide and N-alkyl phthalimide. These are formed in order when reacted with different reagents like KOH and haloalkanes.

Chemistry Multiple Choice Questions on “Biomolecules – Proteins – 1”.

1. Proteins are polymers of ______
a) α-amino acids
b) β-amino acids
c) γ-amino acids
d) δ-amino acids
Answer: a
Clarification: Amino acids contain amino (NH₂) and carboxyl (COOH) functional groups. α-amino acids contain the NH₂ group on the carbon adjacent to the COOH group. Proteins on hydrolysis yield only α-amino acids.

2. If the basic formula of an α-amino acid is R-CH(NH₂)-COOH, where R is the side chain, what is the primary point of distinction between any two proteins?
a) Number of amino groups
b) Number of carboxyl groups
c) The side chain R
d) Relative positions of amino, carboxyl groups and R
Answer: c
Clarification: α-amino acids are the constituents of proteins. Different proteins are formed by polymerisation of different α-amino acids, which are formed due to the difference in the side chain substituted group R, which may be as simple as hydrogen or as complex as imidazole.

3. Which of the following amino acids is optically inactive?
a) Glycine
b) Alanine
c) Lysine
d) Valine
Answer: a
Clarification: Glycine is the only naturally occurring α-amino acid that is optically inactive. This is because the α-carbon of glycine has two hydrogen atoms attached to it.

4. What is the one letter code for tyrosine?
a) T
b) Y
c) R
d) S
Answer: b
Clarification: Tyrosine is a natural amino acid first obtained from cheese, hence the name. It has a 4-hydroxyphenylmethyl side chain. Its three-letter symbol is Tyr and its code is Y.

5. Which of the following amino acids are aromatic in nature?
a) Methionine
b) Isoleucine
c) Proline
d) Histidine
Answer: d
Clarification: Amino acids can be aromatic only if their side chain consists of an aromatic compound. Methionine has a sulphur containing straight chain. Isoleucine has an isobutyl side group. Proline has a complex cyclic side chain. Histidine has a basic aromatic side group, imidazole.

6. Which of the following is a non-essential amino acid?
a) Threonine
b) Glutamine
c) Phenylalanine
d) Valine
Answer: b
Clarification: Threonine, phenylalanine and valine along with seven other amino acids cannot be produced by the human body and must be obtained through diet. These are essential amino acids. Other amino acids like glutamine are synthesized by the human body and are therefore called non-essential amino acids.

7. Which of the following is a neutral amino acid?
a) Glycine
b) Lysine
c) Arginine
d) Histidine
Answer: a
Clarification: Neutral amino acids contain equal number of amino and carboxyl groups. Lysine, arginine and histidine contain two NH₂ groups are one COOH group, and are hence basic amino acids.

8. Cysteine is a/an ________ amino acid.
a) acidic
b) essential
c) aromatic
d) sulphur containing
Answer: d
Clarification: Cysteine is a semi-essential, neutral amino acid. Its side chain is HS-CH₂ (thiol), which is aliphatic and contains sulphur. Its symbol is Cys and its code is C.

9. Which of the following amino acids contains only one amino group?
a) Leucine
b) Lysine
c) Asparagine
d) Glutamine
Answer: a
Clarification: Lysine, asparagine and glutamine all contain two amino (NH₂) groups, of which one is a part of the side chain. Leucine has a purely alkyl side chain and the only amino group it has is one the α-carbon.

10. What is the one letter code for asparagine?
a) A
b) P
c) N
d) S
Answer: c
Clarification: For the sake of simplicity, each amino acid has been given an abbreviation which is either a three-letter symbol or a one-letter code. Asparagine is represented as Asn or N.

11. Which of the following is incorrect regarding tryptophan?
a) It is an essential amino acid
b) It is a basic amino acid
c) It has an aromatic side chain
d) It is a non-polar amino acid
Answer: b
Clarification: Tryptophan is an essential amino acid with bicyclic aromatic side chain, indole. It cannot be synthesized by the human body and must be taken for nitrogen balance in the body. It has one amino and carboxyl group each and is a neutral amino acid.

12. The structure shown below is ______

a) Side chain of histidine
b) Side chain of tryptophan
c) Side chain of proline
d) Proline
Answer: d
Clarification: Proline is a neutral aliphatic amino acid, with cyclic pyrrolidine side chain, making it non-polar. It is non-essential as the human body can synthesize it.

13. Identify the amino acid with the formula HOOC-CH₂-CH₂-CH(NH₂)-COOH.
a) Glutamic acid
b) Aspartic acid
c) Glutamine
d) Asparagine
Answer: a
Clarification: Glutamic acid is an amino substituted dicarboxylic acid which can be synthesized by the body. Its IUPAC name is 2-aminopentanedioic acid. Since it has two COOH groups (one more than the number of NH₂ groups), it is acidic in nature.

14. Amino acids behave like carboxylic acids
a) True
b) False
Answer: b
Clarification: Generally, amino acids behave like salts rather than simple amines or carboxylic acids. This is due to the presence of both acidic (COOH) and basic (NH₂) groups in the same molecule.

15. Amino acids can show amphoteric behaviour.
a) True
b) False
Answer: a
Clarification: In an aqueous solution of amino acid, COOH loses a proton and NH₂ accepts a proton, giving rise to a dipolar zwitter ion. This is neutral but contains both positive and negative terminals. In this form, amino acids react both with acids and bases, and hence show amphoteric behaviour.

Chemistry Question Papers for IIT JEE Exam on “Chemicals in Food”.

1. Which of the following chemicals are added to increase the shelf-life of foods?
a) Food colour
b) Sweeteners
c) Artificial flavours
d) Antioxidants
Answer: d
Clarification: Chemicals are added to food for improving their aesthetic, to increase their shelf-life or to add nutritive value. Food colours, flavours and sweeteners simply enhance the appearance and taste of food. Antioxidants are used for preservation.

2. Which of the following artificial sweeteners can be only used in soft drinks?
a) Aspartame
b) Alitame
c) Sucralose
d) Saccharin
Answer: a
Clarification: Aspartame is an artificial sweetening agent which is unstable at higher temperatures. It is a compound formed from aspartic acid and phenylalanine. It is 100 times as sweet as cane sugar.

3. Alitame is ______ times sweeter than cane sugar.
a) 50
b) 100
c) 600
d) 2000
Answer: d
Clarification: Alitame is a compound consisting of one hydroxy group, three keto groups and a tetra-methyl sulphur containing cyclic group. It is a very strong sweetener and provides uncontrollable sweetness to food.

4. Which of the following is not a preservative?
a) Dulcin
b) Potassium metasulphite
c) Sodium benzoate
d) Sorbic acid salts
Answer: a
Clarification: Preservatives are chemicals that are added to food to prevent their spoilage and preserve their nutritive value and taste. Salt, sugar and oils are the most common examples. Potassium metasulphite is used in jams. Dulcin is an artificial sweeter.

5. Which of the following is not an antioxidant?
a) BHT
b) BHA
c) Saccharin
d) Sulphur dioxide
Answer: c
Clarification: Antioxidants prevent food spoilage by itself reacting with oxygen and protecting the food from oxidation. Butylated hydroxy toluene and butylated hydroxy anisole increase the shelf lives of food like butter. SO₂ is also used in wines, fruits and sugar syrups for preservation.

Chemistry Question Papers for IIT JEE Exam,

Chemistry Multiple Choice Questions on “Amorphous and Crystalline Solids”.

1. In polar molecular solids, the molecules are held together by ________
a) dipole-dipole interactions
b) dispersion forces
c) hydrogen bonds
d) covalent bonds
Answer: a
Clarification: Molecular solids are solids that are collections of molecules held together by intermolecular forces. In polar molecules such as HCl, So₂, etc., the molecules are held together by dipole-dipole interactions.

2. Diamond is an example of _______
a) solid with hydrogen bonding
b) electrovalent solid
c) covalent solid
d) glass
Answer: c
Clarification: The solids in which constituent particles are attached to each other by covalent bonds are called covalent solids. Diamond, graphite, silicon, SiC, AIN, quartz are examples of covalent solids.

3. Silicon is found in nature in the forms of ________
a) body-centered cubic structure
b) hexagonal-closed packed structure
c) network solid
d) face-centered cubic structure
Answer: c
Clarification: Silicon due to its catenation property form network solid. Catenation is the ability of an atom to form bonds with other atoms of the same element. The compounds of silicon are reactive and not stable.

4. Which one of the following are the dimensions of cubic crystal?
a) a =b ≠ c
b) a = b = c and α = β ≠ γ = 90
c) a = b = c and α = β = γ = 90
d) a ≠ b = c and α = β ≠ γ = 90
Answer: c
Clarification: The dimensions of a cubic crystals are a = b = c, α = β = γ = 90.

5. Which of the following is not a crystal system?
a) Cubic
b) Trigonal
c) Triclinic
d) Hexaclinic
Answer: d
Clarification: Hexaclinic is not a crystal system. Their crystal system are cubic, tetragonal, rhombohedral or trigonal, orthorhombic or rhombic, monoclinic, triclinic and hexagonal.

6. In face-centred cubic cell, a unit cell is shared equally by __________
a) four unit cells
b) two unit cells
c) one unit cell
d) six unit cells
Answer: d
Clarification: The unit cell in which atoms are present at corners as well as faces of unit cell is known as face-centred cubic unit cell. In face-centred cubic cell, a unit cell is shared equally by six unit cells.

7. The unit cell with a≠b≠c and α=β=γ=90 refers to __________ crystal system.
a) hexagonal
b) trigonal
c) triclinic
d) orthorhombic
Answer: d
Clarification: In orthorhombic crystal system, all three axes are unequal in length and all are perpendicular to one another. It is also called as rhombic crystal system. Topaz, barite are some examples of orthorhombic crystals.

8. Which is the most unsymmetrical crystal system?
a) Triclinic crystal system
b) Cubic crystal system
c) Hexagonal crystal system
d) Trigonal crystal system
Answer: a
Clarification: Most unsymmetrical crystal system is triclinic in which all three axes are unequal in length none is perpendicular to another. Triclinic unit cells has the least symmetrical shape of all unit cells. Turquoise is an example of triclinic crystal.

9. In the simple cubic cell, each corner atom is shared by __________
a) eight unit cells
b) one unit cell
c) two unit cells
d) six unit cells
Answer: a
Clarification: The unit cell in which the constituent atoms are present only at the corner is known as simple cubic cell. It is also referred to as a primitive cubic cell. In the simple cubic cell, each corner atom is shared by eight different unit cells.

10. The points which shows the position of atoms in crystal are called as _________
a) crystal lattice
b) crystal parameters
c) bravais lattice
d) lattice point
Answer: d
Clarification: The point at which the atoms may be present on the unit cell is termed as lattice point. It shows the position of atoms in crystal.

11. The unit cell with a≠b≠c and α=γ=90, β≠90 refers to __________crystal system.
a) cubic
b) tetragonal
c) monoclinic
d) triclinic
Answer: c
Clarification: In monoclinic crystal system, all the three axes are unequal in length and two axes are perpendicular to each other. Gypsum and borax are examples of monoclinic crystals.

12. Which type of solid crystals will conduct heat and electricity?
a) Ionic
b) Covalent
c) Molecular
d) Metallic
Answer: d
Clarification: Metallic crystals consist of metal cations surrounded by a sea of mobile valence electrons. These electrons are capable of moving through the entire crystal. The metallic crystals conduct heat and electricity due to the presence of these mobile electrons in them.

13. Which is not a characteristic of crystalline solids?
a) They undergo a clean cleavage
b) They are true solids
c) They are isotropic
d) They have sharp melting points
View Answer

Answer: c
Clarification: Amorphous solids are isotropic that is they have identical properties in all directions, whereas crystalline solids are anisotropic that is they have different properties in different directions.

14. Which of the following is a characteristic of amorphous solid?
a) They are true solids
b) They have sharp melting points
c) They undergo clear cleavage
d) They are isotropic
Answer: d
Clarification: Amorphous solids are isotropic that is they have identical properties in all directions. The remaining options are the characteristics of crystalline solids.

15. Solids are classified as ___________
a) crystalline and ionic solids
b) metallic and amorphous solids
c) molecular and covalent solids
d) crystalline and amorphous solids
Answer: d
Clarification: Based on their crystal structures, solids are classified as crystalline and amorphous solids. In crystalline solids, the constituent particles are arranged in a regular manner. In amorphous solids, the constituent particles are not arranged in any regular manner.

16. Quartz is an example of ___________
a) molecular solids
b) ionic solids
c) covalent solids
d) metallic solids
Answer: c
Clarification: Quartz is a common example of covalent solids. In covalent solids, the constituent particles are attached to each other by covalent bonds. Diamond, graphite, silicon are other examples of covalent solids.

17. Solid carbon dioxide is an example of _________
a) metallic crystal
b) covalent crystal
c) ionic crystal
d) molecular crystal
Answer: d
Clarification: Solid CO₂ is an example of molecular crystal. These solids have molecules as their constituent particles. These solids may be bonded by vander waals’ forces or by dipole-dipole attraction or by strong hydrogen bonds. H₂, Cl₂, I₂ are some examples of molecular solids.

Chemistry Online Quiz for Schools on “Colligative Properties and Determination of Molar Mass – 2”.

1. Why is ‘raising of viscosity’ of a solution after addition of solute, not considered to be a colligative property?
a) The resultant viscosity depends on the nature of the solute
b) The resultant viscosity depends on the amount of solute
c) The resultant viscosity depends on the nature of solvent
d) The resultant viscosity depends on the amount of solvent
Answer: a
Clarification: A colligative property is identified by the fact that it has no dependence on the nature of the particles of solute. However, in the case of viscosity it really depends on the solute that is added to the solvent. Thus, the change in viscosity after addition of a non-volatile solute cannot be considered to be a colligative property.

2. At 70°C the vapor pressure of pure water is 31 kPa. Which of the following is most likely the vapor pressure of a 2.0 molal aq. glucose solution at 70°C?
a) 30.001 kPa
b) 29.915 kPa
c) 28.226 kPa
d) 32.392 kPa
Answer: b
Clarification: Given, P⁰_water = 31 kPa
Concentration of solution, c = 2 molal = 2 moles of glucose/kg of water
From law of relative lowering of vapor pressure, ΔP/P⁰ = X₂, where X₂ is the mole fraction of glucose in the solution.
Mass of water = 1 kg = 1000 g
Molecular weight of water = 18 g/mole
Moles of water = 1000/18 = 55.556 moles
X₂ = 2/(2 + 55.556) = 0.035
ΔP = 31 kPa x 0.035 = 1.085 kPa
Final pressure = 31 kPa – 1.085 kPa = 29.915 kPa.

3. 117 g of NaCl is added to 222 g of water in a saucepan. At what does temperature does water boil at 101.325 kPa? Ebullioscopy constant for water = 0.52 K kg mol^-1 and b.p. = 100°C
a) 98.3°C
b) 102.8°C
c) 104.7°C
d) 101.5°C
Answer: c
Clarification: Given,
Weight of solvent, w₁ = 222 g
Weight of solute, w₂ = 117 g
K_b = 0.53 K kg mol^-1
Now, addition of a non-volatile solute causes elevation in boiling point, ΔT_b
ΔT_b = (k_b x 1000 x w₂)/(M₂ x w₁)
On substituting, ΔT_b = (0.52 x 1000 x 117)/(58.5 x 222) = 4.7°C
New boiling temperature = 100 + 4.7 = 104.7°C.

4. Boiling point of chloroform is 61°C. After addition of 5.0 g of a non-volatile solute to 20 g chloroform boils at 64.63°C. If k_b = 3.63 K kg mol^-1, what is the molecular weight of the solute?
a) 320 g/mol
b) 100 g/mol
c) 400 g/mol
d) 250 g/mol
Answer: d
Clarification: Given,
b.p. of chloroform = 61°C
New b.p. after addition = 64.63°C
Mass of solute, w₂ = 5.0 g
Mass of solvent, w₁ = 20 g
K_b = 3.63 K kg mol^-1
From these, ∆T_b = 64.63 – 61 = 3.63°C
Using ΔT_b = (k_b x 1000 x w₂)/(M₂ x w₁)
M₂ = (k_b x 1000 x w₂)/(ΔT_b x w₁)
M₂ = (3.63 x 1000 x 5)/(3.63 x 20) = 250 g/mol.

5. Pure CS₂ melts at -112°C. 228 grams of propylene glycol crystals is mixed with 500 grams of CS₂. If k_f of CS₂ = -3.83 K kg mol^-1 what is the depression in freezing point?
a) -23°C
b) -135°C
c) -20°C
d) -100°C
Answer: a
Clarification: Given,
k_f = -3.83 k kg mol^-1
Mass of solute, w₂ = 228 g
Mass of solvent, w₁ = 500 g
Molar mass of solute, M₂ = 76 g/mole
Moles of solute = w₂/M₂ = 228/76 = 3 moles
Molality of the solution, m = Number of moles of solute/mass of solvent (kg)
m = 3 moles/0.5 kg = 6 molal
We know, ΔT_f = k_f x m
ΔT_f = -3.83 x 6 = -23°C.

6. What are colligative properties useful for?
a) Determining boiling and melting temperature
b) Determining molar mass
c) Determining equivalent weight
d) Determining van’t Hoff factor
Answer: b
Clarification: Colligative properties serve the purpose for determining molar masses of unknown compounds. Colligative properties are not used to determine boiling and melting temperatures as it would result in an incorrect value upon the addition of a solute. Equivalent weight can only be determined if the molar mass is known and the van’t Hoff factor is determined in a similar manner.

7. A pair of solution bears the same osmotic pressure. What is this pair of solutions called?
a) Hypertonic
b) Hypotonic
c) Isotonic
d) Osmolarity
Answer: c
Clarification: Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions. Hypertonic solutions are those in which the concentration of solute outside the membrane and a lesser concentration within the membrane. Osmolarity is a type of concentration expressed in number of solute particles per liter.

8. A cell with lots of salt inside it is placed in a vessel containing just water. Which process takes place?
a) Dialysis
b) Filtration
c) Shriveling
d) Osmosis
Answer: d
Clarification: Osmosis is a mass transfer process due to which water molecules move from a region of higher water potential to lower water potential down the potential gradient. Dialysis and filtration are processes using the concept of diffusion of impurities. Shriveling is the shrinking of a cell. In this case, the cell swells up.

9. What is a necessary condition for osmosis to take place?
a) Semi-permeable membrane
b) Same concentration of solvent
c) High temperature
d) Pressure greater than osmotic pressure
Answer: a
Clarification: A semi permeable membrane is a must condition since it facilitates the blocking of solute particles from diffusing through and allows only water molecules to pass through. Obviously, water is the solvent in osmosis hence the concentration of solvent cannot be same if osmosis has to occur. If pressure greater than osmotic pressure is applied, reverse osmosis takes place. Meaning, water molecules will flow from region of lower water potential to higher water potential.

10. Which is the most appropriate method for determining the molar masses of biomolecules?
a) Relative lowering of vapor pressure
b) Elevation of boiling point
c) Depression in freezing point
d) Osmosis
Answer: d
Clarification: Osmotic pressure method has the greatest advantage over other methods that even for very dilute concentrations it gives a large magnitude. This would not be true in case other methods. It is highly useful for biomolecules since they are unstable at extremely high and low temperatures thus eradicating the method using boiling and freezing points.

Chemistry Online Quiz for Schools,

Chemistry Multiple Choice Questions on “Surface Chemistry – Adsorption”.

1. A finely divided substance is more effective as an adsorbent.
a) True
b) False
Answer: a
Clarification: Adsorption is a surface phenomenon. It is dependent on the amount of the adsorbent exposed to the adsorbate. A finely divided substance provides a large surface area and hence, provides more sites where adsorption can take place.

2. What is the process called when the molecules of a substance are retained at the surface of a solid or a liquid?
a) Absorption
b) Adsorption
c) Sorption
d) Desorption
Answer: b
Clarification: Adsorption is the process which involves the accumulation of the molecules of a substance in higher concentration on the surface of a solid or a liquid. For example, gasses are adsorbed on the surface of charcoal.

3. Which of the following forces is involved in physical adsorption?
a) Gravitational force
b) Magnetic force
c) Van der Waals force
d) Electromagnetic force
Answer: c
Clarification: In physical adsorption, the molecules of the adsorbate stick to the surface of the adsorbent due to very weak forces which is known as Van der Waals force. Van der Waals force is similar to the forces that cause condensation of gas into liquid.

4. In physisorption, the adsorbent does not show specificity towards a particular gas.
a) True
b) False
Answer: a
Clarification: In physisorption, there is no chemical bond between the adsorbent and adsorbate. They are held together by Van der Waals forces. Since Van der Waals forces are universal and almost same for all gases, the adsorbent does not show specificity to a particular gas.

5. Which of the following statements is true with respect to the extent of physisorption?
a) Increases with increase in temperature
b) Decreases with increase in surface area
c) Decreases with increase in the strength of Van der Waals forces
d) Decreases with increase in temperature
Answer: d
Clarification: Physisorption is an exothermic process. According to Le-Chatelier’s principle, an exothermic reaction is favoured by a decrease in temperature. Therefore, the extent of physisorption decreases on increasing temperature.

6. Which of the following can result in a transition from physisorption to chemisorption?
a) Decrease in temperature
b) Increase in temperature
c) Decrease in pressure
d) Increase in surface area
Answer: b
Clarification: Physisorption occurs due to Van der Waals forces. On increasing the temperature, the adsorbate can split into atoms and form chemical bonds with the adsorbent and hence, cause chemisorption.

7. Which of the following statements is incorrect with respect to physisorption?
a) It is reversible
b) It is spontaneous
c) ΔH d) ΔS > 0
Answer: d
Clarification: Physisorption is reversible as the molecules of the adsorbate are held to the adsorbent by weak Van der Waals forces which can be broken easily. Physisorption is exothermic and so, ΔH is negative. Adsorption is a spontaneous process. In adsorption, the movement of the adsorbed molecule is restricted. As a result the entropy change (ΔS) is negative.

8. Which of the following statements is true with respect to the types of adsorption?
a) Chemisorption is stronger than physisorption
b) Physisorption is stronger than chemisorption
c) They are both equal
d) They cannot be compared
Answer: a
Clarification: Chemisorption is stronger than physisorption. Chemisorption involves the forming of a chemical bond between the adsorbent and the adsorbate whereas in physisorption, the molecules are held together by weak Van der Waal’s forces. Therefore, chemisorption is stronger.

9. Which of the following is an example of sorption?
a) Sponge in water
b) Cotton dipped in ink
c) Water on silica gel
d) Oxygen on metal surface
Answer: b
Clarification: Sorption refers to the process where adsorption and absorption occur at the same time. Cotton dipped in ink is one of the cases where both adsorption and absorption, that is, sorption occurs.

10. Which of the following statements is not true with respect to chemisorption?
a) Depends on nature of adsorbate and adsorbent
b) Has a large heat of adsorption
c) Forms a unimolecular layer
d) Occurs at low temperature
Answer: d
Clarification: Chemisorption is a kind of adsorption that involves a chemical reaction between the adsorbent and adsorbate, resulting in the formation of a chemical bond between the two. Since a bond is to be formed, chemisorption is highly dependent on the reactants(adsorbate and adsorbent) and has a large heat of adsorption. Also, since it involves bond formation, it cannot be multi-layered. Chemisorption is favoured to occur at high temperature.