Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Amines Chemical Reactions – 1”.

1. Amines are generally ______ in nature.
a) electrophilic
b) acidic
c) basic
d) neutral
Answer: c
Clarification: Amines behave as nucleophiles due to the presence of unshared pair of electrons on N atom. This also makes them proton acceptor, and react with acids to form salts.

2. Which of the following is associated with decrease in pK_b value of amines?
a) Increase in acidic strength
b) Increase in basic strength
c) Better proton donation
d) Better electron acceptor
Answer: b
Clarification: The basic strength of amines is expressed in terms of dissociation constant K_b. Greater the K_b value, stronger the base. It is more commonly expressed as pK_b=-logK_b. Hence smaller the value of pK_b, more is the basic strength of the amine.

3. The equilibrium constant of the reaction of amines with _____ is taken as a measure of its basic character.
a) an acid
b) a base
c) water
d) a Lewis base
Answer: c
Clarification: For the reaction,
RNH₂ + H₂O ↔ RNH₃⁺ + OH^–, the dissociation constant is,
K_eq = ([RNH₃⁺][OH^–])/([RNH₂][H₂O])
(Rightarrow) K_eq[H₂O] = ([RNH₃⁺][OH^–])/[RNH₂]
(Rightarrow) K_b = ([RNH³⁺][OH^–])/[RNH₂]

4. What is the product formed when ethanamine reacts with HBr?
a) NH₄Br
b) CH₃NH₃Br
c) CH₃CH₂NHBr
d) CH₃CH₂NH₃Br
Answer: d
Clarification: Since amines are basic in nature, they react with acids to form salts of ammonium. Ethanamine (CH₃CH₂NH₂) reacts with HBr in a reversible reaction to form ethylammonium bromide through addition.

5. The reaction between methylamine and hydrogen iodide results in the formation of a _______
a) colourless liquid
b) dark coloured gas
c) white solid
d) yellow liquid
View Answer

Answer: c
Clarification: Methylamine and hydrogen iodide are allowed to mix at very cold temperatures for about 2 hours. The resulting product is allowed to evaporate and methanaminium iodide (CH₃NH₃I) is obtained in the form of a white powder.

6. Which of the following is not produced on the reaction of methylammonium chloride with sodium hydroxide?
a) HCl
b) CH₃NH₂
c) H₂O
d) NaCl
Answer: a
Clarification: Amine salts on treatment with a base like NaOH gives back the parent amine along with a water molecule and a salt. For example, CH₃NH₃Cl with NaOH gives methanamine , water and sodium chloride.

7. Identify X and Y respectively in the following reaction.

a) OH^–, HCl
b) HCl, OH^–
c) H₂O, OH^–
d) HCl, H₂O
Answer: b
Clarification: Aniline undergoes addition in the presence of HCl to give anilinium chloride (salt) which on action with OH^– ions (from a base like NaOH) gives back aniline.

8. What is the correct order of pK_b values of the following amines?
a) Methanamine > Ethanamine > Benzenamine
b) Benzenamine > Ethanamine > Methanamine
c) Ethanamine > Methanamine > Benzenamine
d) Benzenamine > Methanamine > Ethanamine
Answer: d
Clarification: The higher pK_b value means lower basic strength. This means that benzenamine has the lower basic strength among methanamine and ethanamine. This is because of the electron withdrawing nature of the aryl group. Also in primary amines, the increase in size of alkyl group increases +I effect leading to high electron density on N atom and as a result higher basicity.

9. The treatment of N,N-dimethylaniline with acetic acid gives no reaction.
a) True
b) False
Answer: b
Clarification: N,N-Dimethylaniline reacts with acetic acid to form a salt, N,N-dimethylanilinium acetate. This is due to the basic nature of amines.

10. If the pK_b value of ammonia is 4.75, predict the pK_b value of methanamine?
a) 3.38
b) 4.70
c) 8.92
d) 9.38
Answer: a
Clarification: Aliphatic amines are stronger bases than ammonia and hence have a much lower pK_b than ammonia. This is because of the electron donating nature of alkyl groups which increase the negative charge on nitrogen atom, making it less prone to electron acceptance.

11. If the K_b values of ammonia, methylamine and ethylamine are x, y and z respectively, identify the correct relation between x, y and z from the following.
a) x > y
b) y c) x > z
d) x > y > z
Answer: b
Clarification: Larger the value of K_b, stronger is the base. Since ethylamine is a stronger base than methylamine which is stronger than ammonia, due to the effect of alkyl groups, the K_b vale of methylamine will be less than that of ethylamine (y

12. Consider three gaseous alkylamines A, B and C or 1°, 2° and 3° respectively. What will be the correct order of their basicity?
a) A > B > C
b) C > B > A
c) B > A > C
d) B > C > A
Answer: b
Clarification: In the gaseous state, the solvation effect is missing and hence the expected order of basicity will be 3° > 2° > 1°. This is because as the number of alkyl groups increase, the +I effect strengthens and more negative charge is accumulated on the N atom. This makes the unpaired share mire available for sharing with the proton of the acid.

13. The basic strength of alkylamines does not depend on which of the following?
a) Number or alkyl groups
b) Size of alkyl groups
c) Physical state of the amine
d) Presence of an aromatic ring
Answer: d
Clarification: Since only basicity of alkylamines is in question the presence of aromatic ring is not considered. As the size and number of alkyl groups increases, the stability of ammonium ion (formed from the amine) increases due to dispersal of more positive charge by the +I effect of alkyl groups. The physical state, gaseous or aqueous, also determines the basicity as the hydration effect comes into play.

14. What is the correct order of basicity of aliphatic amines purely on the basis of solvation effect of the ammonium cation?
a) 1° > 2° > 3°
b) 3° > 2° > 1°
c) 2° > 1° > 3°
d) 2° > 3° > 1°
Answer: a
Clarification: In aqueous phase, the greater the size of the ion, lesser will be the hydration and less stabilised is the ion. Since the stability of ions is directly proportional to the basic strength of amines, primary amines will be the most basic and tertiary amines will be the least basic. This order is completely opposite to that based on the inductive effect of alkyl groups.

15. Stearic hinderance of alkyl groups has an effect on the basic character of amines.
a) True
b) False
Answer: a
Clarification: When the alkyl group is small, there is no stearic hinderance to hydrogen bonding. But when the alkyl group size or number increases, there will be hinderance to formation of hydrogen bonds, and this affects the order of basic strength of amines.

Chemistry Multiple Choice Questions on “Biomolecules – Enzymes”.

1. Which of the following best describes a particular enzyme?
a) Chemical catalyst
b) Fibrous protein
c) Highly selective
d) Can be used for various reactions
Answer: c
Clarification: Enzymes are biological catalysts produced by living cells. They differ from other catalysts in being highly selective and specific. Almost all enzymes are globular proteins. They are very specific for a particular reaction and a particular substrate.

2. Enzymes are generally named after the ________
a) compound on which they work
b) compound which they form as product
c) medium in which they act
d) place from where they are derived
Answer: a
Clarification: Enzymes are generally name after the compound or class of compounds on which they work. For example, the catalyst that hydrolyses the reaction of starch (amylum) to glucose is called as amylase.

3. The enzyme which catalyses the conversion of proteins to amino acids is ______
a) invertase
b) urease
c) nuclease
d) protease
Answer: d
Clarification: Proteases are enzymes that helps in protein catabolism by hydrolysis of peptide bonds. Pepsin and trypsin are the most common proteases, found in the human stomach and pancreas respectively. Both of them are major digestive enzymes.

4. Which of the following is a substrate specific enzyme?
a) Maltase
b) Carboxylase
c) Hexokinase
d) Carbonic anhydrase
View Answer

Answer: a
Clarification: Substrate specific enzymes are those which can act only on one particular compound to give a product(s). For example, maltase acts only on maltose to break the glycosidic linkage between the two glucose units.

5. Enzymes are regarded as ______
a) biocatalysts
b) messengers
c) inhibitors
d) antibodies
Answer: a
Clarification: Enzymes are biological catalysts produced by living cells which catalyse the biochemical reactions in living organisms. Chemically, enzymes are naturally occurring simple or conjugate proteins.

6. Enzymes are basically ______
a) polysaccharides
b) sugars
c) polypeptides
d) pyrimidine bases
Answer: c
Clarification: Almost all enzymes are globular proteins, which are nothing but very long chains of amino acids residues (>100) and higher molecular mass, or polypeptides.

7. Cellulose is not digestible by humans due to the absence of which of the following enzymes?
a) Amylase
b) Urease
c) Cellulase
d) Invertase
Answer: c
Clarification: The enzyme cellulase hydrolyses cellulose into glucose and thereby digests it. However, the human stomach does not produce any enzyme capable of digesting cellulose, and hence it cannot be digested.

8. Enzymes reduce the magnitude of activation energy for a reaction.
a) True
b) False
Answer: a
Clarification: For example, the activation energy for acid hydrolysis of sucrose is 6.22kJ/mol, while the activation energy is only 2.15kJ/mol when it is hydrolysed in the presence of sucrase.

9. The hydrolysis of lactose can be catalysed only by the enzyme lactase. Also, lactase is only able to work on lactose and no other compound.
a) True
b) False
Answer: a
Clarification: Enzymes are highly specific in nature. Almost all biochemical reactions ae controlled by its own specific enzyme.

10. The prosthetic groups which get attached to the enzyme at the time of reaction are called _____
a) cofactors
b) coenzymes
c) messengers
d) inhibitors
Answer: b
Clarification: Most active enzymes are associated with some non-protein components required for their activity. These are called prosthetic groups and they may be cofactors or coenzymes.

11. Albinism is caused by the deficiency of which enzyme?
a) Phenylalanine hydroxylase
b) Streptokinase
c) Prolidase
d) Tyrosinase
Answer: d
Clarification: Deficiency of tyrosinase results in insufficient production of melanin. This causes a condition of white skin and hair called albinism. It can be prevented by including the required supply of enzyme through diet.

12. Identify the correct statement about enzymes.
a) Enzymes increase the activation energy of a reaction
b) Enzymes need to be used in excess compared to the reagent to catalyse the reaction
c) Enzymes work only at their optimum temperature and pH
d) The activity of enzymes cannot be affected by other compounds
Answer: c
Clarification: Enzymes decrease the activation energy of a reaction. Extremely small quantities of enzymes can increase the rate of reaction by thousands. Enzyme action can be inhibited by compounds known as enzyme inhibitors.

Chemistry Multiple Choice Questions on “Solid State – Crystal Lattices and Unit Cells”.

1. Which of the following is regarded as the ‘repeatable entity’ of a 3D crystal structure?
a) Unit cell
b) Lattice
c) Crystal
d) Bravais Index
Answer: a
Clarification: Unit cell is the smallest entity of a crystal lattice which, when repeated in space (3 dimensions) generates the entire crystal lattice. Lattice comprises of the unit cells which hold all the particles in a particular arrangement in 3 dimensions. Crystal is a piece of homogenous solid and Bravais indices are used to define planes in crystal lattices in the hexagonal system.

2. Which of the following unit cells has constituent particles occupying the corner positions only?
a) Body-centered cell
b) Primitive cell
c) Face centered cell
d) End-centered cell
Answer: b
Clarification: According to classification of unit cells, a primitive unit cell is one which has all constituent particles located at its corners. BCC has one particle present at the center including the corners. FCC has an individual cell shared between the faces of adjacent cells. End centered cells have cells present at centers of two opposite faces.

3. What is the coordination number of a body-centered unit cell?
a) 6
b) 12
c) 8
d) 4
Answer: c
Clarification: Coordination number of a unit cell is defined as the number of atoms/ions that surround the central atom/ion. In the case of BCC, the central particle is surrounded by 8 particles hence, 8.

4. Which of the following arrangements of particles does a simple cubic lattice follow?
a) ABAB
b) AABB
c) ABCABC
d) AAA
Answer: d
Clarification: Simple cubic lattice results from 3D close packing from 2D square-packed layers. When one 2D layer is placed on top of the other, the corresponding spheres of the second layer are exactly on top of the first one. Since both have the same, exact arrangement it is AAA type.

5. If a crystal lattice has 6 closed-pack spheres, what the number of tetrahedral voids in the lattice?
a) 12
b) 6
c) 36
d) 3
Answer: a
Clarification: For a crystal lattice, if there are N close-packed spheres the number of tetrahedral voids are 2N and number octahedral voids are N. For N=6, number of tetrahedral voids = 2 × 6 = 12.

6. Which of the following possess anisotropic nature within their structure?
a) Hair wax
b) Snowflakes
c) Polythene
d) Crystal glass
Answer: b
Clarification: Crystalline solids possess anisotropic nature within their structure. Anisotropy is the directional dependence of a property. Meaning, a property within the crystal structure will have different values when measured in different directions. Snowflake is a crystalline solid whereas the rest are amorphous solids.

7. Identify the dimensional relation for the unit cell illustrated below.

a) a = b = c
b) a = b ≠ c
c) a ≠ b ≠ c
d) a ≠ b = c
Answer: c
Clarification: The given figure represents an orthorhombic unit cell. Experimentally, it is determined that for orthorhombic unit cells a ≠ b ≠ c. All sides are unequal. It results from extension of cube along two pairs of orthogonal sides by two distinct factors.

8. A compound is formed by atoms of elements A occupying the corners of the unit cell and an atom of element B present at the center of the unit cell. Deduce the formula of the compound.
a) AB₂
b) AB₃
c) AB₄
d) AB
Answer: d
Clarification: The description is of a BCC. For BCC, each atom at the corner is shared by 8 unit cells. One atom at the center wholly belongs to the corresponding unit cell.
Therefore, total number of atoms of A present=(frac{1}{8}) x 8=1
Total number of atoms of B present=1
Therefore, A:B=1:1 implying the formula of the compound is AB.

9. Atoms of element X form a BCC and atoms of element Y occupy 3/4^th of the tetrahedral voids. What is the formula of the compound?
a) X₂Y₃
b) X₃Y₂
c) X₃Y₄
d) X₄Y₃

Answer: a
Clarification: The number of tetrahedral voids form is equal to twice the number of atoms of element X. Number of atoms of Y is 3/4^th the number of tetrahedral voids i.e. 3/2 times the number of atoms of X. Therefore, the ratio of numbers of atoms of X and Y are 2:3, hence X₂Y₃.

10. What is the total volume of the particles present in a body centered unit cell?
a) 8πr³
b) (frac{8}{3})πr³
c) (frac{16}{3})πr³
d) (frac{32}{3})πr³

Answer: b
Clarification: Since particles are assumed to be spheres and volume of one sphere is (frac{4}{3})πr³, total volume of all particles in BCC = 2 x (frac{4}{3})πr³=(frac{8}{3})πr³ since a BCC has 2 particles per cell.

11. If the aluminum unit cell exhibits face-centered behavior then how many unit cells are present in 54g of aluminum?
a) 1.2042 x 10²⁴
b) 5.575 x 10²¹
c) 3.011 x 10²³
d) 2.4088 x 10²⁴
Answer: c
Clarification: Atomic mass of Al = 27g/mole (contains 6.022 x 10²³ Al atoms)
Since it exhibits FCC, there are 4 Al atoms/unit cell.
If 27g Al contains 6.022 x 10²³ Al atoms then 54g Al contains 1.2044 x 10²⁴atoms.
Thus, if 1 unit cell contains 4 Al atoms then number of unit cells containing 1.2044 x 10²⁴ atoms=(1.2044 x 10²⁴ x 1)/4 = 3.011× 10²³ unit cells.

12. What is the radius of a metal atom if it crystallizes with body-centered lattice having a unit cell edge of 333 Pico meter?
a) 1538.06 pm
b) 769.03 pm
c) 288.38 pm
d) 144.19 pm
Answer: d
Clarification: For body-centered unit cells, the relation between radius of a particle ‘r’ and edge length of unit cell ‘a’ is given as (frac{sqrt{3}}{4})a=r
On substituting the values we get r = (frac{sqrt{3}}{4}) x 333 pm = 144.19 pm is the radius of the metal atom.

13. How many parameters are used to characterize a unit cell?
a) Six
b) Three
c) Two
d) Nine
Answer: a
Clarification: A unit cell is characterized by six parameters i.e. the three common edge lengths a, b, c and three angles between the edges that are α, β, γ. These are referred to as inter-axial lengths and angles, respectively. The position of a unit cell can be determined by fractional coordinates along the cell edges.

14. What is each point (position of particle) in a crystal lattice termed as?
a) Lattice index
b) Lattice point
c) Lattice lines
d) Lattice spot
Answer: a
Clarification: Each point of the particle’s position is referred to as ‘lattice point’ or ‘lattice site’. Every lattice point represents one constituent particle which may be an atom, ion or molecule.

15. If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.
a) 2923 kg/m³
b) 5846 kg/m³
c) 8768 kg/m³
d) 1750 kg/m³

Answer: b
Clarification:
Given,
Edge length (a) = 500 pm = 500 x 10^-12 m
Atomic mass (M) = 110 g/mole = 110 x 10^-3 kg/mole
Avogadro’s number (N_A) = 6.022 x 10²³/mole
z = 4 atoms/cell
The density, d of a metal is given as d=(frac{zM}{a^3N_A})
On substitution, d=(frac{4 times 110 times 10^{-3}}{(500 times 10^{-12})^3 times 6.022 times 10^{23}})=5846 kg/m³.

Chemistry Multiple Choice Questions on “Electrochemical Cells”.

1. An electrochemical cell can only convert electrical energy to chemical energy.
a) True
b) False
Answer: b
Clarification: An electrochemical cell can convert electrical energy to chemical energy and can also convert electrical energy to chemical energy. There are two types of electrochemical cells- Galvanic cell and Electrolytic cell.

2. An electrochemical cell generally consists of a cathode and an anode. Which of the following statements is correct with respect to the cathode?
a) Oxidation occurs at the cathode
b) Electrons move into the cathode
c) Usually denoted by a negative sign
d) Is usually made up of insulating material
Answer: b
Clarification: Cathodes are usually metal electrodes. It is the electrode where reduction takes place. The cathode is the positive electrode in a galvanic cell and a negative electrode in an electrolytic cell. Electrons move into the cathode.

3. When equilibrium is reached inside the two half-cells of the electrochemical cells, what is the net voltage across the electrodes?
a) > 1
b) c) = 0
d) Not defined
Answer: c
Clarification: A half-cell is half of an electrochemical cell (electrolytic or galvanic), where either oxidation or reduction occurs. At equilibrium, there is no transfer of electrons across the half cells. Therefore, the potential difference between them is nil.

4. Which of the following is not a generally used electrolyte in the salt bridges used to connect the two half-cells of an electrochemical cell?
a) NaCl
b) KNO₃
c) KCl
d) ZnSO₄
Answer: d
Clarification: A salt bridge is a device used to connect the oxidation and reduction half-cells of a galvanic cell (a type of electrochemical cell). Strong electrolytes are generally used to make the salt bridges in electrochemical cells. Since ZnSO₄ is not a strong electrolyte, it is not used to make salt bridges.

5. When no current is drawn through an electrochemical cell, the sum of the electrode potentials of the two electrodes is called cell emf. True or False?
a) True
b) False
Answer: a
Clarification: Emf of a cell is equal to the maximum potential difference across its electrodes, which occurs when no current is drawn through the cell. It can also be defined as the net voltage between the oxidation and reduction half-reactions.

6. Which of the following statements is correct regarding Electrochemical cells?
a) Cell potential is an extensive property
b) Cell potential is an intensive property
c) The Gibbs free energy of an electrochemical cell is an intensive property
d) Gibbs free energy is undefined for an electrochemical cell
Answer: b
Clarification: Cell potential is an intensive property as it is independent of the amount of material present. Gibbs free energy is defined for an electrochemical cell and is an extensive property as it depends on the quantity of the material.

7. Which of the following factors does not affect the electrode potential of an electrode?
a) Nature of the electrode (metal)
b) Temperature of the solution
c) Molarity of the solution
d) Size of the electrode
Answer: d
Clarification: Electrode potential is the tendency of an electrode to accept or to lose electrons. Electrode potential depends on the nature of the electrode, temperature of the solution and the concentration of metal ions in the solution. It doesn’t depend on the size of the electrode.

8. Why are the saturated solutions of electrolytes for the salt bridge prepared in agar-agar jelly or gelatin?
a) The jelly acts as an electrolyte
b) It helps the electrolytes to mix with the contents of the half cells
c) It helps maintain the electrical polarity between the two half-cell solutions
d) It keeps the electrolyte in semi-solid phase and prevents it from mixing with the two half-cell solutions
Answer: d
Clarification: The salt bridge connects the two half-cell solutions to complete the circuit of the electrochemical cell. The electrolytes of the salt bridge are generally prepared in agar-agar or gelatin so that the electrolytes are kept in a semi-solid phase and do not mix with the half-cell solutions and interfere with the electrochemical reaction.

9. Which of the following is not a characteristic feature of a salt bridge?
a) Salt bridge joins the two halves of an electrochemical cell
b) It completes the inner circuit
c) It is filled with a salt solution (or gel)
d) It does not maintain electrical neutrality of the electrolytic solutions of the half-cells
Answer: d
Clarification: A salt bridge is a junction that connects the anodic and cathodic compartments in a cell or electrolytic solution. It maintains electrical neutrality within the internal circuit, preventing the cell from rapidly running its reaction to equilibrium.

10. Which of the following is not a type of electrochemical cell?
a) Voltaic cell
b) Photovoltaic cell
c) Electrolytic cell
d) Fuel Cell
Answer: b
Clarification: A Voltaic or Galvanic cell is a type of electrochemical cell that converts chemical energy into electrical energy. Photovoltaic cells are used to convert light energy into electrical energy. An Electrolytic cell is a type of electrochemical cell that converts electrical energy into chemical energy. A fuel cell is an electrochemical cell that converts the chemical energy of a fuel and an oxidizing agent into electricity.

11. What is the direction of flow of electrons in an electrolytic cell?
a) Anode to cathode externally
b) Anode to cathode internally
c) Cathode to anode externally
d) Cathode to anode in the solution
Answer: a
Clarification: An electrolytic cell is a type of electrochemical cell. An electrolytic cell converts electrical energy into chemical energy. Electrons flow from anode to cathode through the external supply in an electrolytic cell. In the solution, only ions flow and not the electrons.

12. Which of the following is a not a secondary cell?
a) Nickel-cadmium cell
b) Lead storage cell
c) Mercury cell
d) Leclanche cell
Answer: d
Clarification: A secondary battery (a series of cells) is one which can be charged, discharged into a load, and recharged many times. Nickel-cadmium cell, Lead storage cell and Mercury cell are examples of secondary cells. Leclanche cell is an example of a primary cell.

13. Which of the following statements regarding primary cells is false?
a) Primary cells cannot be recharged
b) They have low internal resistance
c) They have an irreversible chemical reaction
d) Their initial cost is cheap
Answer: b
Clarification: Primary cells cannot be used again and again. Since there is no fluid inside, these cells are also known as dry cells. The internal resistance is high and the chemical reaction is irreversible. Their initial cost is cheap.

14. What is the observation when the opposing external applied potential to an electrochemical cell is greater than the cell’s potential?
a) The electrochemical cell behaves like an electrolytic cell
b) The electrochemical cell stops functioning
c) Only oxidation reactions occur in the cell
d) Only reduction reactions occur in the cell
Answer: a
Clarification: In an electrochemical cell, when an opposing externally potential is applied and increased slowly, the reaction continues to take place. When the external potential is equal to the potential of the cell, the reaction stops. Once the externally applied potential is greater than the potential of the cell, the reaction goes in the opposite direction and the cell behaves like an electrolytic cell.

15. Which of the following conditions are satisfied when the cell reaction in the electrochemical cell is spontaneous?
a) ΔG° > 0
b) E°_cellc) E°_cell = 0
d) ΔG° Answer: d
Clarification: For all spontaneous chemical reactions, the change in Gibbs free energy (ΔG°) is always negative. For a spontaneous reaction in an electrolytic cell, the cell potential (E°cell) should be positive.

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

1. Which of the following colloidal system represents a gel?
a) Solid in liquid
b) Solid in gas
c) Liquid in solid
d) Liquid in gas
Answer: c
Clarification: A gel is a colloidal system in which the dispersed phase is a liquid and the dispersion medium is a solid. Toothpaste, jam, cheese, rubber and gelatin (animal protein) are some of the examples of colloidal systems which are gels.

2. How are colloidal solutions of gold prepared by different colours?
a) Different diameters of colloidal gold particles
b) Variable valency of gold
c) Different concentration of gold particles
d) Impurities produced by different methods
Answer: a
Clarification: Colloidal solutions of gold prepared by different methods are of different colours because of different diameters of colloidal gold particles. The colour of colloidal solutions depends upon the size of the colloidal particles.

3. What are the dispersed phase and dispersion medium in alcohol respectively?
a) Alcohol, water
b) Solid, water
c) Water, alcohol
d) Solid, alcohol
Answer: d
Clarification: A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcohol. For example: A colloidal solution of cellulose nitrate in ethyl alcohol is an alcohol.

4. What is the range of the size of colloidal particles?
a) 1 to 100 nm
b) 10 to 100 pm
c) 1 to 100 µm
d) 1 to 10 mm
Answer: a
Clarification: A colloid is typically a two-phase system consisting of a continuous phase (the dispersion medium) and dispersed phase (the particles or emulsion droplets). The particle size of the dispersed phase typically ranges from 1 nanometre to 1 micrometre.

5. By using what can the colloidal particles can be separated from particles of true solution?
a) Parchment paper
b) An ultracentrifuge machine
c) An electrolyte
d) Ordinary cloth
Answer: a
Clarification: Centrifugation is a process which involves the use of the centrifugal force for the sedimentation of heterogeneous mixtures with a centrifuge used in industry and in laboratory settings. This process is used to separate two immiscible liquids. Particles of colloids are big enough to be blocked by parchment paper or animal membrane.

6. What isthe order of diameter of colloidal particles?
a) 10^-3 m
b) 10^-6 m
c) 10^-15 m
d) 10^-7 m
Answer: d
Clarification: Colloidal state of matter is, therefore, a state in which the size of the particles is such (1 to 1000 nm) that they can pass through filter paper but not through animal or vegetable membrane. Thus, every substance can be brought into the colloidal state by adopting suitable methods.

7. Dust is a colloid.
a) True
b) False
Answer: a
Clarification: Yes, the statement is true. Dust is a colloid if suspended in the air. It consists of a solid in a gas which is present in the atmosphere and it comes under the category of aerosols (contains small particles of liquid or solid dispersed in a gas).

8. When hit by light, what happens to a colloidal mixture?
a) Absorbed
b) Reflected
c) Diffracted
d) Passes through
Answer: c
Clarification: When light strikes a colloidal mixture, it is reflected off the large particles and spreads out. It is so because the colloidal particles move rapidly and randomly. This is what happens when light hits a mixture.

9. Under which category is colloidal system?
a) Homogeneous mixture
b) Heterogeneous mixture
c) Suspensions
d) True solution
View Answer

Answer: b
Clarification: Colloidal sols form heterogeneous mixtures consisting of particles of dispersed phase and the dispersion medium. The dispersed particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas.

10. What is the colloidal solution of a gas in liquid called?
a) Aerosol
b) Gel
c) Foam
d) Solution
Answer: c
Clarification: Depending upon whether the dispersed phase and the dispersion medium are solids, liquids or gases, eight types of colloidal systems are possible. The colloidal solution wherein gas is the dispersed phase and liquid is the dispersion medium is called foam.

Chemistry Multiple Choice Questions on “P-Block Elements – Nitric Acid”.

1. Which of the following is not an oxo-acid of nitrogen?
a) Hyponitric acid
b) Hyponitrous acid
c) Nitrous acid
d) Nitric acid
Answer: a
Clarification: Hyponitric acid does not exist. The rest three mentioned are commonly occurring oxoacids of nitrogen. Hyponitrous acid, H₂N₂O₂ is an isomer tautomer of nitramide, with the structure of the former being HON = NOH. Nitrous acid, HNO₂ is usually formed in the atmosphere prior conversion to nitric acid. It is highly unstable.

2. Which of the following is true regarding nitric acid?
a) It is a strong reducing agent
b) It is a weak oxidizing agent
c) Its basicity is unity
d) It is non-planar in gaseous state

Answer: b
Clarification: Nitric acid is a very weak reducing agent since it has a polar O – H bond. This breaks to donate the H⁺ ion which is why it is a strong oxidizing agent and a strong acid. Since there is only one cleavable O – H bond, the basicity of nitric acid is unity (one HNO₃ molecule can donate only 1 H⁺ ion). It exists as a planar molecule in vapor phase.

3. Which of the following reactions best represents lab scale preparation of nitric acid?
a) 3HNO₂ → HNO₃ + H₂O + 2NO
b) NO₂ + O₂ → NO₃
c) NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃
d) 3NO₂ + H₂O → 2HNO₃ + NO

Answer: c
Clarification: The most appropriate lab scale preparation method of nitric acid, HNO₃ is using an alkali nitrate salt and react it with concentration nitric acid in a glass retort. Nitrous acid being highly unstable decomposes into nitric acid. The other two sets of reaction represent the industrial process of manufacturing nitric acid i.e. Ostwald’s process.

4. What is the name of the industrial process to manufacture nitric acid?
a) Contact process
b) Haber-Bosch process
c) Solvay process
d) Ostwald’s process

Answer: d
Clarification: Ostwald’s process is the name of the industrial process to manufacture nitric acid in bulk. It involves the oxidation of ammonia which forms nitric oxide. This is then reacted with more oxygen to produce nitrogen dioxide. Subsequently, nitrogen dioxide is dissolved in water to produce adequate concentrations of nitric acid. Contact process is used to produce sulfuric acid. Solvay is used to obtain sodium carbonate and Haber-Bosch to obtain ammonia.

5. What is the catalyst used in the industrial manufacture of nitric acid?
a) Powdered iron (III) oxide
b) Vanadium (V) oxide
c) Zinc-mercury amalgam
d) Platinum-Rhodium gauze sheet
Answer: d
Clarification: Pt-Rh gauze sheet is widely used as the catalyst in ammonic oxidation, the first step of Ostwald’s process. Fe₂O₃ is used in Haber’s process; V₂O₅ in contact process and Zn (Hg) is used in Clemmensen reduction of aldehydes.

6. What is the nitric acid – water composition by mass, respectively, for the components to form an azeotrope?
a) 70% – 30%
b) 68% – 32%
c) 30% – 70%
d) 32% – 68%
Answer: b
Clarification: Experimentally, it is determined that nitric acid and water form a constant boiling azeotrope at 68% – 32% by mass composition, respectively. Here, it becomes impossible to separate water and nitric acid by distillation methods. Thus, concentrated sulfuric acid is used for dehydration and removal of water.

7. Which of these gases is released upon treating zinc with diluted and then concentrated nitric acid?
a) Nitrogen dioxide and nitrous oxide
b) Nitric oxide and nitrous oxide
c) Nitrous oxide and nitrogen dioxide
d) Nitrous oxide and nitric oxide
Answer: c
Clarification: The products released depend on the concentration of nitric acid. In case of zinc metal, diluted nitric acid treatment release nitrous oxide and concentrated nitric acid causes the release of nitrogen dioxide.

8. What product(s) is/are formed when aluminum metal is treated with concentrated nitric acid?
a) Al (NO₃) ₃
b) Al (NO₂) ₃ + H₂
c) Al₂O₃
d) Al₄O₃
Answer: c
Clarification: Aluminum does not dissolve in nitric acid. This is because treatment with nitric acid results in the formation of a tough oxide layer. This oxide layer prevents it from further reacting with the oxide. Hence, the compound formed is Al₂O₃ i.e. aluminum (III) oxide.

9. Which reagent is predominantly used in pickling of stainless steel?
a) Iodic acid
b) Nitric acid
c) Phosphoric acid
d) Sulfuric acid
Answer: b
Clarification: Pickling of stainless steel is the process of removal of a thin layer of the alloyed metal from the surface. The common reagent used is nitric acid along with calculated amounts of hydrofluoric acid.

10. How many moles of nitric acid is required to convert 1 mole of sulfur to sulfuric acid?
a) 10
b) 4
c) 48
d) 20
Answer: c
Clarification: 1 mole of sulfur, S₈ requires 48 moles of concentrated nitric acid. The reaction is given by S₈ + 48HNO₃ → 8H₂SO₄ + 48NO₂ + 16H₂O. 10, 4 and 20 moles of concentrated nitric acid is required to produce iodic acid, carbon dioxide and phosphoric acid from 1 mole of iodine, carbon and phosphorus, respectively.