# Free Chemistry MCQs for CUET Exam Preparation

Are you getting ready for the Common University Entrance Test (CUET)?
Do you need help with the chemistry section of the exam?

Well, you’re in luck! We have free chemistry MCQs that are made explicitly for the CUET exam. Our MCQs cover different topics like organic chemistry, physical chemistry, and inorganic chemistry.
Using our MCQs, you can test yourself and determine which areas to improve. This will help you do better in the CUET exam.

Our MCQs were created by experts who know everything there is to know about chemistry. We ensure that our MCQs are always up-to-date and follow the latest exam pattern and trends. We also use advanced algorithms to provide our high-quality MCQs to give you the best learning experience possible.
Everyone should have access to education. That’s why we offer our MCQs for free to everyone. Whether you’re a student preparing for the CUET exam or want to learn about chemistry, our platform has something for you.

Aside from our MCQs, we offer other resources like study materials, video lectures, and practice tests. These resources are designed to be attractive, interactive, and informative. They will help you improve your knowledge and understanding of chemistry.

If you want to increase your chances of success in the CUET exam and take the first step towards your dream university, visit our platform today. Check out our free chemistry MCQs and other resources.

### 1. Choose the incorrect interaction that is present when substrates bind to the active site of enzymes

• (A) Van-der Waal’s interaction.
• (B) Ionic bonding
• (C) Metallic bonding
• (D) Dipole-dipole interaction

(C) In the first step, before drugs interact with enzymes, the active site of enzyme molecules holds the substrate to form an enzyme-substrate complex.
The substrate can bind to the active site through some interactions, such as hydrogen bonding, ionic bonding, dipole-dipole interaction, and van der Waal’s interaction. No such metallic interaction is present that aids in the process of substrate binding.

### 2. Starch is a

• (A) natural polymer.
• (B) semi-synthetic polymer.
• (C) synthetic polymer.
• (C) synthetic polymer.

(A)Starch is a natural polymer found naturally in plants and animals. It is made of the monomer glucose, joined together by glycosidic linkages. So, the correct option is A.

### 3. Polythene and Buna-S are

• (A)copolymer and homopolymer, respectively.
• (B)copolymers.
• (C)homopolymer and copolymer, respectively.
• (D)homopolymers.

(C) Polymers made from one type of monomer are said to be homopolymers. Polythene is made from one type of monomer unit: ethene. So, polythene is a homopolymer.
Polymers synthesised from two different types of monomers are called copolymers. An example of a copolymer is buna-s. It is synthesised from the addition polymerization of 1,3-butadiene and styrene.

### 4. Bakelite is an example of thermosetting plastic because

• (A) on heating, it undergoes extensive crosslinking, which makes it non-recyclable
• (B) on heating, it undergoes extensive vaporisation which makes it non-recyclable.
• (C) on heating it transforms into liquid form, which makes it recyclable.
• (D) on heating, it undergoes extensive crosslinking, which makes it recyclable.

(A) Polymers are large molecules made by combining subunits called monomers.
Polymers found in nature, that is, in animals and plants, are called natural polymers. Polymers synthesised in laboratories or industries are termed synthetic polymers.
A thermosetting polymer (also known as plastic) hardens irreversibly upon heating. Bakelite is a thermosetting plastic because it is a cross-linked and heavily branched molecule. On heating, it undergoes extensive crosslinking, which makes it hard and non-recyclable. In such plastics, the intermolecular forces are much stronger since these are covalently bonded three-dimensional solids.

### 5. In reaction with sodium amalgam and water, glucose is reduced. The product of this reduction is

• (A) gluconic acid.
• (B) sorbitol.
• (C) saccharic acid.
• (D) glucose oxime.

(B)Upon reduction with sodium amalgam, glucose gives a hexahydric alcohol known as sorbitol. The hydrogen adds to the carbonyl functional group of glucose and converts the aldehydic functional group into alcohol. Therefore, the product obtained from the given reduction is sorbitol.

### 6. Which among the following is not true about amino acids?

• (A)All amino acids are optically active.
• (B)They are soluble in water.
• (C)They have high melting points.
• (D)Amino acids are crystalline solids.

(A) Amino acids are major constituents of our body and are the monomeric units of proteins. Amino acids are colourless crystalline solids.
They are readily soluble in water but slightly soluble in alcohol due to their polar nature. Amino acids have high melting points because of the presence of strong intermolecular dipole-dipole interactions or electrostatic force of attraction among their molecules.
This type of attraction arises due to the formation of zwitter-ions (from amino acids).
Note that glycine is the only amino acid that is optically inactive. All other amino acids are optically active. In glycine, the carbon atom attached to amino and carboxyl groups is achiral because of the two hydrogen atoms attached to it.
So, glycine is optically inactive. Therefore, concerning amino acids, option (A) is false.

### 7. Amino acids can behave as

• (A) an acid.
• (B) a base.
• (C) both an acid and base.
• (D) a neutral compound.

(C) Amino acids are biomolecules that contain carboxylic and amino groups, that is, $$-\mathrm{COOH} \text { and }-\mathrm{NH}_2$$ respectively. The number of these functional groups is different in different amino acids, along with their relative positions. They behave like salts in an aqueous solution. The group is acidic and it loses while amino groups are basic and accept . This leads to the formation of a dipolar ion called zwitterion which can be shown as follows.

Thus, amino acids can behave as both an acid and base.

Question-8-

### 8. Which of the following compounds is glycine?

• (A)
• (B) >
• (C)
• (D)

(A) Glycine is termed as the simplest α-amino acid with a hydrogen atom as the side chain. Its structure is shown below.

### 9. Which of the following is an example of water-in-oil type of emulsion?

• (A) ice cream
• (B) mayonnaise
• (C) milk
• (D) butter

(D) Water-in-oil is an emulsion type in which water droplets are the dispersed phase, that is, they are dispersed in oil, which is the continuous phase for this kind of emulsion. They are also called oily emulsions.
Among the given options, ice-cream is an oil-in-water type emulsion as the fat particles are in dispersed phase. They are dispersed among the continuous phase, which is water and sugar with air bubbles.
Mayonnaise is an oil-in-vinegar emulsion. Oil droplets are in the dispersed phase, which are spreading throughout the water-based continuous phase, that is, vinegar. Milk is also an example of oil-in-water emulsion. The dispersed phase in milk is liquid fat and the continuous phase is water.
Butter is the example of water-in-oil emulsion. The dispersed phase for butter is water droplets, which are dispersed in fat (continuous phase). Therefore, butter is considered a water-in-oil emulsion.

### 10. An emulsifying agent is used in oil-in-water emulsions because_____.

• (A)They help in stabilization by the formation of the interfacial film.
• (B)They dissolve the oil droplets in water.
• (C)They completely convert the oil into vapor.
• (D)They completely dissolve the water into oil.

(A) An emulsion is a colloidal mixture of two or more liquids that are independently immiscible. The types of emulsions include O/W (oil dispersed in ) and W/O ( dispersed in oil).
As these are immiscible, an emulsion is highly unstable and separates into two different layers. As a result, an agent that stabilizes and keeps the particles suspended is added, which is an emulsifier or emulsifying agent. Emulsifiers are a part of surfactants that stabilize an emulsion by increasing its kinetic stability. They are amphiphilic, that is, they possess both, hydrophilic and hydrophobic parts, which tends to bring about a decrease in interfacial tension between two immiscible liquids.
For instance, an emulsifying agent such as soap is added in O/W type emulsion to create an interfacial film between the oil particles and to further stabilize the emulsion.

### 11. Which of the following gels does not exhibit elasticity?

• (A) Gelatin
• (B) Silica gel
• (C) Agar
• (D) Starch

(B)A gel is a type of colloid in which the dispersed phase is liquid and the dispersed medium is solid. There are two types of gels.
Elastic gel: These gels possess elasticity. When force is applied on these gels, their shape changes, and they regain their original shape when the force is removed. The dehydration of these gels gives solid mass. The addition of water followed by heating and cooling again reverts it back to its original texture. For example: agar-agar, gelatin, starch and others.
Non-elastic gel: These gels are rigid and do not possess elasticity. The dehydration of these gels gives rigid, solid mass.
The addition of water followed by heating and cooling again does not give the original gel. For example, silica gel, ferric hydroxide gel and others.

### 12. The gold number of P is 0.20, Q is 15, R is 5 and S is 20. Which colloid has the highest protecting power?

• (A)P
• (B)Q
• (C)R
• (D)S

(A)The gold number is defined as the least amount (in mg) of a protective colloid needed such that the coagulation of 10 ml of a standard hydro gold sol is prevented on the addition of 1 ml of a 10% NaCl solution.
A higher magnitude of gold number implies that more protective colloid is needed for the prevention of coagulation of the colloid. Thus, gold number is inversely related to the protecting power of a colloid. Here, gold number of colloid P is the least. Therefore, P will have the highest protecting power.

### 13. The oxidation states of the central atoms in $$\left[\operatorname{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}$$ and $$\left[\operatorname{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}$$ are

• (A)+3 and +3 respectively.
• (B)+2 and +3 respectively.
• (C)+2 and +2 respectively.
• (D)+3 and +2 respectively.

(A)In coordination complexes, the total charge on the coordination sphere is equal to the summation of the oxidation states of the ligands and metal ions. In the complex $$\left[\operatorname{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}$$ the central atom is cobalt (Co). The total charge on the coordination sphere is +2 The oxidation state of $$\mathrm{NH}_3$$ is zero, as it a neutral ligand and the oxidation state of $$\mathrm{Cl}^{-}$$is -1. Let the oxidation state of Co be x. Therefore,
\begin{aligned} & x+5 \times 0+(-1)=+2 \\ & x \quad=+3 \text {, } \\ \end{aligned}
In the complex
$$\left[\mathrm{Fe}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}$$
the central atom is iron (Fe). The total charge on the coordination sphere is 0. The oxidation state of ethylenediamine is zero, as it a neutral ligand and the oxidation state of $$\mathrm{Cl}^{-}$$i is -1. Let the oxidation state of Fe be y. Therefore,
\begin{aligned} & y+2 \times 0+2(-1)+(-1)=0 \\ & y=+3 \\ & \end{aligned}
Therefore, the oxidation states of the central atoms in $$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}$$ and $$\left[\mathrm{Fe}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}$$ are +3 and +3 respectively.

### 14. The IUPAC name of $$\left[\operatorname{Pt}\left(\mathrm{NH}_3\right)_3(\mathrm{Br})\left(\mathrm{NO}_2\right) \mathrm{Cl}\right]^{+} \mathrm{Cl}^{-}$$ is

• (A)triammine chloridobromidonitro platinum(IV) chloride
• (B)triammine bromido chlorido nitro platinum(IV) chloride
• (C)triammine bromidonitrochlorido platinum(IV) chloride
• (D)triammine nitrochloridobromido platinum(IV) chloride

(B)The given compound can be written as $$\left[\operatorname{Pt}\left(\mathrm{NH}_3\right)_3(\mathrm{Br})\left(\mathrm{NO}_2\right) \mathrm{Cl}\right]^{+} \mathrm{Cl}^{-}$$ So, the complex ion with metal (Pt) is a cation. Thus, the metal’s name is retained as it is, that is, platinum. The ligands are named before the metal ion.
The ligands and their names to be used in this complex are $$\mathrm{NH}_3$$ (amine), Br (bromido), $$\mathrm{NO}_2$$ (nitro) and Cl (chlorido). Here, only $$\mathrm{NH}_3$$ is a neutral ligand. Thus, amine comes first. The ligands Br, $$\mathrm{NO}_2$$ and Cl are all monodentate ligands with a charge of -1 Hence, alphabetically, bromido comes first, followed by chlorido and lastly, nitro. Now, the oxidation number of the metal ion is to be calculated.
It can be observed that the complex cation is attached to one counter anion, that is, chloride ion. The charge on the chloride anion is -1. Hence, in order to maintain the electroneutrality of the complex, the complex cation must possess a charge of +1. Let the oxidation number (O.N.) of Pt be x.
$$\begin{array}{ll} x+0+(-1)+(-1)+(-1) & =+1 \\ x=+4 \end{array}$$
So, the oxidation number of Pt in this complex is +4, which is written as IV in Roman numeral. Therefore, the name of the given coordination compound is triamminebromidochloridonitroplatinum(IV) chloride.

### 15. Which of the following is biologically significant chelate?

• (A)Carbon monoxide
• (B)Ammonia
• (C)Cyanide ion
• (D)EDTA

(D)Chelating agents are substances that have two or more donor atoms in order to bind to a single metal ion. Among the given ligands, EDTA (ethylenediaminetetraacetic acid) is a multidentate ligand and it is used to remove heavy metals such as aluminum, calcium, mercury, copper, iron from the body. The structure of EDTA is shown below.
Carbon monoxide, ammonia, and cyanide ion are monodentate ligands and they can donate only a single pair of electrons to a metal atom. The structures of ammonia, cyanide ion and carbon dioxide are shown below.

Therefore, EDTA is a biologically significant chelate.

### 16. Which of the following options is correct?

• (A) Lanthanide complexes are used in calibration of UV-visible spectrophotometer.
• (B) Nd is used in laser machining.
• (C) Gd is used in MRI.
• (D)All of the above

(D)Lanthanoide series consists of 15 chemical metallic elements with atomic number 57 to 71 from lanthanum to lutetium.
Lanthanide complexes can be used in calibration of various spectroscopic instruments such as UV-visible spectrophotometer because the absorption bands are sharp and f orbitals are deep inside the atom.
An element of lanthanide series, that is, Nd (neodymium) is used in construction and working of laser machines.
Gd (gadolinium) is widely used as a contrast agent in magnetic resonance imaging (MRI) machines. Therefore, all the given statements are correct.

### 17. The radius of thorium (atomic number 90) in its +4 oxidation state is 99 pm. Which of the following values will be closest to the atomic radius of californium in +4 oxidation state (atomic number 98)?

• (A) 96 pm
• (B) 92 pm
• (C) 88 pm
• (D) 86 pm

17. (D) 17. Thorium and californium are the part of actinoid series. The ionic radii of actinoid elements decrease gradually while moving along the actinoid series from left to right. A steady decrease in the ionic radii with an increase in the nuclear charge is known as actinoid contraction.
While moving from left to right in the actinoid series, the nuclear charge increases by one, which is not compensated due to poor shielding of 5f-orbital. Therefore, with an increase in the atomic number, the inward pull experienced by 5f-electrons increases, which results in a steady decrease in the size.
The contraction in ionic radii is greater in case of actinoids as compared to lanthanoids due to poor shielding of 5f-electrons. Therefore, out of the given options, the least value of ionic radii in picometer (pm) will be attributed to californium.

### 18. Haemoglobin is a coordination complex. The central atom of haemoglobin is

• (A)cobalt
• (B)iron
• (C)magnesium
• (D)chromium

18. (B) 18. Haemoglobin is a coordination complex. It helps in carrying oxygen to various parts of the body. So it acts as an oxygen carrier. Single haemoglobin unit contains four haeme units. Each haeme unit of haemoglobin contains a coordination complex of iron in which iron is coordinated with six ligands.
So, its coordination number is six. Central iron is attached with five nitrogen atoms. Four of the five nitrogen atoms are from the porphyrin ring and the fifth nitrogen is from histidine amino acid residue. The sixth coordination site is occupied by a oxygen molecule.

### 19. The color of copper(II) salts are usually

• (A)blue
• (B)red
• (C)crimson
• (D)orange

19. (A) 19. Copper is a transition metal and it has partly filled d-orbital. The electron configuration of $$\mathrm{Cu}^{2+}$$ is $$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^9$$ Copper (II) is one of the most reactive transition metals, and it also forms a wide range of salts. Iron and copper are the most common salt-forming transition metals. When transition metals bond with ligands or other ions due to the presence of strong field or inductive effect of ligands and the different symmetry of d-orbitals, making these d-orbitals non-degenerative, which means they can split into different energy levels.
In Copper(II) complexes, Cu has 9 electrons (unpaired) in its outermost shell; hence, it can show d-d transition, which means the electron can absorb light from the visible region and gets excited. This electron can transit from one energy level to another in the d-orbital. This is known as the d-d transition.
Due to the absorption of energy from the visible range, these complexes show distinct colors. Copper(II) complexes are generally blue or green, as it absorbs light from the red region of the spectrum and gives its complementary color, which is blue to green.

### 20. Which of the following proteins is used in textile industry?

• (A) Gelatin
• (B) Casein
• (C) Glue
• (D) Trypsin

20. (B) 20. Gelatin is a protein which is used in capsules and photographic plates. Casein is a protein which is used in the textile industry. It is a milk protein. It is used in the textile industry for manufacturing artificial silk and wool. Glue is a protein which is used as an adhesive and in sizing paper. Trypsin is an enzyme which catalyzes the reactions in which proteins convert into amino acids.

### 21. Which of the following reagent is used for the concentration of bauxite?

• (A)Silica
• (B)Sodium hydroxide
• (C)Hydrogen chloride
• (D)Sodium bicarbonate

(B) The three major steps involved in the extraction of a metal from its ore involve concentrating the ore, isolating the metal from its ore, and purifying the extracted metal. Aluminium is extracted from bauxite ore.
Approximately 50% of the ore is aluminium and the remaining is zinc, silica, iron oxides etc.
The concentration of bauxite ore $$\left(\mathrm{Al}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O}\right)$$ involves the removal of unwanted materials from it.
The process used for the leaching of bauxite is called Bayer’s process. The reagent used is a strong base, sodium hydroxide (NaOH). After this, processes such as filtration, drying, washing and heating are carried out to obtain pure alumina. The reactions involved are shown below.
\begin{aligned} & \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right] \\ & 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]+\mathrm{CO}_2 \rightarrow \mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{H}_2 \mathrm{O}+2 \mathrm{NaHCO}_3 \\ & \mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{H}_2 \mathrm{O} \stackrel{1470 \mathrm{~K}}{\longrightarrow} \mathrm{Al}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{O}\end{aligned}

### 22. Which of the following combination gives cast iron?

• (A)Pig iron and scrap iron
• (B)Pig iron and wrought iron
• (C)Wrought iron and carbon steel
• (D)Rust and carbon steel

(A) Cast iron is extremely hard and brittle. It is prepared by melting pig iron with scrap iron in the presence of coke by using a hot air blast.
Cast iron has a lower carbon content as compared to pig iron. It contains lower content of carbon (2-4%) with other impurities such as sulfur (S), phosphorus (P), silicon (Si), and manganese (Mn). Cast iron is used in the preparation of a pure form of iron called wrought iron. It is also used for railway sleepers, toys, casting stoves, gutter pipes etc.
Wrought iron contains less than 0.05% of carbon content and 99.95% of iron content. Carbon steel contains 0.6-1.4% of carbon content and 99.4-98.6% of iron content. Pig iron contains a very high carbon content, typically 3.5-4.5%.

### 23. Among all the following, which one is an aluminium alloy?

• (A)Steel
• (B)Nichrome
• (C)Babbitt metal
• (D)Duralumin

(D) An alloy is a substance that is formed by mixing two or more metals. Sometimes, an alloy can be formed by mixing metallic and non-metallic elements. Steel is an alloy of iron (85-88%), carbon (less than 1.2%), and other elements such as chromium, manganese, sulphur, phosphorus, nickel, and copper (10.5%). It is an extremely popular alloy, due to its low cost and strength. It is widely used in construction, in making everyday household items etc. Nichrome mainly consists of nickel (80-85%) and chromium (15-20%). It also contains some other elements that are mainly used to make resistance wire. Babbitt metal is an alloy of tin (90%), antimony (7%) and copper (3%). It is mainly used in making plain bearings (as bearing surfaces) because of its low measure of friction with steel. Duralumin is an alloy of aluminium (95%), copper (4%), magnesium (less than 1%) and manganese (0.5%). Because of its high strength and resistance to corrosion, it is used in aircraft, railroad cars and boats.

### 24. Which of the following statements is incorrect regarding colligative properties?

• (A)The solution should be very dilute.
• (B)The solute should be non-volatile.
• (C)The solvent should be non-volatile.
• (D)Association should not occur.

(C) Colligative properties of a solution depend on the ratio of the number of solute particles to the solvent particles in a solution, but not on the nature of the chemicals present in it.
The experimental value and calculated value of the colligative properties resemble each other when the solution is very dilute in nature, non-volatile and when the solution does not dissociate or associate.
Dilute solution of a non-volatile solute exhibits properties that only depend on the solute particles irrespective of their nature. In case of association or dissociation occurring in the solution, the number of particles increases and decreases accordingly. Hence, the experimental value differs from the calculated value.

25. The freezing point of an aqueous solution is $$-0.186^{\circ} \mathrm{C}$$ and the freezing point of the solvent is as the solvent is water. The freezing point depression can be expressed as
\begin{aligned} \Delta T_f & =T_{f(\text { solvent })}-T_{f(\text { solution })} \\ & =K_f \times m \end{aligned}
Here, $$\Delta T_b$$ is the depression in the freezing point,$$K_f$$ represents the molal depression constant and m is the molality of the solute.
$$\begin{array}{cl} T_{\text {solvent }}-T_{\text {solution }} & =K_f \times m \\ 0-(-0.186) & =0.93 \times m \\ 0.186 & =0.93 \times m \\ m & =0.2 \text { molal } \end{array}$$
The expression for the elevation in boiling point is \begin{aligned} \Delta T_b & =T_{b(\text { solvent })}-T_{b(\text { solution })} \\ & =\quad K_b \times m \end{aligned}
Here,$$\Delta T_b$$ represents the elevation in the boiling point, $$K_b$$ is the molal elevation constant and m is the molality of the solute. Now, putting the values,
\begin{aligned} \Delta T_b & =K_b \times m \\ & =0.64 \times 0.2 \\ & =0.128 \mathrm{~K} \end{aligned}

### 26. The freezing point of seawater is

• (A)4 Degree Celsius
• (B)2 Degree Celsius
• (C)10 Degree Celsius
• (D)0 Degree Celsius

(D) For a liquid, the freezing point is the temperature at which that particular liquid can turn into a solid. For pure water, the freezing point is 0 Degrees Celsius but the sea water remains below 0 Degree Celsius because of the presence of salts in seawater that causes freezing point depression.
Freezing point depression can be said as the difference in the temperature between the actual freezing point of a pure solvent and the freezing point observed after the addition of a solute.

### 27. Methanol and ammonia are soluble in water. The reason for this is attributed to

• (A)the non-polar nature of water.
• (B)the absence of H-bonding in methanol and ammonia with water molecule.
• (C)the presence of H-bonding in methanol and ammonia with water molecule.
• (D)Both (A) and (B) are correct.

(C) Methanol and ammonia are soluble in water and the reason for their solubility is attributed to the fact that both the molecules are polar in nature and the atoms in both the molecules possess electronegativity difference.
Thus, the polar molecules will dissolve easily in polar solvent like water.
The molecules that have electronegative elements like O, F or N attached with hydrogen atoms show hydrogen bonding. Thus, methanol, ammonia and water, all show hydrogen bonding. Thus, these molecules are easily dissolved in water due to the H-bond.

### 28. The solubility of a solid substance can increase with the __________.

• (A)increase in pressure.
• (B)decrease in pressure.
• (C)increase in temperature.
• (D)decrease in temperature.

(C) The solubility of a solute can be defined as the maximum concentration that can get dissolved in a particular quantity of solvent at any given temperature.
From the definition of solubility, it is clear that solubility depends on temperature. Solubility is directly proportional to the temperature, which means that with the increase in temperature, solubility also increases. When a solute is added to any solvent, in order to dissolve that solute- kinetic energy of that solvent, molecules must overcome the intermolecular forces of that particular solute so that the solute molecules can be broken apart and get dissolved.
Solubility increases with temperature mainly because with the increase in temperature, the average kinetic energy of solvent molecules increases and as a result, it can break apart more solute molecules by overcoming the intermolecular attraction force of solute molecules.
The solubility of a solid substance is not much affected by the change in pressure, although the solubility of gaseous substances increases with the increase in partial pressure and decreases with the decrease in partial pressure.

### 29. How much charge is required for the reduction of 1 mol of $$\mathrm{Fe}^{+2}$$ to Fe?

• (A)96500 C
• (B)193000 C
• (C)289500 C
• (D)386000 C

(B) The reaction at the cathode is shown below.
$$\mathrm{Fe}^{+2}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$$
One Faraday is the charge required to deposit one gram equivalent of a substance. One Faraday is equal to
$$96500 \mathrm{C} \mathrm{mol}^{-1}$$
According to the reaction, for the conversion of 1 mol of $$\mathrm{Fe}^{+2}$$ to Fe, it requires 2 mole of electrons or 2 F of electricity. Thus, the charge (Q) required for the reduction of 1 mol of $$\mathrm{Fe}^{+2}$$ to Fe is calculated as \begin{aligned} Q & =2 \mathrm{~F} \\ & =2 \times 96500 \mathrm{C} \\ & =193000 \mathrm{C} \end{aligned}

### 30. What is the correct order for the conductivity of the following materials— silver (Ag), copper (Cu), iron (Fe) and gold (Au)?

• (A) Au>Ag>Fe>Cu
• (B) Au>Ag>Cu>Fe
• (C) Ag>Cu>Au>Fe
• (D) Ag>Fe>Au>Cu

(C) Electrical conductivity is the property of a material to conduct electricity or heat. It is determined by the number of electrons in the outermost shell and the nature and structure of the materials. Additionally, it depends on the temperature—it decreases with an increase in temperature. The electronic configurations of silver (Ag), copper (Cu), iron (Fe), and gold (Au) are shown below.
\begin{aligned} & \mathrm{Ag}:[\mathrm{Kr}] 4 d^{10} 5 s^1 \\ & \mathrm{Cu}:[\mathrm{Ar}] 3 d^{10} 4 s^1 \\ & \mathrm{Au}:[\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^1 \\ & \mathrm{Fe}:[\mathrm{Ar}] 3 d^{10} 4 s^2 \end{aligned}
Here, iron has two electrons in its outermost shell, but Ag, Cu, and Au have one electron each in their valence shells. So, Ag, Cu, and Au are more conductive than Fe. However, down the periodic table, atomic size increases, but due to lanthanide contraction, the atomic radii of Au are less than that of Ag.
Hence, the outermost electron of silver will easily generate mobile electrons. Therefore, the correct order for the conductivity of the given materials is. Ag>Cu>Au>Fe

### 31. An electrolytic cell is a device that uses electrical energy to drive ____.

• (A)spontaneous reversible reaction.
• (B)non-spontaneous redox reaction.
• (C)spontaneous redox reaction.
• (D)non-spontaneous oxidation reaction.

(B)An electrolytic cell is an electrochemical cell that is used to provide electrical energy to drive non-spontaneous redox reactions. This cell converts electrical energy into chemical energy.

### 32. Which of the following statement is incorrect?

• (A)Primary cells discharge rapidly.
• (B)Primary cell is called dry cell.
• (C)Primary cell has high internal resistance.
• (D)Alkaline batteries are an example of the primary cell.

(A) Batteries or cells are basically parallel electrochemical cells. There are mainly two types of cells, primary and secondary. Primary cells cannot be charged and are for single use, but secondary cells are rechargeable but get discharged quickly.
Primary cells have high density. Consequently, they discharge slowly. They do not contain any liquid or fluid and are known as dry cells. Internal resistance is higher for these cells. Primary cells are cheaper than secondary cells. Some examples of primary cells are alkaline batteries and dry cells.

### 33. Which of the following elements is not capable of forming clathrates with quinol?

• (A)Helium
• (B)Krypton
• (C)Neon
• (D)Both (A) and (C)

(D) Clathrates or cage compounds are formed when atoms of noble gases are trapped in the cavities of a huge crystal lattice of certain organic and inorganic compounds.
In clathrates, gas molecules penetrate the cavities of the crystal structure composed of water molecules. Clathrates formed with quinol molecules are quite large in size. Both helium and neon atoms are very small to be retained in such cavities. So, they escape from such cavities. Therefore, helium and neon are not capable of forming clathrate compounds with quinol.

### 34. Which of the following elements is used as an inert protective gas for welding metals?

• (A)Calcium
• (B)Iron
• (C)Helium
• (D)Potassium

(C) Inert gases include argon, neon, krypton, xenon, helium and others. Helium is commonly used to enhance heat input, and thus welding speed.
It is used as an inert protective gas for welding metals like aluminum and in rocket propulsion (to pressurize fuel tanks, particularly those for liquid hydrogen because only helium remains a gas at liquid-hydrogen temperature).

### 35. The product formed in the following reaction is

• (A)
• (B )
• (C)
• (D)

(A) When alkyl halide is heated with dry $$\mathrm{Ag}_2 \mathrm{O}$$ ether is formed as the final product. This reaction follows $$\mathrm{S}_{\mathrm{N}} 2$$ a mechanism. So, on heating chloropropane with dry $$\mathrm{Ag}_2 \mathrm{O}$$ dipropyl ether is formed as shown below.

### 36. Which of the following ethers is widely used as an anesthetic?

• (A)Methyl ethyl ether
• (B)Diphenyl ether
• (C)Dipropyl ether
• (D)Diethyl ether

(D) Diethyl ether is a solvent used in the production of plastics and synthetic dyes, and it is a part of starting fluids. It is a powerful anesthetic whose actions are followed by muscle relaxants and analgesics.
It takes a long time to induct and is extremely flammable and explosive. Because of its characteristics, it is used as an anesthetic.

### 37. The reaction of an aldehyde with hydrazine and KOH gives an alkane. This reaction is called

• (A)Wolff-Kishner reduction
• (B)Clemmensen reduction
• (C)Cannizzaro reaction
• (D)Aldol reduction

(A) The reaction of an aldehyde with hydrazine in the presence of a strong base at 453-473 K gives alkane. This reaction is called the Wolff-Kishner reduction.

### 38. Which of the following tests is used to detect the presence of keto group?

• (A)Hinsberg test
• (B)Lucas test
• (C)Sodium nitroprusside test
• (D)All of the above

(C) Hinsberg test is used to differentiate between primary, secondary and tertiary amines. Lucas test is used to differentiate between primary, secondary and tertiary alcohols. Sodium nitroprusside test is used to detect the presence of a keto group in any compound. The appearance of the red color marks the presence of the keto group. \begin{aligned} & \mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{OH}^{-} \rightarrow \mathrm{CH}_3 \mathrm{COCH}_2^{-}+\mathrm{H}_2 \mathrm{O} \\ & {\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}+\mathrm{CH}_3 \mathrm{COCH}_2^{-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}_2 \mathrm{CH}_3 \mathrm{COCH}_2\right]^{3-}} \end{aligned}
The final complex appears to be red in color.

### 39. What is the IUPAC name of the following compound?

• (A)5-bromo-2-methylbenzoic acid
• (B)3-bromo-6-methylbenzoic acid
• (C)4-bromo-1-methylbenzoic acid
• (D)1-bromo-4-methylbenzoic acid

(A) Carboxylic acids are the organic compounds where the functional group is Naming of an organic compound should be done based on the IUPAC rules as per which numbering will be in alphabetical order.
In this scenario, the position where the carboxylic group is attached to the ring is marked as 1 and methyl group positioned carbon is methyl and where bromine is substituted is marked as 5. So, the name is 5-bromo-2-methylbenzoic acid.

### 40. The compounds given below are

• (A)functional group isomers.
• (B)chain isomers.
• (C)tautomers.
• (D)Both (B) and (C)

(A) The compounds that have the same molecular formula, but different functional groups exhibit functional isomerism. The given isomers are functional group isomers. This is because in the given compounds, the molecular formula is the same and only the functional groups are different. One compound is nitrite and the other is nitro. Chain isomerism occurs due to the difference in the arrangement of carbon atoms constituting the chain, that is, a straight or branched chain of carbon atoms, which is not the case here. Also, the two structures obtained upon the transfer of a hydrogen atom from one atom to another atom within the structure are termed tautomers. In the given structures, the transfer of hydrogen atoms does not take place. So, compounds (I) and (II) are not tautomers.

### 41. The correct structural formula of 2-nitropropane and ethyl nitrite are

• (A)CH3Ch3(NO2)CH3 and CH3Ch3NO2
• (B)CH3CH(NO2)CH3 and CH3Ch3NO2
• (C)CH3Ch3(NO2)CH3 and CH3Ch3ONO
• (D)CH3CH(NO2)CH3 and CH3Ch3ONO

(C) A nitro group has one nitrogen and two oxygen atoms. It is an ambident group with two different bonding atoms that form a bond between an oxygen atom and a nitrogen atom.
If the nitro group attacks through the nitrogen atom, it is considered a nitro, whereas if the nitro group attacks through one of the oxygen atoms, it is considered a nitrite. So, the structural formula of 2-nitropropane is CH3Ch3(NO2)CH3 and that of ethyl nitrite is CH3Ch3ONO

### 42. 2X + Y→2Z is a second-order reaction. What will be the molecularity of this reaction?

• (A)2
• (B)1.5
• (C)3
• (D)5

(C) Molecularity can be defined as the summation of stoichiometric coefficient of the reactants that react together to form a product.
It means that it is the total number of reactants (molecules, atoms, ions and others) that takes part in a reaction, where the reaction must be elementary in order to calculate the molecularity. 2X + Y→2Z, For this reaction, 2 moles of X and one mole of Y react to form 2 moles of Z. Hence, molecularity is (2+1) = 3

### 43.The given reaction is Y→Z If the concentration of Y is increased two times, then the rate of the reaction becomes eight times. Calculate the order of the reaction.

• (A)2
• (B)1
• (C)3
• (D)0

(C) For the given reaction, the general rate of the reaction can be represented as $$r=k[\mathrm{Y}]^n \cdots \cdots(1)$$ Here, k is the rate constant of the reaction and n is the order of the reaction. If the concentration of Y is increased two times, the rate of the reaction becomes eight times. So, the new rate of the reaction can be expressed as $$8 \times r=k[2 \mathrm{Y}]^n \cdots \cdots(2)$$ Now, divide equation (2) by equation (1). \begin{aligned} \frac{8 \times r}{r} & =\frac{k[2 \mathrm{Y}]^n}{k[\mathrm{Y}]^n} \\ 8 & =(2)^n \\ (2)^3 & =(2)^n \\ n & =3 \end{aligned} Therefore, the order of the reaction is 3.

### 44. Consider that the X→Z reaction follows a zero-order reaction. The time required to consume all the reactants is

• (A)K/X
• (B)X/2K
• (C)X/K
• (D)2K/X

(C) It is known that a zero order reaction is independent of reactant concentrations and there is no change in the rate with the increase or decrease of the concentration of the reactants. For zero-order reaction, $$-\frac{d[X]}{d t}=k$$ So, x=kt Here, x is the concentration of the reactant, k is the rate constant and t is the time. Now, the time required to complete the reaction is calculated as follows. x=kt t=x/k

### 45. For the reaction 2P+Q→R+S, the activation energy of the backward reaction is greater than the forward reaction. This reaction is

• (A)Endothermic
• (B)Endothermic
• (C)Independent of temperature.
• (D)Neither exothermic nor endothermic.

(B) Activation energy is the minimum energy requirement or barrier that needs to be met or overcome in order for a reaction to occur.
If the activation energy of the reverse reaction or the product formation is higher, that means the formation of the product is favorable, and the reaction needs less energy to proceed. Hence, it is an exothermic reaction.

### 46. Calculate the number of electrons shared by each oxygen atom with the central atom in carbon dioxide.

• (A)8
• (B)2
• (C)6
• (D)4

(B) According to the octet rule, an element can gain or lose one or more than one electron to achieve stability by possessing eight electrons in the outermost shell, which is the electronic configuration for a noble gas.
Helium is an exception to the octet rule as it has only two valence electrons. Carbon has 4 electrons and oxygen has 6 electrons in their valence shells. In CO₂ during bond formation, each oxygen atom shares 2 electrons with the central atom, that is, carbon.
Hence, carbon in carbon dioxide has 8 electrons in its outermost shell, which means that it follows the octet rule. The Lewis dot structure of CO₂ is shown below.

### 47. In PCl₅, phosphorus does not follow the octet rule as it has ____ electrons after bonding.

• (A)10
• (B)7
• (C)6
• (D)9

(A) In PCl₅ the central atom (phosphorus) is bonded with five chlorine atoms. Phosphorus and chlorine have 5 electrons and 7 electrons respectively in their outermost shells. In the case of chlorine, it needs only one electron to complete its octet. Hence, each chlorine atom takes one electron from phosphorus and shares one of its electrons to form an electron pair in order to form a bond. Now, phosphorus has 10 electrons (5 electrons of its own and one from each of the five chlorine atoms). The Lewis dot structure of PCl₅ is shown below.

### 48. The electronegativity difference between two atoms X and Y is 1.9. These two atoms form

• (A)electrovalent bonding
• (B)non-polar covalent bonding.
• (C)metallic bonding
• (D)polar covalent bonding.

(A) Ionic bonding or electrovalent bonding is formed between the atoms with a greater electronegativity difference and with a greater difference in the ionization energy. In ionic bonding, one of the bonding atoms completely donates electrons and becomes a positive ion.
The other atom that accepts electrons will possess a negative charge. If the electronegativity difference is greater than 1.7, the bond will be ionic. If the electronegativity difference between atoms lies between 0.6 to 1.7, they form a polar covalent bonding, and if it lies between 0.0 to 0.5, they form a non-polar covalent bonding.
Metallic bonding can not be relate with the electronegative difference. Since the electronegativity difference between X and Y is 1.9, they form an ionic bonding.

### 49. If the number of valence electrons in an oxygen atom is 6, what is the formal charge on the oxygen atom in

• (A)1
• (B)-1
• (C)0
• (D)-2

(C) In h3O, the central atom (oxygen) is bonded to two hydrogen atoms and has two lone pairs. The Lewis structure of h3O is shown below. According to the Lewis dot structure, the number of bonding and non-bonding electrons are 4. The formal charge on the oxygen atom is calculated by the formula given below. $$\text { Formal charge on } \mathrm{O}=\text { Valence electrons }- \text { Non }- \text { bonding electrons }-\frac{\text { Bonding electrons }}{2}$$ Substitute the known values in the formula given above. \begin{aligned} \text { Formal charge on } \mathrm{O} & =6-4-\frac{4}{2} \\ & =0 \end{aligned} Therefore, the formal charge on oxygen in h3O is 0.

### 50. A coordinate covalent bond can be formed between

• (A)Bronsted acid and Bronsted base.
• (B)Lewis acid and Lewis base.
• (C)Arrhenius acid and Arrhenius base.
• (D)any acid and any base.

(B) A coordinate covalent bond is formed by donating electrons from one electron-rich species to another electron-deficient species.
The electron-rich species act as Lewis bases and the electron-deficient species act as Lewis acids. For example, a coordinate covalent bond exists between $$\mathrm{NH}_3 \text { and } \mathrm{BF}_3$$ in the addition compound $$\left[\mathrm{NH}_3: \mathrm{BF}_3\right]$$ Here, NH3 is an electron-rich species or Lewis base because of the presence of a lone pair of electrons on nitrogen.
It donates electrons to the electron-deficient species BF3 and forms a coordinate covalent bond.