CUET Chemistry Question Paper 2025, Download Previous Year Question Paper with Answers

Unit

Title

I

Solid State

II

Solutions

III

Electrochemistry

IV

Chemical Kinetics

V

Surface Chemistry

VI

General Principles and Processes of Isolation of Elements

VII

p-Block Elements

VIII

d and f Block Elements

IX

Coordination Compounds

X

Haloalkanes and Haloarenes

XI

Alcohols, Phenols, and Ethers

XII

Aldehydes, Ketones, and Carboxylic Acids

XIII

Organic Compounds Containing Nitrogen

XIV

Biomolecules

Hormones –Elementary idea (excluding structure).

XV

Polymers

XVI

Chemistry in Everyday Life

Title

Link

CUET Chemistry question paper 2024

CUET 2024 chemistry question paper - Set A

CUET 2024 chemistry question paper - Set B

CUET 2024 chemistry question paper - Set C

CUET 2024 chemistry question paper - Set D

Title

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CUET Chemistry question paper 2023

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Section:

CHEMISTRY

ItemNo:

1

QuestionType:

MCQ

Question:

The molecularity of the following elementary reaction is $\mathrm{NH}_4 \mathrm{NO}_2 \rightarrow \mathrm{~N}_2+2 \mathrm{H}_2 \mathrm{O}$

A:

0

B:

1

C:

2

D:

3

Correct Answer

(1) Unimolecular.

Explanation

Molecularity refers to the number of reactant molecules involved in an elementary reaction. In the given reaction:

There is only one reactant molecule involved in the decomposition reaction of ammonium nitrite (NH₄NO₂), so this reaction is unimolecular.

Section:

CHEMISTRY

ItemNo:

2

QuestionType:

MCQ

Question:

Which of the following is not the characteristic of physisorption?

A:

It arises because of van der Waals forces.

B:

It is not specific in nature

C:

Enthalapy of adsorption is high

D:

It results into multi molecular layers on adsorbent surface under high pressure

Correct Answer

Enthalpy of adsorption is high.

Explanation

Physisorption, or physical adsorption, occurs due to weak Van der Waals forces and is not specific to any particular substance. The enthalpy of adsorption in physisorption is generally low (not high) because the forces involved are weak compared to chemisorption. Additionally, physisorption can result in multilayer adsorption on the adsorbent surface, especially under high pressure. Therefore, the statement about high enthalpy of adsorption is incorrect for physisorption.

Section:

CHEMISTRY

ItemNo:

3

QuestionType:

MCQ

Question:

Which is an emulsion ?

Ans:

Smoke

B:

Hair Cream

C:

Paint

D:

Cheeze

Correct Answer

Hair Cream.

Explanation

An emulsion is a mixture of two immiscible liquids (like oil and water) where one is dispersed in the other. Hair cream is an example of an emulsion because it consists of oil and water mixed together with emulsifiers to form a stable product.

The other options are not emulsions:

• Smoke is a colloidal mixture of solid particles in a gas.
  
  

• Paint is typically a suspension or dispersion, not an emulsion.
  
  

• Cheese is a dairy product, but it's not considered an emulsion.

Section:

CHEMISTRY

ItemNo:

4

QuestionType:

MCQ

Question:

Among the following statements, choose the correct statements,

A. In Ionic solid, ions are the constituent particles.

B. Ionic solids are soft.

C. Ionic solid are electrical insulators in the solid state.

D. Ionic solid conducts electricity in molten state.

E. Ionic solids have low melting and boiling points.

A:

A,C,D

B:

A,D,E

C:

A,B,C

D:

A,C,E

Correct Answer

A, C & D only

Explanation

A. In Ionic solid, ions are the constituent particles.
This statement is correct because ionic solids are made up of positively and negatively charged ions.

B. Ionic solids are soft.
This statement is incorrect. Ionic solids are generally hard and brittle because of the strong electrostatic forces between ions.

C. Ionic solids are electrical insulators in the solid state.
This statement is correct because, in the solid state, ions are fixed in place and cannot move to conduct electricity.

D. Ionic solids conduct electricity in the molten state.
This statement is correct because, in the molten state, the ions are free to move, allowing ionic solids to conduct electricity.

E. Ionic solids have low melting and boiling points.
This statement is incorrect. Ionic solids typically have high melting and boiling points due to the strong electrostatic forces between the ions.

Section:

CHEMISTRY

ItemNo:

5

QuestionType:

MCQ

Question:

Which of the following is a positively charged Sol?

A:

Starch

B:

Gum

C:

Gold Sol

D:

Blood

Correct Answer

Gold Sol.

Explanation

A Sol is a type of colloidal dispersion where solid particles are dispersed in a liquid. When a positively charged sol is formed, the solid particles have a positive charge. Gold Sol is an example of a positively charged sol where gold particles are dispersed in a liquid.

• Starch and Gum are not colloidal sols in the context of charge; they typically form gels or pastes.
  
  

• Blood is a complex biological fluid and is not considered a colloidal sol with a net charge in the same sense as Gold Sol.

Section:

CHEMISTRY

ItemNo:

6

QuestionType:

MCQ

Question:

The metal refined by Van Arkel method is

A:

Ni

B:

Zr

C:

Cu

D:

Sn

Correct Answer

Zr (Zirconium)

Explanation

The Van Arkel method is used for refining certain metals, particularly zirconium (Zr) and titanium (Ti). In this method, the impure metal is reacted with iodine to form a volatile compound, which is then decomposed to yield pure metal. It is commonly used for the refinement of metals like zirconium, not for nickel (Ni), copper (Cu), or tin (Sn).

Section:

CHEMISTRY

ItemNo:

7

QuestionType:

MCQ

Question:

Rate constant ' k ' for a certain reaction is $\mathrm{k}=2.3 \times 10^{-5} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$. Order of the reaction is:

A:

0

B:

1

C:

2

D:

3

Correct Answer

2

Explanation

The given rate constant, k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹, has units that correspond to the second-order reaction.

For a reaction of order n, the units of the rate constant k are:

• For zero-order: k has units of mol L⁻¹ s⁻¹
  
  

• For first-order: k has units of s⁻¹
  
  

• For second-order: k has units of L mol⁻¹ s⁻¹
  
  

• For third-order: k has units of L² mol⁻² s⁻¹
  
  

Since the units of k in the given question are L mol⁻¹ s⁻¹, this corresponds to a second-order reaction.

Section:

CHEMISTRY

ItemNo:

9

QuestionType:

MCQ

Question:

Among the following statements related to ionic conductance, choose the correct statements.

A. Ionic conductance depends on the nature of electrolyte

B. Ionic conductance is due to the movements of electrons

C. Ionic conductance is also called electronic conductance

D. Ionic conductance depends on temperature

E. Ionic conductance also depends on the nature of solvent Choose the correct answer from the options given below:

A:

A,B,C

B:

B,C,D

C:

B,C,E

D:

A,D,E

Correct Answer

A, D and E only

Explanation

A. Ionic conductance depends on the nature of electrolyte.
This statement is correct. Ionic conductance varies with the type of electrolyte (e.g., strong or weak electrolytes).

B. Ionic conductance is due to the movements of electrons.
This statement is incorrect. Ionic conductance is due to the movement of ions, not electrons.

C. Ionic conductance is also called electronic conductance.
This statement is incorrect. Ionic conductance refers to the conduction of electricity through the movement of ions, whereas electronic conductance refers to the flow of electrons.

D. Ionic conductance depends on temperature.
This statement is correct. Ionic conductance generally increases with temperature, as the ions move more freely at higher temperatures.

E. Ionic conductance also depends on the nature of solvent.
This statement is correct. The solvent affects the dissociation of the electrolyte and the mobility of ions, which in turn affects ionic conductance.

Section:

CHEMISTRY

ItemNo:

10

QuestionType:

MCQ

Question:

The artificial sweetener used only for cold food is

A:

Alitame

B:

Sucralose

C:

Aspartame

D:

Saccharin

Correct Answer

Aspartame.

Explanation

Aspartame is an artificial sweetener that is commonly used in various food products, but it is heat-sensitive and breaks down at high temperatures. Therefore, it is typically used in cold food and beverages, not in hot cooking or baking.

The other options are:

• Alitame: A sweetener that is heat-stable and can be used in cooking.
  
  

• Sucralose: A heat-stable sweetener, suitable for cooking and baking.
  
  

• Saccharin: Another heat-stable sweetener, used in both cold and hot food.

Section:

CHEMISTRY

ItemNo:

11

QuestionType:

MCQ

Question:

$\Lambda_m^{\circ}$ for $\mathrm{NaCl}, \mathrm{HCl}$ and NaOAc are $126.4,425.9$ and $91.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. Calculate $\Lambda^{\circ}$ for HOAc

A:

$390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

B:

$643.3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

C:

$461.3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

D:

$208.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Correct Answer

390.5 S cm² mol⁻¹.

Explanation

Given values:

- $\Lambda_m^{\circ}(\mathrm{NaCl})=126.4 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

- $\Lambda_m^{\circ}(\mathrm{HCl})=425.9 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

- $\Lambda_m^{\circ}(\mathrm{NaOAc})=91.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

$\Lambda_m^{\circ}(\mathrm{HOAc})=91.0+425.9-126.4$

$\Lambda_m^{\circ}(\mathrm{HOAc})=390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Section:

CHEMISTRY

Item No:

12

Question Type:

MCQ

Question:

The products formed at cathode and anode by electrolysis of aqueous NaCl solution respectively are

A:

$\mathrm{Na}, \mathrm{Cl}_2$

B:

$\mathrm{Na}, \mathrm{O}_2$

C:

$\mathrm{H}_2, \mathrm{Cl}_2$

D:

$\mathrm{H}_2, \mathrm{O}_2$

Correct Answer

$\mathrm{H}_2, \mathrm{Cl}_2$

Explanation

Explanation: During the electrolysis of aqueous NaCl solution, the products formed at the cathode and anode are:

- At the cathode (reduction): The reduction reaction involves the formation of hydrogen gas $\left(\mathrm{H}_2\right)$ because water is reduced more easily than sodium ions.

$2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \rightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$

- At the anode (oxidation): Chloride ions $\left(\mathrm{Cl}^{-}\right)$are oxidized to form chlorine gas $\left(\mathrm{Cl}_2\right)$.

$2 C l^{-} \rightarrow C l_2+2 e^{-}$

Thus, the products formed at the cathode and anode are hydrogen gas $\left(\mathrm{H}_2\right)$ and chlorine gas $\left(\mathrm{Cl}_2\right)$, respectively.

Section:

CHEMISTRY

ItemNo:

13

QuestionType:

MCQ

Question:

Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution

A:

$0.278 \times 10^{-3} \mathrm{M}$

B:

$0 \cdot 278 \mathrm{M}$

C:

$2.78 \times 10^{-3} \mathrm{M}$

D:

$2.78 \mathrm{M}$

Correct Answer

0.278 M.

Explanation

Given:

- Mass of $\mathrm{NaOH}=5 \mathrm{~g}$

- Volume of solution $=450 \mathrm{~mL}=0.450 \mathrm{~L}$ (since $1 \mathrm{~L}=1000 \mathrm{~mL}$ )

Step 1: Calculate moles of NaOH

The molar mass of NaOH is:

$\text { Molar mass of } \mathrm{NaOH}=23+16+1=40 \mathrm{~g} / \mathrm{mol}$

Now, calculate the moles of NaOH :

$\text { moles of } \mathrm{NaOH}=\frac{\text { mass }}{\text { molar mass }}=\frac{5 \mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}=0.125 \mathrm{~mol}$

Step 2: Calculate the molarity

Now, use the molarity formula:

$M=\frac{0.125 \mathrm{~mol}}{0.450 \mathrm{~L}}=0.278 \mathrm{M}$

Section:

CHEMISTRY

ItemNo:

14

QuestionType:

MCQ

Question:

Atoms of element B form a hep lattice and those of element A occupy $2 / 3 \mathrm{rd}$ of tetrahedral voids. What is the formula of the compound formed by the elements $A$ and B?

A:

$\mathrm{A}_3 \mathrm{~B}_4$

B:

$\mathrm{A}_4 \mathrm{~B}_3$

C:

$\mathrm{A}_2 \mathrm{~B}_3$

D:

$\mathrm{A}_3 \mathrm{~B}_2$

Correct Answer

$\mathrm{A}_3 \mathrm{~B}_2$

Explanation

Step 1: Number of atoms in one unit cell of B (hcp structure)

In an hcp lattice, the number of atoms per unit cell is 6.

Step 2: Number of tetrahedral voids in one unit cell of B

In the hcp structure, there are 8 tetrahedral voids per unit cell.

Step 3: Number of voids occupied by A

Since A occupies $2 / 3$ of the tetrahedral voids, the number of voids occupied by $A$ in the unit cell of $B$ is:

$\text { Number of voids occupied by } A=\frac{2}{3} \times 8=\frac{16}{3} \approx 5.33$

Thus, the number of atoms of A that occupy the voids is 5.33 per unit cell.

Step 4: Determine the ratio of A to B

From the above, we know that there are 6 atoms of $B$ and 5.33 atoms of $A$ in one unit cell.

The simplest ratio of A to B is approximately 3:2 (since 5.33 atoms of A corresponds to 3 atoms when rounded).

Thus, the formula of the compound is: A3B2

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SOLID STATE

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POLYMERS

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CHEMISTRY IN EVERYDAY LIFE

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