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CUET Chemistry Previous Year Question Papers: The Common University Entrance Test (CUET) 2025 is just two weeks away, as the Chemistry examination is scheduled for May 8, 2025. Previous Year Question papers offer invaluable insights related to question papers, exam patterns, difficulty levels of the Chemistry Exam, and essential topics frequently repeated in the examination.
Latest: CUET Previous Year Question Papers | CUET General Aptitude Test Syllabus 2025
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The chemistry CUET exams' Previous Year Question papers are available on the NTA website; candidates can download and practise to get better marks in the exam. CUET Chemistry 2025 evaluates candidates' conceptual understanding and analytical skills across 16 key physical, organic, and inorganic chemistry units. The format of CUET 2025 is a computer-based test that follows a multiple-choice question format with negative marking for incorrect answers. Scoring well in CUET 2025 will open doors to various top CUET participating universities 2025 for the candidates.
The CUET 2025 exam syllabus is vast, and the candidates should be well-versed with the syllabus to tackle all the challenges set forth by the CUET 2025 Chemistry question paper. The detailed CUET Chemistry syllabus is given below.
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 |
To ensure that you have enough resources for practicing, we bring you the CUET 2024 Chemistry question papers to give you a comprehensive coverage of the full syllabus in short and precise manner. Candidates are advised to go through the below mentioned links for reference.
Title | Link |
CUET Chemistry question paper 2024 | |
Question papers work as a great resource in the journey of preparation for CUET 2025. Below we have attached the link of CUET 2023 Chemistry Question Paper, to help students understand the difficulty level, exam pattern and variety of questions.
Title | Download Link |
CUET Chemistry question paper 2023 |
Here we provide you with the CUET 2025 Chemistry previous year questions with explanations so that you have a better knowledge of the nature of questions being asked and enhance your chances of scoring well in the exam.
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:
|
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. B. Ionic solids are soft. C. Ionic solids are electrical insulators in the solid state. D. Ionic solids conduct electricity in the molten state. E. Ionic solids have low melting and boiling points. |
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.
|
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:
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. 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. |
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:
|
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: - $\Lambda_m^{\circ}(\mathrm{NaCl})=126.4 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ - $\Lambda_m^{\circ}(\mathrm{HCl})=425.9 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ - $\Lambda_m^{\circ}(\mathrm{NaOAc})=91.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ Now, calculate $\Lambda_m^{\circ}(\mathrm{HOAc})$ :$\Lambda_m^{\circ}(\mathrm{HOAc})=91.0+425.9-126.4$ $\Lambda_m^{\circ}(\mathrm{HOAc})=390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ |
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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Regular practice with previous year papers and a strategic preparation plan will strengthen subject knowledge and improve time management skills. Students must follow the tips shared by experts as follows:
Understand the syllabus: Familiarize yourself with the CUET exam syllabus 2025 to know what topics will be covered, allowing you to focus on relevant areas.
Verify clarification: If you have any doubts or questions, don't hesitate to ask your teachers or classmates for clarification.
Review class notes and textbooks: Go through your class notes and the recommended books to reinforce the concepts taught in the course.
Practice previous year papers: Solve the CUET question paper PDF download to understand the CUET exam pattern and identify frequently asked questions.
Create a study schedule: Plan a structured study schedule that allocates sufficient time to cover all topics systematically.
Pen Down Short notes: Prepare concise notes summarizing key points for quick revision during the CUET last-minute preparations.
Practice problem-solving: Chemistry involves calculations and problem-solving, so practice numerical questions to improve your skills.
Stay calm and focused: During the exam, remain composed and read the questions carefully before answering to ensure accurate responses.
Use online resources: Utilise online resources like video lectures, interactive quizzes, and study forums to enhance your understanding.
Preparing through mock tests is one of the most effective preparation mechanisms for any examination. As the CUET 2025 Chemistry syllabus is vast and requires extensive preparation, experts at Careers 360 have designed a set of mock tests to give the best possible practice for the candidates. The links to the mock tests are given below:
Title | Link |
SOLID STATE | |
SOLUTIONS | |
ELECTROCHEMISTRY | |
CHEMICAL KINETICS | |
SURFACE CHEMISTRY | |
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS | |
P-BLOCK ELEMENTS | |
D AND F BLOCK ELEMENTS | |
COORDINATION COMPOUNDS | |
HALOALKANES AND HALOARENES | |
ALCOHOLS, PHENOLS, AND ETHERS | |
ALDEHYDES, KETONES, AND CARBOXYLIC ACIDS | |
ORGANIC COMPOUNDS CONTAINING NITROGEN | |
BIOMOLECULES | |
POLYMERS | |
CHEMISTRY IN EVERYDAY LIFE |
The format of CUET 2025 chemistry question papers is includes multiple-choice questions. The overall CUET 2025 exam contains 50 questions per section.
The CUET Chemistry 2025 question paper consists of three sections: Section A, Section B, and Section C namely as Language test, domain specific section and general test respectively
The overall difficulty level of the CUET exam is moderate to difficult, but the difficulty level changes as per the subjects and years.
Yes, some practical-based questions may be included in Section C to evaluate experimental knowledge.
The use of calculators is usually not allowed in the CUET exam 2025, unless specified.
Focus on understanding concepts, practise CUET chemistry previous year question papers, and manage time during preparation.
CUET previous year question papers with answers pdf will be available from NTA website.
Yes, the question papers of CUET examination are generally based on the prescribed syllabus for the course.
CUET previous year question papers with answers pdf download link available on the NTA website.
Hello,
You cannot get admission to engineering courses in Delhi University (DU) through CUET scores .
DU offers engineering courses under its Faculty of Technology, and admissions to these courses are based on JEE Main scores, not CUET.
Hope it helps !
Hello,
Delhi University (DU) offers various engineering courses under its Department of Technology.
Admission to these courses is primarily based on the Joint Entrance Examination (JEE) Main scores. However, some courses may also consider CUET (Common University Entrance Test) scores for admission.
The exact number of seats filled through CUET scores can vary each year and are not publicly disclosed.
Hope it helps !
Admissions for CUET aren't solely based on 2 Non-Med Percentage along with JEE Mains percentile. The eligibility criteria for CUET require students to have scored at least 50% marks in their Class 12th exam for general candidates and 45% for reserved categories.
Additionally, CUET has its own exam pattern, which includes multiple-choice questions divided into three sections:
- Section 1: Language proficiency(English/Hindi/regional languages)
- Section 2: Domain-specific subjects
- Section 3: General Aptitude
It's also important to note that while JEE Mains is a separate entrance exam, some universities may consider both CUET and JEE Mains scores for admission to certain programs. However, the specific admission criteria may vary depending on the university and course.
To confirm the admission criteria for your desired course, I recommend checking the official websites of the participating universities or contacting them directly.
Yes, if the CUET UG application form does not ask for the 10th marksheet upload and only requires a photograph and signature, your application should still be considered valid. Since you are currently appearing for the 12th exams, the system may not require additional documents at this stage. However, double-check the official guidelines or contact CUET support to confirm.
A PG (Postgraduate) degree in Forensic Science, specifically an M.Sc. in Forensic Science, is a two-year program that provides specialized knowledge and skills in the scientific analysis and application of techniques for collecting and analyzing evidence to solve crimes.
You can refer to following link for the paper
CUET forensic science question paper
GOOD luck!!
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