CBSE Class 12 Chemistry Case Study Questions with Solution - PDF Download

CBSE Class 12 Chemistry Case Study Questions with Solution – PDF Download

This article gives students important questions on case study-based that will help them do well on the CBSE Class 12 Chemistry Board Exam on February 27, 2024.

There are different parts to the Chemistry examination, and each part has a particular number of questions and marks. It is very important for students to know the paper format so they can study well.

SectionNumber of Questions & TypeMarks per Question
Section A16 (MCQs)1 mark
Section B5 (Short answer questions)2 marks
Section C7 (Short answer questions)3 marks
Section D2 (Case Study-Based Questions)4 marks
Section E3 (Long Answer Type Questions)5 marks

This section has important case study-based questions for Class 12 Chemistry, along with their answers. Each question is worth 4 marks, and students are provided with an internal choice.

  • The following questions are case-based questions. Each question has an internal choice and carries 4 marks each. Read the passage carefully and answer the questions that follow.

Q.1 Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, DNA was first identified in the late 1860s by Swiss chemist Friedrich Miescher. Then, in the decades following Miescher’s discovery, other scientists–notably, Phoebus Levene and Erwin Chargaff–carried out a series of research efforts that revealed additional details about the DNA molecule,including its primary chemical components and the ways in which they joined with one another. Without the scientific foundation provided by these pioneers, Watson and Crick may never have reached their groundbreaking conclusion of 1953: that the DNA molecule exists in the form of a three-dimensional double helix. Chargaff, an Austrian biochemist, as his first step in this DNA research, set out to see whether there were any differences in DNA among different species. After developing a new paper chromatography method for separating and identifying small amounts of organic material, Chargaff reached two major conclusions: (i) the nucleotide composition of DNA varies among species. (ii) Almost all DNA, no matter what organism or tissue type it comes from maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is similar to the amount of thymine (T), and the amount of guanine (G) approximates the amount of cytosine (C). In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. This conclusion is now known as “Chargaff’s rule.” Chargaff’s rule is not obeyed in some viruses. These either have single- stranded DNA or RNA as their genetic material. Answer the following questions: a. A segment of DNA has 100 adenine and 150 cytosine bases.

  • What is the total number of nucleotides present in this segment of DNA?

Ans. A = 100 so T = 100 C=150 so G = 150

Total nucleotides = 100+100+150+150 =500

  •       A sample of hair and blood was found at two sites. Scientists claim that the samples belong to the same species. How did the scientists arrive at this conclusion?

Ans. They studied the nucleotide composition of DNA. It was the same so they concluded that the samples belong to the same species.

  • The sample of a virus was tested and it was found to contain 20% adenine, 20% thymine, 20 % guanine and the rest cytosine. Is the genetic material of this virus (a) DNA- double helix (b) DNA-single helix (c) RNA? What do you infer from this data?

Ans. A = T = 20%

But G is not equal to C so double helix is ruled out.

The bases pairs are ATGC and not AUGC so it is not RNA. The virus is a single helix DNA virus.

Q.2 Henna is investigating the melting point of different salt solutions. She makes a salt solution using 10 mL of water with a known mass of NaCl salt. She puts the salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts. She repeats each experiment.

Assuming the melting point of pure water as O degree C, answer the following questions:

  • One temperature in the second set of results does not fit the pattern.

Which temperature is that? Justify your answer.

  • Why did Henna collect two sets of results?
  • In place of NaCl, if Henna had used glucose, what would have been the melting point of the solution with 0.6 g glucose in it?

Ans. The melting point of ice is the freezing point of water. We can use the depression in the freezing point property in this case.

  • 3rd reading for 0.5 g there has to be an increase in depression of freezing point and therefore decrease in freezing point so also decrease in melting point when amount of salt is increased but the trend is not followed in this case.
  • two sets of reading help to avoid error in data collection and give more objective data.
  • Read the passage carefully and answer the questions that follow.

Q.3 Early crystallographers had trouble solving the structures of inorganic solids using X-ray diffraction because some of the mathematical tools for analyzing the data had not yet been developed. Once a trial structure was proposed, it was relatively easy to calculate the diffraction pattern, but it was difficult to go the other way (from the diffraction pattern to the structure) if nothing was known a priori about the

arrangement of atoms in the unit cell. It was important to develop some guidelines for guessing the coordination numbers and bonding geometries of atoms in crystals. The first such rules were proposed by Linus Pauling, who considered how one might pack together oppositely charged spheres of different radii. Pauling proposed from geometric considerations that the quality of the “fit” depended on the radius ratio of

the anion and the cation. If the anion is considered as the packing atom in the crystal, then the smaller cation fills interstitial sites (“holes”). Cations will find arrangements in which they can contact the largest number of anions. If the cation can touch all of its nearest neighbour anions then the fit is good. If the cation is too small for a given site, that coordination number will be unstable and it will prefer a lower coordination structure. The table below gives the ranges of cation/anion radius ratios that give the best fit for a given coordination geometry.

  • The radius of Ag+ ion is 126pm and of I ion is 216pm. The coordination number of Ag+ion is:





  • A solid AB has square planar structure. If the radius of cation A+ is 120pm, calculate the maximum possible value of anion B
    • 240 pm
    • 270 pm
    • 280 pm
    • 290 pm
  • A “good fit” is considered to be one where the cation can touch:
  • all of its nearest neighbour anions.
    • most of its nearest neighbour anions.
    • some of its nearest neighbour anions.
    • none of its nearest neighbour anions.

Ans. (i) 6

The radius of Ag+ ion is 126pm and of I ion is 216pm. The coordination number of Ag+ ion is:

ρ = rcation/ranion = 126/ 216 = 0.58

Radius ratio lies in the range 0.414 – 0.732, so has coordination number 6 or 4 according to the table.

Since none of the options is 4, so the answer is 6

  • 290 pm

Square planar means ratio ratio is between 0.414 – 0.732

If radius of cation is 120 pm then anion should be in the range ρ = rcation/ranion

0.414 = 120/ x so x = 289.8 = 290 pm

0.732 = 120/ x so x = 163.9 = 164 pm

  • all of its nearest neighbour anions
  • Read the passage given below and answer the questions that follow.

There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix.

The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1:1 for small nuclei (Z < 20). Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food you eat, and yes, in your body.

The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive. In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaged chemicals.

The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first order kinetics. It has a half – life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate when the person died. All living organisms consume carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years. This is all because of the fact that nuclear reactions naturally occur in living organisms.

  • Why is Carbon -14 radioactive while Carbon -12 not? (Atomic number of Carbon: 6)
  • Researchers have uncovered the youngest known dinosaur bone, dating around 65 million years ago. How was the age of this fossil estimated?
  • Which are the two most common radioactive decays happening in the human body?

Ans. (a) Ratio of neutrons to protons is 2.3: 1 which is not the stable ratio of 1:1

  • Age of fossils can be estimated by C-14 decay. All living organisms have C-14 which decays without being replaced back once the organism dies.
  • carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium.
  • Read the passage given below and answer the following questions:

Some colloids are stable by their nature, i.e., gels, alloys, and solid foams. Gelatin and jellies are two common examples of a gel. The solid and liquid phases in a gel are interdispersed with both phases being continuous. In most systems, the major factor influencing the stability is the charge on the colloidal particles. If a particular ion is preferentially adsorbed on the surface of the particles, the particles in suspension will repel each other, thereby preventing the formation of aggregates that are larger than colloidal dimensions. The ion can be either positive or negative depending on the particular colloidal system, i.e., air bubbles accumulate negative ions, sulphur particles have a net negative charge in a sulphur sol, and the particles in a metal hydroxide sol are positively charged. Accumulation of charge on a surface is not an unusual phenomenon-dust is attracted to furniture surfaces by electrostatic forces. When salts are added to lyophobic colloidal systems the colloidal particles begin to form larger aggregates and a sediment forms as they settle. This phenomenon is called flocculation, and the suspension can be referred to as flocculated, or colloidally unstable. If the salt is removed, the suspension can usually be restored to its original state; this process is called deflocculation or peptization. The original and restored colloidal systems are called deflocculated, peptized, or stable sols. Why does a small amount of salt have such a dramatic effect on the stability of a lyophobic colloidal system? The answer lies in an understanding of the attractive and repulsive forces that exist between colloidal particles. Van der Waals forces are responsible for the attractions, while the repulsive forces are due to the surface charge on the particles. In a stable colloid, the repulsive forces are of greater magnitude than the attractive forces. The magnitude of the electrical repulsion is diminished by addition of ionized salt, which allows the dispersed particles to aggregate and flocculate. River deltas provide an example of this behaviour. A delta is formed at the mouth of a river because the colloidal clay particles are flocculated when the freshwater mixes with the salt water of the ocean

  • Gelatin is a                                               colloidal system.
  1. Solid in solid
  • Solid in gas
  • Liquid in solid
  • Liquid in gas

(ii).Colloidal solutions are stable due to:

  1. presence of charges on the colloidal particles
  • formation of aggregates by colloidal particles
  • preferential adsorption on the surface
  • preferential absorption on the surface
  • Settling down of colloidal particles to form a suspension is called:
  1. flocculation
  • peptization
  • aggregation
  • deflocculation
  • When Van der Waals forces are greater than forces due to the surface charge on the particles,
    • flocculation occurs.
  • the colloid is stable.
  • peptization takes place.
  • deflocculation occurs.

Ans. (i) C,

(ii) C

(iii) A

(iv) A

Case study-based questions play a pivotal role in scoring well in Class 12 Chemistry. The importance lies in:

  • Effective tackling of such questions.
  • Developing an efficient time management strategy.
  • Cultivating holistic subject comprehension essential for future pursuits.

Students find chemistry difficult, and they should keep practicing, especially with questions based on case studies, if they want to do well on the final Board Examination of 2023-24

Access the detailed case study-based questions and answers for CBSE Class 12 Chemistry here.


  • Mayank Agrawal

    Mayank Agrawal is a passionate blogger, web developer, and Android developer with a knack for storytelling and building user-friendly experiences. He enjoys weaving words into engaging narratives for his blog and crafting intuitive web and mobile applications that users love. While his skills encompass both front-end and back-end development, his true passion lies in crafting engaging Android applications that solve real-world problems and improve people's lives.

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