CBSE Class 12 Chemistry Case Study Based Important Questions with Solutions, Download PDF Here

CBSE 12th Chemistry Case Study Based Questions: CBSE Board exam for class 12 Chemistry will have different types of questions, like MCQs, assertion-based questions, short-answer type questions, long-answer type questions, and case study based questions. To help the students of CBSE Class 12 Chemistry with case study-based questions, we have provided some important case study-based questions and answers. Practising these questions will help students score the best marks in board exams. 

The questions typically involve interpreting graphs, analyzing experimental setups, evaluating chemical reactions, or understanding industrial processes, thereby moving beyond rote memorization to a deeper level of comprehension and application.

CBSE Class 12 Chemistry Paper Pattern

CBSE Class 12 Chemistry exam will be conducted for 70 marks. The time duration for students to attempt the paper will be 3 hours. All questions are compulsory. Use of log tables and calculators is not allowed.

Important Case Study Based Questions for Class 12th Chemistry with Solutions

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. 

a. 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 

b. 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. 

c. 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.

Q2. Dependence of the rate of reaction on the concentration of reactants, temperature, and other factors is the most general method for weeding out unsuitable reaction mechanisms. The term mechanism means all the individual collisional or elementary processes involving molecules (atoms, radicals, and ions included) that take place simultaneously or consecutively to produce the observed overall reaction. For example, when hydrogen gas reacts with bromine, the rate of the reaction was found to be proportional to the concentration of H2 and to the square root of the concentration of Br2 . Furthermore, the rate was inhibited by increasing the concentration of HBr as the reaction proceeded. These observations are not consistent with a mechanism involving bimolecular collisions of a single molecule of each kind. The currently accepted mechanism is considerably more complicated, involving the dissociation of bromine molecules into atoms followed by reactions between atoms and molecules: It is clear from this example that the mechanism cannot be predicted from the 8 overall stoichiometry.(source: Moore, J. W., & Pearson, R. G. (1981). Kinetics and mechanism. John Wiley & Sons.)

a. Predict the expression for the rate of reaction and order for the following:

H2 + Br2 🡪 2 HBr

What are the units of rate constant for the above reaction?

b. How will the rate of reaction be affected if the concentration of Br2 is tripled?

OR

What change in the concentration of H2 will triple the rate of reaction?

c. Suppose a reaction between A and B, was experimentally found to be first

order with respect to both A and B. So the rate equation is:

Rate = k[A][B]

Which of these two mechanisms is consistent with this experimental finding? Why?

Mechanism 1

A → C + D (slow)

B + C → E (fast)

Mechanism 2

A + B → C + D (slow)

C → E (fast)

Q3. Amines are basic in nature. The pKb value is a measure of the basic strength of an amine. Lower the value of pKb , more basic is the amine. The effect of substituent on the basic strength of amines in aqueous solution was determined using titrations. The substituent “X” replaced “-CH2 ” group in piperidine ( compound 1) and propylamine CH3CH2CH2NH2 , (compound 2).

(source: Hall Jr, H. K. (1956). Field and inductive effects on the base strengths of amines. Journal of the American Chemical Society, 78(11), 2570-2572.) Study the above data and answer the following questions:

a. Plot a graph between the electronegativity of the substituent vs pKb value of the corresponding substituted propyl amine (given that pKa + pKb =14). Is there any relation between the electronegativity of the substituent and its basic strength?

b. The electronegativity of the substituent “C6H5CON” is 3.7, what is the expected pKa value of compound C6H5CONHCH2CH2NH2? 2 1 10 (i) 9.9 (ii) 9.5 (iii) 9.3 (iv) 9.1

c. The pKa value of the substituted piperidine formed with substituent “X” is found to be 8.28. What is the expected electronegativity of “X” (i)3.5 (ii)3.4 (iii)3.8 (iv) 3.1

OR

What is the most suitable pKa value of the substituted propylamine formed with substituent “X” with electronegativity 3.0 (i)10.67 (ii)10.08 (iii)10.15 (iv)11.10

Q5. Read the following paragraph and answer the questions:

An ideal solution of two liquids is a solution in which each component obeys Raoult's law which states that the vapour pressure of any component in the solution depends on the mole fraction of that component in the solution and the vapour pressure of that component in the pure state. However, there are many solutions which do not obey Raoult's law. In other words, they show deviations from ideal behaviour which may be positive or negative. However, in either case, corresponding to a particular composition, they form a constant boiling mixtures called azeotropes.

(i) The mole fraction of Ethyl alcohol in its solution with Methyl alcohol is 0.80. The vapour pressure of pure Ethyl alcohol at this temperature is 40mm of Mercury. What is its vapour pressure in the solution if the solution is ideal?

(ii) Why do a solution of Phenol and Aniline exhibit negative deviation from ideal behaviour?
(iii) Write and example for maximum boiling azeotrope.
(iv) Why pure Ethyl alcohol cannot be obtained from rectified spirit even by fractional distillation?
(v) When two liquids A & B are mixed the volume of the resulting solution is found to be slightly greater than sum of the volumes of A & B. Identify the type of deviation exhibited by the solution.

Q6. Colligative properties of a solution depend upon the number of moles of the solute dissolved and do not depend upon the nature of the solute. However, they are applicable only to dilute solutions in which the solutes do not undergo any association or dissociation. For solutes undergoing such changes, Van't Hoff introduced a factor, called Van't Hoff factor (i). This has helped not only to explain the abnormal molecular masses of such solutes in the solution but has also helped to calculate the degree of association or dissociation.

(i) What is Van’t Hoff factor (i) for a compound undergoing tertramerization in an organic solvent?
(ii) Arrange the following in the increasing order of freezing point 0.1M Al2(SO4)3, 0.1M KCl, 0.1M Glucose, 0.1M K2SO4
(iii) The molar mass of Sodium Chloride determined by elevation of boiling point method is found to be abnormal. Why?
(iv) What is the elevation of boiling point of a solution of 13.44g of CuCl2 in 1kg of water? (Kb for water = 0.52Kkg/mol-1, molar mass of CuCl2 = 134.4g/mol)

(v) Equimolal solutions of NaCl and BaCl2 are prepared in water. Freezing pint of NaCl is found to be -2 0C. What freezing point do you expect for BaCl2 solution?

The important case study-based questions with answers for CBSE Class 12 Chemistry can be accessed from the below-given PDF link:

Importance of Solving Case Study Questions for Class 12 Chemistry

• Case study-based questions in Class 12 are pivotal for good scores. Practicing these questions equips students to tackle these types of questions effectively.
• Solving case study-based questions helps students develop an efficient time management strategy.
• Excelling in Case study-based questions cultivates a holistic subject comprehension essential for future pursuits.

Among all the subjects, chemistry is considered one of the toughest because of the nature of the questions. Thus, this subject requires consistent practice. To score well in the final exams, students are advised to solve good number of questions. 

Also Check:

• CBSE Class 12 Chemistry Sample Paper 2025-26
• CBSE Class 12 Syllabus 2025-26
• CBSE Class 12 Skill Subject Syllabus

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CBSE Class 12 Chem Term 1 Sample Paper 2025

Important Name Reactions of Class 12 Chemistry with Previous Year Questions

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CBSE Class 12 Preparation Tips: TOP 10 Ways to Score High in CBSE Class 12th Board Exam

Read

the

passage

carefully

and

answer

the

questions

that

follow.

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.