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CONTENTS
Blueprint (iv)
Time Management (v)Topper Tips (x)
Concept Maps (Chapter-wise)
(A scientifically proven way to visually revise)
Phase 1
Previous Years
Phase 2
Includes 50% new
Competency Qs
Phase 3
Includes 50% new
Competency Qs
1
Introducing Official CBSE Papers (Solved)
Sample Paper 1
(8th Sep. 2023 Practice Paper)
23
Sample Paper 2
(31st Mar. 2023-24 Sample Paper)
35
Sample Paper 3
(CBSE 2022-23)
47
Sample Paper 4
(CBSE Term 1 & 2 2021-22)
59
Sample Paper 5
(CBSE Compartment 2021)
74
Sample Paper 6
(CBSE 2019-20)
87
Sample Paper 7
(CBSE 2018-19)
96
Fresh Papers (Solved)
Sample Paper 8
107
Sample Paper 9
120
Sample Paper 10
132
Sample Paper 11
143
Fresh Papers (Self Assessment)
Sample Paper 12
156
Sample Paper 13
162
Sample Paper 14
167
Sample Paper 15
172
Sample Paper 16
177
Self-evaluation Charts (Sample Paper 12-16)
182
Evaluate your answers using step-wise
marks breakdown & know your final score.
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(iv)
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–
–
–
16Q
Short Answer
Case/ Data-based
Long Answer
TOTAL
5Q
–
–
–
(1 internal choice)
5Q
–
–
7Q
–
–
(1 internal choice)
7Q
–
–
–
(3 marks)
Section C
2Q
–
(1 internal choice)
per case-study
2Q
–
–
–
–
(4 marks)
Section D
3Q
(3 internal choices)
3Q
–
–
–
–
–
(5 marks)
Section E
70 marks
15 marks
8 marks
21 marks
10 marks
4 marks
12 marks
Total Marks
Note:- 2023-24 Chemistry sample paper contains competency-based questions in the form of Stand-alone, Diagram, Statement, Assertion/ Reason,
Table-based and Case-based questions.
–
4Q
Assertion/
Reason
Very Short Answer
12Q
(2 marks)
(1 mark)
MCQ
Section B
Section A
(Question Paper Design)
Blueprint
Time
Management
for the 3 hours paper
CHEMISTRY
Section
Questions
Time to be Spent
Total Time
(maximum)
Reading Time (Mandatory): 15 min
SECTION A
(1 mark)
12Q
(MCQs)
4Q
(A/ R)
SECTION B
(2 marks)
5Q
(VSA)
SECTION C
(3 marks)
7Q
SECTION D
(4 marks)
SECTION E
(5 marks)
2Q
(SA)
(Case-based)
3Q
(LA)
~ 1.5 minutes
(per Question)
30 minutes
~ 2 minutes
(4 minutes extra given*)
(per Question)
~ 3 minutes
(per Question)
~ 4 minutes
20 minutes
(5 minutes extra given*)
35 minutes
(per Question)
(7 minutes extra given*)
~ 10 minutes
25 minutes
(per Question)
(5 minutes extra given*)
~ 15 minutes
50 minutes
(per Question)
(5 minutes extra given*)
Revision Time
20 minutes
TOTAL TIME:
180 minutes
~ means approximate time (1 minute +/– is okay)
* Extra time for competency-based Qs is suggested.
IMPORTANT:
1.
The mandatory 15 minutes Reading Time should be used to skim through the
paper and decide which questions to attempt first.
2. Revision time is a must to have (at the end) to achieve three things:
– Attempt the questions you have left or are not 100% sure about
– Check if any question (sub-part) is left unattempted
– Double check, if the correct options are picked in the Objective section
(v)
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FAQs
1. If a student fails in Pre-board
examination, does that mean
one cannot appear in the Board
examination?
Pre-Boards help students to know how
well they are prepared for the Board
examination. A student cannot be detained
from appearing in the Board examination
if otherwise eligible.
2. Which subjects are mandatory and
what happens if I fail in any?
You have to choose a language (Hindi
Elective or Hindi Core or English Elective
or English Core) as a compulsory subject.
Other than 1st, you have to choose a 2nd
subject from any one Academic Electives
(I.P., Computer Science or any other
language out of 34 Regional languages).
You have to choose any three compulsory
Electives from a pool of Academic or
skill subjects (Physics, Chemistry and
Mathematics/Biology) as 3rd, 4th and 5th
subject. You can choose any optional
Elective from a pool of Languages,
Academic and Skill subjects as 6th subject.
3. When will CBSE provide a datesheet
for 2024 boards?
CBSE will most probably conduct Class
10 and 12 board exams for the 2023-24
academic session from February 15th, 2024.
The examination period is estimated to
span approximately 55 days, concluding on
10th April, 2024.
you just have to understand the topics and
concepts well (basics). They are not that
difficult to answer once you understand
the style and nature of the questions
asked.
6. Do examiners deduct marks for
exceeding the word limit and
spelling mistakes, especially in the
language papers?
No marks are deducted for exceeding the
word limit. Marks for spelling mistakes and
other errors are deducted in the Language
Papers.
7. Will questions be asked from the
Board’s sample paper?
Sample question papers help you know
the design, pattern and types of questions
that can be asked. Questions in the
examination may be from any part of the
syllabus. So, prepare thoroughly from the
entire syllabus.
8. Is it compulsory to write the
answers in the same sequence as in
the question paper?
No, you may attempt those questions
in the beginning which you know best.
Make sure that you write correct question
number to each answer.
9. Whom should I reach out to in
case of examination, admit card or
results-based issues?
For detailed information it is always
suggested to reach out to your school
authorities first.
4. What is the use of the CBSE Marking
Scheme that is always provided by
along with CBSE Sample Papers?
However, if you want to escalate, we have
compiled a comprehensive list of all the
important CBSE contacts.
Marking scheme provides an ideal
answer that CBSE expects you to write.
It also helps you to understand the exact
breakdown of marks for each step. It is a
bit technical in nature but you should refer
to our Self-evaluation charts to understand
this better.
5. I really have no clue what
competency-based questions are.
Am I going to struggle?
CBSE committed that atleast 50% of such
questions will come in the board paper and
now the sample paper proves the same. To
solve competency-based questions (CBQs),
Scan the QR code to access the
Important CBSE Contacts list.
(ix)
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Topper Tips
Limit yourself to a fixed time per Question
All your months of hard work culminates into your performance in those 3 hours.
One technique that you should apply in your exam paper is to constrain yourself to
a certain number of minutes per question type. If you cross that time, you mark that
question and move to the next one. This way you can attempt the whole paper in
time and then get back to the questions that you marked for the final attempt.
Read 3-5 marks Questions Twice
Sometimes in a rush to start solving the paper, you misinterpret or miss out on
sub-parts of the questions. Even worse, you end up providing a 2m answer to a 3m
question or vice versa (which is a waste of time). Always, read the questions twice to
ensure you understand what is being asked and how much you need to write to get
full marks.
Key Competency-based Qs
Look for specific terms like Analyse, Identify, Suggest, Observe, etc, during the reading
time. These questions are designed to test your analytical, creative and critical
thinking skills, in which you must always use the concepts learnt.
Self Analyse for Mistakes
After solving numericals, formula-based questions or chemical reactions, read and
analyse them for any mistakes. You may lose half to 1 mark easily for silly mistakes in
formulas, calculations, diagram labellings or units.
Focus on Graphical Representations & Diagrams
In questions like draw/plot a comparsion graph and answer, be extra cautions while
creating these answers. Catching mistakes like using the same systems of unit in
comparision/calculation at the right time helps a lot in scoring maximum marks.
Dedicate Rough Work Space
Using a rough space apart from your answer sheet is the best way to do calculations
or work. Keep that space limited to the right margin of the answer sheet or to another
page altogether.
(x)
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Homogeneous mixture of two or
more components.
• Component present in lesser
amount is called “solute”.
• Component present in greater
amount is called “solvent”.
INTRODUCTION
• Effect of temperature: For
endothermic reactions, solubility
increases and for exothermic
reactions, solubility decreases
with rise in temperature.
• No effect of pressure is there.
SOLUBILITY OF SOLID
IN A LIQUID
800 ml
200
400
800
LIQUID - LIQUID
SOLUTION
150 ml
40
80
120
• Don’t obey Raoult’s law
• P1 ≠ P10 X1
Non Ideal Solution
• Attraction of A-B < A-A , B-B.
• ΔHmix = +ve
• ΔVmix = +ve
Eg: Ethanol & Acetone
• Min. boiling Azeotropes
Show +ve Deviation
• Attraction A-B > A-A, B-B.
• ΔHmix = -ve
• ΔVmix = -ve
• Max. boiling Azeotropes
Eg: Acetone & Chloroform
Show -ve Deviation
Eg: C6H6 & C6H5CH3 , C6H14 & C6H16
• Obeys Raoult’s law
• ΔHmix = 0
• ΔVmix = 0
Ideal Solution
2
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250 ml
50
100
150
200
Solute
Salt
SOLUTIONS
Strength
•
•
•
•
•
Solvent
Solution
Salt-Water
SOLUBILITY OF
GAS IN LIQUID
• Exist in single phase.
• Particle size is less than 1nm.
• Dilute solution – Large amount of
solvent.
• Concentrated solution – Small
amount of solvent.
• Aqueous solution – Solvent is water.
• Non-aqueous solution – Solvent other
than water.
PROPERTIES & TYPES
OF SOLUTION
Maximum amount that can be dissolved
in a specific amount of solvent at a
specific temperature.
SOLUBILITY
The pressure exerted by
the vapour in equilibrium
with the liquid/solution at
a particular temperature.
Factors:
• Weaker the intermolecular
forces greater is the
vapour pressure.
• Temp increases with the
increase in vapour pressure.
Vapour Pressure
• If the density of solution is approx. 1
then, Molarity < Molality.
Mass of solute x 1066
• ppm = Mass of solution x 10
• Molarity depends on temperature
but molality does not.
Important Points
Chapter 1
Concept MAPS
1
temperature
Aquatic species are comfortable
in cold water
Solubility ∝ P
Scuba divers suffer from bends.
Climbers suffer from anoxia
Solubility of gases in liquid
decreases with increase in
temperature.
• Solubility ∝
• P = K HX
Henry’s Law
It is the amount of
solute present per unit
solution.
Water
EXPRESSING CONCENTRATION
OF SOLUTION
Moles of solute
Vol. of solution in L
Molarity (M)
Moles of solute
mass of solvent in kg
Molality (m)
XA =
nA
nA + nB
XA + XB ………Xi = 1
Here, n = no. of moles
Mole
fraction (XA)
Mass% = Mass of solute x 100
Mass of solution
Mass %
RAOULT’S LAW
Azeotropes
x1 = 1
x2 = 0
p°1
P2
Mole fraction
x2
p total
1
P
P2
= P1 +
x1 = 0
x2 = 1
I
II
III
• P1 = P1oX1 (Non-volatile solute)
• P2 = Po2X1
• P = P1 + P2 (Volatile solute)
Vapour pressure
p°2
X
0%
Y 100%
P
of both
constituents
Total
vapor
Pressure
fro
’s
Raoult
law
n fro
iatio
dev
w
100%
0%
a
lt’s L
aou
mR
Positive Azeotrope
P1º – P1
Pº – P1
W2 x M1
= X2 or 1
=
M2 x W1
P1º
P1º
• Lowering of vapour pressure
is not a colligative property.
• R. L. V. P. is a colligative property.
P1º = V. P. of pure solvent
P1 = V. P. of solution
W2 = Mass of solute
M1 = Molar mass of solvent
W1 = Mass of solvent
M2 = Molar mass of solute
Relative lowering
of vapour pressure
Negative Azeotrope Negative
ion
iat
dev
itive
Pos
w
’s La
lt
aou
mR
• A liquid mixture that has a constant
boiling point and whose vapour has
the same composition as the liquid.
• +ve deviation are known as min.
boiling azeotrope. Eg: C2H5OH + H2O
• –ve deviation are known as max.
boiling azeotrope. Eg: HNO3 + H2O
3
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ΔTb x W1
1000 x W2 x Kb
Two solutions have the same
osmotic pressure (π)
Eg: Saline solution
(0.9%) & blood
Isotonic Solution
• M2 =
Tbº = Boiling point of solution
Tb = Boiling point of solvent
Kb = Molal Elevation Constant
m = Molality
• ΔTb = Kbm
• If m = 1
ΔTb = Kb
• ΔTb = Tºb – Tb
Elevation in
b.p. (boiling point)
• π of solution1 > π of
solution2
• RBC shrinks in hypertonic
solution.
Hypertonic
w.r.t solution1
• ΔTf = Kfm
• If m = 1
ΔTf = Kf
• Used to calculate M2
for normal molecules
ΔTf = Tf – Tf º
Tf = Freezing point of solvent
Tfº = Freezing point of solution
Kf = Molal Depression Constant
m = Molality
1000 x W2 x Kf
• M2 =
ΔTf x W1
Depression in
f.p. (freezing point)
For Normal Molar Mass
• π of solution1 < π of
solution2
• RBC swells or bursts in
hypotonic solution.
Hypotonic solution
w.r.t to solution1
• External pressure used to
stop osmosis
• π = nRT
V
• Used to calculate M2 for
macromolecules.
W RT
• M2 = 2
πV
π = Osmotic pressure
V = Volume
R = Gas constant
T = Temperature
Osmotic Pressure
Depends on no. of solute particle
COLLIGATIVE
PROPERTIES
Sample Paper
12
Self-Assessment
jlk;u foKku
CHEMISTRY
Time Allowed : 3 Hours
Maximum Marks : 70
General Instructions: Same instructions as given in the Sample Paper 1.
SECTION - A
16 Marks
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mark. There is no internal choice in this section.)
1. The major product of acid catalysed
425
dehydration of 1-methylcyclohexanol is:
(b) 1-methylcyclohexene
(c) 1-cyclohexylmethanol
1
2. KMnO4 is coloured due to:
L (S cm2/mol)
(a) 1-methylcyclohexane
(d) 1-methylenecyclohexane
HCl
375
325
275
225
KCl
175
125
75
0
0.1
(a) d-d transitions
0.3
(b) charge transfer from ligand to metal
(a) 100 Scm2/mol
(b) 115 Scm2/mol
(c) unpaired electrons in d orbital of Mn
(c) 150 Scm2/mol
(d) 125 Scm2/mol
(d) charge transfer from metal to ligand
1
3. Which radioactive isotope would have the
longer half- life 15O or 19O?
(Given rate constants for 15O and 19O
are 5.63 × 10–3 s–1 and k = 2.38 × 10–2 s–1
respectively.)
(a)
15
(b)
19
O
O
(d) None of the above, information given is
insufficient
1
4. The molar conductivity of CH3COOH at
infinite dilution is 390 Scm2/mol. Using the
graph and given information, the molar
conductivity of CH3COOK will be:
1
5. For the reaction, A + 2B → AB2, the order
w.r.t. reactant A is 2 and zero w.r.t. reactant
B. What will be change in rate of reaction if
the concentration of A is doubled and B is
halved?
(a) increases four times
(b) decreases four times
(c) increases two times
(d) no change
(c) Both will have the same half-life
156
0.2
c (M)½
1
6. Which of the following observations is
shown when benzene diazonium chloride
reacts with aniline?
(a) p-Aminoazobenzene (yellow dye)
(b) p-Hydroxyazobenzene (orange dye)
(c) p-Aminoazobenzene (orange dye)
(d) p-Hydroxyazobenzene (yellow dye)
1
Chemistry Class XII
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7. Which of the following is correct regarding
Column - I
Column - II
(Compound)
(Oxidation state of Co)
(A)
[Co(NCS)(NH 3 ) 5 ]
(SO3)
(I)
+4
(B)
[Co(NH3)4Cl2)]SO4
(II)
0
(C)
Na4[Co(S2O3)3]
(III)
+2
(D)
[Co2(CO)8]
(IV)
+3
Lucas reagent?
(a) conc. HCl and an anhy. ZnCl2
(b) conc. HNO3 and an anhy. ZnCl2
(c) conc. HCl and hydrous ZnCl2
(d) conc. HNO3 and hydrous ZnCl2
1
8. Which of the following statements is not
correct for amines?
(a) Most alkyl amines are more basic than
ammonia solution.
(b) pKb value of ethylamine is lower than
benzylamine.
(c) CH3NH2 on reaction with nitrous acid
releases NO2 gas.
(d) Hinsberg’s reagent reacts with secondary
amines to form sulphonamides.
1
Code:
(A)
(a)(I)
(b)(IV)
(c)(III)
(d)(IV)
(B)
(II)
(III)
(I)
(I)
CH3
CHO
(i) 'X', CS 2
(ii) H2O
formed in the following reaction?
(a) CrO3
(b) CrO2Cl2
(c) Alkaline KMnO4
COOK
OH
+
+ (CH3CO)2O H
Salicylic acid
(a) 2-Acetoxybenzoic acid
(b) 3-Acetoxybenzoic acid
(c) 4-Acetoxybenzoic acid
(d) 2-methoxybenzoic acid
(D)
(III)
(I)
(II)
(II)
12. Identify 'X' in the reaction given below.
9. What is the IUPAC name of the product
(C)
(IV)
(II)
(IV)
(III)
1
10. Arrhenius equation can be represented
graphically as follows:
(d) Anhydrous AlCl3
1
In the following question, a statement of
assertion (A) is followed by a statement of
reason (R). Mark the correct choice as:
(a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
(b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
13. Assertion (A):An ether is more volatile than
ln k(s–1)
an alcohol of
molecular mass.
Reason (R):
1 (K–1)
T
(II) Ea/R
(b) (I) A
(II) Ea
(c) (I) ln A
(II) - Ea/R
(d) (I) A
(II) -Ea
Ethers are polar in nature.
1
14. Assertion (A):Proteins are found to have two
different types of secondary
structures viz a-helix and
b-pleated sheet structure.
The (I) intercept and (II) slope of the graph
are:
(a) (I) ln A
comparable
Reason (R):
The
secondary
structure
of proteins is stabilized by
hydrogen bonding.
1
15. Assertion (A):
Magnetic moment values of
actinides are lesser than the
theoretically predicted values.
11. Match the compounds (given in Column I)
with the oxidation state of cobalt present in
it (given in column II) and assign the correct
code.
1
Reason (R):
Actinide elements are strongly
paramagnetic.
Sample Paper 12
1
157
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16. Assertion (A):
Tertiary amines are more
basic
than
corresponding
secondary
and
primary
amines in gaseous state.
Reason (R):
Tertiary amines have three
alkyl groups which cause +I
1
effect. SECTION - B
10 Marks
(This section contains 5 questions with internal choice in one question.
The following questions are very short answer type and carry 2 marks each.)
17. A first-order reaction takes 69.3 min for 50%
completion. What is the time needed for 80%
of the reaction to get completed?
(Given: log 5 = 0.6990, log 8 = 0.9030, log 2 =
0.3010)
2
18. Account for the following:
(A) There are 5 –OH groups in glucose
(B) Glucose is a reducing sugar
OR
What happens when D – glucose is treated
with the following reagents?
(A) Bromine water
Calculate Λm for all concentrations and draw
a plot between Λm and c1/2. Find the value of
20. The rate constant of a first order reaction
increases from 2 × 10–2 to 8 × 10–2 when the
temperature changes from 300 K to 320 K.
Calculate the energy of activation (Ea). (log 2
= 0.301, log 3 = 0.4771, log 4 = 0.6021)
increasing order of their boiling points:
CH3CHO, CH3CH2OH, CH3OCH3,
19. The conductivity of NaCl at 298 K has been
determined at different concentrations and
the results are given below:
2
21. (A)Arrange the following compounds in the
2
(B) HNO3
2
Λºm.
CH3CH2CH3.
(B) Write the structure of the product formed
in the following reaction:
O
Concentration / M
10–4 × k/S m-1
0.001
1.237
0.010
11.85
0.020
23.15
Convert the following:
0.050
55.53
(A) Ethanol to 3-Hydroxybutanal
0.100
106.74
+ C2H5
C
Cl
Anhydrous
AlCl3
OR
(B) Benzoic acid to m-Nitrobenzyl alcohol
SECTION - C
2
21 Marks
(This section contains 7 questions with internal choice in one question.
The following questions are short answer type and carry 3 marks each.)
22. Write the equations for the following
reaction (Any two):
(A) Salicylic acid is treated with acetic
anhydride in the presence of conc. H2SO4
Complex
Stability
Constant (K)
[Cu(NH3)4]2+
4.5 × 1011
(B) Tert butyl chloride is treated with sodium
ethoxide.
[Cu(CN)4]2–
2.0 × 1027
[Ag(NH3)2]+
1.6 × 107
(C) Phenol is treated with chloroform in the
[Co(NH3)6]3+
5.0 × 1033
[Ag(CN)2]–
5.4 × 1018
23. Observe the table related to stability
[Ni(NH3)6]2+
6.1 × 1018
constant of some complex compounds.
[Ni(en)3]2+
4.6 × 1018
Answer the questions based on the table
and related concepts.
[Fe(CN)6]3–
1.2 × 1031
Stability constant of some complexes are
given:
[Fe(CN)6]4–
1.8 × 106
[Cd(NH3)4]2+
1.0 × 107
presence of KOH
158
3
Chemistry Class XII
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(A) Why stability constants of cyanides
complexes higher than complexes with
NH3?
(B) Why is [Cu(NH3)4]2+ more stable than
[Cd(NH3)4]2+?
(C) Why is [Ni(en)3]2+ more stable than
3
[Ni(NH3)4]2+?
(A) State Henry’s law and explain why are
the tanks used by scuba divers filled with
air is diluted with helium (11.7% helium,
56.2% nitrogen and 32.1% oxygen)?
(B) Assume that argon exerts a partial
pressure of 6 bar. Calculate the solubility
of argon gas in water. (Given: Henry’s law
constant for argon dissolved in water, KH
3
= 40Kbar)
25. Give the structures of products A, B and C in
the following reactions:
LiAIH
NH
C4H9Br is treated with aq. KOH solution.
The rate of this reaction depends upon the
concentration of the compound ‘A’ only.
When another optically active isomer ‘B’ of
this compound was treated with aq. KOH
solution, the rate of reaction was found to
be dependent on concentration of compound
24. Answer the following questions:
KCN
4
→ A 
→B
(A) CH3CH2Br 
26. Compound ‘A’ with the molecular formula
HNO
2

→C
0.° C
and KOH both.
Write down the structural formula of both
compounds ‘A’ and ‘B’. Out of these two
compounds, which one will be converted to
3
the product with inverted configuration?
27. tert-Butyl bromide reacts with aq. NaOH
by SN1 mechanism while n-butyl bromide
3
reacts by SN2 mechanism. Why?
28. (A)The CFSE of [CoCl6]3- is 18000 cm-1.
Calculate the CFSE for [CoCl4]- complex.
(B) A complex of the type [M(AA)₂X₂]n+ is
known to be optically active. What does
NaOH + Br
3
2
→ A 
→B
(B) CH3COOH 
∆
this
CHCl 3 + Alc.KOH

→C
complex? Give one example of such a
3
indicate about the structure of the
3
complex.
SECTION - D
8 Marks
(The following questions are case-based questions. Each question has an internal choice and carries
4 (1 + 1 + 2) marks each. Read the passage carefully and answer the questions that follow.)
29. 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.
Melting point in °C
Mass of the
salt used
(in g)
Readings
Set 1
Reading
Set 2
1
0.3
–1.9
–1.9
2
0.4
–2.5
–2.6
3
0.5
–3.0
–5.5
4
0.6
–3.8
–3.8
S.No.
5
0.8
–5.1
–5.0
6
1.0
–6.4
–6.3
Assuming the melting point of pure water as
0°C, answer the following questions:
(A) One temperature in the second set of
results does not fit the pattern. Which
temperature is that? Justify your answer.
OR
Why did Henna collect two sets of
results?
1
(B) 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?
1
(C) What is the predicted melting point if
1.2 g of salt is added to 10 ml of water?
Justify your answer.2
Sample Paper 12
159
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30. Strengthening
the Foundation: Chargaff
Formulates His "Rules"
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.
(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?
1
(B) A sample of hair and blood was found at
two sites. Scientists claim that the samples belong to same species. How did the
scientists arrive at this conclusion?
1
(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?
OR
How can Chargaff ’s rule be used to infer
that the genetic material of an organism
is double- helix or single- helix?
SECTION - E
2
15 Marks
(The following questions are long answer type and carry 5 marks each.
Two questions have an internal choice.)
31. Answer the following (Any five):
(A) Why are all copper halides known except
that copper iodide?
(B) Why is the Eo(V3+/V2+) value for vanadium
comparatively low?
(C) Why HCl should not be used for
potassium permanganate titrations?
(D) Explain the observation. "At the end of
each period, there is a slight increase in
the atomic radius of d-block elements."
(E) What is the effect of pH on dichromate
ion solution?
(F) Why do transition elements show
variable oxidation states?
(G) What happens when (NH4)2Cr2O7 is
heated? 5
32. (A)What is the process that involves the
spontaneous transfer of electrons
between substances and is used in
batteries to generate electrical energy?
160
(B) Write the reaction occurring at anode and
cathode and the products of electrolysis
of aq. KCl.
(C) What is the pH of HCl solution when the
hydrogen gas electrode shows a potential
of -0.59 V at standard temperature and
pressure?
OR
(A) Molar conductivity of substance “A” is
5.9 × 103 S/m and “B” is 1 x 10–16 S/m.
Which of the two is most likely to be
copper metal and why?
(B) What is the quantity of electricity in
Coulombs required to produce 4.8 g of
Mg from molten MgCl2? How much Ca
will be produced if the same amount of
electricity was passed through molten
CaCl2? (Atomic mass of Mg = 24 u,
atomic mass of Ca = 40 u).
Chemistry Class XII
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(C) What is the standard free energy
change for the following reaction at
room temperature? Is the reaction
spontaneous?
Sn(s) + 2Cu2+(aq) → Sn2+ (aq) + 2Cu(s)
5
33. A hydrocarbon (a) with molecular formula
C5H10 on ozonolysis gives two products
(b) and (c). Both (b) and (c) give a yellow
precipitate when heated with iodine in
presence of NaOH while only (b) give a silver
mirror on reaction with Tollen’s reagent.
(A) Identify (a), (b) and (c).
(B) Write the reaction of (b) with Tollen’s
reagent
(C) Write the equation for iodoform test for
(c).
(D) Write down the equation for aldol
condensation reaction of (b) and (c).
OR
An organic compound (a) with molecular
formula C2Cl3O2H is obtained when (b)
reacts with Red P and Cl2. The organic
compound (b) can be obtained on the
reaction of methyl magnesium chloride with
dry ice followed by acid hydrolysis.
(A) Identify (a) and (b)
(B) Write down the reaction for the
formation of (a) from (b). What is this
reaction called?
(C) Give any one method by which organic
compound (b) can be prepared from its
corresponding acid chloride.
(D) Which will be the more acidic compound
(a) or (b)? Why?
(E) Write down the reaction to prepare
methane from the compound (b).
5
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Sample Paper
Sample Paper 12
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SELF EVALUATION CHART
Sample Paper - 12
Question
Marks
(Type)
Topic
(Chapter Name)
Q1
1m
(MCQ)
Chemical reactions
(Alcohols, Phenols and
Ethers)
(b) 1-methylcyclohexene
Q2
1m
(MCQ)
Colour in Coordination
Compounds
(Coordination
Compounds)
(b) charge transfer from ligand to metal
Q3
1m
(MCQ)
Half-life of a reaction
(Chemical Kinetics)
(a) 15O
Q4
1m
(MCQ)
Kohlrausch law of
independent migration
of ions
(Electrochemistry)
(b) 115 Scm2/mol
Q5
1m
(MCQ)
Dependence of Rate on
Concentration
(Chemical kinetics)
(a) increases four times
Q6
1m
(MCQ)
Chemical reactions of
amines
(Amines)
(a) p-Aminoazobenzene (yellow dye)
Q7
1m
(MCQ)
Lucas Test
(Alcohol, Phenol and
Ethers)
(a) conc. HCl and anhy. ZnCl2
Q8
1m
(MCQ)
Properties of amines
(Amines)
(c) CH3NH2 on reaction with nitrous acid releases
NO2 gas.
Q9
1m
(MCQ)
Q10
1m
(MCQ)
Q11
1m
(MCQ)
Oxidation state of
coordination compounds
(d) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(Coordination
Compounds)
Q12
1m
(MCQ)
Preparation of aldehyde
(Aldehydes, Ketones and (b) CrO2Cl2
Carboxylic Acids)
Q13
1m
(A-R)
Properties of Ethers
(Alcohols, Phenols and
Ethers)
(b) Both (A) and (R) are true but (R) is not the correct
explanation of (A).
1
Q14
1m
(A-R)
Structure of Proteins
(Biomolecules)
(b) Both (A) and (R) are true but (R) is not the correct
explanation of (A).
1
182
Full Marks
(Breakdown)
Chemical properties of
carboxylic acid
(a) 2-Acetoxybenzoic acid
(Aldehydes, Ketones and
Carboxylic acids)
Arrhenius equation
(Chemical kinetics)
(c) (I) ln A (II) - Ea/R
Your
Performance
1
1
1
1
1
1
1
1
1
1
1
1
Chemistry Class XII
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Q15
1m
(A-R)
Magnetic properties of
Actinoids
(The d- and f-Block
Elements)
(b) Both (A) and (R) are true but (R) is not the correct
explanation of (A)
1
Q16
1m
(A-R)
Chemical properties of
amines
(Amines)
(a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
1
Q17
2m
(VSA)
First order reaction
(Chemical Kinetics)
✔ Calculate the rate constant (1/2m)
✔ Write the formula of integrated rate law (1m)
✔ Calculate the time needed for the reaction to get
2
completed (½m)
Q18
2m
(VSA)
Q18
(OR)
Q19
Q20
Chemical reactions of
glucose
(Biomolecules)
✔ (A) Write reaction of D-glucose with bromine
water. (1m)
✔ (B) Write reaction of D-glucose with nitric acid.
Fehlings reagent. (1m)
(1m)
✔ Write formula of molar conductivity (½m)
✔ Draw a plot using calculation (1m)
✔ Find Λ using a graph (1/2m)
2
OR
2
Effect of concentration
on molar conductivity
(Electrochemistry)
2m
(VSA)
Effect of temperature on
activation energy
(Chemical Kinetics)
✔ Write the formula of activation energy (1/2m)
✔ Calculate the energy of activation (11/2m)
2
Properties of aldehydes,
ketones and carboxylic
acids
(Aldehydes, ketones and
carboxylic acids)
✔ (A) Write the correct order of arrangement. (1m)
✔ (B) Write the structure of the product formed.
2
✔ (A) Write the reaction of conversion (1m)
✔ (B) Write the reaction of conversion (1m)
2
2m
(VSA)
Q21
(OR)
Q23
✔ (A) Write reaction for the acetylation of glucose.
(1m)
✔ (B) Write reaction for reduction of glucose by
2m
(VSA)
Q21
Q22
Chemical reactions of
glucose
(Biomolecules)
Properties of aldehydes,
ketones and carboxylic
acids
(Aldehydes, ketones and
carboxylic acids)
3m
(SA)
Properties of alcohols,
phenols and ethers
(Alcohols, Phenols and
Ethers)
3m
(SA)
Applications of
Coordination
Compounds
(Coordinate compounds)
0
m
(1m)
✔ Write
the reaction in the form of chemical
equation in each case. (1m + 1m + 1m)
✔ (A) Give reason why stability constant of
cyanides is higher (1m)
✔ (B) Write the reason why [Cu(NH ) ] is more
stable. (1m)
✔ (C) Write the reason why (Ni(en) ] is more
3 4
stable. (1m)
3
2
3
2+
2+
3
✔ (A) Write about the Henry's Law. (1m)
Q24
3m
(SA)
Henry’s law
(Solutions)
Write the reason why tanks are diluted with
Helium. (1m)
✔ (B) Calculate the mole fraction of Argon in water.
3
(1m)
Self Evaluation Chart Sample Paper 12
183
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Q25
3m
(SA)
Chemical properties of
amines
(Amines)
✔ (A) Identify ‘A’, ‘B’ and ‘C’ (1/2m + 1/2m + 1/2m)
✔ (B) Identify ‘A’, ‘B’ and ‘C’ (1/2m + 1/2m + 1/2m)
3
Q26
3m
(SA)
Chemical properties of
haloalkanes
(Haloalkanes and
Haloarenes)
✔ Identify the compound ‘A’ and ‘B’. (1m + 1m)
✔ Which of the two will be converted to product
3
3m
(SA)
Chemical properties of
haloalkanes
(Haloalkanes and
Haloarenes)
3m
(SA)
Crystal Field Theory
(Coordination
Compounds)
Q27
Q28
with inverted configuration (1m)
✔ Give reason for the reaction of tert-Butyl bromide
with NaOH via S 1. (1m)
✔ Give reason for the reaction of n-butyl bromide
with NaOH via S 2. (1m)
✔ Give proper reaction for explanation. (1m)
N
N
✔ (A) Calculate CFSE. (1m)
✔ (B) Write about the structure
of the complex.
(1m)
Give one example of complex. (1m)
3
3
✔ (A) Give the justification. (1m)
Q29
Q30
Q31
184
4m
(CBQ)
4m
(CBQ)
5m
(LA)
Colligative properties
(Solutions)
Nucleic Acids
(Biomolecules)
Chemical properties of
d- and f- block elements
(The d- and f- block
elements)
OR
Give two reasons mentioning why Henna
collected two sets of results. (1/2m + 1/2m)
(B) Calculate the melting point of the solution.
(1m)
(C) Calculate the predicted melting point. (2m)
✔
✔
✔ (A) Write the number of nucleotides present.
(1m)
✔ (B) Mention the method that scientists used to
reach the conclusion. (1m)
✔ (C) Identify the genetic material of the virus out
of the options (a),(b) and (c). (1m)
Write the inference. (1m)
OR
Mention the inference of the genetic material
by Chargaff's rule. (2m)
✔ (A) State the reason. (1m)
✔ (B) State the reason. (1m)
✔ (C) State the reason. (1m)
✔ (D) Explain the observation of the trend. (1m)
✔ (E) Write the effect of the pH. (1m)
✔ (F) State the reason. (1m)
✔ (G) Write the reaction. (1m)
(Any five)
4
4
5
Chemistry Class XII
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Electrolytic cells and
electrolysis
(Electrochemistry)
Q32
Electrolytic cells and
electrolysis
(Electrochemistry)
Chemical properties
(Aldehydes, Ketones and
Carboxylic acids)
Q33
Write the formula for calculation of pH. (1m)
Calculate the pH. (1m)
5
✔ (A) Identify the copper metal out of 'A' and 'B'.
5m
(LA)
Q32
OR
✔ (A) Write the name of the process. (1m)
✔ (B) Write the reaction occuring at the cathode
and anode. (1/2m + 1/2m)
✔ (C) Write the Nernst equation. (1m)
(1/2m)
Write the reason why you chose. (1/2m)
(B) Calculate the quantity of electricity required.
(1m)
Calculate the weight of Ca produced. (1m)
(C) Calculate the free energy change. (1m)
Say whether reaction is spontaneous or
nonspontaneous. (1m)
✔
OR
✔
5
✔ (A) Identify (a), (b), and (c). (1/2m + 1/2m + 1/2m)
✔ (B) Write the reaction in the form of chemical
equation. (1/2m)
✔ (C) Write the reaction in the form of chemical
equation. (1/2m)
✔ (D) Write the reaction in the form of chemical
5
equation. (1/2m)
Write the product in the reaction.
(1/2m + 1/2m + 1/2m + 1/2m)
5m
(LA)
Q33
(OR)
✔ (A) Identify the compounds 'a' and 'b'
(1/2m + 1/2m)
✔ (B) Write the reaction in the form of chemical
equation. (1/2m)
Write the named reaction. (1/2m)
Chemical properties
(Aldehydes, Ketones and
(C) Write the method. (1m)
Carboxylic acids)
(D) Choose the acidic compound out of 'A' and 'B'.
(1/2m)
State the reason why you chose. (1/2m)
(E) Write the reaction in the form of chemical
equation. (1m)
✔
✔
OR
5
✔
TOTAL
Self Evaluation Chart Sample Paper 12
70
185
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SOLUTIONS
SAMPLE PAPER - 12
SECTION - A
1. (b) 1-methylcyclohexene
Explanation: According to Saytzeff rule i.e.,
highly substituted alkene is major product.
Here dehydration reaction takes place, alkene
is formed due to the removal of a water
molecule.
7. (a) conc. HCl and an anhy. ZnCl2
Explanation: Lucas reagent is conc. HCl and
anhy. ZnCl2
8. (c)CH3NH2 on reaction with nitrous acid
releases NO2 gas.
Explanation: The gas evolved when
methylamine reacts with nitrous acid is
nitrogen.
2. (b) charge transfer from ligand to metal
Explanation: The Mn atom in KMnO4 has +7
oxidation state with electron configuration
[Ar]3d04s0 Since no unpaired electrons are
present, d−d transitions are not possible. The
molecule should, therefore, be colourless.
Its intense purple colour is due to L→M (ligand
to metal) charge transfer 2p(L) of O to 3d(M) of
Mn.
3. (a)
15
O
Explanation: The rate constant for the decay
of O–15 is less than that for O–19 . Therefore,
the rate of decay of O–15 will be slower and
will have a longer half life .
Related Theory

Amines are basic in nature as they react with acids
to form salts. As the nitrogen atom of amines
possesses a lone pair of electrons, it behaves as
Lewis base. The basic strength of amines increases
with increase in value of Kb or decrease in the value
of pKb. The reactions are shown below.
+
–
CH3—NH3 X (Salt)
CH3—NH2 + HX
+
NH2
–
NH3Cl
+ HCl
Related Theory

The time in which the concentration of reactants is
reduced to half of its initial concentration is called
half-life of the reaction. It is denoted as t1/2.
4. (b) 115 Scm2/mol
Explanation: Λ°CH3COOK = Λ°CH3COOH +ΛΛ°KCl –
Λ
Λ°HCl = 390 + 150 – 425
= 115 S cm2/mol
Aniline
Anilinium chloride
9. (a) 2-Acetoxybenzoic acid
Explanation: The IUPAC name of the product is
2-Acetoxybenzoic acid and the reaction is
COOK
COOK
OH
+
+(CH3CO) 2O H
OCOCH3
+ CH3COOH
5. (a) increases four times
Explanation: Rate = [A]2
If [A] is doubled then Rate’
= [2A]2 = 4 [A]2
= 4 Rate
Salicylic acid
(Aspirin)
10. (c) (I) ln A (II) - Ea/R
Explanation: Arrhenius equation is
6. (a) p-Aminoazobenzene (yellow dye)
Explanation: Benzene diazonium chloride
reacts with aniline to form p-Aminoazobenzene.
+
–
N NCl + H
Acetylsalicylic acid
+
NH2 H
ln k =
p – Aminoazobenzene
(yellow dye)
–
NH2 + Cl + H2O
RT
+ ln A
By comparing the above equation with straight
line equation i.e., y = mx + c
ln k =
N N
- Ea
So,
m (slope) =
- Ea
RT
+ ln A
- Ea
R
C (intercept) = ln A
Sample Paper 12
1
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11. (d) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
hydrogen bonding in alcohols. H is bonded to
the electronegative O atom in alcohols. As a
result, the H atom of one alcohol's OH group
joins forces with the O atom of the second
alcohol's OH group to form a hydrogen bond.
Explanation:
(A) [Co(NCS)(NH3)5](SO3)
Let oxidation state of Co is x.
x – 1 + 5 × 0 = + 2
14. (b)Both (A) and (R) are true but (R) is not the
x=+2+1
correct explanation of (A).
= + 3
Explanation:
(B) [Co(NH3)4Cl2](SO4)
Let oxidation state of Co = x.
Caution
x=4

(C) Na4[Co(S2O3)3]
Let oxidation state of Co = x.
x + 3 × (–2) = – 4
correct explanation of (A).
x=–4+6=+2
Explanation: The magnetic moment is less as
(D) [Co2(CO)8]
the 5f electrons of actinides are less effectively
Let oxidation state of Co = x.
x–8×0=0
x=0
It must be noted that proteins are polymers of
a-amino acids. Therefore, hydrolysis of proteins
yields a-amino acids only.
15. (b)Both (A) and (R) are true but (R) is not the
x–6=–4
and
stabilise the a-helix structure.
x –2=+2
H-bonding
chains are two important mechanisms that
⇒ x + 4 × 0 + 2 × (–1) = +2
⇒
Intrachain
minimising steric hindrance between side
shielded which results in quenching of orbital
contributions , they are strongly paramagnetic
due to presence of unpaired electrons.
12. (b) CrO2Cl2
16. (a)Both (A) and (R) are true and (R) is the
Explanation: This is known as Etard reaction.
The chromyl chloride oxidises the methyl
group to a chromium complex in CS2. This
complex on hydrolysis gives benzaldehyde.
13. (b)Both (A) and (R) are true but (R) is not the
correct explanation of (A).
Explanation: Even though they have the same
molecular formula, ethers are more volatile
than alcohols. This is due to intermolecular
correct explanation of (A).
Explanation: On the basis of the +I group, the
basicity order of amines in the gaseous phase
can be described. Alkyl groups drive electrons
towards
nitrogen
because
they
release
electrons, which increases the availability of
the unshared pair for sharing the basic nature
which increases with increase in number of
alkyl groups.
SECTION - B
18. (A)Acetylation
t=
[R ]
2.303
log o
t
[R]
2.303
100
log
0.01
20
t = 230.3 log 5
t = 160.9 min
2
(log 5 = 0.6990)
acetic
gives
glucose
pentaacetate
which confirms the presence of five –OH
groups. Since it exists as a stable compound,
five –OH groups should be attached to
different carbon atoms
CHO
(CHOH)4
CH4OH
Glucose
CHO
Acetic
anhydride
O
—
k=
with
(CH—O—C—CH3)4
O
—
For first order reaction,
glucose
—
= 0.01 min–1
anhydride
of
—
0.693
1
k=
=
69.3
100
—
0.693
k
—
17. Half life t½ =
CH2—O—C—CH3
Glucose penta-acetate
Chemistry Class XII
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OR
(B) Glucose reduces Fehling's reagent
H—
O
—
—
C
(A)
(CHOH)4
+ 2Cu2+ + 2H2O
—
(CHOH)4
CH2OH
CH2OH
H—C—OH
Gluconic acid
(B)
COOH
Oxidation
CH2OH
— — — —
H—C—OH
HO—C—H
(CHOH)4
—
(CHOH)4
—
C
O
—
—
—
HO—
—
CHO
CH2OH
—
H—C—OH
COOH
Saccharic
acid
+Cu2O + 4H+
H—C—OH
19. H—C—OH
1 Scm–1
100 Sm–1
=1
(unit conversion factor)
CH2OH
Λm =
1000×K
(S cm2
Molarity
Concentration
(M)
K (S m–1)
10–3
1.237 × 10–4
1.237 × 10–4
1000×1.237×10–4
10–2
11.85 × 10–4
11.85 × 10–4
1000×11.85×10–4
2 × 10–2
23.15 × 10–4
23.15 × 10–4
1000×23.15×10–4
5 × 10–2
55.53 × 10–4
55.53 × 10–4
1000×55.53×10–4
10–1
—
Br2 water
—
— — — —
H—C—OH
HO—C—H
COOH
—
—
CHO
K (S cm–1)
c 1/2 (M1/2)
mol–1)
10–3
10–2
2×10–2
5×10–1
= 123.7
0.0316
= 118.5
0.100
= 115.8
0.141
= 111.1
0.224
–4
106.74 × 10–4 106.74 × 10–4 1000×106.74×10 = 106.7
10–1
124
0.316
0.032, 123.7
cm(Scm2 mol–1)
122
120
0.1, 118.5
118
0.14, 115.8
116
114
112
0.22, 111.1
110
108
0.32, 106.7
106
0
0.05
0.1
0.2
0.15
-1 1/2
0.25
0.3
0.35
c1/2 (mol L )
Λº = Intercept on Λm axis = 124.0 S cm2 mol–1, which is obtained by extrapolation to zero concentration.
Sample Paper 12
3
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20. Given: k2 = 8 × 10–2, k1 = 2 × 10–2, T1 = 300 K,
CH3CH2CH3 < CH3OCH3 < CH3CHO <
T2 = 320 K
log
⇒ log
⇒
⇒
k2
k1
8×10−2
2×10−2
log4 =
Ea =
Ea
1 1
=
 − 
2.303R  T1 T2 
(B)
O
Et
EtCOCl
AlCl3
Ea
 1
1 
−
=

2.303 × 8.314  300 320 
OR
Ea
20
×
19.147 300 × 320
(A)
CH3
19.147×0.6021×300×320
20
CH2
OH
P.C.C
CH3CHO
OH
–
Ethanal
Ethanol
CH3
CH
CH2
CHO
OH
3-hydroxy butanal
[Q log4 = 0.6021]
19.147 × 0.6021 × 4800 kJ mol −1
=
1000
⇒
CH3CH2OH
(B)
COOH
COOH
HNO3(conc.)
H2SO4 (conc.), D
Ea = 55.3364 kJ mol–1
CONH2
+NH3
NO2
D
NO2
Benzoic acid
21. (A)CH3CH2CH3 have weak Van der Waals
LiAlH4
forces thus lowest boiling point and in
CH3CHO there are more dipole-dipole
interactions than CH3OCH3, thus having a
greater boiling point. Therefore the order
will be:
CH2OH
CH2
NH2
HNO2
NO2
NO2
m-Nitro benzyl alcohol
SECTION - C
22. (A) Aspirin is formed
COOH
OH
+ (CH3CO) 2O
stability constant more stable the complex.
Hence, (Cu(NH3)4]2+ more stable than
[Cd(NH3)4]2+.
H+
(C)[Ni(en)3]2+ is more stable than (Ni(NH3)6]2+
because en is a bidentate ligand. Hence, it
shows chelation effect which is responsible
for its high stability.
Salicylic acid
COOH
OCOCH3
+ CH3COOH
Acetylsalicylic acid
[Aspirin]
24. (A)Henry’s law: The partial pressure of the
gas in vapour phase (P) is proportional
to the mole fraction of the gas (X) in the
solution.
The pressure in underwater is high, so the
solubility of gases in blood increases. When
the diver comes to surface the pressure
decreases so does the solubility causing
bubbles of nitrogen in blood, to avoid this
situation and maintain the same partial
pressure of nitrogen in underwater too, the
dilution is done.
sodium ethoxide
→ (CH3)2C = CH2
(B) (CH3)3CCl 
2-methylpropene
(C) o-Hydroxybenzaldehyde will be formed
OH O
OH
CHCl3
H
3KOH
(Any two)
23. (A)The stability constant of cyanides complex
is higher than complex with NH3 because
cyanide is a strong field ligand.
(B)The numerical value of the stability
constant is a measure of the stability of
the complex. Greater the magnitude of the
4
Related Theory

The solubility of gas in a liquid is directly proportional
to the partial pressure of the gas present above the
surface of liquid. This is called Henry’s law.
Chemistry Class XII
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(B) 27. tert-butyl bromide reacts with aq. NaOH
P = KH X
by SN1 mechanism because it produces a
Mole fraction of argon in water
X=
=
halide group. The carbocation then reacts
with the nucleophile OH–. The primary halide
6
n-butylbromide, on the other hand, is unable
40×103
to create a stable carbocation, therefore it
X = 1.5 × 10–4
25. (A)
stable carbocation after the cleavage of the
P
KH
KCN
CH3CH2Br
undergoes the SN2 process, which is a oneLiAIH4
CH3CH2C N
Bromoethane
'A'
step substitution involving OH– attack and
simultaneous leaving X– to form n-butyl
alcohol.
Propane nitrile
CH3
HNO2
0.º C
CH3CH2CH2NH2
CH3CH2CH2OH
'B'
CH3
CH3
Br
C
CH3
Ionization
–
Br
slow
'C'
1-propanamine
Propan - 1- ol
CH3
CH3
CH3COOH
Ethanoic acid
CH3
C
'A'
NH2
CHCl3+ alc. KOH
0.º C
fast
OH
CH3
CH3CH2CH2
C
'C'
Methyl carbylamine
HO
Br
H H
reaction proceeds through the SN1 mechanism.
Since the SN1 mechanism is favoured by 3°
alkyl halides. ‘A’ should be
H
CH2CH2CH3
HO
C
H H
n-Butyl alcohol
28. (A) Given, CFSE of [CoCl6]3– is 18000 cm–1
Then, CFSE of tetrahedral complex [CoCl4]–
is given by
CH3
Br
Transition state
C4H9Br + KOH(aq) → C4H9OH + KBr
depends upon the concentration of ‘A’ only, the
–
C
H
HO
n-Butyl bromide
alcohol.
In the case of compound ‘A’ rate of reaction
CH2CH2CH3
+
26. Alkyl halide on reaction with aq. KOH gives
Dt =
4
× 18000 cm−1
9
= 8000 cm–1
CH3
C
C
tert-Butyl alcohol
–
CH3
CH3
CH3
CH3
CH3N C
'B'
Methanamine
CH3
–
OH
C
NaOH + Br2
Ethanamide
CH3NH2
C
tert-Butyl carbocation
(stable)
(B)
NH3
CH3
(B)The given complex is optically active,
Br
so it must be octahedral in nature. This
In the case of compound ‘B’ which is an
means that ‘A’ must be a bidentate ligand.
optically active isomer of ‘A’ rate of reaction
The complex can have the structure
depends upon the concentration of ‘B’ as well
[Pt(en)2Cl2]2+. Only its cis isomer can show
as KOH. Therefore, the reaction occurs by
optical isomerism.
SN2 mechanism which is favoured by 1°∘ and
2°∘ alkyl halides.
The compound 'B' is
CH3
CH
CH
Br
CH3
]2+
en
Cl
Pl
en
en
en
Cl
Pt
]2+
Cl
Cl
Non- Superimposable isomers of [Pt(en)2Cl2]2+
Sample Paper 12
5
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SECTION - D
29. The melting point of ice is the freezing point of
water. We can use the depression in freezing
point property in this case.
(A) 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 on this case.
OR
Two sets of reading help to avoid error in
data collection and give more objective
data.
(B) ∆Tf (glucose) = 1×K f ×
0.6×1000
180×10
...(i)
(C) Depression in freezing point is directly
proportional to molality (mass of solute
when the amount of solvent remains same)
0.3 g depression is 1.9°C
0.6 g depression is 3.8°C
1.2 g depression is 3.8 × 2 = 7.6°C
30. (A) A = 100 and T = 100
C = 150 and G = 150
Total nucleotides = 100 + 100 + 150 + 150
= 500
(B) They studied the nucleotide composition of
DNA. It was the same so they concluded
that the samples belong to same species.
(C) A = T = 20%
But G is not equal to C so double helix is
ruled out.
0.6×1000
∆Tf (NaCl) = 2×K f ×
58.5×10
3.8 = 2×K f ×
The bases pairs are ATGC and not AUGC so
it is not RNA.
0.6×1000
58.5×10
The virus is a single helix DNA virus.
OR
Divide equation (ii) by (ii)
According to Chargaff's rule, all double
helix DNA will have the same amount of
A and T as well as C will be same amount
as G. If this is not the case then the helix is
single stranded.
∆T (glucose)
58.5
f
=
2 × 180
3.8
∆Tf (glucose) = 0.62
Freezing point or Melting point = – 0.62 °C
SECTION - E
31. (A) Cu2+ oxidizes iodide ion to iodine.
(B) The low value for V is related to the stability
of V2+ (half-filled t2g level)
(C) Permanganate titrations in presence of
hydrochloric acid are unsatisfactory since
(G) (NH4)2Cr2O7
hydrochloric acid is oxidised to chlorine.
(D) The d-orbital is full with ten electrons and
shield the electrons present in the higher
s-orbital to a greater extent resulting in
increase in size.
(E) The
chromates
interconvertible
and
in
dichromates
aqueous
are
solution
depending upon pH of the solution.
Increasing the pH (in basic solution) of
dichromate ions a colour change from
orange to yellow is observed as dichromate
ions change to chromate ions.
6
(F) The variability of oxidation state of
transition elements is due to incompletely
filled d-orbitals and presence of unpaired
electrons, i.e., (ns) and (n – 1) d electrons
have approximate equal energies.
Heat
N2 + H2O + Cr2O3
Nitrogen
Chromic oxide
32. (A)The process that involves the spontaneous
transfer of electrons between substances
and are used in batteries is Oxidationreduction (redox) reaction.
(B) KCl(aq) → K+(aq) + Cl–(aq)
Cathode: H2O(l) + e– → ½ H2(g) + OH–(aq)
Anode: Cl–(aq) → ½ Cl2(aq) + e–
Net reaction:
KCl(aq) + H2O(l) → K+(aq) + OH–(aq) + ½ H2(g)
+ ½ Cl2(g)
Chemistry Class XII
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(C) Given, potential of hydrogen gas electrode
= −0.59 V
Electrode reaction: H+ + e– → 0.5 H2
Applying Nernst equation,
E
H+ 


 H2 
E
E
∴∴
H 


 H2 
+
H+ 


 H2 
n
[H2]
−0.59
−0.59
pH
= E°  H+  –


 H2 
[H ]1/2
0.059
log 2 +
n
[H ]
=0V
33. (A) (a) is an alkene CH(CH3) = C(CH3)2
(b) is an aldehyde with –CH3 group
CH3CHO
(c) is a methyl ketone O = C(CH3)2
(B) CH3CHO + [Ag(NH3)2]+ + OH– →
CH3COO– + Ag + NH3 + H2O
(C) CH3COCH3 + NaOH + I2 → CHI3
+ CH3COONa
= –0.59 V
=1
= 1 bar
= 0 – 0.059 ( – log [H+] )
= −0.059pH
= 10
OR
(A) “A” is copper, metals are conductors thus
have high value of conductivity.
(B) Mg2+ + 2e–  Mg
1 mole of magnesium ions gains two moles
of electrons or 2F to form 1 mole of Mg
24 g Mg requires 2 F electricity
4.8 g Mg requires 2 × 4.8/24 = 0.4 F =
0.4 × 96500 = 38600C
2+
–
Ca + 2e → Ca
2 F electricity is required to produce 1 mole
= 40 g Ca
0.4 F electricity will produce 8 g Ca
(C) F = 96500C, n = 2,
Sn2+(aq) + 2e– → Sn(s) ; E° = – 0.14V
Cu2+(aq) + e– → Cu+(aq), ; E° = 0.15 V
E°cell = E°cathode – E°anode
= 0.15 – (–0.14) = 0.29V
ΔG° = –nFE°cell
= –2 x 96500 x 0.29 = –55970 J/mol
(D) CH3COCH3 + CH3CHO
↓ Ba(OH)2
(CH3)2C(OH)CH2COCH3 + CH3CH(OH)CH2CHO
+ (CH3)2C(OH)CH2CHO
+ CH3CH(OH)CH2COCH3
↓ heat
(CH3)2C = CHCOCH3 + CH3CH = CHCHO +
(CH3)2C = CHCHO
+ CH3CH = CHCOCH3
OR
(A) (a): CCl3COOH (b): CH3COOH
(i) Red P/Cl
2→
(B) CH3COOH 
CCl3COOH,
(ii) H O
2
Hell Volhard Zelinsky reaction
(C) CH3COCl
HO
2
→
CH3COOH
(D) (a) will be more acidic due to presence of
3 Cl groups (electron withdrawing groups)
which increase acidity of carboxylic acid.
(i)NaOH,CaO (ii)heat
→ CH4 + Na2CO3
(E) CH3COOH
Sample Paper 12
7