ISC Class 12 Sample Paper 2027 Chemistry Paper 1 (Theory)

FOR ISC CLASS 12 EXAM PREPARATION

ISC Class 12 2027
Sample Paper ·
Chemistry Paper 1
(Theory)
EXAM YEAR TYPE SUBJECT

ISC Class 12 2027 Sample Paper Chemistry Paper 1 (Theory)

Notes · Sample Papers · Previous Year Papers · Mock Tests

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CHEMISTRY
PAPER 1
(THEORY)
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Maximum Marks: 70
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Time Allotted: Three Hours
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Reading Time: Additional Fifteen Minutes
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Instructions
1. You are allowed an additional fifteen minutes for only reading the
question paper.
2. You must NOT start writing during the reading time.
3. This question paper has 10 printed pages.
4. It is divided into four sections and has twenty one questions in all.
5. Answer all questions.
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6. Section A has fourteen subparts. Each question carries 1 mark.
7. While attempting Multiple Choice Questions in Section A, you are
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required to write only ONE option as the answer.
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8. Section B has ten questions. Each question carries 2 marks.
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9. Section C has seven questions. Each question carries 3 marks.
10. Section D has three questions. Each question carries 5 marks.
11. Internal choices have been provided in two questions each in Sections B,
C and D.
12. The intended marks for questions are given in brackets [ ].
13. All working, including rough work, should be done on the same sheet as,
and adjacent to the rest of the answer.
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14. Balanced equations must be given wherever possible and diagrams where
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they are helpful.
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15. When solving numerical problems, all essential workings must be shown.
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las 16. In working out problems, use the following data:
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ag Gas constant R = 1·987cal deg-1mol-1 = 8·314JK-1mol-1
=0·0821dm3atmK-1mol-1, 1 l atm = 1dm3atm = 101·3 J.
1 faraday = 96500 coulombs, Avogadro’s number = 6·023×1023.

Instruction to Supervising Examiner
Kindly read aloud the Instructions given above to all the candidates present in
the examination hall.

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a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 1 of 21

Note: The Specimen Question Paper in the subject provides a realistic format of the
Board Examination Question Paper and should be used as a practice tool. The questions for the
Board Examination can be set from any part of the syllabus, though the format of the Board
Examination Question Paper will remain the same as that of the Specimen Question Paper.
The weightage allocated to various topics, as given in the syllabus, will be strictly adhered to.

SECTION A - 14 MARKS

Question 1
(A) Fill in the blanks by choosing the appropriate word(s) from those given in the [4×1]
brackets.
[methyl alcohol, ethyl alcohol, paired, one, Cannizzaro, d-d, salicylaldehyde,
Perkin’s, unpaired, f-f, salicylic acid, Kolbe’s, two, 193000, 96500,
Reimer-Tiemann]
(i) Formaldehyde reacts with conc. alkali to give sodium formate and
_________ due to the absence of α - hydrogen atom. The reaction is
known as _________ reaction.
(Recall)
(ii) Electrochemical equivalent (Z) is the quantity of substance deposited or
discharged by a charge of ________ coulomb, whereas the equivalent
conductance (E) is the amount of substance deposited or discharged by a
charge of _________ coulombs. (Understand)
(iii) The colour of transition metal ions is due to ________ electron(s) in
d-subshell and _________ transition. (Analyse)
(iv) The product formed when sodium phenoxide is treated with CO2 at 400K,
under 4-7 atm pressure and then acidified with dilute HCl is _________.
The reaction is known as _______ reaction. (Recall)
(B) Select and write the correct alternative from the choices given below. [7×1]
(i) The correct order of freezing point of 0·5M solution of urea, AlCl3, KCl
and MgCl2 is: (Evaluate)
(a) AlCl3 > MgCl2 > KCl > urea
(b) Urea > KCl > MgCl2 > AlCl3
(c) MgCl2 > KCl > AlCl3 > urea
(d) KCl > MgCl2 > AlCl3 > urea
(ii) Glucose molecules react with (x) number of molecules of phenyl
hydrazine to yield glucosazone. The melting point of glucosazone is
2060C. The value of (x) is: (Recall)
(a) one
(b) two
(c) three
(d) four

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2 ISC SPECIMEN QUESTION PAPER 2027

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a complexes with ammonia.
(iii) Cobalt (III) chloride forms several octahedral
Which of the following will NOT give a test for chloride ions with silver
nitrate solution at 250C? (Understand)
(a) CoCl3.6NH3
(b) CoCl3.5NH3
(c) CoCl3.4NH3

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(d) CoCl3.3NH3

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(iv) The following data was obtained for a hypothetical reaction
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2A+B →
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2AB

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l as (mol L-1) (mol L-1) (mol L-1 sec-1)
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2
0∙5
1∙0
0∙5
0∙5
1∙8 × 10-4
3∙6 × 10-4
3 1∙0 1∙0 3∙6 × 10-4
The rate law expression for this reaction is: (Apply)
1 1
(a) rate = k[A] [B]
(b) rate = k[A]1[B]0
(c) rate = k[A]2[B]1
(d) rate = k[A]0[B]1
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(v) Which of the following reactions will NOT give primary amine?

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(Understand)
(a) C2H5NC →
𝐿𝑖𝐴𝑙𝐻4

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_________
(b) C2H5CN →
𝐿𝑖𝐴𝑙𝐻4 g
a _________
𝐵𝑟2 /𝐾𝑂𝐻
(c) C2H5CONH2 → _________
𝐵𝑟2 /𝐾𝑂𝐻
(d)CH3CONH2 → _________
(vi) Given below are two statements marked Assertion and Reason. Read the
two statements carefully and select the correct option.
Assertion: When an ether containing a primary alkyl group and tertiary
alkyl group is treated with hydrogen iodide, tertiary alkyl
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iodide is formed.

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Reason: The reaction takes place via SN1 mechanism, and the more stable
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tertiary carbocation is formed easily which combines with
s e iodide ion to form tertiary alkyl iodide. (Analyse)
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g la (a) Both Assertion and Reason are true, and Reason is the correct a
a explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

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(vii) Given below are two statements marked Assertion and Reason. Read the
two statements carefully and select the correct option.
Assertion: Magnesium metal can liberate H2 gas from an aqueous
solution of HCl.
Reason: Magnesium has positive value of standard reduction electrode
potential. (Analyse)
(a) Both Assertion and Reason are true, and Reason is the correct
explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.
(C) Read the passage carefully and answer the questions that follow. [3×1]
The hydroxy derivatives of aliphatic hydrocarbons are called alcohols. Ethanol
is perhaps the most important aliphatic alcohol. During COVID-19 pandemic
alcohol-based sanitizers became an essential protective tool worldwide, reducing
the spread of the virus. Nowadays, it is also used as a substitute for petrol in
internal combustion engines.
The boiling points of alcohols are considerably higher than those of the
corresponding ethers due to the presence of intermolecular hydrogen bonding in
alcohols and it increases with increase in molecular mass.
For isomeric alcohols the boiling point follows the order:
Primary (1o) alcohol > Secondary (2o) alcohol > Tertiary (3o) alcohol, which is
evident from the following data:

Compound Propan-1-ol Butan-1-ol Butan-2-ol 2-methyl-propan-2-ol
Boiling 370K 391K 371K 356K
point
Aliphatic alcohols behave as weak acids and ionize to a small extent. Their acidic
strength is even less than that of both water and phenol. Phenols are hydroxy
derivatives of arenes and have varied application in manufacturing of dyes,
plastics, drugs, explosives, etc. At different concentrations, phenols are used both
as a disinfectant and an antiseptic.
(i) The boiling point of 2-methyl-propan-2-ol is much lower than that of
butan-1-ol, although both have the same molecular formula. Give a reason
to explain such observation. (Evaluate)
(ii) Although alcohols are weaker acids than water, phenol is more acidic than
alcohols. Justify. (Evaluate)
(iii) In what form (i.e. concentration), does phenol act as a disinfectant or an
antiseptic? (Recall)

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4 ISC SPECIMEN QUESTION PAPER 2027

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SECTION B – 20 MARKS
Question 2 [2]
The standard reduction electrode potential (Eo) of some metals is given below:
Metal Eo
Ag+/Ag +0∙80V
Mg2+ / Mg ­2∙37V
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Li+/Li ­3∙05V
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Cu2+/Cu +0∙34V
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Fe2+/Fe ­0∙44V
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Arrange these metals in the increasing order of their reducing power. Justify the order. g
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Question 3 [2]
An alkene [A] with molecular formula, C5H10 on ozonolysis and subsequent hydrolysis
gives a mixture of two compounds, [B] and [C]. Compound [B] gives positive Fehling’s
test and also reacts with iodine and NaOH solution. Compound [C] does not give
Fehling’s test but forms iodoform.
Identify compounds [A], [B] and [C]. Write the iodoform reaction either with compound
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[B] or [C]. (Understand)

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Question 4
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[Cr(NH3)6]3+ ag
(i) Answer the following questions with respect to the coordination complex ion

(a) What is the hybridisation state of the central metal atom? (Understand)
(b) Comment on the magnetic nature of the complex ion. (Analyse)
OR
(ii) Observe the diagram below of splitting of d-orbital in octahedral field and answer
the following questions.

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∆ or 10Dqa
+0.60∆o
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o

g
-0.40∆o
a
a

(a) Identify the electronic configuration of d5 ion, if ∆o < P. (Understand)
(b) Comment on the relationship between crystal field stabilisation energy (∆o)
and the strength of the ligand. (Analyse)

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Question 5 [2]
Rohan, a patient of thyroid disorder, was given certain amount of radioactive Iodine-131
as a part of his treatment.
What amount of Iodine-131 will be left in Rohan’s body after 48 days, if it is known that
the half-life period of radioactive iodine-131 is 12 days? Assume that no amount of the
isotope was eliminated through biological processes. (Apply)

Question 6 [2]
How can the following conversions be carried out? (Write the chemical equations):
(i) Aniline to chlorobenzene (Recall)
(ii) Chloroethane to n-butane (Recall)

Question 7 [2]
Substantiate the following statements with reasons.
(i) Cobalt is attracted in a magnetic field whereas Zinc is not. (Understand)
(Atomic number of Co = 27 and Zn =30)
(ii) The melting points and boiling points of Zn, Cd and Hg are low. (Understand)
(Atomic number of Zn = 30, Cd = 48 and Hg = 80)

Question 8 [2]
Write the Nernst equation and calculate the emf of the following cell at 298K.
Mg(s) / Mg2+ (1x10-3M) // Cu2+(1x10-4M) / Cu(s)
Given: E ° (Mg2+ /Mg) = -2∙373V, E ° (Cu2+ /Cu) = +0∙337V (Apply)

Question 9 [2]
(i) Illustrate the following reactions with an example in each case:
(a) Coupling reaction (Recall)
(b) Balz-Schiemann reaction (Recall)
OR
(ii) An organic compound [A] with molecular formula C3H7NO forms compound [B]
on heating with Br2 and KOH. Compound [B] on heating with CHCl3 and alcoholic
KOH produces a pungent smelling compound [C]. Compound [B] on reacting with
C6H5SO2Cl forms compound [D] which is soluble in alkali.
Write the structures of compounds [A], [B], [C] and [D]. (Understand)

Question 10 [2]
In general, it is observed that the rate of a reaction becomes double with every 10oC rise
in temperature. If this generalisation holds true for a reaction, calculate the value of
activation energy when temperature changes from 295 K to 305 K. (Apply)

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6 ISC SPECIMEN QUESTION PAPER 2027

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Question 11 a [2]
The students of Class XII were on a study trip. They visited a nearby lake and took some
water samples which were rich in sodium chloride salt. The boiling point of lake water
was found to be 373∙032 K.
If 500 g of lake water sample was used which contained 0∙45 g of NaCl, what is the
observed molecular weight of NaCl in the lake sample? Assume that NaCl ionises
completely in lake water.
(Given: kb of water = 0∙52K kg mol-1, boiling point of pure water = 373 K) (Analyse)
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Question 12 se g l a
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[3]

agchemistry teacher and asked to perform different reactions.
Two students
by their
Rehan and Ananya were given the same compound, alkyl halide (C H Cl) 2 5

•
Rehan was instructed to use alcoholic KCN as a reagent and Ananya was
instructed to use alcoholic AgCN.
• Rehan observed a compound was formed that could be converted to a carboxylic
acid on hydrolysis but Ananya could not obtain the same acid.
• The teacher explained that the difference was due to the bonding behaviour of
cyanide ions in the reagents.

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Write the chemical equations of the reactions and the IUPAC names of the compounds

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formed by Rehan and Ananya. (Understand)

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Question 13
las [3]

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(i) For a given first order reaction, it takes 5 minutes for the initial concentration of
0∙6 moles litre-1 to become 0∙2 moles litre-1. Calculate the rate constant for this
reaction. (Apply)
(ii) Find the overall order of the reactions which have the following rate law
expression: (Understand)
(a) Rate = k[A] 1/2[B] 3/2
(b) Rate = k[A]3/2[B]-1
(c) Rate = k[A]1[B]2
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c. oQuestion 14 m .co
s e [3]

s em (i) Compounds [X] and [Y] are functional isomers of each othera with molecular
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a (Understand)
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(a) Draw the structural formula of both the isomers.
(b) Which of the compounds will have a lower boiling point and why?
(Understand)
(ii) Methanol does not give yellow precipitate of iodoform when heated with iodine
and alkali, but ethanol gives iodoform test with the same reagents. Explain.
(Understand)
OR

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(iii) How will the following be converted? (Give chemical equations) (Recall)
(a) Diethyl ether from sodium ethoxide
(b) Propan-2-ol from Grignard’s reagent
(c) Phenol from Cumene

Question 15 [3]
(i) Complete and balance the following chemical equation.
KMnO4 + H2SO4 + H2S → _____ + _____ + _____ + _____ (Understand)
(ii) Why do transition metals form alloys? (Understand)
(iii) Potassium dichromate (K2Cr2O7) acts as a powerful oxidising agent in acidic
medium. Explain. (Analyse)

Question 16 [3]
(i) Identify the non-reducing sugar which on hydrolysis gives two reducing
monosaccharides. (Understand)
(ii) Which structure of protein is normally unaffected during denaturation of protein?
(Recall)
(iii) Name the vitamins whose deficiency cause the following diseases. (Recall)
(a) Rickets
(b) Night blindness

Question 17 [3]
(i) Three electrolytic cells (X), (Y) and (Z) containing solutions of AgNO3, CuSO4
and ZnSO4 respectively are connected in series. A steady current of 1∙5 ampere is
passed through these electrolytic cells until 1∙45g of silver is deposited at the
cathode of cell (X).
(Atomic weight of Ag =108, Cu = 63∙5 and Zn = 65∙3)
(a) How much of charge is given to the electrolyte solution? (Apply)
(b) What mass of copper and zinc is deposited at the respective cathode?
(Analyse)
OR
(ii) The molar conductivities at infinite dilution for NaI, CH3COONa and
(CH3COO)2Mg are 126∙9, 91∙0 and 187∙8 ohm-1 cm2 mol-1 respectively at 25oC.
Find out the molar conductivity of MgI2 at infinite dilution by using Kohlrausch’s
law of independent migration of ions. (Apply)

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8 ISC SPECIMEN QUESTION PAPER 2027

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Question 18 a [3]
Write the chemical equations for the following conversions: (Recall)
(i) Nitrobenzene to benzene
(ii) Aniline to bromobenzene
(iii) Methyl cyanide to ethylamine

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SECTION D – 15 MARKS
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emcompound [A] with molecular formula C Cl O H is obtained when glas
Question 19 [5]

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a [B] reacts with red phosphorous and Cl . The organic compound [B] a
(i) An organic 2 3 2

cang
compound
afollowed
2
be obtained on the reaction of methyl magnesium bromide with carbon dioxide
by acid hydrolysis. Upon partial reduction, compound [B] forms
compound [C] with molecular formula C2H4O, which gives yellow precipitate of
iodoform on heating with iodine in presence of sodium hydroxide. Compound [C]
also reacts with dilute NaOH to form compound [D].
(a) Identify the compounds [A], [B], [C] and [D]. (Understand)
(b) Write down the reaction for the formation of [A] from [B]. What is the
reaction called?
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(c) Write the chemical test to distinguish between compound [C] and acetone.
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(d) Which will be more acidic, compound [A] or [B]? Why? (Evaluate)

(ii)
aOR
Identify the compounds [A], [B] and [C] in the following reactions: (Analyse)

SnCl2/HCl CH3MgBr Acidified
(a) CH3C≡N [A] [B] [C]
H2O H2O K2Cr2O7

H2O [O] PCl5
(b) H – C ≡ C – H [A] [B] [C]
2+
Hg / H2SO4 K2Cr2O7 + H2SO4

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(iii) Write chemical equations to convert the following: om
c. (Recall)
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s e (b) Benzaldehyde to cinnamic acid
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a Question 20 [5]
(i) The compound [PtCl2(NH3)2] also known as cis-platin, is used in chemotherapy.
It also demonstrates how coordination geometry impacts medical science.
(a) Write its IUPAC name and the type of isomerism it exhibits. (Recall)
(b) What is the geometry of the compound? (Recall)
(ii) What type of structural isomers are [Pt(NH3)3Br]NO3 and [Pt(NH3)3(NO3)]Br?

m .
How can they be distinguished by using a chemical test? (Analyse)

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(iii) A coordination compound CoCl3.4H2O precipitates silver chloride when treated
with silver nitrate. The molar conductance of its solution corresponds to a total of
two ions. Write the structural formula of the compound and name it. (Evaluate)

Question 21 [5]
(i) Answer the following:
(a) 300 ml of aqueous solution of protein contains 1∙85g of protein, the osmotic
pressure of the solution at 25oC is found to be 3∙05 × 10-3atm. Calculate the
molar mass of the protein. (Apply)
(b) Phenol (C6H5OH) associates in benzene to form a dimer. A solution of 2·5g
of phenol in 120g of benzene lowers its freezing point by 0∙85K.
Calculate the degree of association of phenol.
(kf for benzene = 5∙12 K kg mol-1, molecular mass of phenol = 94 g mol-1)
(Apply)
(c) The molecular weight of potassium chloride and sucrose is determined by
the depression of freezing point method. Compared to their theoretical
molecular weight, what will be their observed molecular weights when
determined by the above method? Justify your answer. (Evaluate)
OR
(ii) (a) Arrange the following aqueous solutions in increasing order of osmotic
pressure. Give reasons in support of your answer. (Analyse)
(1) 6∙0g per litre of urea solution (molecular weight of urea = 60g mol-1)
(2) 72∙0g per litre of glucose solution (molecular weight of
glucose = 180g mol-1)
(3) 5∙85g per litre of sodium chloride solution (molecular weight of
sodium chloride = 58∙5g mol-1)
(b) Calculate the mole fraction (x) of ethanol and water if 92g of ethanol is
dissolved in 540g of water.
(Atomic mass of C = 12, O = 16 and H = 1) (Apply)
(c) What type of azeotropic mixture will be formed by a solution of chloroform
and acetone? Justify your answer based on strength of intermolecular
interactions that develops in the solution. (Evaluate)

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10 ISC SPECIMEN QUESTION PAPER 2027

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CHEMISTRY PAPER
(THEORY)
ANSWER KEY
SECTION A - 14 MARKS

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Question 1
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(A)
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(i)
s em
Methyl alcohol, Cannizzaro
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(ii) One, ag
a
(iii) Unpaired, d-d
(iv) Salicylic acid, Kolbe’s
(B) [7×1]
(i) (b) Urea > KCl > MgCl2 > AlCl3
(ii) (c) three
(iii) (d) CoCl3.3NH3
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(iv) (b) k[A]1[B]0 .co
LiAlH4
s em
laare true, and Reason is the correct explanation
(v) (a) C2H5NC → __________

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(vi) (a) Both Assertion and Reason
for Assertion.
(vii) (c) Assertion is true and Reason is false.
(C) [3×1]
(i) The boiling point of 2-methyl propan-2-ol is much lower than that of
butan-1-ol. This is due to the fact that with branching the surface area
decreases. Consequently, boiling point also decreases.
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(ii) The greater acidic character of phenol as compared to alcohols can be
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explained on the basis of resonance. The oxygen attracts the electron pair of
O-H bond strongly towards itself, weakens the O-H bond and therefore
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a
A 0·2% of phenol acts as an antiseptic, whereas its 1% solution is used as ag
ag disinfectant.

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SECTION B – 20 MARKS

Question 2 [2]
Increasing order of reducing power of metals is Ag < Cu < Fe < Mg < Li
1
Reducing power =
Reduction potential

i.e., as the reduction potential (value of metal) decreases, reducing power of metal increases.

Question 3 [2]
CH3
|
[A] CH3–CH=C–CH3 or 2-methyl but-2-ene

[B] CH3CHO or acetaldehyde or ethanol

[C] CH3–C=O or propanone or acetone
|
CH3
Iodoform reaction with compound [B]
Δ
CH3CHO + 3I2 + 4NaOH → CHI3 + 3NaI + HCOONa + 3H2O

Or
Iodoform reaction with compound [C]
Δ
CH3–C=O + 3I2 + 4NaOH → CHI3 + 3NaI + CH3COONa + 3H2O
|
CH3

Question 4 [2]
(i) For the coordination complex ion [Cr(NH3)6]3+
(a) Type of hybridisation; d2sp3
(b) Magnetic behaviour; paramagnetic
Reason: There are three unpaired electrons.
OR
(ii) (a) t 32g eg 2

(b) If the value of CFSE (Δ0) is more, ligand is strong. (i.e. low spin complex)

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2 ISC SPECIMEN ANSWER KEY 2027

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Question 5 a [2]
Given: t1/2 = 12 days, t = 48 days
Total time 48
Number of half lives = = =4
Half life period 12

Amount left after 4 half lives
1 4
A = Ao x ( )
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2

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1
A = Ao x
16
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16
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i.e. ; of the initial amount
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a s ag [2]
Question 6 gl
a
(i) Aniline to chlorobenzene
𝐶𝑢2 𝐶𝑙2 /𝐻𝐶𝑙
C6H5N2Cl → C6H5Cl + N2
or
𝐶𝑢/𝐻𝐶𝑙
C6H5N2Cl → C6H5Cl + N2

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(ii) Chloroethane to n-butane

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𝑒𝑡ℎ𝑒𝑟
C2H5Cl + 2Na + ClC2H5 → C2H5–C2H5 + 2NaCl
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n-butane
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Question 7 ag [2]
(i) Cobalt has d7 electrons in the outer shell and therefore, has three unpaired electrons.
Therefore, it will be attracted in a magnetic field.
Zinc has completely filled d10 configuration and therefore, will not be attracted by the
magnetic field.
(ii) In Zn, Cd and Hg, all the electrons in d-subshell are paired. Hence, the metallic bonds
present in them are weak. Therefore, they have low melting and boiling points.
m
m .co
.co
Question 8
s e m [2]

s emE = E – log [Mg2+
(aq) ]
g l a
a
0∙059

l a
o
cell (Nernst Equation)
cell

ag
2+ ]
n [𝐶𝑢(aq)
E =Eo
–E
cell
o
cathode
o
anode
= 0∙337V – (–2∙373 V)
Eocell = + 2∙71 V
0.059 1 x 10−3
Ecell = 2∙71 – 2 log1 x 10−4 or
= 2∙71 – 0∙0295log10 or
= 2∙71 – 0∙0295 x 1
Ecell = 2∙6805 V

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. c s e
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e l a
l a s 3 ISC SPECIMEN ANSWER KEY 2027
a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 13 of 21

Question 9 [2]
(i) (a) Coupling reaction
alkali
N2+ . Cl− + OH →
273 – 278K
N=N OH + HCl
p-hydroxyazobenzene
(or any other correct example)
(b) Balz-Schiemann reaction
N2+ . Cl− N2+ . BF4− F
273 – 278K heat
+ HBF4
Fluorobenzene
(or any other correct example)
OR
(ii) [A] C2H5CONH2
[B] C2H5NH2
[C] C2H5N≡C
[D] C6H5SO2–N–C2H5
|
H

Question 10 [2]
Given: T1 = 295K, T2 = 305K
k
log 2 = 2
k1
k2 a E 1 1
logk = 2∙303R [ − ] or
1 T1 T2
E 1 1
log2 = 2∙303 xa8∙314 [ − ] or
295 305
a E 10
0·3010 = 19∙147 x or
295 x 305
0∙3010 x 19∙147 x 295 x 305
Ea =
10
= 151854∙8 J K-1 mol-1 or
Ea = 151∙854 kJ mol-1

Question 11 [2]
Given: ΔTb = 0∙032 K, w = 0∙45g, W = 500g
Kb = 0∙52 K kg mol-1, NaCl → Na+ + Cl- , i = 2
i x 1000 x kb x w
M(obs) =
ΔTb x W
2 x 1000 x 0∙52 x 0∙45
M(obs) = 0∙032 x 500
M(obs) = 29∙25 g mol-1

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4 ISC SPECIMEN ANSWER KEY 2027

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 14 of 21

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a
SECTION C – 21 MARKS
Question 12 [3]
Rehan’s Reaction:
C2H5Cl + KCN → C2H5CN + KCl
(alc) Propane nitrile
Ananya’s Reaction:
C2H5Cl + AgCN →
m
.co
C2H5NC + AgCl
m
.co
(alc) Ethyl carbylamine
e m
e m l a s
Question 13
l as ag [3]

ag
(i) Given; t = 5 minutes, a = 0∙6 moles litre-1
(a – x) = 0∙2 moles litre-1
2∙303 a
K= log10
t (a−x)
2∙303 0∙6
= log10 (0∙2)
5
2∙303
= log 3
5
2∙303
= x 0 ∙ 477
t
m
.co
K = 0∙2197 min-1

m
(ii) The overall order of the following reactions are as follows:
(a) Rate = k[A]1/2[B]3/2
s e
1 3
overall order of reaction = 2 + 2 = 2
g la
(b) Rate = k[A]3/2[B]-1
a
3 1
overall order of reaction = – 1 =
2 2
1 2
(c) Rate = k[A] [B]
overall order of reaction = 1 + 2 = 3

Question 14 [3]

m
.co
(i) (a) Functional isomers
m
c. o (b) Compound [Y] will have lower boiling point as it is an ether which mcannot form
[X] = CH3CH2OH and [Y] = CH3-O-CH3

s e
s em associated molecules by intermolecular hydrogen bond.
g l a
la (ii) a
ag
OH
|
Methanol does not have CH3-CH- group and therefore, does not give iodoform test
Δ
CH3OH + I2 + NaOH → No reaction
OH
|
Ethanol contains CH3-CH- group give iodoform test
Δ

om .
CH2CH2OH + I2 + NaOH → CHI3 (Iodoform)

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e l as
l a s 5 ISC SPECIMEN ANSWER KEY 2027
a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 15 of 21

OR
(iii) (a) Diethyl ether from sodium ethoxide
330K
C2H5ONa + BrC2H5 → C2H5-O-C2H5 + NaBr
Diethyl ether
(b) Propan-2-ol from Grignard’s reagent
H H H
dry ether H2 O,H+
CH3-C=O + CH3MgBr → CH3-C-OMgBr → CH3-C-OH
-Mg(OH)Br
CH3 CH3
Propan-2-ol
(c) Phenol from Cumene
CH3
CH3-CH-CH3 CH3-C-O-O-H OH
H+ ,H2 O
+ O2 → → + CH3CO-CH3

Cumene Phenol

Question 15 [3]
(i) 2KMnO4 + 3H2SO4 + 5H2S → K2SO4+2MnSO4+8H2O+5S
(ii) Atoms of transition metal can easily take place in the crystal lattice of another metal in
the molten state and are miscible with each other and thus form alloys.
(iii) Potassium dichromate (K2Cr2O7) acts as a powerful oxidising agent in acidic medium.
In the presence of dilute sulphuric acid, K2Cr2O7 liberates nascent oxygen and
therefore, acts as an oxidising agent.
Or
K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O + 3[O]

Question 16 [3]
(i) Sucrose
(ii) Primary structure
(iii) (a) Rickets – Vitamn D
(b) Night blindness – Vitamin A

Question 17 [3]
(i) Cell [X] contains AgNO3
The reaction may be represented as follows
AgNO3 ⇔ Ag+ + NO3-
Ag+ + e- ⇔ Ag (at cathode)
(a) 1 mole or 108 g of Ag is deposited by 96500 coulombs
96500 x 1∙45
1∙45 g of Ag is deposited by = 1295∙6 coulombs
108

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6 ISC SPECIMEN ANSWER KEY 2027

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(b) In cell [Y] the electrode reaction is: a
Cu2+ + 2e- → Cu
2 x 96500 coulombs of current will deposit 63∙5 g of Cu
63∙5 x 1295∙6
1295∙6 coulombs of current will deposit = 0∙426 g of Cu
2 x 96500

In cell [Z] the electrode reaction is:
Zn2+ + 2e- → m
.co
Zn

o m m
c
2 x 96500 coulombs of current will deposit 65∙3 g of Zn

m . of current will deposit s e
a
65.x 1295∙6
e
1295∙6 coulombs
s
= 0∙438 g of Zn
l
ag
2 x 96500

g l a OR
a ∧ (NaI) = 126.9 ohm cm mol
(ii) Given; o
m
-1 2 -1

∧om (CH3COONa) = 91∙0 ohm-1 cm2 mol-1
∧om (CH3COO)2Mg = 126∙9 ohm-1 cm2 mol-1
The molar conductivity at infinite dilution for MgI2 may be calculated as:
∧om (MgI2) = ∧om (CH3COO)2Mg + 2∧om (NaI) – 2∧om (CH3COONa)
= 187∙8 + 2(126∙9) – 2(91∙0)
m
= 259∙6 ohm-1 cm2 mol-1
m .co
s e
g la
Question 18
(i) Nitrobenzene to benzene
a [3]

Sn/HCl HNO2 +HCl H3 PO2 +H2 O
C6H5NO2 + 6[H] → C6H5NH2 → C6H5N2+ Cl- → C6H6
273 – 278K
(or any other correct reaction)
(ii) Aniline to bromobenzene
NH2 N2+ Cl- Br

m
+HNO2 +HCl Cu2 Br2 +HBr

m
→
273 – 278K
→ + N2
c o
(or any other correct.reaction)
. c o em
e m(iii) Methylcyanide to ethylamine
la s
l as CH3CN + 4[H] →
Na/alcohol
CH3CH2NH2
ag
ag
or H2/Ni

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e l a
l a s 7 ISC SPECIMEN ANSWER KEY 2027
a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 17 of 21

SECTION D – 15 MARKS

Question 19 [5]
(i) (a) Identify the compounds A, B, C and D
[A] Cl3C-COOH or trichloroacetic acid
[B] CH3COOH or acetic acid
[C] CH3CHO or acetaldehyde
OH
|
[D] CH3–C–CH2–CHO or 2-hydroxybutan-al
|
H
(b) 𝑅𝑒𝑑 𝑃
CH3COOH + 3Cl2 → Cl3C-COOH + 3HCl
[B]
Hell-Volhard-Zelinsky (HVZ) reaction
(c) Compound [C] acetaldehyde gives Tollen’s reagent test (silver mirror), Fehling’s
solution test (Red ppt of Cu2O), Schiff’s reagent test (pink colour), whereas
acetone does not response to these tests. (or any other correct test)
(d) Compound [A] i.e. trichloroacetic acid will be more acidic due to the electron
withdrawing tendency of Cl atoms (-I effect). It stabilises the carboxylate anion by
dispersing the negative charge and therefore, strengthening the acid. Compound
[B] i.e. acetic acid is less acidic due to the absence of -I effect.
OR
(ii) (a) [A] CH3CHO
[B] CH3CHOHCH3
[C] CH3COCH3
(b) [A] CH3CHO
[B] CH3COOH
[C] CH3COCl
(iii) (a) Acetone to Propene
CH3 CH3 OH
LiAlH4 443K
C=O → C → CH3–CH=CH2 + H2O
H2SO4
CH3 CH3 H

(b) Benzaldehyde to cinnamic acid
(I)CH3 COONa
C6H5CHO + (CH3CO)2O → C6H5CH=CHCOOH + CH3COOH
H2O, boil
Cinnamic acid

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8 ISC SPECIMEN ANSWER KEY 2027

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Question 20 a [5]
(i) (a) diamminedichloridoplatinum (II)
Cl NH3 Cl NH3

Pt Pt

Cl NH3 H3N Cl

m
cis-diamminedichlorido trans- diamminedichlorido

.co
platinum(II) platinum(II)
m
.co m
Geometrical isomerism

m s e
e
(b) Square planar geometry
s l a
l a ag
(ii) Ionisation isomers

ag
[Pt(NH3)3Br]NO3 → [Pt(NH3)3Br]+ + NO−3
Aqueous solution of this isomer when heated strongly with conc. H2SO4 gives brown
fumes of NO2. It also gives brown ring test with freshly prepared FeSO4 solution and
conc. H2SO4.
[Pt(NH3)3NO3]Br → [Pt(NH3)3NO3]+ + Br −
Aqueous solution of this isomer when reacts with AgNO3 solution gives yellow
precipitate of AgBr which is sparingly soluble in excess of NH4OH.
(iii) Structural formula of compound is
[Co(H2O)4Cl2]Cl → [Co(H2O)4Cl2]+ + Cl− m
AgNO3 + Cl− → AgCl + NO− 3
m .co
(aq) ppt
s e
Formula:
g la
a
H2O
Cl H2O
Cl
Co

Cl H2O
H2O
IUPAC Name – tetraaquadichlorocobalt(III)chloride

m
c. o
Question 21 [5]
m
(i) (a) Given; π = 3∙05 x 10-3 atm, T = 298K

m .co 300
V = 1000 = 0.3Lit, w = 1.85 g
s e m
s e wxRxT
g l a
la
π = CRT =
g
mxV
a
a 1∙85 x 0∙0821 x 298
m = 3∙05 x 10−3 x 0∙3
m = 49466 g mol-1

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e l a
l a s 9 ISC SPECIMEN ANSWER KEY 2027
a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 19 of 21

(b) Given; kf = 5∙12 k kg mol-1, ΔTf = 0∙85k,
m = 94 g mol-1, w = 2∙5 g, W = 120 g
n=2
1000 x kf x w
m(observed) = ΔTf x W
1000 x 5∙12 x 2∙5
m(obs) = 0∙85 x 120
= 125∙49 g mol-1
M(normal) 94
(i) = M = 125∙49 = 0∙749
(observed)
1−i 1−0∙749
α (degree of association) = 1 = 1
1− 1−
n 2
0∙251
= 0∙5 = 0∙502 or 50∙2%
(c) The observed molecular mass of potassium chloride will be half of its theoretical
molecular mass. The depression in freezing point is a colligative property which
depends upon the number of solute particles present in solution. KCl is a strong
electrolyte, it furnishes two ions KCl → K+ + Cl-.
The observed molecular mass of sucrose as determined by depression in freezing
point method will be same as the theoretical molecular mass.
OR
(ii) (a) Osmotic pressure is a colligative property which depends upon the number of
solute particles.
Osmotic pressure ∝ No. of solute particles
wt. of urea 6
(1) Moles of urea = mol. wt. of urea = 60 = 0∙1 moles
1 mole of urea = 6∙023 x 1023 x 0∙1 = 6∙023 x 1022 particles
wt. of glucose 72
(2) Moles of glucose = mol. wt. of glucose = 180 = 0∙4 moles = 6∙023 x1023 x 0∙4
= 24∙092 x 1022 particles
wt. of NaCl 5∙85
(3) Moles of sodium chloride = mol. wt. of NaCl = 58∙5 = 0∙1 moles
1 mole of NaCl = 2 x 6∙023 x 1023 particles, NaCl → Na+ + Cl-
0∙1 mole of NaCl = 2 x 6∙023 x 1023 x 0∙1 = 12∙046 x 1022 particles
Therefore, increasing order of their osmotic pressure is:
Urea < NaCl < glucose
(b) Given: wt. of C2H5OH = 92 g, wt. of H2O = 540 g
mol. wt. of C2H5OH = 46 g mol-1, mol. wt. of H2O = 18 g
92
No. of moles of C2H5OH (n) = 46 = 2 moles
540
No. of moles of H2O (N) = 18 = 30 moles
n 2 2
Mole fraction of C2H5OH = n+N = 2+30 = 32
X(C2 H5OH) = 0.0625
N 30 30
Mole fraction of water (H2O) = n+N = 2+30 = 32
X(H2 O) = 0∙9375

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10 ISC SPECIMEN ANSWER KEY 2027

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a of maximum boiling
(c) The mixture of chloroform and acetone is an example
azeotrope. It shows negative deviation from Raoult’s Law because of the increase
in intermolecular forces of attraction between chloroform and acetone, since they
form hydrogen bonds between them.

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m .co s e m
s e l a
g l a ag
a

m
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s e
g la
a

m
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m .co s e m
s e g l a
g la a
a

o m m .
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e l a
l a s 11 ISC SPECIMEN ANSWER KEY 2027
a g
a g For more Question Papers, Sample Papers, Notes & Syllabus visit Page 21 of 21