CBSE Sample Papers for Class 11 Chemistry Set 9 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 9 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

All questions are compulsory.
This question paper contains 37 questions.
Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.
Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.
Questions 28 – 34 in Section C are long-answer I type questions carrying 3 marks each.
Questions 35-37 in Section D are long-answer II type questions carrying 5 marks each.
There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.
Use log tables, if necessary. Use of calculator is not allowed.

Section – A

Question 1.
Which of the following statements is correct about the reaction given below? [1]
4 Fe(s) + 3o₂(g) → 2 Fe₂o₃ 9g)
(A) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.
(B) Total mass of reactants = total mass of product, therefore, law of conservation of mass.
(C) Amount of Fe₂O₃ can be increased by taking any one of the reactants (iron or oxygen) in excess.
(D) Amount of Fe₂O₃ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in
Answer:
(A) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.

Explanation:
According to law of conservation of mass,
Total mass of reactants = Total mass of products

Question 2.
Which of the following statements about a compound is incorrect ? [1]
(A) A molecule of a compound has atoms of different elements.
(B) A compound can not be separated into its constituent elements by physical methods of separation.
(C) A compound retains the physical properties of its constituent elements.
(D) The ratio of atoms of different elements in a compound is fixed.
Answer:
(A) A molecule of a compound has atoms of different elements.

Explanation:
Compound is formed when two or more elements are chemically joined. So, when the elements are joined, the atoms lose their individual properties.

Question 3.
A measured temperature on Fahrenheit scale is 200°F. What will this reading be on Celsius scale? [1]
(A) 40°C
(B) 94°C
(C) 93.3°C
(D) 30°C
OR
Which of the following polynuclear hydrocarbon is carcinogenic in nature?
(A) 1, 2-Benzanthracene
(B) 3-methyl-cholanthrene
(C) 1, 2 Benzpyrene
(D) All of the above
Answer:
(C) 93.3°C

(D) All of the above

Explanation:
All of the above hydrocarbons are carcinogenic in nature.

Question 4.
PCl₅ , PCl₃ and Cl₂ are at equilibrium at 500 K in a closed container and their concentrations are 0.8 × 10^-3 mol L^-1, 1.2 × 10^-3 mol L^-1 and 1.2 × 10^-3 mol L^-1 respectively. The value of Kc for the reaction
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) will be: [1]
(A) 1.8 × 10³ mol L^-1
(B) 1.8 × 10^-3
(C) 1.8 × 10^-3 L mol^-1
(D) 0.55 × 10⁴
Answer:
(A) 1.8 × 10³ mol L^-1

Explanation:
k_c = \(\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\) = \(\frac{1.2 \times 10^{-3} \times 1.2 \times 10^{-3}}{0.8 \times 10^{-3}}\)
= 1.8 × 10^-3

Question 5.
Which of the following arrangements represent increasing oxidation number of the central atom? [1]
(A) CrO₂^– , CIO₃^–, CrO₄^2-, MnO₄^–
(B) CIO₃^– , CrO₄^2-, MnO₄^–, CrO₂^–
(C) CrO₂^– , CIO₃^–, MnO₄^–, CrO₄^2-
(D) CrO_r^2- , MnO₄^–, CrO₂^–, CIO₃^–
Answer:
(A) CrO₂^– , CIO₃^–, CrO₄^2-, MnO₄^–

Explanation:
Oxidation number of central atom in the given arrangement:

Question 6.
How many isomers are possible in disubstituted benzene? [1]
(A) 2
(B) 3
(C) 4
(D) 1
Answer:
(B) 3

Explanation:
There are three isomers possible in di-substituted benzene, i.e., ortho, para and meta.

Question 7.
Ethyne on passing through red hot iron tube at 873K undergoes cyclic polymerisation. Three molecules of ethyne polymerise to form: [1]
(A) ethane
(B) ethene
(C) benzene
(D) ethanol
Answer:
(C) benzene

Question 8.
Which of the following is the correct IUPAC name? [1]
(A) 3-Ethyl-4,4-dimcthylheptane
(B) 4,4-Dimethyl-3-ethylheptane
(C) 5-Ethyl-4,4-dimethylheptane
(D) 4/4-Bis(methyl)-3-ethylheptane
OR
Electronegativity of carbon atoms depends upon their state of hybridisation. In which of the following compounds, the carbon marked with asterisk is most electronegative?
(A) CH₃—CH₂—*CH₂—CH₃
(B) CH₃ —*CH—CH—CH₃
(C) CH₃—CH₂—C = *CH
(D) CH₃—CH₂—CH = *CH₂
Answer:
(A) 3-Ethyl-4,4-dimcthylheptane

Explanation:
Because, while writing IUPAC name, the alkyl groups are written in alphabetical order. Thus, lower locant 3 is assigned to ethyl. Prefix, di, tri, and tetra are not included in alphabetical order.

Explanation:
Electronegativity increases as the state of hybridization changes from sp³ to sp² and sp² to sp. Thus, sp hybridised carbon has the highest electronegativity.

Question 9.
The correct IUPAC name of the following alkane is [1]

(A) 3,6-Diethyl-2-methyloctane
(B) 5-Isopropyl-3-ethyloctane
(C) 3-Ethyl-5-isopropyloctane
(D) 3-Isopropyl-6-ethyloctane
Answer:
(A) 3,6-Diethyl-2-methyloctane

Explanation:

Question 10.
Which of the following reactions involve homogeneous equilibrium or heterogeneous equilibrium? [1]
(i) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
(ii) C(s) + CO₂(g) → 2CO(g)
OR
What happens to the pH of the buffer solution on dilution?
Answer:
(i) Homogeneous equilibrium
(ii) Heterogeneous equilibrium

No change occurs on dilution.

Question 11.
Why cannot the motion of an electron around the nucleus be determined accurately ? [1]
OR
What are Transuranic Elements?
Answer:
The motion of an electron around the nucleus cannot be determined accurately because in order to study the position of electron, high energy radiation is needed which changes the velocity of electron, Therefore the position and velocity cannot be determined accurately.

The elements which follow uranium [i.e., Neptunium to Uub (Z = 112) of actinoids] in the periodic table are called transuranic or man-made elements. These element do not occur in nature because their half- life periods are so short.

Question 12.
Why electron affinities of Be and Mg are positive ? [1]
Answer:
Be and Mg belong to group 2. They have fully filled s-orbitals and the additional electron cannot be placed in the much higher energy p-orbitals of valence shell.
This would be energetically unfavourable and thus Be and Mg have positive electron affinity.

Question 13.
Describe in brief Lothar Meyer’s classification of elements. [1]
OR
What is meant by periodic classification of elements?
Answer:
Lothar Meyer (1830 – 1895) was a German chemist. He was one of the pioneers in developing the first periodic table of chemical elements.
Lothar Meyer used physical properties such as atomic volume, melting point, boiling point, etc. to arrive at his table of elements. He showed that when the properties of the elements such as atomic volume, melting point, boiling point, etc., are plotted as a function of their atomic weights, the elements with similar properties occupied almost similar positions.

By periodic classification of the elements, we mean the arrangement of the elements in such a way that the elements with similar physical and chemical properties are grouped together. Various scientists made contributions, but however, the contributions
made by Mendeleev are of great significance and he gave a periodic table called as Mendeleev’s Periodic Table which was older and replaced by long form of periodic table.

Question 14.
Assertion (A): Simple distillation can help in separating a mixture of propan-l-ol (boiling point 97°C) and propanpne (boiling point 56°C).
Reason (R): Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation. [1]
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
Answer:
(A) Both A and R are correct and R is the correct explanation of A.

Question 15.
Assertion (A): Energy of resonance hybrid is equal to the average of energies of all canonical forms. [1]
Reason (R): Resonance hybrid cannot be presented by a single structure.
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
Answer:
Option (D) is correct.

Explanation:
Canonical structures always have more energy than resonance hybrid. Resonance hybrids are always more stable than any of the canonical structures. The delocalisation of electrons lowers the orbitals energy and gives stability.

Question 16.
Why entropy of steam is more than that of water at its boiling point ? [1]
OR
Write the IUPAC name of:
CH₃—CH₂—CH₂—CH₂—COOH
Answer:
Entropy of steam is more than that of water at its boiling point because in steam, the molecules can move freely so there is more disorder i.e. more entropy.

Pentanoic acid.

Read the passage given below and answer the questions given below it:
The branch of science dealing with the relations between energy, heat, work and accompanying changes in the nature and behaviour of various substances around us is called thermodynamics. The main aim of the study of chemical thermodynamics is to learn (i) transformation of energy from one form into another form, (ii) utilization of various forms of energy and (iii) changes in the properties of systems produced by chemical or physical effects. The laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature do not change with time for a system in equilibrium state.

Answer the questions from (17) to (20):

Question 17.
A system which can neither exchange matter nor energy with the surroundings is called _________ [1]
(A) Open system
(B) Isolated system
(C) Closed system
(D) Ideal system
Answer:
(B) Isolated system

Explanation:
In an isolated system, there is no exchange of energy or matter between the system and the surroundings.

Question 18.
Which of the following is intensive property? [1]
(A) Molarity
(B) Temperature
(C) Density
(D) All of these
Answer:
(D) All of these

Explanation:
An intensive property, is a physical property of a system that does not depend on the system size or the amount of material in the system.

Question 19.
Temperature and heat are __________ [1]
(A) Extensive properties
(B) Intensive properties
(C) Intensive and extensive properties respectively
(D) Extensive and intensive properties respectively
Answer:
(C) Intensive and extensive properties respectively

Explanation:
An extensive property is a property that depends on the amount of matter in a sample, e.g., Mass, volume and heat are examples of extensive properties. An intensive property is a property of matter that depends only on the type of matter in a sample and not on the amount, e.g., Pressure and temperature.

Question 20.
In adiabatic process, [1]
(A) q > 0
(B) q = 1
(C) q = 0
(D) q < 0
Answer:
(C) q = 0

Explanation:
Adiabatic process is the change occurring within a system as a result of transfer of energy to or from the system in the form of work only, i.e., no heat is transferred.

Section – B

Question 21.
(a) What is a nucleophile?
(b) Reduction of alkyl halides takes place in which conditions? [2]
Answer:
(a) Nucleophile (nucleus-loving) is a chemical species that donates an electron pair to an electrophile (electron-loving). Hence, a nucleophile should have either a negative charge or an electron pair to donate.

(b) Alkanes can be prepared by the reduction of alkyl halides (except fluorides) with zinc and dilute hydrochloric acid.

Question 22.
Glycine is an amino acid. It exists in the form Zwitter ion as

What are the conjugate acid and conjugate base of this zwitter ion? [2]
Answer:

Question 23.
How do you express the bond strength in terms of bond order ? [2]
Answer:
Bond order is directly proportional to the bond strength.
Bond order ∝ bond strength
Greater the bond order of molecule or ion, greater will be the bond strength. Because, bond enthalpy increases with increase in bond order. Hence, bond strength is also high.

Question 24.
A proton is moving with kinetic energy 5 × 10^-27 J. What is the velocity of the proton ? [2]
OR
Calculate the mass of the photon with wavelength of 5 pm.
Answer:
∵ Mass of proton = 1.67 × 10^-27 kg
K.E. = 5 × 10^-27 J

Given λ = 5 pm = 5 × 10^-12 m
Velocity of photon = Velocity of light
= 3.0 × 108 m/s.
∵ λ = \(\frac{h}{m v}\)
or m = \(\frac{h}{\lambda v}\) = \(\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{5 \times 10^{-12} \mathrm{~m} \times 3.0 \times 10^8 \mathrm{~m} / \mathrm{s}}\)
= 4.417 × 10^-31 kg

Commonly Made Error:
Some students do not convert pm to m.

Answering Tip:
Always check the compatibility of units.

Question 25.
Which of the following compounds are aromatic according to Huckel’s rule ?

OR
How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m- nitrochlorobenzene [2]
Answer:
(A) = Has 8π electrons, does not follow Huckel rule. Orbitals of one carbon atom are not in conjugation. It is non-aromatic.
(B) = Follow Huckel’s rule, has 6p delocalised electrons. Hence, is aromatic.

(i) Benzene canbe converted into p-nitrobromobenzene as:

(ii) Benzene can be converted into m-nitrochloroben- zene as:

Question 26.
Draw formulas for the first five members of each homologous series beginning with the following compounds.
(i) H-COOH
(ii) CH₃COCH₃
Answer:
(i) H-COOH: Methanoic acid
CH₃-COOH: Ethanoic acid
CH₃-CH₂-COOH: Propanoic acid
CH₃-CH₂-CH₂-COOH: Butanoic acid
CH₃-CH₂-CH₂-CH₂-COOH: Pentanoic acid

(ii) CH₃COCH₃: Propanone
CH₃COCH₂CH₃: Butan-2-one
CH₃COCH₂CH₂CH₃: Pentan-2-one
CH₃COCH₂CH₂CH₂CH₃: Hexan-2- one
CH₃COCH₂CH₂CH₂CH₂CH₃: Heptan-2-one

Question 27.
Complete the following reactions:

Answer:

Section – C

Question 28.
Calculate:
(i) Mass in gram of 5.8 mol N₂O
(ii) Number of moles in 8.0 g of O₂
(iii) Molar mass if 11.2 L at STP weigh 8.5 g
OR
The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molality. (The density of acid solution is 1.02 g cm^-3).
Answer:
(i) Molecular mass of N₂O = 2(14) + 16 = 44
Mass = Molecular mass × No. of moles
= 44 × 5.8
= 255.2 g

(ii) Number of moles of
O₂ = \(=\frac{\text { Mass of } \mathrm{O}_2}{\text { Molecular mass of } \mathrm{O}_2}\)
= \(\frac{8.0}{32}\)
= 0.25 mol

(iii) ∵ 11.2 L at STP weigh = 8.5 g
∴ 22.4 L at STP will weigh = \(\frac{8.5}{11.2}\) × 22.4
= 17 g mol^-1

Molarity = 1.35 M
Molecular mass of H₂SO₄ = 2 + 32 + 64 = 98
Mass of H₂SO₄ in 1 L solution = 1.35 × 98 = 132.3 g
Mass of 1 L solution = 1000 × 1.02
= 1020 g (‘ Density = 1.02 g cm^-3)
Mass of water in solution = 1020 – 132.3
= 887.7 g
= 0.888 kg
Number of moles of solute
Molality = \(\frac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)
= \(\frac{1.35}{0.888}\)
= 1.52 m

Question 29.
(a) Name the dipositive ion represented by the electronic configuration: 1s², 2s²2p⁶, 3s², 3p⁶.
(b) The bromine atom possess 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ? [3]
Answer:
(a) Electronic configuration = 1s², 2s²2p⁶, 3s², 3p⁶
No. of electrons = 18
Atomic number = 18 + 2 = 20
∴ Ion is Ca²⁺.

(b) Br = 2, 8, 18, 7
= 1s², 2s²2p⁶, 3s²3p⁶3d¹⁰, 4s²4p⁵
Since, effective nuclear charge decreases as the distance of the orbitals increases from the nucleus. Therefore, 4p electrons experience the lowest effective nuclear charge.

Question 30.
(a) Arrange the following species in increasing order of their ionic size :
N^3-, Na⁺, F^–, O^2-
(b) How does atomic radius vary in a period and in a group? How do you explain the variation ?
Answer:
(a) The species N^3- , O^2- , F^–, Na⁺ in their increasing order of ionic radii:
Na⁺ < F^– < O^2- < N^3-
Ionic radius for anions is more than cations and it increases as the anionic charge increases.

(b) Across the period: Atomic radii decrease on account of the following facts:
Nuclear charge increases progressively by one unit but the additional electron goes to the same principal shell. As a result, the electron cloud is pulled closer to the nucleus by the increased effective nuclear charge.

Down the group: Atomic radii increases.

Reason: Nuclear charge increases with increase in atomic numbers but at the same time, there is progressive increase in principal energy levels.

Commonly Made Error:
Students get confuse with periodic properties and give wrong answers.

Answering Tip:
Understand the variations in the periodic properties of the periodic table.

Question 31.
(i) What do you understand by the following : [3]
(a) Markovnikov’s rule
(b) Huckel’s rule
(ii) Complete the following and name the products :

OR
(a) Write chemical reactions to illustrate the following :
(i) Kolbe’s reaction
(ii) Wurtz reaction
(b) What type of hydrocarbons are present in high octane gasoline ?
Answer:
(i) (a) Markovnikov’s Rule : This rule states that, ‘The negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms, e.g.

(b) Huckel’s Rule : This rules states that all planar cyclic conjugated polyenes containing (4n + 2) π electrons where, n = 0, 1, 2, ……….. are aromatic in nature.

Commonly Made Error:
Some students confuse Markovnikov’s with anti-Markovnikov’s rule.

Answering Tip:
Always write the equation whenever possible.

(ii)

(a) (i) Kolbe’s reaction : In this reaction, an aqueous solution of sodium or potassium salt of carboxylic acid on electrolysis gives alkane having even number of carbon atoms at the anode.

(ii) Wurtz reaction : In this reaction, alkyl halides on treatment with sodium in dry ether gives higher alkanes, preferably containing even number of carbon atoms.

Commonly Made Error:
Students often make mistake of either writing only equations or only illustrations.

Answering Tip:
Always understand that “Illustrate” means explain with suitable examples.

(b) Branched chain aliphatic and/or aromatic hydrocarbons.

Question 32.
(i) What is the main difference between electromagnetic wave theory and Planck’s quantum theory ? [3]
(ii) Which rule is violated in the following orbital diagram:

Answer:
(i) Electromagnetic wave theory: Energy is radiated or absorbed continuously.
Planck’s quantum theory: Energy is radiated or absorbed not continuously but discontinuously in the form of small packets called quanta or photons.
(ii) Hund’s rule of maximum multiplicity is being violated.

Question 33.
Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre ? [3]
Answer:
W = ?
ΔU = ?
V_i = 1 litre
V_f = 5 litre
External pressure (P_ex) = 0
∴ Work done W = – P_ex(V_f – V_i)
= – 0(5 – 1)
= 0

For isothermal expansion q = 0
∴ According to first law of thermodynamics
ΔU = q + W
= 0 + 0 = 0

Question 34.
Justify that the reaction:
2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)
is a redox reaction. Identify the species oxidised/ reduced, which acts as an oxidant and which acts as a reductant. [3]
Answer:
First of all, assign oxidation number to each of the species in the given reaction.

So, it is concluded that in this reaction, copper is reduced from +1 oxidation state to zero oxidation state and sulphur is oxidised from -2 oxidation state to +4 oxidation state.
Hence, this reaction is a redox reaction.
Further, Cu₂O is used to oxidise sulphur. So, Cu(I) is an oxidant and sulphur of Cu₂S is used to reduce copper both in Cu₂S itself and Cu₂O to decrease its oxidation number, so, sulphur of Cu₂S is a reductant.

Section – D

Question 35.
Which hybrid orbitals are used by carbon atoms in the following molecules? [5]
(a) CH₃ – CH₃
(b) CH₃ – CH = CH₂
(c) CH₃ – CH₂ – OH
(d) CH₃ – CHO
(e) CH₃ – COOH
OR
Use the molecular orbital energy level diagram to show that N₂ would be expected to have a triple bond, F₂, a single bond and Ne₂, no bond.
Answer:

Since, the bond order of Ne₂ is zero, so, Ne₂ is unstable, i.e., it does not exist as a molecule.

Question 36.
How do you account for the formation of ethane during chlorination of methane? [5]
Answer:
Chlorination of methane proceeds through a free radical chain mechanism. The whole reaction takes place in the given three steps.
Step 1: Initiation : The reaction begins with the homolytic cleavage of Cl – Cl bond as:

Step 2: Propagation : In the second step, chlorine free radicals attack methane molecules and break down the C-H bond to generate methyl radicals as:

These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and CH₃Cl are the major products formed, other higher halogenated compounds are also formed as:

Step 3: Termination : Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:

Hence, by this process, ethane is obtained as a by¬product of chlorination of methane.

Question 37.
(a) What is meant by homogeneous equilibrium? Give an example.
(b) The value of K_c = 4.24 at 800 K for the reaction.
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
Calculate equilibrium concentrations of CO₂, H₂, CO and H₂O at 800 K, if only CO and H₂O are present initially at concentrations of 0.10 M each.
(c) For which equilibrium, the vapour pressure is constant at a given temperature. [5]
Answer:
(a) Homogeneous Equilibrium: Equilibrium in a system having only one phase is called homogeneous equilibrium.
e.g., N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

(b) For the reaction,

Let x mole L^-1 of each of the product be formed.
At equilibrium, arrange in the same order under the reactants and products of the chemical reaction. where x = amount of CO₂ and H₂ at equilibrium. Hence, equilibrium constant can be written as
k_c = \(\frac{x^2}{(0.1-x)^2}\) = 4.24
x² = 4.24 (0.01 + x² – 0.2x)
x² = 0.0424 + 4.24x² – 0.848x
3.24x² – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
For quadratic equation ax² + bx + c = 0

The value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.
Therefore, the equilibrium concentrations are:
[CO₂] = [H₂] = x = 0.067 M
[CO] = [H₂O] = 0.1 – x
= 0.1 – 0.067
= 0.033 M