CBSE Sample Papers for Class 12 Chemistry Set 11 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 11 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

• There are 35 questions in this question paper with internal choices.
• Section A consists of 18 multiple-choice questions carrying 1 mark each.
• Section B consists of 7 very short answer questions carrying 2 marks each.
• Section C consists of 5 short answer questions carrying 3 marks each.
• Section D consists of 2 case-based questions carrying 4 marks each.
• Section E consists of 3 long answer questions carrying 5 marks each.
• All questions are compulsory.
• Use of log tables and calculators are not allowed.

SECTION – A (18 Marks)
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mart There is no internal choice in this section.)

Question 1.
The charge required for the reduction of 1 mol of Cr₂O₇^2- ions into Cr³⁺ is: [1]
(a) 6 × 96500 coulomb
(b) 2 × 96500 coulomb
(c) 3 × 96500 coulomb
(d) 4 × 96500 coulomb
Answer:
(a) 6 × 96500 coulomb

Explanation:
When Cr₂O₇^2- is reduced to Cr³⁺, the oxidation state changes from +6 to +3. Each chromium atom accepts three electrons. Hence, two chromium atoms will require 6 electrons total.
Cr₂O₇^2- → 2Cr³⁺ + 6e^–
Thus, the charge required to reduce one mole of Cr₂O₇^2- is 6 Faraday and 1 Faraday is equal to 96500 C therefore, 6 × 96500 C of charge is required for the reduction process.

Related Theory:
One Faraday is defined as the charge in coulombs (C) of 1 mole of electrons. Faraday’s constant is approximately 96485 C mol^-1. We can calculate 1F by multiplying the charge on one electron (1.602 × 10-19) by Avogadro’s number (6.022 × 1023).

Question 2.
Arrange the following compounds in the correct decreasing order of reactivity towards neucleophilic substitution reaction. [1]
(I) CH₃CHO
(II) HCHO
(III) CH₃COCH₃
(IV) C₆H₅CHO
(a) (ll) > (l) > (lll) > (IV)
(b) (l) > (lll) > (ll) > (IV)
(c) (lll) > (l) > (ll) > (IV)
(d) (IV) > (I) > (II) > (III)
Answer:
(a) (II)>(I)>(III)>(IV)

Explanation:
In nucleophilic acyl substitution reactions, the C=O group remains in the final reaction product. The overall transformation replaces a group originally attached to the C=O (e.g. the Z group), with a nucleophile.

Question 3.
The rate constant for first-order reaction is shown in the following graph: [1]

The half-life for this reaction is:
(a) 69.3 sec
(b) 6.63 sec
(c) 0.693 sec
(d) 0.069 sec
Answer:
(a) 69.3 sec

Explanation:
For a first order reaction:
t_½ = 0.693/ k
t_½ = 0.693/10^-2
t_½ = 0.693 × 102 = 69.3 sec

Related Theory:
In a chemical reaction, the half-life of a species is the time it takes for the concentration of that substance to fall to half of its initial value.

Question 4.
Which of the following is correct about in-bonding in nucleotide? [1]
(a) A-T, G-C
(b) A-G, T-C
(c) G-T, A-C
(d) A-A, T-T
Answer:
(a) A-T G-C

Explanation:
Adenine pairs with thymine with 2 hydrogen bonds. Guanine pairs with cytosine with 3 hydrogen bonds. Guanine and cytosine bonded base pairs are stronger than thymine and adenine bonded base pairs in DNA.

Question 5.
Which of the following solutions shows positive deviation from Raoult’s law? [1]
(a) Acetone + Aniline
(b) Acetone + Ethanol
(c) Water + Nitric acid
(d) Chloroform + Benzene
Answer:
(b) Acetone + Ethanol

Explanation:
Molecules of ethanol in pure state are hydrogen bonded. But when acetone is added to the ethanol, molecules of acetone
get in between the molecules of ethanol and break some of the hydrogen bond, which weaken the intermolecular attractive forces resulting in increase in vapour pressure of mixture than ethanol in pure state.

Question 6.
The correct IUPAC name of the given compound is: [1]

(a) Benzene 1,4-diol
(b) Hydroquinone
(c) p-Dihydroxy benzene
(d) m-Dihydroxy benzene
Answer:
(a) Benzene 1, 4 -diol

Related Theory:
The International Union of Pure and Applied Chemistry (IUPAC) have given certain rules to name the organic compounds known as the nomenclature system. It is a set of logical rules devised to circumvent problems caused by arbitrary nomenclature.

Question 7.
Lanthanoid contraction is due to increase in: [1]
(a) shielding of 4f-electro ns
(b) effective nuclear charge
(c) size of f-electrons
(d) atomic radius
Answer:
(a) Shielding of 4f electrons

Explanation:
The lanthanide contraction is caused by a poor shielding effect of the 4f electrons. With the increase in the atomic number, the atomic radius decreases because the elements in row 3 have 4f electrons. These electrons are not effectively shielded resulting in the greater and effective nuclear charge. Thus, the size decrease beyond lanthanum in the periodic table and is known as lathanoid contraction.

Question 8.
Identify the end product (C) in the following sequence: [1]

(a) C₂H₅CH₂NH₂
(b) C₂H₅CONH₂
(c) C₂H₅COOH
(d) C₂H₅NH₂ + HCOOH
Answer:
(c) C₂H₅COOH

Explanation:

Question 9.
Which of the following alcohol is most soluble in water? [1]
(a) Propanol
(b) Butanol
(c) Pentanol
(d) Hexanol
Answer:
(a) Propanol

Explanation:
Propanol is most soluble in water because the lower alcohols are highly soluble in water due to the presence of -OH group in the alcohols which forms H-bond with itself and molecular association takes place which causes the increase in the boiling point of the corresponding alcohols due to the increase in the number of carbon atoms and thus high temperature is required to break this association of bonds and thus, the solubility in water increases. The extent of the hydrogen bonding in alcohols depends on the number of the carbon atoms attached in the chain and takes place as shown below:

Question 10.
Dissachrides that are reducing in nature are: [1]
(a) sucrose and lactose
(b) sucrose and maltose
(c) lactose and maltose
(d) sucrose, lactose and maltose
Answer:
(c) lactose and maltose

Explanation:
Reducing disaccharides like lactose and maltose have only one of their two anomeric carbons involved in the glycosidic bond, while the other is free and can convert to an open-chain form with an aldehyde group.

Related Theory:
The carbon derived from the carbonyl carbon of the open chain form of the carbohydrate molecule.

Question 11.
Find the asymmetric carbon atom in the following molecules: [1]

(a) (I), (II), (III), (IV)
(b) (I), (III), (IV)
(c) (II), (III), (IV)
(d) (I), (II), (III)
Answer:
(d) (I), (II), (III)

Explanation:
The four substituents linked to carbon in molecules (I), (II), and (III) are all distinct, making them asymmetric molecules.

Question 12.
Which of the following elements has the lowest boiling point?
(a)

(b)

(c)

(d)

Answer:
(d)

Explanation:
The lowest boiling point is for tertiary butyl bromide.

Question 13.
An optically active alkyl using the S_N² mechanism is used to: [1]
(a) racemisation
(b) none of these
(c) inversion of configuration
(d) retention of configuration
Answer:
(c) inversion of configuration

Explanation:
In the S_N² mechanism, the nucleophile attacks from the back, inverting the conformation as a result.

Question 14.
The following two optically active stereo-structures of CH₃-CH(OH)COOH are possible: [1]
(a) mesomers
(b) enantiomers
(c) diastereomers
(d) atropisomers
Answer:
(c) diasteromers

Explanation:
Two possible stereo-structures of CH₃CHOH/COOH, which are optically active, are called diastereomers.

In the following questions, 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

Question 15.
Assertion: Order of the reaction can be zero, fractional and even negative.
Reason: Order of a reaction is usually a whole number. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
The order of the reaction is usually a whole number. However, the order of a reaction can be zero, fractionaL and can also be negative for inhibitors.

Question 16.
Assertion: Mixture of ortho and para nitro phenols are separated by steam distillation.
Reason: Ortho nitrophenol is steam volatile and has intramolecular hydrogen bonding. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
Steam distillation is a separation process which purifies isolate temperature- sensitive materiaLs, such as naturaL aromatic compounds. In steam distillation, dry steam is passed through the pLant materiaL. These vapours undergo condensation and coLLection in receivers. Since, the intramolecular hydrogen bonding is present in the molecule of ortho and para nitro phenols they can be separated through steam distillation process.

Related Theory:
Intermolecular Hydrogen Bonding occurs when the hydrogen bonding is between H-atom of one molecule and an atom of the electronegative element of another molecule. For example.
(1) Hydrogen bond between the molecules of hydrogen fluoride.
(2) Hydrogen bond in alcohol or water molecules.

Question 17.
Assertion: Fe³⁺ ion is more paramagnetic than Fe²⁺.
Reason: Fe³⁺ ion contains five unpaired electrons whereas Fe²⁺ has four unpaired electrons. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
The atomic number of Fe is 26 and thus Fe³⁺ ion contains five unpaired electrons whereas Fe²⁺ has four unpaired electrons which can be explained with the electronic configurations of the two ions.

Atomic number of Fe = 26
Electronic Configuration of Fe, with 26 electrons
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Fe³⁺ means (26 – 3) electrons = 23 electrons
Due to the fact that partially filled orbital configuration is stable enough.
We get,
Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4d⁰ 3d⁵

Question 18.
Assertion: In glucose all the six C atoms are present in straight chain.
Reason: Glucose reacts with HI to give n-Hexane. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Glucose on prolonged heating with the HI gives the n-hexane and n-hexane has 6 carbon atoms connected with each other in the straight chain thus, this concludes that all the 6 carbon atoms in the glucose molecule are also connected linearly in the form of a straight chain

Related Theory:
The two more reactions are frequently used to describe the structure of glucose. These are:
1. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five-OH groups in the structure of glucose.

2. Glucose gets oxidised to six carbon carboxylic acid on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.

Section – B (14 Marks)
(The following questions are very short answer type with internal choice in two questions and carry 2 marks each.)

Question 19.
(A) Write any four factors on which solubility of a gas in liquid depends.
(B) On what factors vapour pressure of a liquid depends? [2]
Answer:
(A) Four important factors are:

1. Nature of gas.
2. Nature of liquid
3. Temperature
4. Pressure

(B) Nature of the liquid and temperature

Question 20.
Convert the following:
(A) Benzene to AniLine
(B) Aniline to Phenol
OR
What happens when:
(A) aniline reacts with cone. H₂SO₄.
(B) aniline reacts with acetic anhydride. [2]
Answer:
(A)

(B)

OR

(A) Aniline reacts with conc. H₂SO₄ to give the sulphanilic acid.

(B) Aniline reacts with acetic anhydride to give the acetanilide.

Question 21.
Alkyl halides are often dehydrohalogenated using sodium methoxide in methanol, sodium ethoxide in ethanol, potassium tert- butoxide in tert-butyl alcohol, or dimethyl sulfoxide, (CH₃)₂SO. a chemical process that creates an alkene or an alkyne when a hydrogen atom and a halogen atom in a molecule are separated from nearby atoms. On the basis of the above data, answer the following questions:
(A) Why dehydrohalogenation is called elimination reactions?
(B) Write the correct order towards dehydrohalogentaion of 1°, 2°, and 3° alkylhalides. [2]
Answer:
(A) The dehydrohalogenation reaction of alkyl halides are β-elimination reactions because the hydrogen atom removed is actually α β- hydrogen.

(B) 3⁰ > 2⁰ > 1⁰

Question 22.
An aromatic compound X with molecular formula C₈H₈O forms an orange red precipitate with 2, 4-DNP reagent and gives yellow precipitate on heating with Iodine in the presence of NaOH. It neither reduces Tollen’s reagent or Fehling’s solution nor does it decolourise bromine water or Bayer’s reagents. On drastic oxidation with chromic acid gives a carboxylic acid Y having the molecular formula C₇H₆O₂. Identify the compound X and Y and give the reactions involved. [2]
Answer:

Question 23.
Complete and balance the following equations:
(A) Cu + SO₄^2- + H⁺ →
(B) Zn + NO₃^– + OH^– →
OR
Why do the transition elements have higher enthalpies of atomisation? In 3d series (Sc to Zn), which element has the lowest enthalpy of atomisation and why? [2]
Answer:
(A) Cu + SO₄^2- + H⁺; → Cu²⁺ + SO₂ + 2H₂O
(B) Zn + NO₃^– + OH^– → 4ZnO₂^2- + NH₃ + 2H₂O

OR

Transition metals have high effective nuclear charge, greater number of valence electrons and some unpaired electrons. They, thus have strong metal-metal bonding. Hence, transition metals have high enthalpies of atomisation. In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell electronic configuration of Zn is 3d¹⁰4s². Due to the absence of unpaired electrons in ns and (n-1)d shells, the interatomic electronic bonding is the weakest in zinc. Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition elements.

Question 24.
(A) Write down the IUPAC name of the following complex:
[Co(NH₃)₅ (NO₂)](NO₃)₂
(B) Write the formula for the following complex: Potassium tetracyanidonickelate(II) [2]
Answer:
(A) IUPAC name of [Co(NH₃)₅(NO₂)](NO₃)₂:
Pentaamminenitrocobalt(III)nitrate.

(B) Formula of the complex is K₂ [Ni(CN)₄].

Question 25.
Give an example of the following ligands. [2]
(A) Tetradentate ligands
Answer:

(B) Hexadendate ligands
Answer:

Section – C (15 Marks)
(The following questions are short answer type with internal choice in two questions and carry 3 marks each.)

Question 26.
Answer the following related to manganate and permanganate ions.
(A) Geometry of both
(B) Colour exhibited by both the ions.
(C) Oxidation state of Mn in both the ions [3]
Answer:
(A) Both are tetrahedral in shape

(B) Colour of manganate ion is green and color of permanganate is purple.

(C) Oxidation state of MnO₄^2- is +6 and oxidation state of MnO₄^– is +7.

Question 27.
Complete the following reactions:

OR
(A) Arrange 1°, 2° and 3° alcohols in correct order of reactivity of different types of alcohol towards hydrogen halide. Explain how you did so.
(B) What will you obtain butanol from catalytic reduction of butanal? Give the chemical equation and IUPAC name of the product.
(C) Why alcohols have high boiling points? [3]
Answer:

OR

(A) 3° alcohol > 2° alcohol > 1° alcohol.

Explanation: Reaction of hydrogen
halides (HX) with alcohols depends on the degree of carbocation generated due to heterocyclic cleavage of HX bond. If alcohol is primary then reaction will proceed through S_N², and for tertiary alcohol it will proceed through S_N¹ mechanism. For secondary alcohol it can go through both path (S_N¹ or S_N²). S_N¹ being ionic in nature takes place very fast while S_N² is molecular in nature and hence is slow. Hence, we can sum up the above information as – rate of reaction is 3° alcohol > 2° alcohol > 1° alcohol.

(B) Catalytic reduction of butanal will give Butan-1-ol

(C) Alcohols have high boiling point because of the intermolecular hydrogen bonding with the water molecule due to which the
bonds become so strong that it cannot be broken easily.

Question 28.
The conductivity of 0.0001028 mol L^-1 acetic acid is 4.95 × 10^-5 Scm^-1. Calculate its dissociation constant if λ°_m for acetic acid (CH₃COOH) is 390.5 S cm² mol^-1. [3]
Answer:
Given k = 4.95 × 10^-5 Scm^-1
M = C = 0.0001028 molL^-1

Question 29.
(A) What is the difference between double salts and complexes?
(B) Explain [FeF₆]_3- is paramagnetic and outer orbital complex.
(C) What type of ligand is cyanide (CN^–) ion? [3]
Answer:
(A) The difference between double salt and complex salt are:

Double Salt	Complex Salt
1. These exit only in solid state and dissociate into constituent species in their solution.	They retain their identity in solid as well as in solution state.
2. They lose their identity in dissolved state.	They do not lose their identity in dissolved state.
3. In double salt the metal atom or ion exhibit normal valency.	In co-ordination compounds, the number ofnegative ions or molecules surrounding the central metal atom is beyond its normal valency.

(B)

Here F^– is a weak field ligond and it obeys Hund’s rule of leaving maximum number of unpaired electrons (n = 5) and it uses one 4s, three 4p and two 4d orbitais which are sp³d² hybridised, So it is an outer orbital complex.

(C) Ambidentate ligand

Question 30.
Carry out the following conversions: [3]
(A) Aniline to iodobenzene
Answer:

(B) Nitrobenzene to benzoic acid
Answer:

(C) Aniline to p-bromoaniline
Answer:

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.)

Question 31.
Solution is a homogeneous mixture of two or more substances in same or different physical phases. The substances forming the solution are called components of the solution. On the basis of number of components a solution of two components is called binary solution. In a binary solution, solvent is the component which is present in large quantity while the other component is known as solute. [If water is used as a solvent, the solution is called aqueous solution and if not, the solution is called non-aqueous solution.]

Depending upon the amount of solute dissolved in a solvent we have the following types of solutions: Unsaturated solution a solution in which more solute can be dissolved without raising temperature is called an unsaturated solution. Saturated solution a solution in which no solute can be dissolved further at a given temperature is called a saturated solution. Liquid – Vapour phase diagrams of binary solutions are drawn to illustrate the composition in the liquid and vapour phases of the two volatile components. The die hansbeswishow his for the binary mixture of benzene and toluene, which form a nearly ideal binary solution: (A) as a function of Pressure, at 23°C and (B) as a function of temperature at P = 1 atm. The x-axis gives x(benzene).

(A) Identify the regions as vapour (V), liquid (L) or vapour + liquid (V + L), on graphs B. [1]
(B) State Henry’s law. [1]
(C) In an aqueous solution, CH₃OH has a mole fraction of 0.02 and a density of 0.99 g cm^-3. Find out the molarity and molality of it.
OR
How will you determine the relative molecular mass of a solute on the basis of depression of freezing point? [2]
Answer:
(A)

(B) Henry’s Law states that at a constant temperature, the soLubiLity of a gas is directly proportional to the pressure of the gas. In other words, the partial pressure of the gas in vapour phase (P) is proportionaL to the mole fraction of the gas (X) in the solution.
P = K_H.X
Here, K_H is the Henry’s law constant.
Hence, the given statement is Henry’s law.

(C) Let x mole of CH₃OH and y mole of water be present in solution.
Mole fraction of CH₃OH = \(\frac{x}{x+y}\) = 0.02

OR

The molecular mass of a solute dissolved in a particular solvent can be determined with the help of depression of freezing point for the solution as follows.
If W_A g of a solute are dissolved in W_B g of a solvent, the molality m of the solution is given by,
m = \(\frac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)
= \(\frac{W_A / M^{\prime}}{W_{-} / 1000}\)
Or m = \(\frac{1000 \times W_A}{W_B \times M^{\prime}}\)
Where, M’ is the molecular mass of the solute. Putting the above value of m in equation
T_f = K_f × m
we get
∆T_f = \(\mathrm{K}_f \times \frac{1000 \times \mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}} \times \mathrm{M}^{\prime}}\)
M’ = \(\frac{1000 \times \mathrm{K}_f \times \mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}} \times \Delta \mathrm{T}_f}\)

Question 32.
Amino acids publishes contributions from all fields of amino acid and protein research: analysis, separation, synthesis, biosynthesis, cross linking amino acids, racemization/ enantiomers, modification of amino acids as phosphorylation, methylation, acetylation, glycosylatior. and no enzymatic glycosylation, new roles for amino acids in physiology and pathophysiology, biology, amino acid analogues and derivatives, polyamines, radiated amino acids, peptides, stable isotopes and isotopes of amino acids. Applications in medicine, food chemistry, nutrition, gastroenterology, nephrology, neurochemistry, pharmacology, excitatory amino acids are just some of the topics covered. Fields of interest include: Biochemistry, food chemistry, nutrition, neurology, psychiatry, pharmacology, nephrology, gastroenterology, microbiology.
(A) How glycine can be considered as a non-essential amino acid? [1]
(B) Give one of the examples of denaturation of proteins. [1]
(C) (i) What is meant by invert sugar?
(ii) What is difference between nucleotide and nucleoside?
OR
(i) Name the disesaes which are caused by the deficiency of enzymes.
(ii) Which type of secondary structure is found in keratin and fibroin? [2]
Answer:
(A) Glycine is non-essential amino acid because it is synthesised in our body and need not to be taken in the diet essentially.

(B) Coagulation of egg white when the egg is boiled.

(C) (i) A liquid sweetener called invert sugar is created by combining water and table sugar (sucrose). It results in a thick, sweet syrup that contains 50% fructose and 50% glucose when the links between the sugars in sucrose are broken.
(ii) Nucleoside is formed by base and sugar. And nucleotide is formed by base, sugar and phosphate group.

OR

(i) Phenylketonuria and albinism
(ii) A prevalent motif in the secondary structure of proteins, the alpha helix (-helix) is a right-hand helix conformation in which each backbone NH group hydrogen bonds to the backbone C=O group of the amino acid positioned four residues earlier in the protein sequence.

Section – E (15 Marks)
(The following questions are long answer type and carry 5 marks each. Two questions have an internal choice.)

Question 33.
(A) For the reaction:
2A + B → A2B
The rate= k [A].[B] with k = 2.0 × 10^-6 mol^-2 L^-2 s^-1. Calculate the initial rate of the reaction when [A] = 0.1 molL^-1, [B] = 0.2 molL^-1. calculate the rate of the reaction after [A] is reduced to 0.06 molL^-1
(B) (i) Mention the factors that affect the rate of a chemical reaction.
(ii) What is average rate of the reaction?
(iii) What are pseudo first order reactions?
OR
(A) Why cannot be molecularity of any reaction be equal to zero?
(B) Consider the reaction given below:
2NO_(g) + O_2(g) → 2NO_2(g)

Concentration of NO (mot/L)	Concentration of O2 (mol/L)	Rate of formation of NO2/mol L-1s-1
(1) 0.60	0.60	0.768
(2) 0.30	0.30	0.096

What information can you derive from this?
(C) What is the unit of rate constant for second order reaction?
(D) Can molecularity of the reaction be equal to zero?
(E) How the rate of the reaction can be determined mathematically? [5]
Answer:
(A) 2A + B → A₂B
Initial rate = k [A].[B]²
Given, [A] = 0.1 M, [B] = 0.2 M and
k = 2.0 × 10^-6 M^-2s^-1 (mol^-2L²s^-1).
Then, Initial rate = 2 × 10^-6 × 0.1 × (0.2)²
= 8 × 10^-9 Ms^-1 (molL^-1 s^-1)
Now, according to quatation,
Concentration of A left = [A] = 0.06M
Concentration of B left = [B] = [0.2 M – 0.02 M]
= 0.18 M
Rate = k [A].[B]²
= 2 × 10-6 × 0.06 × (0.18)²
= 3.89 × 10-9 Ms^-1 (molL^-1s^-1)

(B) 1. The factors which are responsible for the effect in chemical reaction’s rate are:

• Temperature.
• Presence of a catalyst.
• The concentration of reactants (pressure in case of gases).

2. The average rate of reaction is defined as the ratio of the change in the concentration of the reactants or the products of a chemical reaction to the time interval. Following is the mathematical representation of the average rate:
Average rate of reaction = Ax/At

3. A pseudo first-order reaction can be defined as a second-order or bimolecular reaction that is made to behave like a first-order reaction. This reaction occurs when one reacting material is present in great excess or is maintained at a constant concentration compared with the other substance

OR

(A) Molecularity is the number of reacting species taking part in an elementary reaction, as a reaction to happen the least number of molecules taking part is one, molecularity cannot be zero. This example shows that the rate of any reaction decreases with decrease in concentration of reactants during the course of the reaction.

(B) Rate of a reaction is directly proportional to concentration of reactants taking part in reaction, as suggested by the following rate expression:
Rate = k [A]^x [B]^y
So, as the concentration of reactants decrease upon progress of reaction, rate of reaction also decreases. So, decreasing the concentration decreases the rate of product formation hence, the rate of the reaction also increases.

(C) Unit of rate constant for second order reaction is moL^-1Ls^-1.

(D) No moLecuLarity of the reaction cannot be equal to zero.

(E) Rate of the reaction is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period At.

Question 34.
Answer the following:
(A) Arrange the following in increasing order of their solubility in water: Acetaldehyde, Formaldehyde, Butyraldehyde and Benzaldehyde.
(B) What is Fehling’s solution?
(C) Observe the following graph of boiling point vs molecular weight:
What do you understand about the trend comparing aldehydes, carboxylic acids and primary alcohols. [5]

Answer:
(A) For aldehydes as the carbon chain increases in length, solubility in water decreases for aliphatic aldehydes
– Formaldehyde < Acetaldehyde < Butyraldehyde < Benzaldehyde.

(B) Fehling’s solution is alkaline solution of CuSO₄ along with some Rochelle salt.

(C) The boiling point is in the following order:
carboxylic acids > alcohols > aldehydes
Carboxylic acids have higher boiling points than alcohols due to the presence of dimer formation due to intramolecular hydrogen bonding which increases the strength of the vander waals dispersion forces.
In alcohols there is presence of intermolecular hydrogen bonding which is stronger as compared to dipole – dipole interactions in aldehydes.

Question 35.
When a conductivity cell was filled with 0.02 M KCl, it had a resistance of 82.4 ohm at 25°C and when filled with 0.005 N K₂SO₄ it had a resistance of 326 ohm. Calculate:
(A) Cell constant
(B) Specific conductivity, and
(C) Equivalent conductivity of 0.005 N K₂SO₄ solution. The specific conductance of 0.02 M KCl; is 0.002765 Ohm^-1 cm^-1.
OR
(A) Describe the elctrolysis of copper sulphate solution using copper electrode.
(B) In an electrolysis experiment, two series-
connected cells were subjected to a current for five hours. A solution of gold salt is present in the first cell, while a solution of copper sulphate is present in the second cell. The first cell received 9.85g of gold. Find the quantity of copper deposited on the cathode in the second cell if the oxidation -number of gold is +3. Calculate the current’s magnitude in amperes as well. [5]
Answer:
(A) Calculation of cell constant:
Resistance of 0.02 M KCl solution = 82.4 ohm
Specific conductivity of KCl = 0.002768 ohm^-1 cm^-1
∴ Conductance of 0.02 M KCl solution
= \(\frac{1}{82.4}\) ohm^-1
∴ Specific conductivity = Cell constant × Conductance
Specific conductivity
∴ Cell constant = \(\frac{\text { Specific conductivity }}{\text { Conductance }}\)
= \(\frac{0.002768}{1 / 82.4}\) = 0.2281 cm^-1

(B) Calculation of specific conductivity:
Resistance of 0.005 N K₂SO₄ solution = 326 ohm
∴ Conductance of 0.005 N K₂SO₄ solution
= \(\frac{1}{326}\) ohm^-1
Specific conductivity of 0.005N K₂SO₄ solution
= Cell constant × Conductance
= 0.2281 × \(\frac{1}{326}\)
= 6.997 × 10^-4 ohm^-1 cm^-1

(C) Calculation of equivalent conductivity: Equivalent conductivity is given by
∧_eq = \(k \times \frac{1000}{\text { Normality of solution }}\)
Therefore, the equivalent conductivity of K₂SO₄ solution
∧_eq = \(\frac{6.997 \times 10^{-4} \times 1000}{0.005}\)
= 139.94 ohm^-1 cm² eq^-1

OR

(A) CuSO₄ ⇌ Cu⁴⁺ + SO^2-₄
At cathode, copper is deposited.
Cu²⁺ + 2e^– → Cu
At anode, the copper of the electrode is oxidised to Cu²⁺ ions or SO₄^2- ions dissolve equivalent amount of copper of the anode.
Cu → Cu²⁺ + 2e^–
or Cu + SO^2-₄ → CuSO₄ + 2e^–
Thus, during electrolysis, copper is transferred from anode to cathode.

(B) We know that,
Mas of Au deposited Eq. mass of Au Mass of Cu deposited Eq. mass of Cu