CBSE Sample Papers for Class 12 Chemistry Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 3 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 CFSE for octahedral [CoCl₆]^4_– is x cm^-1.
The CFSE for tetrahedral [CoCl₄]^2- will be:
(a) x cm^-1
(b) less than x cm^-1
(c) greater than x cm^-1
(d) none of the above
Answer:
Less than x cm^-1

Explanation:
The crystal field splitting in the tetrahedral field is intrinsically smaller than in the octahedral field. For most purposes the relationship may be represented as ∆_t = – ∆₀

Question 2.
Fe³⁺ ion is more stable than Fe²⁺ because:
(a) more the charge on the atom, more the stability.
(b) configuration of Fe²⁺ is 3d⁶ while Fe³⁺ is 3d⁵
(c) Fe²⁺ has a larger size than Fe³⁺.
(d) Fe³⁺ ions are colored hence more stable.
Answer:
(b) configuration of Fe²⁺ is 3d⁶ while Fe³⁺ is 3d⁵

Explanation:
3d⁵ configuration is more stable due to singly occupied half-filled orbitals.

Question 3.
Identify A, B, C and D:

(a) A = C₂H₄, B = C₂H₅OH, C = C₂H₅NC, D = C₂H₅CN
(b) A = C₂H₅OH, B = C₂H₄, C = C₂H₅CN, D = C₂H₅NC
(c) A = C₂H₄OH, B = C₂H₄OH, C = C₂H₅CN, D = C₂H₅NC
(d) A = C₂H₅OH, B = C₂H₄, C = C₂H₅NC, D = C₂H₅CN
Answer:
(a) A = C₂H₄, B = C₂H₅OH, C = C₂H₅NC, D = C₂H₅CN

Explanation:

Question 4.
Schiff’s base is:
(a) a substituted imine
(b) a substituted amine
(c) a substituted nitrile
(d) none of the above
Answer:
(a) a substituted imine

Explanation:
Schiffs bases are aldehyde or ketone-like compounds in which the carbonyl group is replaced by an imine or azomethine group.

Question 5.
Complete the given analogy:
Methanoic acid: Rubber industry:: Sodium benzoate: ………………………..
(a) Food preservative industry
(b) Perfume industry
(c) Soap industry
(d) Leather industry
Answer:
(a) Food preservative industry

Explanation:
Sodium benzoate is best known as a preservative used in processed foods and beverages to extend shelf life.

Question 6.
Identify the final product of the reaction sequence.

(a) Benzophenone
(b) Acetophenone
(c) Diphenyl
(d) Methyl salicylate
Answer:
(b) Acetophenone

Explanation:

Question 7.
Match the items of Column I with items of Column II:

Column-I (Distinguish)	Column-II (Test)
(A) Methanol and Ethanol	(i) Lucas reagent
(B) Phenol and Cyclohexanol	(ii) Sodium metal
(C) Methanol and Diethyl ether	(iii) lodoform Test
(D) n-Butyl alcohol and tert-Butyl alcohol	(iv) Ferric chloride

(a) (A) – (iv), (B) – (i), (C) – (iii), (D) – (ii)
(b) (A) – (ii), (B) – (iii), (C) – (i), (D) – (iv)
(c) (A) – (iii), (B) – (iv), (C) – (ii), (D) – (i)
(d) (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i)
Answer:
(c) (A) – (Hi), (B) – (iv), (C) – (ii), (D) – (i)

Explanation:
(A) Methanol and Ethanol by Iodoform test,
(B) Phenol and Cyclohexanol by Ferric chloride,
(C) Methanol and Diethyl ether by Sodium metal,
(D) n-Butyl alcohol and tert-Butyl alcohol by Lucas Reagent

Question 8.
Reaction of toluene with a halogen in presence of lewis acid is an example of:
(a) Electrophilic elimination
(b) Nucleophilic substitution
(c) Electrophilic substitution
(d) Free radical addition
Answer:
(c) Electrophilic substitution

Explanation:
Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. Electrophile Cl⁺ attacks to electron rich benzene ring and replaces hydrogen. So, the reaction is the electrophilic substitution reaction.

Question 9.
Conductivity of 0.00241 M acetic acid is 7.896 x 10″s S cm”1. What is its molar conductivity?
(a) 18.5 S cm² mol^-1
(b) 28.6 S cm² mol^-1
(c) 19.2 S cm² mol^-1
(d) 32.76 S cm² mol^-1
Answer:
(d) 32.76 S cm² mol^-1

Explanation:
∧°_m = \(\frac{k \times 1000}{M}\)
= \(\frac{7.896 \times 10^{-5} \mathrm{Scm}^{-1} \times 1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}}{0.00241 \mathrm{molL}^{-1}}\)
= 32.76 S cm² mol^-1

Question 10.
Which of the following is a secondary allylic alcohol? [1]
(a) But-3-en-2-ol
(b) But-2-en-2-ol
(c) Prop-2-enol
(d) Butan-2-ol
Answer:
(a) But-3-en-2-ol

Explanation:

Question 11.
Distinguish between phenol and benzyl alcohol? [1]
(a) Add neutral FeCl₃. Phenol gives violet colour whereas benzyl alcohol does not.
(b) Fehling’s test
(c) Tollen’s test
(d) Iodoform test
Answer:
(a) Add neutral FeCl₃. Phenol gives violet colour whereas benzyl alcohol does not.

Explanation:
Ferric chloride can be used to distinguish between phenol and benzyl alcohol because, phenols react with neutral ferric chloride solution to give blue, violet or green colouration whereas alcohols do not react.

Question 12.
Which of the following cannot be made by using Williamson Synthesis? [1]
(a) Methoxybenzene
(b) Benzyl p-nitrophenyl ether
(c) tert-Butyl methyl ether
(d) Di tert-butyl ether
Answer:
(d) Di tert – butyl ether

Explanation:
In Williamson’s synthesis, primary alkyl halide is always taken otherwise alkene is the major product. Hence, di-tert- butyl ether cannot be made.

Question 13.
For water and nitric acid mixture which of the given graph is correct? [1]
(a)

(b)

(c) Both of these
(d) None of these
Answer:
(b)

Explanation:
Two liquids HNO₃ (A) and water (B) form a maximum boiling azeotrope when mixed in the ratio of 68% and 32% respectively because the forces between the particles in the mixture are stronger than the mean of the forces between the particles in the pure liquids.

Question 14.
Water- HCl mixture:
(I) shows positive deviations
(II) forms minimum boiling azeotrope
(III) shows negative deviations
(IV) forms maximum boiling azeotrope
(a) (I) and (II)
(b) (II) and (III)
(c) (III) and (IV)
(d) (II) and (III)
Answer:
(c) III and IV

Explanation:
Negative deviations from Raoult’s Law: New stronger forces must exist in the mixture than in the original liquids. These are cases where the molecules break away from the mixture less easily than they do from the pure liquids.

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: One molar aqueous solution has always higher concentration than one molal solution.
Reason: One molar solution contains less solvent than one molal solution. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
for aqueous solution,
density = \(\frac{\text { mass of solution }}{\text { Volume of solution }}\) = 1
∴ \(\frac{\text { mass of solvent }}{\text { Volume of solution }}\) < 1
∴ Molarity < Molality

Question 16.
Assertion: Fehling solution oxidises acetaldehyde to acetic acid but not benzaldehyde to benzoic acid. Reason: The C-H bond of CHO group in benzaldehyde is stronger than in acetaldehyde. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Fehling solution is weak oxidising agent and benzaldehyde is weak reducing agnet. Hence, it does not react with Fehling solution. Due to +R effect of benzene ring, the electron density in the carbonyl group of benzaldehyde increases. This in turn, increase the electron density in the C-H bond of aldehyde group. As a result, the C-H bond becomes stronger.

Question 17.
Assertion: The primary valency is ionizable and are satisfied by positive ions.
Reason: Central atom/ions in coordination compounds also acts as Lewis acid. [1]
Answer:
(d) (A) is false but (R) is true

Explanation:
The primary valency in a coordination compound is the number of negative ions which are equivalent to the charge on the metal ion.

Question 18.
Assertion: Electrolysis of NaCl solution gives chlorine at anode instead of O₂.
Reason: Formation of oxygen at anode requires overvoltage. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Generally, the one with the lower value is preferred but in this case chlorine is produced at anode. This is because formation of oxygen at anode requires overvoltage.

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.
Give the structures of (i) and (ii) in the following reactions:
(A)
Answer:

(B)
Answer:

OR

Give reasons for the following:
(A) Aniline does not undergo Fridel-Crafts reaction.
Answer:
Aniline does not undergo Friedel craft’s reactions because the reagent AlCl₃ (the Lewis acid which is used as a catalyst in Friedel crafts reaction), being electron deficient acts as a Lewis base, and attacks on the lone pair of nitrogen present in aniline to form an insoluble complex which precipitates out and the reaction does not proceed.

(B) Although -NH₂ is o- and p- directing group yet aniline on nitration gives a significant amount of m-nitroaniline. [2]
Answer:
Although -NH₂ is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline because nitration is carried out in an acidic medium (a mixture of concentrated HNO₃ and concentrated H₂SO₄). In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).

Question 20.
Write the major product in the following reactions: [2]
(A)
Answer:

(B)
Answer:

Question 21.
Give reason for the following: [2]
(A) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic? (atomic number of Ni = 28)
Answer:
Though both [NiCl₄]^2- and [Ni(CO)₄] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of Ligands. Cl^– is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl₄]^2- is paramagnetic.

(B) Chelating complexes are more stable than unchelated complexes.
Answer:
Chelating complex is more stable than unchelated complex because the ligand is attached to the metal ion from many points and form ring like structures. Thus, there is a strong force of attraction between the metal and the ligand.

Question 22.
The given curve is obtained when molar conductivity (∧_m) is plotted against the square root of concentration, c^1/2 for two electrolytes A and B. [2]

(A) How do you account for the increase in the motar conductivity of the electrolyte A on dilution?
Answer:
From the graph, electrolyte A is a strong electrolyte which is completely ionised in solution. With dilution, the ions are far apart from each other and hence the molar conductivity increases.

(B) As seen from the graph, the value of limiting molar conductivity (∧°_m) for electrolyte B cannot be obtained graphically. How can this value be obtained?
Answer:
Electrolyte B is weak electrolyte and to determine the value of limiting molar conductivity for electrolyte B, indirect method based upon Kohlrausch law of independent migration of ions is used.

Question 23.
In the following reaction:
Q + R → Products
The time taken for 99% reaction of Q is twice the time taken for 90% reaction of Q. The concentration of R varies with time as shown in the figure.

What is the overall order of the reaction? Give the units of the rate constant for the same. Write the rate expression for the above reaction.
OR
Rate constant for a first order reaction has been found to be 2.54 × 10^-3 s^-1. Calculate its three fourth life. [2]
Answer:
For reactant Q, t_99% = 2 × t_90%
Therefore, order of reaction w.r.t reactant Q = 1
From the graph, order of reaction w.r.t reactant, R = 0
So, overall order of reaction = 1 + 0 = 1
Unit of rate constant, k = time^-1
Rate expression: Rate = k[Q][R]⁰
or Rate = k[Q]

OR

k = 2.54 × 10^-3 sec^-1
For a first reaction:

Question 24.
(A) In the following pairs of halogen compounds, which would undergo S_N¹ reaction faster? Explain.

Answer:
In the given compounds, (I) is a primary halide while (II) is secondary halide.
We know that due to steric hindrance, secondary halide is more reactive compared to primary halide towards S_N¹ reaction.
Hence, (II) is secondary halide therefore undergoes S_N¹ reaction faster.

(B) Amongst the isomeric dihalobenzenes which isomer has the highest melting point and why? [2]
Answer:
Amongst the isomeric dihalobenzenes para-isomer has the highest melting point. This is due to greater symmetry of para- isomer that first in crystal lattice better as compared to ortho and meta isomers.

Question 25.
Deduce the acidic strength of following in decreasing order. [2]
(A) Acetic acid, monochloroacetic acid, trichloroacetic acid
Answer:
CCl₃COOH > CH₂ClCOOH > CH₃COOH

(B) p-Hydroxybenzoic acid, benzoic acid and p-toluic acid.
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 questions: [3]
(A) Propanamine and N, N-dimethyl- methanamine contains the same number of carbon atoms, even though propanamine has higher boiling point than N,N- dimethylmethanamine. Why?
Answer:
Due to presence of two H-atoms on N-atoms of 1° amines, they undergo extensive intermolecualar H-bonding while 3° amines due to absence of hydrogen atoms on the N-atom, do not undergo hydrogen bonding, propanamine has higer boiling point than N, N-dimethylmethanamine.

(B) Write the reaction involved in Hoffmann bromamide degradation reaction.
Answer:
When an amide is treated with Br2 in aqueous or ethanolic solution of NaOH, degradation of amide results in formation of primary amine.
Primary amine formed contains one carbon less than the number of carbon atoms in amide.
R-CO-NH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

(C) Give one chemical test to distinguish between aniline and ethanamine.
Answer:
Aniline forms diazonium salt with NaNO₂/ HCl which gives orange-red dye with β-naphthol. Ethylamine does not give this test.

Question 27.
A non reducing disaccharide ‘A’ on hydrolysis with dilute acid gives an equimolar mixture of D-(+)-glucose and D- (-) -fructose. [3]

(A) Identify (i). What is the mixture of D-(+)- glucose and D-(-)-fructose known as? Name the linkage that holds the two units in the disaccharide.
Answer:
(i) is sucrose (C₁₂H₂₂O₁₁).
The mixture of D-(+)- glucose and D-(-) -fructose is known as invert sugar. The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic linkage.

(B) α-Amino acids have relatively higher melting points than the corresponding halo acids. Explain.
Answer:
α-amino acids act as zwitter ions, (H₃N⁺– CHR-COO^–) or dipolar ions. Due to this dipolar salt-like structure, they have string dipole-dipole interactions. Therefore, their melting points are higher than the corresponding halo acids which do not exist as zwitter ions.

Question 28.
Answer any three questions: [3]
(A) Give an example of application of coordination compounds in medicinal chemistry.
Answer:
Cisplatin: [PtCl₂(NH₃)₂] is used in cancer chemotherapy.

(B) Give IUPAC name of [Co(en)₃]³⁺
Answer:
Tris (ethylenediamine)cobalt (III)

(C) Draw structure of [Mn₂(CO)₁₀].
Answer:
[Mn₂(CO)₁₀]

(D) Write the hybridization and shape of [Fe(CN)₆]^3-.
(Atomic number of Fe = 26)
Answer:
d² sp³ octahedral shape

Question 29.
(A) Give reason: [3]
(i) Highest oxidation state of a metal exhibited in its oxide or fluoride only.
Answer:
The highest oxidation state of a metal exhibited in its oxide or fluoride only since fluorine and oxygen are the most electronegative elements. The highest oxidation state shown by any transition element is + 8.

(ii) In volumetric analysis of KMnO₄ vs Mohr’s salt, H₂SO₄ and not HCl is added to Mohr’s salt solution to provide acidic medium.
Answer:
KMnO₄ which is the indicator, oxidises HCl to Cl₂ as HCl is a mild reducing agent. Due to which, the purpose of the experiment is not served. So, dilute sulphuric acid is used.

(B) Give the structure of Cr₂O₇^2-.
Answer:
The structure of Cr₂O₇^2- is:

Question 30.
Account for the following:
(A) Of the d4 species, Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidizing.
(B) Cu²⁺ is more stable in aqueous solution though Cu⁺ has d¹⁰ configuration.
(C) The d¹ configuration is highly unstable in ions.
OR
Following are the transition metal ions of 3d series:
Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following giving reason for your choice of answer:
(A) Which ion is most stable in an aqueous solution?
(B) Which ion is a strong oxidizing agent?
(C) Which ion is colourless? [3]
Answer:
(A) Cr²⁺ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr³⁺ (electronic configuration, d³). This d³ configuration can be written as t³_2g configuration, which is a more stable configuration. In the case of Mn³⁺ (d⁴), it acts as an oxidizing agent and gets reduced to Mn²⁺ (d⁵). This has an exactly half-filled d-orbital and is highly stable.

(B) Cu²⁺ has a smaller size than Cu⁺ as it contains 2 electrons less than Cu⁺.
So, due to its small size the charge of the ion remains concentrated and is not spread widely. Due to this accumulation of charge, the polar water molecules can easily make bonds with the Cu²⁺ ions and release a higher amount of energy, thus making the hydration enthalpy very high.

(C) The ions in d¹ configuration tend to lose one more electrons to get into stable d⁰ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

OR

(A) Cr³⁺, half filled t³_2g
(B) Mn³⁺, due to stable d⁵ configuration in Mn²⁺.
(C) Ti⁴⁺, No unpaired electrons so no d-d transition.

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.
Solutions are homogeneous mixtures of two or more than two components. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. If the vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law, then the solution exhibits positive deviation or negative deviation from Raoult’s law. Respectively some liquids on mixing, form azeotropes which are binary mixtures having the same composition in liquid and vapour phase. There are two types of azeotropes called minimum boiling azeotrope and maximum boiling azeotrope shown in the following graph. Some properties of the solution depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. such properities are called colligative properities.

(A) Name the example of the solution which will show positive deviation from Raoult’s law.
(B) What is azeotrope mixture?
(C) (i) If k_f value of H₂O is 1.86 K/kg/mol.
What is the value of ∆T_f for 0.1 m solution of non-volatile solute is?
(ii) What is the reason behind low concentration of oxygen in blood and tissues of people living at high altitudes?
OR
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non¬volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol^-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance? [2]
Answer:
(A) Methanol and acetone
Explanation:
Acetone + Methanol is an example of solution showing positive deviation from Raoult’s law as acetone- methanol attractions are weaker that acetone- acetone and methanol-methanol attractions.

(B) Azeotrope, a mixture of liquids that has a constant boiling point because the vapour has the same composition as the liquid mixture. The boiling point of an azeotropic mixture may be higher or lower than that of any of its components.

(C) (i) ∆T_f = K_f × molality = 1.86 × 0.1 = 0.186 K
(ii) low atmospheric pressure
Explanation:
Because at high altitude, the partial pressure of oxygen is less than partial pressure of oxygen at the ground level.

OR

P° = 0.850 bar, P_s = 0.845 bar
W_B = 0.5 g, M_B = ?
weight of solvent (benzene), W_A = 39.0 g
and molecular weight of benzene, M_A = 78 g

So, molecular mass of solid = 170 g.

Question 32.
Biomolecules, also called biological molecules, are one of numerous substances that are produced bg cells and living organisms. Biomolecules have a wide range of sizes and structures and perform a vast array of functions. The four major types of biomolecules are carbohydrates, lipids, nucleic acids and proteins.

Carbohydrates, which are made up primarily of molecules containing atoms of carbon, hydrogen, and oxygen, are essential energy sources and structural components of all life. They are among the most abundant biomolecules on Earth. Proteins are major structural elements of cells. They also serve as transporters, moving nutrients and other molecules in and out of cells and as enzymes and catalysts for the vast majority of chemical reactions that take place in living organisms. Nucleic acids, namely DNA and RNA, have the unique function of storing an organism’s genetic code.
(A) Why the concentration of zwitter ion is maximum for neutral amino acid at its isoelectric point? [1]
(B) Which compound is formed when glucose is reacted with NH₂OH? [1]
(C) Classify carbohydrates on the basis of taste.
OR
Describe the following nature of glucose with the help of reaction.
(i) Reducing agent
(ii) Oxidation [2]
Answer:
(A) The concentration of zwitter ion is maximum for neutral amino acid at its isoelectric point as the net charge is zero at this point.

(B) Oxime is formed.

(C) Sugars: Carbohydrates which are sweet in taste are called carbohydrates. They are soluble in water and are crystalline in nature. Some of the examples are glucose, sucrose and lactose etc.
Non-sugars: Carbohydrates which are sweet in taste are called non-sugar carbohydrates. They are insoluble in water and are amorphous in nature. Examples of non-sugars are starch, cellulose etc.

OR

(i) Since, D-glucose is oxidized easily, it acts as a reducing agent and also can reduce the Tollen’s reagent as well as Fehling’s solution.

(ii) On oxidation with a mild oxidizing agent such as bromine water, -CHO group converts to carboxylic acid group and a monocarboxylic acid is formed, the product is called D-gluconic acid.

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) Write Etard reaction.
(B) Arrange the following in increasing order of their reactivity towards nucleophilic addition reaction:

(C) (A) and (B) are two functional isomers of compound C₃H₆O. On heating with NaOH and I₂, isomer (B) forms yellow precipitate of iodoform whereas isomer (A) does not form any precipitate. Write the formula of (A) and (B).
(D) Predict the products of following reactions:
(i)
(ii) [5]
Answer:
(A) Etard Reaction:

(B) Nucleophilic addition order: III > I > II.
Explanation: Nucleophiles are negatively entity positivity of carbon is carbonyl group facilitates nucleophilic addition.

(C) The functional isomers are CH₃COCH₃[A] and CH₃CH₂CHO[B]. The isomer [A] has a methyl ketonic group (CH₃CO-) and responds to iodoform test while the isomer [B] does not since this group is not present.

(D)

Question 34.
(A) State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decreases with dilution?
(B) Calculate the E°ceu of the following cell at 298K:
2Al(s) + 3Cu²⁺_(aq) (0.01M) → 2Al³⁺_(aq) (0.01M) + 3Cu_(s)
Given that E_cell = 1.98V
OR
(A) Define electrochemical cell. What happens if external potential applied becomes greater than E°ceu of electrochemical cell?
(B) Conductivity of 2.5 × 10^-4 M methanoic acid is 5.25 × 10^-5 S/cm. Calculate its molar conductivity and degree of dissociation.
(Given: λ°_(H⁺) = 349.5 Scm²/mol and λ°_(HCOO^–) = 50.5 Scm²/mol) [5]
Answer:
(A) Kohlrausch law of independent migration of ions: The law states that limiting molar conductivity of an electrolyte can be represented by the sum of the individual contributions of the anion and cation of the electrolyte.
∧⁰ = ∨₊λ°₊ + ∨_–k°_–
On dilution, the conductivity (k) of the electrolyte decreases as the number of ions per unit volume of solution decreases.

(B) Al(s) | Al³⁺(ag) (0.01M) || Cu²⁺(ag) (0.01 M) | Cu(s)
LHE [(Al(s) → Al³⁺(ag)] + 3e^–)] × 2 (Oxidation at anode)
RHE [Cu²⁺ (aq) + 2e^– → Cu(s)] × 3 (Reduction at cathode)
∴ n = 6

OR

∧_m = \(\)S cm² mol^-1
∧_m = \(\)S cm² mol^-1
= 210 S cm² mol^-1
∧⁰_m = λ⁰(HCOO^–) + λ⁰(H⁺)
= (50.5 + 349.5) S cm² mol^-1
= 400 S cm² mol^-1
α = ∧_m/∧⁰_m
α = 210/400 = 0.525

Question 35.
(A) consider the following data for reaction between A and B to give C:

Determine the order of reaction with respect to A and with respect to B and the overall order of reaction.
(B) If half life for a zero order reaction is 2 years with initial concentration of reactant being 0.004M, what will be its rate constant?
OR
(A) Write integrated rate equation for first order reaction and also identify the units of rate constant.
(B) Draw a graph between time and concentration of reactant for zero order reaction. What does the slope of graph signify?
(C) Give an example of pseudo first order reaction. [5]
Answer:
(A) We know that
Rate = k[A]^x [B]^y
case I:
2.1 × 10^-2 = k[0.1]^x [1.00]^y
case II:
8.4 × 10^-3 = k[0.2]^x [1.00]^y
Eq. (1) ÷ Eq. (2)

So, order of reaction w.r.t. A = 2
Now, from case II:
(8.4 × 10^-3) = (0.2)^z (2.00)^y …… (3)
Eq. (3) ÷ Eq. (1)
\(\frac{8.4 \times 10^{-3}}{2.1 \times 10^{-3}}=\frac{(0.2)^z(2.00)^y}{(0.1)^z(1.00)^y}\)
⇒ 4 = (2)^z × 2^y
⇒ 2^y = 1
⇒ 2^y = 2⁰
So, order of reaction w.r.t. B = 0 And overall order of reactions
n = x + y
⇒ n = 2 + 0 = 2
⇒ n = 2.

(B) Given [RQ] = 0.004 M, t_1/2 = 2 years
k = \(\frac{\left[R_0\right]}{2 \times t_{1 / 2}}\)
⇒ k = \(\frac{[0.004]}{2 \times 2}\)
So, k = 0.001 mol L^-1 yr^-1

OR

(A) Integrated rate equation for first order reaction
k = \(\frac{2.303}{t}\) log\(\frac{\left[R_0\right]}{[R]}\)
where [R_o] = Initial cone, of reactant
and [R] = Final cone, of reactant unit of rate constant for first order reac.
= s^-1 (time^-1)

(B) Plot showing variation in the concentration vs time for a zero-order reaction is represented as follows:
Concentration decrease linearly with a constant slope with time for zero-order reaction.

Slope of graph represents k.

(C) In the reaction,
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆,
water is present in large excess. Hence, it will not appear in the rate law expression. Thus, it is an example of pseudo first order reaction.