Maharashtra Board Class 12 Chemistry Sample Paper Set 5 with Solutions

Maharashtra State Board Class 12th Chemistry Sample Paper Set 5 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Chemistry Model Paper Set 5 with Solutions

General Instructions:

The question paper is divided into four sections.

1. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each. Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
2. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. 15 to Q. No. 26 are Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each. (Attempt any Three).
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each MCQ, correct answer must be written along with its alphabet.
   E.g., (a) ……/(b) ……./(c) ……./(d) ……… Only first attempt will be considered for evaluation.
8. Draw well labeled diagrams and write balanced equations wherever necessary.
9. Given data:
   Atomic mass of C = 12, H = 1, O = 16,
   Atomic number (Z): V = 23, Zn = 30, Eu = 63, Yb = 70, Ar = 1.
   R = 8.314 JK^-1 mol^-1 or 0.083 L bar K^-1 mol^-1
   N_A = 6.022 × 10²³ mol^-1, F = 96500 C

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [10 Marks]

i. Among the known interhalogen compounds, the maximum number of atoms is ____.
(A) 3
(B) 6
(C) 7
(D) 8
Answer:
(D) 8

ii. The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume) ____.
(A) 5.41%
(B) 3.54%
(C) 4.53%
(D) 53.4%
Answer:
(A) 5.41%

Explanation:
M₂ = \(\frac{W_2 R T}{\pi V}\)
\(\frac{W_2}{V}\) = \(\frac{\pi M_2}{R T}\) \(=\frac{7.65 \mathrm{~atm} \times 180 \mathrm{~g} \mathrm{~mol}^{-1}}{0.08205 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 310 \mathrm{~K}}\)
= 54.1 g dm^-3 = 5.41 g cm^-3 = 5.41%

iii. The conjugate base of [Zn(H₂O)₄]²⁺ is
(A) [Zn(H₂O)₄]^2-.
(B) [Zn(H₂O)₃]^2-
(C) [Zn(H₂O)₃OH]⁺
(D) [Zn(H₂O)H]³⁺
Answer:
(C) [Zn(H₂O)₃OH]⁺

iv. Which of the following is NOT present in DNA?
(A) Adenine
(B) Guanine
(C) Thymine
(D) Uracil
Answer:
(D) Uracil

v. Compound having general formula

is called ____.
(A) Diester
(B) Acid anhydride
(C) Hemiacetal
(D) Acetal
Answer:
(D) acetal

vi. Mohr’s salt is _____.
(A) Ferrous ammonium sulphate
(B) Ferrous sulphate
(C) Ammonium sulphate
(D) Ferric sulphate
Answer:
(A) Ferrous ammonium sulphate

Explanation:
Mohr’s salt = FeSO₄(NH₄)₂SO₄.H₂O

vii. The standard potential of the cell in the following reaction is ____.

(\(\mathrm{E}_{\mathrm{Cd}}^0\) = -0.403V, \(\mathrm{E}_{\mathrm{Cu}}^0\) = 0.334 V)
(A) -0.737 V
(B) 0.737 V
(C) -0.069 V
(D) 0.069 V
Answer:
(B) 0.737 V

viii. The packing efficiency of ccp lattice is ____.
(A) 74%
(B) 68%
(C) 52.4%
(D) 26%
Answer:
(A) 74%

ix. Which of the following is γ-isomer of BHC?
(A) DDT
(B) Lindane
(C) Chloroform
(D) Chlorobenzene
Answer:
(B) Lindane

x.

Answer:
(D)

Question 2.
Answer the following: [8 Marks]

i. Define path function. Give examples.
Answer:
The properties which depend on the path are called path functions.
e.g. work (W) and heat (Q).

ii. How is van’t Hoff factor related to degree of ionization?
Answer:
The van’t Floff factor is related to degree of ionization as follows:
i = 1 + α(n – 1)
or
α = \(\frac{i-1}{n-1}\)
where, α = Degree of ionization/dissociation
i = van’t Floff factor
n= Moles of ions obtained from ionization of 1 mole of electrolyte.

iii. Define rate of reaction.
Answer:
The rate of reaction represents a decrease in concentration of the reactant per unit time or increase in concentration of product per unit time.

iv. Why are alkyl halides generally not prepared by free radical halogenation of alkanes?
Answer:
Free radical halogenation of alkanes leads to the formation of mixture of mono and poly halogen compounds. Hence, free radical halogenation of alkanes is not suitable for preparation of alkyl halides.

v. Write structure of optically active alcohol having molecular formula C₄H₁₀O.
Answer:
The optically active alcohol having molecular formula C₄H₁₀O is

vi. Write the position of zinc (Z = 30) in the periodic table and write its electronic configuration.
Answer:
Zinc (Zn) is placed in the 4^th period and group 12 of the modern periodic table.
Electronic configuration of zinc: (Z = 30) is [Ar] 3d¹⁰ 4s²

vii. Write structural formula of synthetic SBR rubber.
Answer:
SBR rubber:

vii. A solid is hard, brittle and electrically nonconductor. Its melt conducts electricity. What type of solid is it?
Answer:
Ionic solids are hard and brittle. They are nonconductors of electricity in solid state. However, they are good conductors when melted. Hence, the given solid is an ionic solid.

Section – B

Attempt any EIGHT of the following questions: [16 Marks]

Question 3.
Write the equation relating boiling point elevation to the concentration of solution.
Answer:
The boiling point elevation is directly proportional to the molality of the solution. Thus,
ΔT_b ∝ m
∴ ΔT_b = K_b m
where, m is the molality of solution. The proportionality constant K_b is called boiling point elevation constant or molal elevation constant or ebullioscopic constant.

Question 4.
Differentiate between ideal and nonideal solutions.
Answer:

Ideal solutions	Non – ideal solutions
i. Ideal solutions obey Raoults law over entire range of concentrations.	Nonideal solutions do not obey Raoult’s law over the entire range of concentrations.
ii. The vapour pressure of ideal solution always lies between vapour pressures of pure components.	The vapour pressures of these solutions can be higher or lower than those of pure components.
iii. Solvent-solute, solute-solute and solvent-solvent molecular interactions are comparable.	Solute-solvent intermolecular attractions may be weaker or stronger than those between solute-solute molecules and solvent-solvent molecules.
iv. e.g. Benzene + toluene	e.g. Ethanol + acetone, Carbon disulphide + acetone, Phenol + aniline, Chloroform + acetone

Question 5.
Calculate the pH of a buffer solution composed of 0.1 M weak base BOH and 0.2 M of its salt BA.
[K_b = 1.8 × 10^-5 for the weak base]
Answer:
Given: [Base] = 0.1 M, [Salt] = 0.2 M, K_b = 1.8 × 10^-5 for the weak base
To find: pH of the buffer so

Formulae:
i. pOH = pK_b + log₁₀\(\frac{[\text { salt }]}{[\text { tase] }}\)
ii. pH + pOH = 14

Calculation:
pOH of basic buffer is given by Henderson-Hasselbalch equation:
pOH = pK_b + log₁₀\(\frac{[\text { salt }]}{[\text { base }]}\)
pK_b = -log₁₀K_b
= -log₁₀(1.8 × 10^-5) = 5 – log₁₀ 1.8
= 5 – 0.2553 = 4.7447
Substitution in the Henderson-Hasselbalch equation gives
pOH = 4.7447 + log_1o\(\frac{0.2}{0.1}\) = 4.7447 + l0g₁₀2
= 4.7447 + 0.3010 = 50457
From formula (ii),
pH = 14 – pOH = 14 – 5.0457 = 8.9543 ≈ 8.95

Ans: The pH of the given buffer solution iS 8.95.

Question 6.
Write a short note on internal energy.
Answer:

1. Every substance is associated with a definite amount of internal energy.
2. The internal energy of a system is made up of kinetic and potential energies of individual particles of the system. The change in internal energy is given as:
   ∆U = U₂ – U₁
   where, U₁ and U₂ are internal energies of initial and final states, respectively.
3. U is a state function and extensive property.
4. A transfer of energy (as heat or work) from the system would change its internal energy.

To know ∆U the energy supplied to or removed from the system need to be monitored.

a. The energy transferred to the system by heating it or performing work on it is added to the system.
b. The energy transferred from the system by cooling or by performing work on the surroundings is removed from the system.

Question 7.
Write features of reversible process.
Answer:
Features of reversible process:

1. The driving and opposing forces differ by an infinitesimal amount.
2. The process can be reversed by an infinitesimal change in conditions.
3. A reversible process proceeds infinitely slowly and takes place in infinite number of steps.
4. At the end of every step of the process, the system attains mechanical equilibrium with the surroundings.

Question 8.
Write the Nernst equation and explain the terms involved.
Answer:
For any general reaction,
aA + bB → cC + dD
Nernst equation is given by

where
n = moles of electrons used in the reaction,
F = Faraday = 96500 C,
T = temperature in kelvin,
R = gas constant = 8.314 J K^-1 mol^-1.

Question 9.
Mention the names of various steps involved in the extraction of pure metals from their ores.
Answer:
The various steps involved in the extraction of pure metals from their ores are as follows:

1. Concentration of ores
2. Conversion of ores into oxides or other desired compounds
3. Reduction of ores to form crude metals
4. Refining of metals

Question 10.
Iron exhibits +2 and +3 oxidation states. Write their electronic configurations. Which will be more stable? Why?
Answer:

1. Electronic configuration of Fe²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
2. Electronic configuration of Fe³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
3. Fe³⁺ is more stable than Fe²⁺. This is due to the presence of half filled ‘d’ orbital in Fe³⁺.

Question 11.
Give reason: Reactions involving Grignard reagent must be carried out under anhydrous conditions.
Answer:

1. Grignard reagents are highly reactive compounds.
2. They react with water or compounds containing hydrogen attached to electronegative element.

Hence, reactions involving Grignard reagent must be carried out under anhydrous conditions.

Question 12.
Convert but-l-ene to n-butyl iodide.
Answer:
But-1-ene to n-butyl iodide:

Question 13.
Classify the following complexes as homoleptic and heteroleptic
[CO(NH₃)₅Cl]SO₄, [Co(ONO)(NH₃)₅]Cl₂, [CoCl(NH₃)(en)₂]²⁺ and [Co(C₂O₄)₃]^3-.
Answer:

1. Homoleptic complexes: [Co(C₂O₄)₃]^3-.
2. Heteroleptic complexes: [Co(NH₃)₅Cl]SO₄, [Co(ONO)(NH₃)₅]Cl₂, [CoCl(NH₃)(en)₂]²⁺

Question 14.
For the reaction 2A + B → products, find the rate law from the following data.

Answer:
From above observations (i) and (ii)
0.15 = (0.3)^x(0.05)^y …. (1)
0.30 = (0.6)^x(0.05)^y … …. (2)
Dividing (2) by (1)
\(\frac{0.30}{0.15}\) = 2 = \(\frac{(0.6)^x(0.05)^y}{(0.3)^x(0.05)^y}\) = \(\left(\frac{0.6}{0.3}\right)^x\) = 2^x
Hence, x – 1
From observation (i) and (iii) separately in the rate law gives
0.15 = (0.3)^x(0.05)^y …….. (3)
1.20 = (0.6)^x(0.2)^y ………. (4)
Dividing (4) by (3)
\(\frac{1.20}{0.15}\) = \(\frac{0.6}{0.3}\left(\frac{0.2}{0.05}\right)^y\) (∵ x = 1)
∴ 8 = \(2\left(\frac{0.2}{0.05}\right)^y\) = 2 × 4^y
∴ 4 = 4^y
Therefore, y = 1

Ans: The rate law is rate = k[A][B].

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
Explain in detail the applications of nanotechnology/nanomaterials in various discipline.
Answer:
Nanochemistry has contributed to number of innovative products in various disciplines because of their unique physical, chemical, optical, structural, catalytic properties, etc.

Following are the few applications:

i. Nanoparticles can contribute to stronger, lighter, cleaner and smarter surfaces and systems. They are used in the manufacturing of scratchproof eyeglasses, transport, sunscreen, crack resistant paints and so on.

ii. Used in electronic devices, e.g. Magnetoresistive Random Access Memory (MRAM).

iii. Nanotechnology plays an important role in water purification techniques.

a. Water contains waterborne pathogens like viruses, bacteria.
b. Cost-effective filter materials coated with silver nanoparticles (AgNps) is an alternative technology and can be used in water purification.
c. Silver nanoparticles act as highly effective antibacterial agent which kills E. coli from water.

iv. Self-cleaning materials:

a. Lotus is an example of self-cleaning. The lotus plant (Nelumbo nucifera) although grows in muddy water, its leaves always appear clean.
b. The plants’ leaves are super-hydrophobic.
c. Nanostructures on lotus leaves repel water which carries dirt as it rolls off. Lotus effect is the basis of self-cleaning windows.

Question 16.
Calculate the amount of work done in the following cases.
i. Oxidation of one mole HCl_(g) at 200 °C according to the reaction 4HCl_(g) + O_2(g) → 2Cl_2(g) + 2H₂O_(g)
ii. Decomposition of one mole of NO at 300 °C for the reaction 2NO_(g) → N_2(g) + O_2(g)
Answer:
i. Oxidation of 1 mole HCl_(g)
Temperature T = 200 °C = 473 K

ii. Decomposition of one mole of NO
Temperature = T = 300 °C = 573 K
To find: Work done

Formula: W : -∆n_gRT

Calculation:

i. The given reaction is for 4 moles of HCl. For 1 mole of HCl, the reaction is given as follows:

ii. The given reaction is for 2 moles of NO. For 1 mole of NO, the reaction is given as follows:

Ans:

i. The work done is +983 J. The work is done on the system.
ii. The work done is O kJ. There is no work done.

Question 17.
Aqueous solution of sodium carbonate is alkaline whereas aqueous solution of ammonium chloride is acidic. Explain.
Answer:
i. Sodium carbonate (Na₂CO₃) is a salt of weak acid H₂CO₃ and strong base NaOH. When dissolved in water, it dissociates completely.
Na₂C0_3(aq) → 2N\(\mathrm{Na}_{(\mathrm{qq})}^{+}\) + \(\mathrm{CO}_{3(\mathrm{q})}^{2-}\)

ii. The Na+ ions of salt have no tendency to react with OH^– ions of water since the possible product of the reaction is NaOH, a strong electrolyte.

iii. On the other hand, the reaction of \(\mathrm{CO}_{3}^{2-}\) ions of salt with the H₃O⁺ ions from water produces unionized H₂CO₃.

iv. As a result of excess OH ions produced, the resulting solution of Na₂CO₃ is alkaline.

v. Similarly, ammonium chloride (NH₄Cl) is salt of strong acid HCl and weak base NH₄OH.
When NH₄Cl is dissolved in water, it dissociates completely as,
NH₄Cl_(aq) → \(\mathrm{NH}_{4(a q)}^{+}\) + \(\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

vi. \(\mathrm{Cl}_{(\mathrm{aq})}^{-}\) ions of salt have no tendency to react with water because the possible product HCl is strong electrolyte.

vii. The reaction of \(\mathrm{NH}_4^{+}\) ions with OH^– ions form unionized NH₄OH. The hydrolytic equilibrium for NH₄Cl is then written as,

viii. Due to the presence of excess of H₃O⁺ ions, the resulting solution of NH₄Cl is acidic.

Question 18.
i. What is the relationship between coefficients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coefficients are the exponents?
Answer:
Coefficients of reactants ¡n a balanced chemical equation may or may not be the same as the exponents in rate law for the same reaction. For elementary reaction, coefficients in a balanced chemical equation are same as the exponents in rate law.

ii. Distinguish between order and molecularity of a reaction.
Answer:

Order	Molecularity
i. It is an experimentally determined property.	It is a theoretical entity.
ii. It is the sum of powers of the concentration terms of reactants those appear in the rate equation.	It is the number of reactant molecules taking part in an elementary reaction.
iii. It may be an Integer, fraction or zero.	IT is an integer.
iv. It is not based on balanced chemical equation.	It is based on balanced chemical equation.

Question 19.
Explain optical isomerism in 2-chlorobutane.
Answer:

1. The stereoisomerism in which the isomers have different spatial arrangements of groups/atoms around a chiral atom is called optical isomerism.
2. 2-chlorobutane has one chiral carbon atom. The spatial arrangement of the four different groups around the chiral atom is different.
3. Structure of 2-chlorobutane and its mirror image can be represented as,
   
4. 2-Chlorobutane cannot superimpose perfectly on its mirror image as shown in the figure.

Hence, 2-chlorobutane exhibits optical isomerism.

Question 20.
Europium and ytterbium behave as good reducing agents in +2 oxidation state. Explain.
Answer:
i. Europium (Eu) and ytterbium (Yb) show +2 oxidation states.
Their electronic configurations are given below:
Eu [Xe] 4f⁷ 6s² : Eu²⁺ = [Xe] 4f⁷
Yb = [Xe] 4f¹⁴ 6s² ; Yb²⁺ = [Xe] 4f¹⁴

ii. It is clear from the configuration of Eu that Eu²⁺ is favoured by its half filled f-subshell. But it can be easily converted into stable Eu³⁺ by loss of an electron. Due to this reason, Eu²⁺ is a good reducing agent.

iii. Similarly, Yb²⁺ ion is stabilized due to completely filled f-subshell. It can be easily converted into Yb³⁺ by loss of an electron. Due to this reason, Yb²⁺ is a good reducing agent.

Question 21.
Write reaction showing conversion of p-bromoisopropylbenzene into p-isopropylbenzoic acid. ‘
Answer:

Question 22.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by direct reaction of xenon and fluorine in different ratios and conditions, such as temperature, electric discharge and photochemical reaction.

Question 23.
Explain in detail free radical mechanism involved during preparation of addition polymer.
Answer:
Free radical mechanism is most common in addition polymerization. It ¡s also called chain reaction which involves three distinct steps. These are as follows:

i. step 1: Chain initiation:

a. The chain reaction is initiated by a free radical which is formed from an initiator (catalyst) such as benzoyl peroxide, acetyl peroxide, tert-butyl peroxide, etc.

b. For example, acetyl peroxide generates methyl radical as shown below:

c. The free radical (say ) so formed attaches itself to the olefin (vinyl monomer) and produces a new radical, made up of two parts, namely, the attached radical and the monomer unit.

Question 24.
i. Give examples of common disaccharides.
ii. Give reason: Hydrolysis of sucrose is called inversion.
Answer:
i. Examples of common disaccharides: Sucrose, maltose and lactose. [1 Mark]

ii. a. Sucrose (C₁₂H₂₂O₁₁) is dextrorotatory (+66.5°). On hydrolysis with dilute acid or an enzyme called invertase, sucrose gives equimolar mixture of D-(+)-glucose and

b. Since the laevoratotion of fructose (-92.4°) is larger than the dextrorotation of glucose (+52.7°), the hydrolysis product has net laevorotation.

Hence, hydrolysis of sucrose is also called inversion of sucrose.

Question 25.
An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (Density of an element = 14.44 g cm^-3)
Answer:
Given: Type of unit cell is bcc.
Edge length (a) = 288 pm
= 2.88 × 10^-8 cm,
Mass of element (x) = 200 g

To find:

i. Number of unit cells in 200 g of element
ii. Number of atoms in 200 g of element

Formulae:

i. Number of atoms in x g of element = \(\frac{x n}{\rho a^3}\)
ii. Number of unit cells in x g of element = \(\frac{x}{\rho a^3}\)

Calculation:

i. For bcc unit cell, n = 2.
Using formula (i),
Number of atoms in 200 g of element
= \(\frac{200 \mathrm{~g} \times 2}{14.44 \mathrm{~g} \mathrm{~cm}^{-3} \times\left(2.88 \times 10^{-8} \mathrm{~cm}\right)^3}\) = 1.16 × 10²⁴

ii. Using formula (ii),
Number of unit cells in 200 g element
= \(\frac{200 \mathrm{~g}}{14.44 \mathrm{~g} \mathrm{~cm}^{-3} \times\left(2.88 \times 10^{-8} \mathrm{~cm}\right)^3}\) = 5.80 × 10²³

Ans:

i. Number of atoms in 200 g element is 1.16 × 1024.
ii. Number of unit cells in 200 g element is 5.80 × 10²³.

Question 26.
What are Freons? State uses and effects of Freons.
Answer:
Freons:

1. These are organic compounds of chlorine and fluorine, chlorofluorocarbons (CFC’s).
2. The most common representative is dichlorodifluoromethane (Freon-12). Others include chlorodifluoromethane (R-22), trichlorofluoromethane (R-11), etc.
3. Lises: Freons are used,
   • as refrigerants in fridge and air-conditioning, propellants in aerosol and solvents.
   • as blowing agents in making foams and packing materials.
4. Effects:
   • Chlorofluorocarbons are responsible for ozone depletion of ozone in stratosphere.
   • Regular large inhalation of freons results in breathing problems, organ damage, loss of consciousness.

Section – D

Attempt any THREE of the following questions:

Question 27.
i. Write the IUPAC names of the following structures:

a.

b.

ii. How will you prepare:
a. Propiophenone from propanenitrile
b. 4-Methoxyacetophenone from anisole
Answer:
i.

ii.

Question 28.
Give valence bond description for bonding in the complex [VCl₄]^– Draw box diagrams for free metal ion. Which hybrid orbitals are used by the metal? State the number of unpaired electrons.
Answer:
i. Oxidation state of vanadium is +3

ii. Valence shell electronic configuration of free metal ion, V³⁺

iii. Number of cl^– ligands is 4. Therefore, the number of vacant metal ion orbitals required for bonding with ligands must be four.

iv. Four orbitals on metal available for hybridisation are one s and three 4p. The complex is tetrahedral

v. The four metal ion orbitals for bonding with Cl^– ligands are derived from the sp³ hybridization.

vi. Four vacant sp³ hybrid orbitals of V³⁺ overlap with four orbitals of Cl^– ions.

vii. Configuration after complex formation would be

viii. The complex has two unpaired electrons.
The structure of [VCl₄]^– is

Question 29.
i. What is aqua regia? How does it react with noble metals?
ii. Give two reactions showing oxidising property of concentrated H₂SO₄.
Answer:
i.

a. A mixture of three parts of conc. hydrochloric acid and one part of conc. HNO₃ is known as aqua regia. (1 MarkJ
b. Noble metals like gold. platinum get dissolved in aqua regia.

ii. Metals and nonmetals both are oxdised by hot, concentrated sulfuric acid which itself gets reduced to SO₂.

Question 30.
i. Construct a galvanic cell from the electrodes Co³⁺| Co and Mn²⁺ | Mn.
\(\mathrm{E}_{\mathrm{C}_0}^{\circ}\) = 1.82,V, \(\mathrm{E}_{\mathrm{M}_n}^{\circ}\) = -1.18 V. Calculate \(\mathrm{E}_{\mathrm{cell}}^{\circ}\).
ii. Derive the relationship between standard cell potential and equilibrium constant of cell reaction.
Answer:
i. Given: \(E_{c_0}^0\) = 1.82 V, \(E_{M n}^{\circ}\) = -1.18 V
To find \(\mathrm{E}_{\mathrm{cell}}^{\circ}\) and cell representation

Formula:
\(E_{\text {cell }}^{\circ}\) = \(E_{\text {Cathode }}^0\) = \(E_{\text {anode }}^{\circ}\)

Calculation:

Ans: The standard cell potential is 3.00 V.

ii. The relation between standard Gibbs energy change of cell reaction and standard cell potential is given by
-∆G° = \(n F E_{\text {cell }}^o\) …… (1)
The relation between standard Gibbs energy change of a chemical reaction and its equilibrium constant as given in thermodynamics is:
∆G° = -RT ln K
Combining equations (1) and (2), we have
\(-n F E_{c e l l}^o\) = -RT ln K
∴ \(\mathrm{E}_{c e l l}^0\) = \(\frac{R T}{n F}\) ln K = \(\frac{2.303 \mathrm{RT}}{n F}\) log₁₀ K = \(\frac{0.0592}{n}\) log₁₀ K at 25°C

Question 31.
i. What is diazotisation? Write diazotisation reaction of aniline,
ii. Draw resonance structures of aryl diazonium salt.
Answer:
i. Aliphcitic/aromatic primary amines react with nitrous acid to form corresponding diazonium salts. This reaction is called as diazotisation.

ii. The resonance structures of aryl diazonium salt can be given as,

Maharashtra Board Class 12 Chemistry Previous Year Question Papers