12th Chemistry Question Paper 2023 Maharashtra Board Pdf

Maharashtra State Board Class 12th Chemistry Question Paper 2023 with Solutions Answers Pdf Download.

Class 12 Chemistry Question Paper 2023 Maharashtra State Board with Solutions

Time: 3 Hrs
Max. Marks: 70

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.
   Only the first attempt will be considered for evaluation.
   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. Given: R = .8.314 J.K^-1mol^-1
   N_A = 6.022 × 10²³
   F = 96500 C

Section-A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: (10)

(i) The relation between radius of sphere and edge length in body centered cubic lattice is given by formula:
(a) \(\sqrt{3 r}\) = 4a
(b) r = \(\frac{\sqrt{3}}{a} \times 4\)
(c) r = \(\frac{\sqrt{3}}{4} a\)
(d) r = \(\frac{\sqrt{2}}{4} \times a\)
Answer:
(c) r = \(\frac{\sqrt{3}}{4} a\)

(ii) The pH ofweakmonoacidicbase is 11.2, its OH^– ion concentration is:
(a) 1.585 × 10^-3 mol dm^-3
(b) 3.010 × 10^-11 mol dm^-3
(c) 3.010 × 10^-3 mol dm^-3
(d) 1.585 × 10^-11 mol dm^-3
Answer:
(a) 1.585 × 10^-3 mol dm^-3

(iii) Which of the following correctly represents integrated rate law equation for a first order reaction in gas phase:
(a) \(\frac{2.303}{t} \times \log _{10} \frac{P_i}{P_i-P}\)
(b) \(\frac{2.303}{t} \times \log _{10} \frac{P_i}{2 P_i-P}\)
(c) \(\frac{2.303}{t} \times \log _{10} \frac{2 P_i}{P_i-P}\)
(d) \(\frac{2.303}{t} \times \log _{10} \frac{P_i-P}{2 P_i}\)
Answer:
(b) \(\frac{2.303}{t} \times \log _{10} \frac{P_i}{2 P_i-P}\)

(iv) The spin only magnetic moment of Mn²⁺ ion is:
(a) 4.901 BM
(b) 5.916 BM
(c) 3.873 BM
(d) 2.846 BM
Answer:
(b) 5.916 BM

(v) The correct formula of a complex having IUPAC nametetraamminedibromoplatinum(IV) bromide is:
(a) [Pt Br(NH₃)₄]Br₂
(b) [PtBr₂(NH₃)₄]Br
(c) [Pt Br₂(NH₃)₄]Br₂
(d) [PtBr(NH₃)₄]Br
Answer:
(c) [Pt Br₂(NH₃)₄]Br₂

(vi) The allylic halide, among the following is:


Answer:
(d) CH₂=CH-CH₂-X

(vii) The product of following reaction is:

Answer:
(b) CH₃-CH=CH-CH₂-CH₂-OH

(viii) Ozonolysis of 2, 3 dimethyl but-2-ene, followed by decomposition by Zn dust and water gives:
(a) acetaldehyde
(b) propionaldehyde and acetone
(c) acetone
(d) acetaldehyde and butyraldehyde
Answer:
(c) acetone

ix) The glycosidic Linkage present in maltose is:
(a) α, ß -1,2-glycosidic linkage
(b) α-,4-glycosidic linkage
(c) ß-1,4-glycosidic Linkage
(d) α-1,6-glycosidic linkage
Answer:
(b) α-,4-glycosidic linkage

(x) The monomer of natural rubber is:
(a) Isoprene
(b) Acrylonitrile
(c) E-Caprolactam
(d) Tetrafluoroethylene
Answer:
(a) Isoprene

Question 2.
Answer the following questions: (8)
i) Write the name of the technique used to know geometry of nanoparticles.
Answer:
X-ray diffraction (XRD)

(ii) Write the name of product formed by the action of LiAlH₄ ether on acetamide.
Answer:
Ethylamine OREthanamine

iii) Write the structure of the product formed when chlorobenzene is treated with sodium metal in the presence of dry ether.
Answer:

iv) Write the chemical composition of cryolite.
Answer:
Na₃AlF₆

v) Write the name of platinum complex used in the treatment of cancer.
Answer:
Cisplation

(vi) Write the SI unit of cryoscopic constant.
Answer:
K Kg mol^-1 OR °C Kg mol^-1

(vii) Write the correct condition for spontaneity in terms of Gibbs energy.
Answer:
∆G < 0 (∆G is negative)

(viii) Calculate molar conductivity for 0.5 M BaCl₂ if its conductivity at 298 K is 0.01 Ω^-1 cm^-1
Answer:
Molar coductivity(Λ_m)
= \(\frac{1000 \times \text { Conductivity (K) }}{\text { Molarity (C) }}\)
= \(\frac{1000 \times 0.01 \Omega^{-1} \mathrm{~cm}^{-1}}{0.5}\)
= \(20 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Section-B
Attempt any EIGHT of the following questions: (16)

Question 3.
Distinguish between lanthanides and actinides.
Answer:
Distinguish between lanthanides and actinides

Lanthanides	Actinides
1. Lanthanoids show a maximum oxidation state of+4	1. Actinoids
2. Lanthanoids do not form complexes easily.	2. Actinoids have a greater tendency to form complexes with ligands such as thio- ethers.
3. All lanthanoids are non-radioactive except promethium.	3. Actinides
4. Lanthanoids do not form oxocations.	4. Actinides
5. Most of the lanthanoids are colourless in nature.	5. The actinoids are coloured ions.

Question 4.
Calculate the mole fraction of solute, if the vapour pressure of pure benzene at certain temperature is 640 mmHg and vapour pressure of solution of a solute in benzene is 600 mmHg.
Answer:
Given: \(P_0^1\) = 64.0 mm Hg, \(P_1\) = 600 mm Hg
Find: Mole fraction of solute = x₂ = ?
Formula:
x₂ = \(\frac{P_1^0-P_1}{P_1^0}\)
x₂ = \(\frac{P_1^0-P_1}{P_1^0}\) = \(\frac{640-600}{640}\)
x₂ = \(\frac{40}{640}\) = 0.0625

Question 5.
Define: Green chemistry. Write two advantages of nanoparticle and nanotechnology.
Answer:
Green chemistry: Green chemistry is the use of chemistry for pollution prevention by environmentally conscious design of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.
Advantages:
(i) Revolution in electronics and computing.
(ii) Energy sector-nanotechnology will make solar power more economical Energy storage devices will become more efficient
(iii) Medical field: Manufacturing of smart drugs, helps “ cure faster and without side effects. Curing of life- threatening diseases like cancer and diabetes.

Question 6.
Explain the following terms:
(i) Substitutional impurity defect
(ii) Interstitial impurity defect.
Answer:
(i) Substitutional impurity defect: In this defect, the foreign atoms are found at the lattice sites in place of host atoms. The regular atoms are displaced from their lattice sites by impurity atoms.
For example: Solid solution of metals (alloys) brass-Cu atoms are replaced by Zn].

(ii) Interstitial impurity defect: In this defect, the impurity atoms occupy interstitial spaces of lattice structure.
For example: In steel, Fe atoms occupy normal lattice sites. The carbon atoms are present at interstitial spaces.

Question 7.
Write the chemical reactions for the following:
(i) Chlorobenzene is heated with fuming H₂S0₄
(ii) Ethyl bromide is heated with silver acetate.
Answer:
(i) Chlorobenzene is heated with fuming H₂S0₄

(ii) Ethyl bromide is heated with silver acetate.

CH₃COOAg + BrC₂H₅ → CH₃COOC₂H₅ + AgBr↓

Question 8.
Define: Acidic buffer solution. Write the relationship between solubility and solubility product for Pbl₂.
Answer:
Acidic buffer solution: A solution containing a weak acid and its salts with strong base is called an acidic buffer solution.
For example: A solution containing weak acid such as CH₃COOH and its salt such as CH₃COOHNa is an acidic buffer solution.
Relationship between solubility and solubility product for Pbl₂.
For Pbl₂,

x= 1, y = 2
Therefore, K_sp = (1)¹(2)²S¹⁺² = 4S³ …(i)
Relation i.e. K_sp = 4S³ give full credit

Question 9.
What is the action of the following reagents on ethyl amine?
(i) Chloroform and caustic potash
(ii) Nitrous acid
Answer:

Question 10.
Calculate standard Gibbs energy change at 25°C for the cell reaction.
Img 4
Answer:
Given: \(\mathrm{E}_{\mathrm{Cd}}^0\) = -0.403 V, \(\mathrm{E}_{\mathrm{Sn}}^0\) = – 0.136 V, n = 2.
Find: \(\Delta G^0\) = ?
\(\mathrm{E}_{\text {cell }}^0\) = \(\mathrm{E}_{\mathrm{Sn}}^0\) – \(\mathrm{E}_{\mathrm{Cd}}^0\) = – 0-136 -(- 0.403) = 0.267 V
\(\Delta G^0\) = – nF\(\mathrm{E}_{\text {cell }}^0\) = -2 × 96500 × 0.267
\(\Delta G^0\) = – 51531 VC = -51531J = – 51.53 KJ

Question 11.
Write chemical reaction for the preparation of glucose from sucrose. Write structure of D-ribose.
Answer:
Preparation of glucose from sucrose:

Question 12.
Define extensive property. Calculate the work done during the expansion of 2 moles of an ideal gas from 10 dm³ to 20 dm³ at 298 K in vacuum.
Answer:
Extensive property: A property which depends on the amount of matter present in a system is called an extensive property. Examples: Mass, volume, etc. When the gas expands in vacuum, there is no opposing force (p = 0).
No work is done when the gas expands freely in vacuum. Hence, W = 0

Question 13.
Write the reactions for the formation of nylon 6, 6 polymer.
Answer:

Question 14.
Draw structures of the following compounds:
(i) chloric acid
(ii) peroxy disulphuric acid
Answer:

Section-C
Attempt any EIGHT of the following questions: (24)

Question 15.
Define Osmosis.
How will you determine molar mass of non-volatile solute by elevation of boiling point?
Answer:
The net spontaneous flow of solvent molecules into the solution or from more dilute solution to more concentrated solution through a semipermeable membrar)e is called osmosis.
We know that,
∆T_b cc m, ∆T_b = K_bm …(i)
Let W_a = Weight of a solvent,
W₂ = Weight of a solute,
M₂ = Molar mass of solute

Moles of solute in W₂g of solvent = \(\frac{W_2}{M_2}\)

Mass of solvent = W₁ = \(\frac{W_1}{1000}\) kg

The expression of molality.
m = \(\frac{\text { Moles of solute }}{\text { Mass of solvent in kg }}\) = \(\frac{W_2 / M_2}{W_1 / 1000}\)
= \(\frac{1000 \mathrm{~W}_2}{\mathrm{M}_2 \mathrm{~W}_1}\)
Substitution of this value of m in Eq. (i) gives
∆T_b = \(\mathrm{K}_b \frac{1000 \mathrm{~W}_2}{\mathrm{M}_2 \mathrm{~W}_1}\)
M = \(\frac{1000 \mathrm{~K}_b \mathrm{~W}_2}{\Delta \mathrm{~T}_b \mathrm{~W}_1}\)

Question 16.
Convert the following:
(i) Ethyl alcohol into ethyl acetate
(ii) Phenol into behzene
(iii) Diethyl ether into ethyl chloride
Answer:
(i) Ethyl alcohol into ethyl acetate

(ii) Phenol into behzene

(iii) Diethyl ether into ethyl chloride

Question 17.
A weak monobasic acid is 10% dissociated in 0.05 M solution. What is percent dissociation in 0.15 M solution?
Answer:
If α₁ and α₂ are the values of degree of dissociation at two different concentrations C₁ and C₂ respectively, then
K_a = α₁²C₁ = α₂²C₂ Therefore α₁²C₁ = α₂²C₂
Data: C₁ = 0.05 M, C₂ = 0.15 M,
Percent dissociation (α₁) = 10%
α₁ = \(\frac{10}{100}\) = 0.1, Formula α₁²C₁ = α₂²C₂
(0.1)² × 0.05 = \(\alpha_2^2\) × 0.15
0.01 × 0.05 = \(\alpha_2^2\) × 0.15
0.0005 = \(\alpha_2^2\) × 0.15
\(\alpha_2^2\) = \(\frac{0.0005}{0.15}\) = 0.0033
\(\alpha_2^2\) = 0.0574 or 5.74 × 10^-2
∴ % dissociation (α₂) = 5.74%

Question 18.
Explain dehydrohalogenation reaction of 2-chlorobutane. Write use and environmental effect of CFC.
Answer:
2-chlorobutane on heating with alcoholic KOH gives mixture of but-l-ene and but-2-ene.
In this reaction elimination of hydrogen and halogen take place, hence this reaction is also known as dehydrohalogenation reaction. Since hydrogen is eliminated from ß-carbon, this reaction is known as ß-elimination reaction (1, 2-elimination), major product 2-butene is formed according to Saytzeff rule.

Uses:

• They are used as refrigerants in fridge and air conditioning
• They are used as propellants in aerosol and solvents.
• They are used as blowing agents in making foams and packing materials.

Environmental effect of CFG: CFC’s are responsible for ozone depletion in stratosphere and causing problems such as breathing problem, organ damage, loss of consciousness, skin cancer etc.

Question 19.
2000 mmol of an ideal gas expanded isothermally and reversibly from 20 L to 30 L at 300 K, calculate the work done in the process (R= 8.314 JK^-1 mol^-1).
Answer:
Given: n = 2000 mmol = 2000 × 10^-3 mol = 2 mol
Data: T = 300 K, V₁ = 20L, V₂ = 30L R = 8.314J / Kmol
Find: W_max = ?
W_max =-2.303 × nRT \(\log _{10} \frac{V_2}{V_1}\)

W_max = -2.303 × 2 × 8.314 × 300 × \(\log _{10} \frac{30}{20}\)
= -2.303 × 2 × 8.314 × 300 × \(\log _{10} 1.5\)
= -2.303 × 2 × 8.314 × 300 × 0.176
W_max = -2.023 J or W_max = -2.023 KJ

Question 20.
What are interstitial compounds? Give the classification of alloys with examples.
Answer:
Interstitial compounds: When small atoms like hydrogen, carbon or nitrogen are trapped in the interstitial spaces within the crystal lattice, the compounds formed are called interstitial compounds. Steel and cast iron are examples of interstitial compounds of carbon and iron.

Classification of alloys:
Ferrous alloys have atoms of other elements distributed randomly in atoms of iron in the mixture. e.g., nickel steel, chromium steel, stainless steel etc.
Non-ferrous alloys are formed by mixing atoms of transition metal other than iron with a non-transition element.
e.g., Brass, which is an alloy of copper and zinc

Question 21.
Draw labelled diagram of H₂ – O₂ fuel cell. Write two applications of fuel cell.
Answer:
Draw labelled diagram H₂ – O₂ fuel cell.

Applications of fuel cell:

• The fuel cells are used on experimental basis in automobiles.
• The fuel cell are used for electrical power in the space programme.
• In space crafts the fuel cell is operated at such a high temperature that the water evaporates at the same rate as it is formed. The vapour is condensed and pure water formed is used for drinking by astronauts.
• In future, fuel cells can possibly be explored as power generators in hospitals, hotels and home.

Question 22.
Explain formation of [CoF₆]² complex with respect to:
(i) Hybridisation
(ii) Magnetic properties
(iii) Inner/outer complex
(iv) Geometry
Answer:
(a) Oxidation state of central metal Co is +3.
(b) Valence shell electronic configuration of \(\mathrm{Co}^{3 \oplus}\) is

(i) Hybridisation – sp³d²
(ii) Magnetic properties – Paramagnetic
(iii) Inner/outer complex – outer complex
(iv) Geometry – Octahedral complex

Question 23.
What is Pseudo first order reaction? Derive integrated rate law equation for zero order reaction.
Answer:
Certain reactions which are expected to be of higher order follow the first order kinetics. These reactions are called as Pseudo-first order reaction.
Integrated rate law for zero order reaction:
For zero order reaction,
A → P
The differential rate law is given by ……(i)
Rate = d[A] = k[A]° = k
Rearrangement ofEq. (i) gives
d[A] = – kdt
Integration between the limits
[A] = [A]₀ at t = 0 and [A] = [A]_t at t = t gives
\(\int_{[A]_0}^{[A]^t} d[A]\) = \(-k \int_0^t d t\)
or [A]_t – [A]₀ = -kt
Hence, kt = [A]₀ – [A]_t …(ii)
k = \(\frac{[A]_0-[A]_t}{t}\)

Question 24.
Explain Aldol condensation of ethanal
Answer:
Aldehydes containing at least one a-hydrogen atom undergo a reaction in presence of dilute alkali (dilute NaOH, KOH or NajCO-j) as catalyst to form p-hydroxy aLdehydes (aldol). This reaction is known as aldol reaction.
Formation of aldol is an addition reaction. Aldol formed from aldehyde having a-hydrogen undergoes subsequent elimination of water molecule on warming, giving rise to a, p-unsaturated aldehyde. Overall reaction is known as aldol condensation.

It is nucleophilic addition-elimination relation.

Question 25.
Explain anomalous behaviour of oxygen in group 16 with respect to:
(i) Atomicity
(ii) Magnetic property
(iii) Oxidation state
Answer:
(i) Atomicity: Oxygen is a diatomic molecule (O₂) while others are polyatomic molecules.
(For example P₄, S₈)
(ii) Magnetic property: Oxygen is paramagnetic while others are diamagnetic.
(iii) Oxidation state: Oxygen shows -2, – 1, and + 2 oxidation states while other elements show, – 2, + 2, + 4.+ 6 oxidation states.

Question 26.
Write chemical reactionsforthefollowing conversions:
(i) Acetic acid into acetic anhydride
(ii) Acetic acid into ethyl alcohol
Write IUPAC name and structure of methylphenylamine.
Answer:

IUPAC name: N-methylaniline or N-methylbenzenamine

Section-D
Attempt any THREE of the following questions: (12)

Question 27.
Show that, time required for 99.9% completion of a first order reaction is three times the time required for 90% completion. Give electronic configuration of Gd (Z = 64). Write the name of nano structured material used in car tyres to increase the life of tyres.
Answer:
Given: For 99.9% completion, if [A]₀ = 100.
[A]_t = 100 – 99.9 = 0.1
For 90% completion, if [A]₀ = 100,
[A]_t = 100 – 90 = 10

= log 10³
= 3 log 10
= 3
t₁ = 3t₂
Electronic configuration of Gd (Z = 64).
Gd (Z = 64) = [Xe] 4f⁷ 5d¹ 6s²
Name of non structured material used in car tyres to increase the life of tyres.
Carbon block.

Question 28.
Derive relationship between ∆H and ∆U for gaseous reaction.
Define: Vulcanization.
What is peptide bond?
Answer:
At constant pressure, ∆H and ∆U are related as
∆H = ∆H + P∆V
For reaction involving gases, ∆V cannot be neglected and
∆H = ∆U + P∆V = ∆H + P(V₂ – V₁)
∆H = ∆U + PV₂-PV₁ ….(1)
where V₁ is the volume of gas phase reactants and V₂ that of the gaseous products.
Applying ideal gas equation PV = nRT.
When nx moles of gaseous reactants produce n₂ moles of gaseous products.
The ideal gas equation give,
PV₁ = n₁RT and PV₂ = n₂RT ….(2)
Substitution of Eq. (2) into Eq. (1) yields
∆H = ∆U + n₂RT – n₁RT
∆H = ∆U + (n₂ – n₁)RT
∆H = ∆U + ∆n_g RT
where ∆n_g is difference between the number of moles of products and those of reactants.

Vulcanization: The process by which a network of cross links is introduced into an elastomers is called vulcanization.
The bond that connects a-amino acids to each other is called peptide bond.
OR
is called as peptide bond.

Question 29.
Silver crystallizes in fee structure. If edge length of unit cell is 400 pm, calculate density of silver (Atomic mass of Ag = 108 ). Write a note on Haloform reaction.
Answer:
Given: M = 108g mol^-1, n = 4 for fcc, N_A = 6.022 × 10²³ atoms mol^-1, a = 400 pm = 400 × 10^-12 m = 4.00 × 10^-8 cm

Haloform reaction:

• This reaction is given by acetaldehyde, all methyl ketones (CH₃ – CO – R) and all alcohols containing CH₃ -(CHOH)- group. When an alcohol or methyl ketone is warmed with sodium hydroxide and iodine, a yellow precipitate is formed.
• Here the reagent sodium hypoiodite is produced in situ.
• During the reaction, sodium salt of carboxylic acid is formed which contains one carbon atom less than the substrate.
• The methyl group is converted into haloform.
• For example: Acetone is oxidized by sodium hypoiodite to give sodium salt of acetic acid and yellow precipitate of iodoform.
  

Question 30.
Define: Distereoisomers.
Give cis and trans isomers of [Co(NH₃)₄Cl₂].
What is reference electrode?
Give reason: Bleaching action of ozone is also called dry bleach.
Answer:
Isomers which are non-superimposable mirror image of each other is called distereoisomers.
Give cis and trans isomers of [Co(NH₃)₄Cl₂]⁺.

Reference electrode: A reference electrode is then defined as an electrode whose potential is arbitrarily taken as zero or is exactly known.
Reason: Bleaching action of ozone is also called dry bleach
Ozone bleaches in absence of moisture so it also known as dry bleach. .

Question 31.
Write Dow process for preparation of Phenol. What is the action of bromine water on phenol?
Give reason: Group 16th elements have lower ionisation enthalpy compared to group 15th elements. Write two uses of dioxygen.
Answer:
Dow process for preparation of Phenol

Reason: Group 16th elements have lower ionisation enthalpy compared to group 15th elements:
The elements of group 16 have lower ionisation enthalpy values compared to those of group 15 in the corresponding periods, owing to extra stable half filled electronic configuration of p-orbitals in elements of group 15.

Uses of dioxygen.
Dioxygen is important for respiration to sustain animal and aquatic life.
• It is used in the manufacture of steel.
• It is used in oxyacetylene flame for welding and cutting of metals.
• Oxygen cylinders are widely used in hospitals, high altitude flying and mountaineering.
• It is used in combustion offiiels.for example, hydrazine in liquid oxygen provides tremendous thrust (energy) in rockets.

Maharashtra Board Class 12 Chemistry Previous Year Question Papers