Maharashtra Board Class 12 Chemistry Sample Paper Set 4 with Solutions

Maharashtra State Board Class 12th Chemistry Sample Paper Set 4 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Chemistry Model Paper Set 4 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): Sc = 21, Ti = 22, V = 23, Cr = 24, Ar = 18.
   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. Aldol condensation is ______.
(A) electrophilic substitution reaction
(B) nucleophilic substitution reaction
(C) elimination reaction
(D) addition – elimination reaction
Answer:
(D) addition – elimination reaction

ii.

The major product of the above reaction is ____.

Answer:
(C) CH₃ – CH – CH₃

iii. For pH > 7, the hydronium ion concentration would be _____.
(A) 10^-7 M
(B) < 10^-7M
(C) >10^-7M
(D) ≥ 10^-7M
Answer:
(B) < 10^-7 M

iv. Formalin is 40% aqueous solution of _____.
(A) methanal
(B) methanoic acid
(C) methanol
(D) methanamine
Answer:
(A) methanal

v. Which of the following 0.1 M aqueous solutions will exert the highest osmotic pressure?
(A) Al₂(SO₄)₃
(B) Na₂SO₄
(C) MgCl₂
(D) KCl
Answer:
(A) Al₂(SO₄)₃

Explanation:
Colligative properties are proportional to the number of solute particles present in the solution.

Osmotic pressure is a colligative property; hence, Al₂(SO₄)3 exerts the highest osmotic pressure.

vi. Which of the following is the first oxidation product of secondary alcohol?
(A) Alkene
(B) Aldehyde
(C) Ketone
(D) Carboxylic acid
Answer:
(C) Ketone

vii. Which of the following have d°s° configuration?
(A) Sc³⁺
(B) Ti⁴⁺
(C) V⁵⁺
(D) All of the above
Answer:
(D) All of the above

viii. Which among the following hydrides is NOT a reducing agent?
(A) H₂O
(B) H₂S
(C) H₂Te
(D) H₂Se
Answer:
(A) H₂O

ix. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 bar from an ihitial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, ∆U of the gas will be ____.
(A) -500 J
(B) +500 J
(C) -1013 J
(D) + 1010 J
Answer:
(A) – 500 J

Explanation:
Since the container is insulated, this is an adiabatic process.
For adiabatic process,
∆U = +W = – P_ext ∆V = -P_ext (V₂ – V₁)
Initial volume (V₂) = 2.5 L = 2.5 dm³
Final volume (V₂) = 4.5 L = 4.5 dm³
External pressure (P_ext) = 2.5 bar
∆U = W = – 2.5 bar × (4.5 dm³ – 2.5 dm³)
= – 5.0 dm³ bar × \(\frac{100 \mathrm{~J}}{1 \mathrm{dm}^3 \mathrm{bar}}\) = – 500 J

Question 2.
Answer the following:

i. Name the Lewis acids and bases in the complex [PtCl₂(NH₃)₂].
Answer:
Lewis acid: Platinum (II) ion, (Pt²⁺)
Lewis bases: Chloride ion (Cl^–), ammonia (NH₃)

ii. State Markownikov’s rule.
Answer:
Markownikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part (-X) of the reagent gets attached to the carbon atom which carries less number of hydrogen atoms.

iii. Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.
Answer:
a. The molar conductivity of the given solution is related to conductivity as ∧ = \(\frac{k}{c}\)
b. The SI units of k are S m^-1 and that of c are mol m^-3. Hence, the SI units of ∧ is S m² mol^-1.

iv. Define ‘sustainable development’.
Answer:
Sustainable development is development that meets the needs of the present, without compromising the ability of future generations to meet their own need.

v. Name the monosaccharide unit(s) present in maltose.
Answer:
Maltose is a disaccharide made of two units of D-glucose.

vi. Write probable electronic configurations of chromium.
Answer:
The probable (expected) electronic configuration of chromium is
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s² or [Ar] 3d⁴ 4s².

vii. For the reaction, N_2(g) + 3H_2(g) → 2NH_3(g), what is the relationship among \(\frac{\mathrm{d}\left[\mathrm{~N}_2\right]}{\mathrm{dt}}, \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}\) and \(\frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}\)?
Answer:
The relationship among

viii. Write the structure of isoprene.
Answer:

Section – B

Question 3.
Write the conditions for maximum work done by the system.
Answer:
Conditions for maximum work from a system:

1. Any change which takes place in a system should be thermodynamically reversible.
2. The change in the system should take place in an infinite number of steps.
3. The driving force of the change should be infinitesimally greater than the opposing force.
4. The system should be in mechanical equilibrium with its surroundings. (Maximum expansion work is obtained when pex is only infinitesimally smaller than P. In effect, the two pressures are the same. This balance of pressures is a state of mechanical equilibrium).

Question 4.
Predict the major product in the following reactions:

i.

Answer:

ii.

Answer:

Question 5.
Why salts of Sc³⁺, Ti⁴⁺, V⁵⁺ are colourless?
Answer:

1. Condensed electronic configurations of Sc³⁺, Ti⁴⁺, V⁵⁺ are: Sc³⁺: [Ar] 3d⁰; Ti⁴⁺: [Ar] 3d°; V⁵⁺: [Ar] 3d°
2. The ions Sc³⁺, Ti⁴⁺ and V⁵⁺ have completely empty d-orbitals i.e., no unpaired electrons are present. Thus, their salts are colourless, as d-d transitions are not possible.

Question 6.
Write a note on Stephen reaction.
Answer:
Nitriles are reduced to imine hydrochloride by stannous chloride in presence of hydrochloric acid which on acid hydrolysis give corresponding aldehydes. This reaction is called Stephen reaction.

Question 7.
An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (-OH) groups must be present in the original compound?
Answer:
In acetylation reaction H atom of an (-OH) group is replaced by an acetyl group (-COCH₃). This results increase in molecular mass by [(12+16+12+3×1)-1], that is, 42 u.
In the given alcohol, increase in molecular mass = 174 u – 90 u = 84 u
∴ Number of -OH groups = \(\frac{84 u}{42 u}\) = 2

Question 8.
Write the chemical reactions involved in manufacture of Nylon 6,6.
Answer:
Mixing of the two monomers forms nylon salt, which upon condensation polymerization under conditions of high temperature and pressure give the polyamide fibre, nylon 6,6.

Question 9.
How many moles of electrons are required for reduction of 2 moles of Zn²⁺ to Zn? How many Faradays of electricity will be required?
Answer:
i. The balanced equation for the reduction of Zn²⁺ to Zn is
\(\mathrm{Zn}_{(a q)}^{2+}\) + 2e^– → Zn_(s)
The equation shows that 1 mole of Zn²⁺ is reduced to Zn by 2 moles of electrons. For reduction of 2 moles of Zn⁺, 4 moles of electrons will be required

ii. Faraday (96500 Coulombs) is the amount of charge on one mole of electrons. Therefore, for 4 moles of electrons, 4 F of electricity will be required.
Ans: For reduction of 2 moles of Zn²⁺, 4 moles of electrons will be required and 4 F of electricity will be required.

Question 10.
Distinguish between crystalline solids and amorphous solids.
Answer:

Crystalline solids	Amorphous solids
i. The constituent particles ore arranged in a regular and periodic manner.	The constituent particles are arranged randomly.
ii They have sharp and characteristic melting point.	They do not have sharp melting point. They gradually soften over a range of temperature.
iii. They are anisotropic, i.e., have different physical properties in different direction.	They are isotropic, i.e., have same physical properties in all directions.
iv. They have long range order.	They have only short range order.
e.g. Ice, NaCl, etc.	Glass, rubber, plastics, etc.

Question 11.
Sucrose decomposes in acid solution to give glucose and fructose according to the first order rate law. The half life of the reaction is 3 hours. Calculate fraction of sucrose which will remain after 8 hours.
Answer:
Given: t_1/2 = 3 hours , t = 8 hours
To find: \(\frac{[A]_{+}}{[A]_0}\)

Formulae:

i. k = \(\frac{0.693}{t_{1 / 2}}\)
II. k = \(\frac{2.303}{t} \log _{10} \frac{[A]_0}{[A]_{+}}\)

Calculation:
As sucrose decomposes according to first order rate law, k = \(\frac{2.303}{t} \log \frac{[A]_b}{[A]_t}\)
From formula (i),

Ans: The fraction of the sample of sucrose that remains after 8 hours is 0.158.

Question 12.
What is the action of following on ethyl bromide:
i. Na in dry ether
ii. Mg in dry ether
Answer:
i. Ethyl bromide yields butane on treatment with sodium metal in dry ether (Wurtz reaction).

ii. Ethyl bromide yields ethyl magnesium bromide on treatment with magnesium metal in dry ether.

Question 13.
How are primary, secondary and tertiary alcohols identified by using Lucas reagent?
Answer:
i. Primary, secondary and tertiary alcohols can be distinguished from each other in the laboratory using Lucas reagent (conc. HCl and ZnCl₂).

ii. The reaction involved is:

iii. Alcohols are soluble in Lucas reagent but the product alkyl chloride is not. Hence, the clear solution becomes turbid when product starts forming.

iv. Tertiary alcohols reacts fast and the reagent turns turbid instantaneously. Secondary alcohols turn the reagent turbid slowly. Primary alcohols turn the reagent turbid only on heating.

Note: Observations and inferences for Lucas reagent test of alcohols:

Test	Observation	Inference
Alcohol + Lucas reagent	Reagent turns turbid on heating.	Primary alcohol is present.
Alcohol + Lucas reagent	Reagent turns turbid slowly, without heating.	Secondary alcohol is present.
Alcohol + Lucas reagent	Reagent turns turbid immediately, without heating.	Tertiary alcohol is present.

Question 14.
Following graph shows experimental determination of solubilities of some ionic solids in water at various temperatures. Comment on the effect of temperature on solubilities of these compounds

Answer:

1. Solubilities of NaBr, NaCI and KCI change slightly with temperature.
2. Solubilities of KNO₃, NaNO₃ and KBr increase appreciably with increasing temperature.

Section – C

Attempt any EIGHT of the following questions:

Question 15.
Derive the relationship between pH and pOH.
Answer:
Relationship between pH and pOH:
The ionic product of water is given as:
K_w = [H₃O⁺][0H^–]
Now, K_w = 1 × 10^-14 at 298 K
Thus, [H₃O⁺][OH^–] = 1.0 × 10^-14
Taking logarithm of both the sides, we write
log₁₀[H₃O⁺] + log₁₀[OH^–] = -14
-log₁₀[H₃O⁺] + {- log₁₀[OH^–]} = 14
Now, pH = -log₁₀[H30+] and
pOH = -log₁₀[OH^–]
∴ pH + pOH = 14

Question 16.
Calculate the work done in the following reaction at 50 °C. State whether work is done on the system or by the system.
Answer:
Given: Oxidation of 1 mole of SO₂
Temperature = T = 50 °C = 323 K
To find: Work done and to determine whether work is done on the system or by the system.
Formula: W = -∆n_gRT
Calculation:
The given reaction is as follows:
SO_2(g) + \(\frac{1}{2} O_{2(g)}\) → SO_3(g)
Now,
∆n_g = (moles of product gases) – (moles of reactant gases)
∆n_g = 1 – 1.5 = -0.5 mol
Hence,
W = -∆n_g RT
= – (-0.5 mol) × 8.314 J K^-1 mol^-1 × 323 K
= +1343 J = +1.343 kJ
Work is done on the system (since W > 0).
Ans: The work done is +1.343 kJ. The work is done on the system.

Question 17.
i. What are ethers?
ii. How are they classified?
Answer:
i. Ethers are compounds which contain an oxygen atom bonded to two alkyl groups or two aryl groups or one alkyl and one aryl group. Ethers are organic oxides and are considered as anhydrides of alcohols. Ethers are represented as R – O – R’ or Ar – O – Ar’ orR-O-Ar

ii. Ethers are classified based on whether the two alkyl/ aryl groups bonded to oxygen atom are same or different.

a. Symmetrical ethers (simple ethers): If the two alkyl groups or aryl groups attached to oxygen atom are same (R – O – R or Ar – O – Ar).
e.g. CH₃ – O – CH₃, C₆H₅ – O – C₆H₅

b. Unsymmetrical (mixed ethers): If the two alkyl groups or aryl groups attached to oxygen atom are different (R – O – R’ or Ar – O – Ar’).
e.g. CH₃ – O – C₂H₅, C₆H₅ – O – C₅H₅

Question 18.
The density of iridium is 22.4 g cm^-3. The unit cell of iridium is fee. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g mol^-1.
Answer:
Given: Type of unit cell is fee.
Density of iridium (p) = 22.4 g cm^-3
Molar mass of iridium = 192.2 g mol^-1
To find: Radius of iridium atom (r)

Formulae:
i. Density (ρ) = \(\frac{n M}{a^3 N_A}\)
ii. For fcc unit cell, r = 0.3535 a

Calculation:
For fee unit cell, n = 4.
Using formula (i),

Using formula (ii),
r = 0.3535 a
r = 0.3535 × 384.9 pm = 135.7 pm
≈ 136 pm

Ans: Radius of iridium atom (r) is 136 pm.

Question 19.
What are zero, one and two dimensional nanoscale systems?
Answer:
i. Zero-Dimensional Nanostructures:
A zero dimensional structure is one in which all three dimensions are in the nanoscale. That is, all three dimensions < 100 nm
e.g. Nanoparticles, quantum dots, nanoshells, nanorings, microcapsules

ii. One-Dimensional Nanostructures:
A one dimensional nanostructure is one in which two dimensions are in the nanoscale. That is, two dimensions < 100 nm
e.g.Nanowires, nanofibres, nanotubes and nanorods

iii. Two-Dimensional Nanostructures:
A two-dimensional nanostructure is one in which one dimension is in the nanoscale. That is, one dimension < 100 nm
e.g. Thin films, layers and coatings

Question 20.
i. What is the difference between a double salt and a coordination complex? Give an example.
ii. What is the shape of a complex in which the coordination number of the central metal ion is 4?
Answer:
i. A double salt dissociates in water completely into simple ions, whereas a coordination complex dissociates in water with at least one complex ion.
e.g. Mohr’s salt, FeSO₄(NH₄)₂SO₄.6H₂O is a double salt while K₄[Fe(CN)₆] is a complex. (2 Marks)

ii. A coordination complex with coordination number four has tetrahedral or square planar structure.

Question 21.
What are the differences between cast iron, wrought iron and steel?
Answer:
Differences between cast iron, wrought iron and steel:

Cast iron	Wrought iron	Steel
i. Hard and brittle	i. Very šoft	i. Neither too hard nor too soft.
ii. Contains 4% carbon.	ii. Contains less than 0.2% carbon	ii. Contains 0.2 to 2%
iii. Used for making pipes, manufacturing automotive parts, pots, pans, utensils, etc.	iii. Used for making pipes, bars for stay bolts, engine bolts and rivets, etc.	iii. Used in buildings infrastructure, tools, ships, automobiles, weapons, etc.

Question 22.
What is the action of following on SO₂?
i. Cl₂
ii. O₂
iii. NaOH
Answer:
i. Action of Cl₂ on SO₂: Sulfur dioxide reacts with chlorine in the presence of charcoal (catalyst) to form suif uryl chloride.

ii. Action of O₂ on SO₂: Sulfur dioxide is oxidised by dioxygen in presence of vanadium (V) oxide to give sulfur trioxide.

iii. Action of NaOH on SO₂: Sulfur dioxide readily reacts with sodium hydroxide solution to form sodium sulfite.

Question 23.
Can isobutyraldehyde undergo Cannizzaro reaction? Explain.
Answer:
Isobutyraldehyde (2-methylpropanal) can undergo Cannizzaro reaction, even though it contains an a-hydrogen atom.

Explanation:

i. The basic requirement for Cannizzaro reaction is non-enolisable aldehyde. In isobutyraldehyde, the enolate ion is unstable due to +I effect of methyl groups and also due to the negative charge on carbon atom.

ii. If the enolate ion was stable, then isobutyraldehyde would have undergone aldol condensation. Also, due to steric effect a good yield of aldol product is not obtained. Thus, it prefers to undergo Cannizzaro reaction.

Question 24.
What is zeroth order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer:
i. Reactions in which rate is independent of the reactant concentration are zeroth order reaction.

ii. Integrated rate law for zero order reactions:
For zero order reaction, A → P
the differential rate law is given by
rate = \(-\frac{\mathrm{d}[\mathrm{~A}]}{[\mathrm{A}]}\) = k[A]₀ = k ………..(1)
Rearrangement of Eq. (1) gives
d[A] = -k dt (∵ [A]⁰ = 1)
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 ………(2)

iii. The unit of rate constant for the zero order reaction is mol dm^-3 s^-1

Question 25.
Write IUPAC names of the following.

i.

Answer:
2-Bromopropane

ii.

Answer:
1-Iodo-2-methylpropane

iii.
CH₃ – CH = CH – CH₂Cl
Answer:
1-Chlorobut-2-ene

Question 26.
Using Raoult’s law, how will you show that ∆P = \(\mathrm{P}_1^0 \mathrm{x}_2\) ? Where, x₂ is the mole fraction of solute in the solution and \(\mathrm{P}_1^0\) vapour pressure of pure solvent.
Answer:

1. Raoult’s law expresses the quantitative relationship between vapour pressure of solution and vapour pressure of solvent.
2. In solutions of nonvolatile solutes, the law is applicable only to the volatile solvent.
3. The law States that, “the vapour pressure of solvent over the solution is equal to the vapour pressure of pure solvent multiplied by its mole fraction in the solution.”
4. Suppose that for a binary solution containing solvent and one nonvolatile solute, Pi is the vapour pressure of solvent over the solution, x₁ and x₂ are the mole fractions of solvent and solute, respectively and \(\mathrm{P}_1^0\) is the vapour pressure of pure solvent, then, P₁ = \(\mathrm{P}_1^0 x_1\).
5. Since, x₁ = 1 – x₂,
   P₁ = \(P_1^0 x_1\) = \(P_1^0\left(1-x_2\right)\) = \(P_1^0-P_1^0 x_2\)
   ∴ \(P_1^0-P_1\) = \(P_1^0 x_2\)
   ∴ ∆P = \(\mathrm{P}_1^0 x_2\) (∵ ∆P is the lowering of vapour pressure)

Section – D

Attempt any THREE of the following questions: [12 Marks]

Question 27.
i. Henry’s law constant for CH₃Br_(g) is 0.159 mol dm^-3 bar^-1 at 25°C. What is solubility of CH₃Br_(g) in water at same temperature and partial pressure of 0.164 bar?
Answer:
Henry’s law constant = K_H = 0.159 mol dm^-3 bar^-1,
Pressure of the gas = P = 0.164 bar
Solubility (S) = K_HP
= 0.159 mol dm^-3 bar^-1 × 0.164 bar = 0.026 mol dm^-3

Ans: Solubility of CH₃Br gas in water is 0.026 mol dm^-3

ii. Write any two statements of first law of thermodynamics.
Answer:
First Law of thermodynamics:

a. The total energy of a system and surroundings remains constant when the system changes from an initial state to final state.
b. Energy of the universe remains constant.
c. The total internal energy of an isolated system is constant.
d. Energy is neither created nor destroyed and can only be converted from one form to another. (Any two correct statements)

Question 28.
i. Using the relationship AG° of cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property?
Answer:
Under standard state conditions, electrical work done in galvanic cell is given by
∆G° = -nF\(E_{\text {cell }}^{\circ}\)
∆G° is an extensive property since its value depends on the amount of substance. If the stoichiometric equation of redox reaction is multiplied by 2, that is, the amount of substances oxidised and reduced are doubled, ∆G° doubles. The moles of electrons transferred also doubles.
The ratio,
\(\mathrm{E}_{\text {cell }}^{\circ}\) = \(-\frac{\Delta G^{\circ}}{n F}\) then becomes, \(E_{c \text { cell }}^{\circ}\) = \(-\frac{2 \Delta G^{\circ}}{2 n F}\) = \(-\frac{\Delta G^{\circ}}{n F}\)
Thus, \(E_{c e l l}^o\) remains the same by multiplying the redox reaction by 2. It means \(E_{c e l l}^o\) is independent of the amount of substance and is an intensive property.

ii. Sketch the diagram of a dry cell.
Answer:

Question 29.
What is the action of benzenediazonium chloride on:

i. phenol in alkaline medium?
Answer:
Benzenediazonium chloride reacts with phenol in mild alkaline medium to give p-hydroxyazobenzene (orange dye).

ii. aniline?
Answer:
Benzenediazonium chloride reacts with aniline in mild alkaline medium to p aminoazobenzene (yellow dye).

Question 30.
i. What is lanthanoid contraction?
Answer:
Lanthanoid contraction: As we move along the lanthanoid series, there is a decrease in atomic and ionic radii. This steady decrease in the atomic and ionic radii is called lanthanoid contraction.

ii. Write four points of distinction between lanthanoids and actinoids.
Answer:

Lanthanoids	Actinoids
i. In lanthanoidš, last differentiating electron occupies 4f orbital.	In actinoids, last differntiating electron occupies 5f orbital.
ii. They are the elements of first inner transition Series.	They are the elements of second inner transition Series.
iii. They are present in period 6.	They are present in period 7.
iv. Lanthcinoids show a maximum oxidation state of +4	Actinoids show oxidation states of +3, +4, +5, +6 and +7.
v. Lanthanoids have less tendency to form complexes.	Actinoids have greater tendency to form complexes with ligands such as thioethers.
vi. All lanthanoids are non-radioactive (except promethium).	All actinoids are radioactive.
vi. Larithanoids do not form oxocations.	Aclinoids forn oxocations such as UO+, PuO+, Np\(\mathrm{O}_2^{+}\).
viii. Most of the lanthanoids are colourless on nature.	Most of the actinoids are coloured ions.

Question 31.
Write the names and structural formulae of oxoacids of chlorine.
Answer:
i.

ii.

iii.

iv.

Maharashtra Board Class 12 Chemistry Previous Year Question Papers