Maharashtra Board Class 12 Chemistry Sample Paper Set 2 with Solutions

Maharashtra State Board Class 12th Chemistry Sample Paper Set 2 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Chemistry Model Paper Set 2 with Solutions

Section – A

Question 1.
Select and write the correct answers to the following questions:

(i) Molecular solids are
(a) Crystalline solids
(b) Amorphous solids
(c) Ionic solids
(d) Metallic solids
Answer:
(a) Crystalline solids

(ii) On which electrode the oxidation reaction takes place?
(a) Anode
(b) Cathode
(c) Salt bridge
(d) None of them
Answer:
(a) Anode

(iii) Isobutylamine is an example of ……….
(a) 2° amine
(b) 3° amine
(c) 1° amine
(d) Quarternary ammonium salt
Answer:
(c) 1° Amine

(iv) Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molaiv conductivities will be (a) 0.54
(b) 11.574
(c) 0.0864
(d) 1.852
Answer:
(a) 0.54

(v) Identify the chiral molecule from the following.
(a) 1-Bromobutane
(b) 1,1-Dibromobutane
(c) 2,3-Dibromobutane
(d) 1,3-Dibromobutane
Answer:
(c) 2,3-Dibromobutane

(vi) A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
Answer:
(b) 2.54 atm

(vii) Formula for the compound sodium hexacyanoferrate (iii) is
(a) [NaFe(CN)₆]
(b) Na₂ [Fe(CN)₆]
(c) Na[Fe(CN)₆]
(d) Na₃ [Fe(CN)₆]
Answer:
(d) Na₃ [Fe(CN₆)]

(viii) Components of Nichrome alloy are
(a) Ni, Cr, Fe
(b) Ni, Cr, Fe, C
(c) Ni, Cr
(d) Cu, Fe
Answer:
(c) Ni,Cr

(ix) Which of the following has highest electron gain enthalpy?
(a) Fluorine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer:
(b) Chlorine

(x) Which of the Na following is a buffer solution?
(a) CH₃COONa + NaCl in water
(b) CH₃COOH + HCl in water
(c) CH₃COOH + CH₃COONa in water
(d) HCl + NH₄Cl in water
Answer:
(c) CH₃COOH + CH₃COONa in water

Question 2.
Answer the following questions:

(i) What is standard cell potential for the reaction?
3Ni_(s) + 2Al³⁺(1M) → 3Ni²⁺ (1M) + 2Al(s) if E°_Ni = – 0.25 V and E°_Al = – 1.66V?
Answer:
The standard cell potential. for the reaction is -1.4 1V.

(ii) Write IUPAC name of

Answer:
2-Bromo-3-methylpent-3-ene

(iii) Name some chain growth polymers.
Answer:
Chain growth polymers are polyacrylonitrile, polyvinylchloride, polythene, etc.

(iv) Which amide does produce ethanamine by Hoffman bromamide degradation reaction?
Answer:
Propanamide (C₂H₅CONH₂) produces ethanamine by Hofmann bromamide degradation reaction.

(v) Give two uses of ClO₂.
Answer:
Uses of chlorine dioxide are as follows:
(a) Bleaching agent for paper pulp and textiles.
(b) In water treatment.

(vi) What is meant by shielding of electrons’ in an atom?
Answer:
(a) The decrease in the force of attraction exerted by
the nucleus on the valence electrons due to the presence of electrons in the inner shells is called the shielding effect.
(b) As a result of the shielding effect, the effective nuclear charge experienced by the valence electron is less than the actual nuclear charge.

(vii) Name a compound where Frenkel defect is found.
Answer:
AgCl

(viii) Write the formula for tetra ammine platinum (ii) chloride.
Answer:
[Pt(NH₃)₄]Cl₂

Section – B

Attempt any Eight of the following questions:

Question 3.
Give the structures of Thiosulphuric acid and Peroxy monosulphuric acid.
Answer:
(i) Thiosulphuric acid, H₂S₂O₃.

(ii) Peroxy monosulphuric acid, H₂SO₅

Question 4.
Explain the role of green chemistry.
Answer:

1. To promote innovative chemical technologies that reduce or eliminate the use or generation of hazardous substances in the design, manufacture, and use of chemical products.
2. The green chemistry helps to reduce capital expenditure to prevent pollution.
3. Green chemistry incorporates pollution, prevention practices in the manufacture of chemicals and promotes pollution prevention and industrial ecology.
4. Green chemistry is a new way of looking at chemicals and their manufacturing process to minimise any negative environmental effects.

Question 5.
How many moles of electrons are passed when 0.8 Ampere current is passed for 1 hour through molten CaCl₂?
Answer:
Calculate: Using formula (i),
Quantity of electricity passed
= I(A) × t(s) = 0.8 × 3600 = 2880 C
Using formula (ii),
Number of moles of electrons passed
\(\frac{Q(C)}{96500(C / \text { mole })}\) = \(\frac{2880 C}{96500(C / \text { mole })}\)
= 0.03 mole
Number of moles of electrons passed through molten CaCl₂ is 0.03 mole.

Question 6.
Write formulae of the following complexes.
(i) Potassium ammine trichloroplatinate (ii)
(ii) Dicyanoaurate (i) ion
Answer:
(i) K[Pt(NH₃)Cl₃]
(ii) [Au (CN)₂]^–

Question 7.
Explain the basic nature of amines with suitable example.
Answer:
Nitrogen atom of amines contains a lone pair of electrons which can be donated. Thus, amines act as bases and nucleophiles.

e.g.,
(i) The reaction of ethylamine with dilute hydrochloric acid results in the formation of ethyl ammonium chloride.

Question 8.
Give one example showing reducing property of ozone.
Answer:
Reaction showing reducing property of ozone is,

Question 9.
What is the enthalpy of atomisation? Give an example.
Answer:
The enthalpy change accompanying the dissociation of all molecules in one mole of gas phase substance into gaseous atoms is called enthalpy of atomisation.
For example: Atomisation of methane molecule.
CH_4(g) → C(g) + 4H_(g); ∆_atmH = 1160 kJ mol^-1

Question 10.
Give uses of potassium permanganate (KMnO₄).
Answer:
Uses of potassium permanganate (KMnO₄):

1. Used as an antiseptic.
2. Used for unsaturation tests in the laboratory.
3. Used in the volumetric analysis of reducing agents.
4. Used for detecting halides in qualitative analysis.
5. Used as a powerful oxidising agent in the laboratory and industry.

Question 11.
Why formic acid is stronger than acetic acid?
Answer:

1. The negatively charged acetate ion (i.e., the conjugate base of acetic acid) gets destabilised by + I effect of – CH₃ group.
2. Lesser is the stabilisation of the conjugate base, weaker is the acid.
3. In formate ion, there is no such destabilisation effect. Thus, formic acid is stronger than acetic acid.

Question 12.
Give the following named reactions.

(i) Wurtz-Fittig reaction
Answer:
Wurtz-Fitting Reaction:

(ii) Fittig reaction
Answer:
Fitting reaction:

Question 13.
Why p-nitrophenol is a stronger acid than phenol?
Answer:
The conjugate base of p-nitrophenol is better resonance stabilised due to six resonance structures compared to the five resonance structures of the conjugate base of phenol The resonance structure has a negative charge on only electgro negative oxygen atoms. Hence, p-nitrophenol is a stronger acid than phenol.

Question 14.
How vapour pressure lowering is related to a rise in boiling point of solution?
Answer:

1. At the boiling point of a liquid, its vapour pressure is equal to 1 atm.
2. In order to reach boiling point the solution and solvent must be heated to a temperature at which their respective vapour pressures attain 1 atm.
3. At any given temperature, the vapour pressure of a solution is lower than that of pure solvent. Hence, the vapour pressure of solution needs a higher temperature to reach 1 atm than that needed for vapour pressure of solvent Therefore, lowering vapour pressure causes a rise in the boiling point of a solution.

Section – C

Attempt any Eight of the following questions:

Question 15.
What is the action of hydrazine on cyclopentanone in presence of KOH in ethylene glycol? How ketones are prepared from nitriles ?
Answer:
Ketones are prepared by reacting nitriles with Grignard reagent in dry ether as solvent followed by acid hydrolysis.

Question 16.
Aluminium crystallises in cubic close packed structure with unit cell edge length of 353.6 Pm. What is the radius of Al atom? How many unit cells are there in 1.00 cm³ of Al?
Answer:
(i) Using formula (i),
r = 0.3535a
∴ r = 0.3535 × 353.6 = 125 pm

(ii) Using formula (ii),
Number of unit cells in volume (V) of metal = \(\frac{\mathrm{V}}{a^3}\)
∴ Number of unit cells in 1.00 cm³ of
Al = \(\frac{1.00}{\left(3.536 \times 10^{-8}\right)^3}\)
= 2.26 × 10²²
(a) Radius of Al atom (r) is 125 pm.
(b) Number of unit cells in 1.00 cm³ of Al is 2.26 × 10²².

Question 17.
Distinguish between S_N1 and S_N2 mechanism of substitution reaction.
Answer:

Factor	SN2	SN1
Kinetics	2nd order	1st order
Molecularity	Bimolecular	Unimolecular
Number of steps	One step	Two step
Bond making and bond breaking	Simultaneous	First the bond in the reactant breaks and then a new bond in product is formed
Transition state	One step, one transition state	Two steps, two transition state
Direction of attack of nucleophile	Only back side attack	Back side attack and front side attak
Nucleophile	Strong Nucleophile favourable.	Weak Nucleophile favourable.

Question 18.
The vapour pressure of water at 20°c is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water?
Answer:
Molar mass of urea (NH₂CONH₂)
= 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol^-1
Molar mass of water = 18g mol^-1
Now, using formula,
\(\frac{P_1^0-P_1}{P_1^0}\) = \(\frac{W_2 M_1}{M_2 W_1}\)
= \(\frac{17 \mathrm{~mm} \mathrm{Hg}-P_1}{17 \mathrm{~mm} \mathrm{Hg}}\)
= \(\frac{2.8 \mathrm{~g} \times 18 \mathrm{~g} \mathrm{~mol}^{-1}}{50 \mathrm{~g} \times 60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
∴ \(\frac{17 \mathrm{~mm} \mathrm{Hg}-P_1}{17 \mathrm{~mm} \mathrm{Hg}}\) = 0.0168
∴ 17 mm Hg – P₁ = 0.2856 mm Hg
∴ 17 mm Hg – P₁ = 0.2856 mm Hg
∴ P₁ = 17mm Hg – 02856 mm Hg = 16.71 mm Hg.
Vapour pressure of the given solution is 16.71 mm Hg.

Question 19.
Write reaction showing conversion of acetaldehyde into acetaldehyde dimethyl acetal.
Answer:

Question 20.
Give valence bond description for the 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) The oxidation state of vanadium is + 3.

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

(iii) The number of Cl^– Ligands is k 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 21.
Obtain the relationship between the rate constant and half-life of a first order reaction.
Answer:
The integrated rate law for the first order reaction is given by the equation
k = \(\frac{2.303}{t}\)log₁₀\(\frac{[\mathrm{A}]_0}{[\mathrm{~A}]_t}\)
Where [A]₀ = initial concentration of the reactant at t = 0
The concentration falls to [A]_t at time t from the start of
the reaction. The concentration of the reactant falls to [A]₀2 at time t₁/2.
Therefore, t = t_1/2
[A]_t = [A]₀/2
So, the equation can be written as

Question 22.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Answer:
Hess’s law of constant heat summation:
Hess’s law of constant heat summation states that, “Overall the enthalpy change for a reaction is equal to sum of enthalpy changes of individual steps in the reaction”.

For example:
The enthalpy changes for a chemical reaction is the same regardless of the path by which the reaction occurs. Hess’s Law is a direct consequence of the fact that enthalpy is state function.’ The enthalpy change of a reaction depends only on the initial and final states and not on the path by which the reaction occurs.

To determine the overall equation of reaction, reactants and products in the individual steps are added or subtracted Like algebraic entities.
Consider the synthesis of NH₃.

1. 2H_2(g) + N_2(g) → N₂H_4(g), \(\Delta_{\underline{r}} \mathrm{H}^0{ }_1\) = + 95.4 kJ
2. N₂H_4(g) + H_2(g) → 2NH_3(g); \(\Delta_{\underline{r}} \mathrm{H}_1^0\) = -187.6 kJ
3. H_2(g) + N_2(g) → 2NH_3(g); ∆_rH° = -92.2 kJ

The sum of the enthalpy changes for steps (1) and (2) is equal to enthalpy change for the overall reaction.

Application of Hess’s Law
The Hesss law has been useful to calculate the enthalpy and the enthalpy changes for the reactions with their enthalpies being not known experimentally.

Question 23.
Write structure of natural rubber and neoprene rubber along with the name and structure of their monomers.
Answer:

Question 24.
What are pseudo-first order reactions? Give one example and explain why it is pseudo-first order.
Answer:
A reaction which has higher order true rate law but experimentally found to behave as first order are called pseudo first order reaction. For example: Hydrolysis of sucrose.

is an example of pesudo first order reaction, becauše water takes part in the reaction the true rate law rate = k[C₁₂H₂₂O₁₁][H₂O] indicates that the reaction must be second order. Similarly to the hydrolysis of ester. [H₂O]. It is constant and the rate law became rate = k[C₁₂H₂₂O₁₁]
Thus, the second order true law is converted, to first order rate law.

Question 25.
Why it is impossible to measure the potential of a single electrode?
Answer:

1. Every oxidation reaction needs to be accompanied by a reduction reaction.
2. The occurrence of only oxidation or only reduction is not possible.
3. In galvanic cell oxidation and reduction occur simultaneously.
4. The potential associated with the redox Can be experimentally measured. For the measurement of potential two electrodes need to be combined together where the redox reaction occurs.
   Hence, it is impossible to measure the potential of a single electrode.

Question 26.
(i) Calculate the standard enthalpy of.
N₂H_4(g) + H_2(g) → 2NH_3(g)
If ∆H°(N – H) = 389 kJ mol^-1, ∆H°(H – H) = 435 kJ mol^-1, ∆H°(N-N) = 159 kJ mol^-1
(ii) The enthalpy change of the following reaction:
CH_4(g) + Cl_2(g) → CH₃Cl_(g) + HCl_(g), ∆_rH° = – 104 kJ.
Calculate C-Cl bond enthalpy. The bond enthalpies are:

Answer:
(i)

(ii)

Section – D

Attempt any Three of the following questions:

Question 27.
(i) In NaOH solution [OH] is 2.87 × 10^-4. Calculate the pH of solution.
(ii) The solubility product of AgBr is 5.2 × 10^-13. Calculate its solubility in mol dm^-3 and g dm^-3 (Molar mass of AgBr = 187.8 g mol^-1)
Answer:
(i) pOH = – log₁₀ [OH^–]

(ii) pH + pOH = 14
Calculation:
From formula (i),
pOH = -log₁₀[OH^–]
∴ pOH = – log₁₀ [2.87 × 10^-4]
= – log₁₀ 2.87 – log₁₀ 10^-4
= -log₁₀ 2.87 + 4
= 4 – 0.4579
pOH = 3.5421
From formula (ii),
pH + pOH = 14
pH = 14 – pOH
= 14 – 3.5421
= 10.4579, pH of the solution is 10.4579

(ii) The solubility product of AgBr is:

The solubility in g dm^-3 = molar solubility in mol dm^-3 × molar mass g mol^-1
S = 7.2 × 10^-7 mol dm^-3 × 187.8 g mol^-1
= 1.35 × 10^-4 g dm^-3

Question 28.
Draw a neat diagram for the Haworth formula of glucopyranose. Give the industrial applications of enzyme catalysis.
Answer:

Some examples of industrial application of enzyme catalysis are:

1. Conversion of glucose to sweet-tasting fructose, using glucose isomerase.
2. Manufacture of new antibiotics, using pencillin G acylase.
3. Manufacture of laundry detergents, using proteases.
4. Manufacture of esters used in cosmetics, using genetically engineered enzyme.

Question 29.
Distinguish between rhombic sulphur and monoclinic sulphur.
Answer:

Rhombic (α-sulphur)	Monoclinic (β-sulphur)
It is a pale yellow coloured soild.	It is bright yellow soild
It forms orthorhombic crystals	It forms needle-shaped monoclinic crystals
Its melting point is 385.8 K	Its melting point is 393 K
Its density is 2.06 g/cm3	Its density is 1.98 g/cm3
It is insoluble in water and soluble in CS2	Soluble in CS2
It is stable below 369 K and transforms to β-sulphur above this temperature	It is stable above 369 K and transforms into α-sulphur below this temperature.
It is prepared by the evaporation of rolls sulphur in CS2.	It is prepared from rhombic sulphur.

Question 30.
Write structural formulae for:
(i) Pentane-1,4-diol
(ii) Cyclohex-2-en-1-oL
(iii) p-Nitrophenol
(iv) Salicylic acid
Answer:

Question 31.
Explain the trends in atomic radii of d-block elements.
Answer:

1. Atomic radii of the elements of the transition series decrease gradually from left to right.
2. As we move across a transition series from left to right the nuclear charge increases by one unit at a time.
3. The last filled electron enters a penultimate (n – 1) d subshell. However, d orbitals in an atom are less penetrating or more diffused and, therefore d electrons offer smaller screening effects.
4. The result is that effective nuclear charge also increases as the atomic number increases along with a transition series. Hence the atomic radii gradually decrease across a transition series from left to right.

Maharashtra Board Class 12 Chemistry Previous Year Question Papers