Maharashtra Board Class 12 Chemistry Sample Paper Set 1 with Solutions

Maharashtra State Board Class 12th Chemistry Sample Paper Set 1 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Chemistry Model Paper Set 1 with Solutions

Section – A

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

(i) A compound forms hep structure, number of Octahedral and Tetrahedral voids in 0.5 mole of substance is respectively.
(a) 3.011 × 10²³, 6.022 × 10²³
(b) 6.022 × 10²³, 3.011 × 10²³
(c) 4.011 × 10²³, 2.011 × 10²³
(d) 6.011 × 10²³, 12.022 × 10²³
Answer:
(a) 3.011 × 10²³, 6.022 × 10²³

(ii) In which reaction mechanism carbocation is formed?
(a) SN₁
(b) SN₂
(c) Both (a) and (b)
(d) None of them
Answer:
(a) SN₁

(iii) The solubility product of a sparingly soluble salt is 5.2 × 10^-13. Its solubility in mol dm × 10^-3 is:
(a) 7.2 × 10^-7
(b) 1.35 × 10^-4
(c) 7.2 × 10^-8
(d) 13.5 × 10^-8
Answer:
(a) 7.2 × 10^-7

(iv) Neutral solutions have the pH of
(a) 8
(b) 7
(c) 9
(d) 14
Answer:
(b) 7

(v) Polonium has the half-life of
(a) 13.8 days
(b) 12 days
(c) 5 days
(d) 102 days
Answer:
(a) 13.8 days

(vi) Most stable oxidation state of titanium is:
(a) +2
(b) +3
(c) +4
(d) +5
Answer:
(c) +4

(vii) Which cell is used as a source of power in flashlights?
(a) Dry cell
(b) Fuel cell
(c) Hydrogen cell
(d) None of them
Answer:
(a) Dry cell

(viii) In which of the following series all the elements are radioactive in nature:
(a) lanthanides
(b) actinides
(c) d-block elements
(d) s-block elements
Answer:
(b) actinides

(ix) Pressure cooker reduces cooking time for food because:
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature.
Answer:
(a) boiling point of water involved in cooking is increased

(x) CH₂OH-CO-(CHOH)₄-CH₂OH is an example of
(a) Aldohexose
(b) Aldoheptose
(c) Ketotetrose
(d) Ketoheptose
Answer:
(d) Ketoheptose

Question 2.
Answer the following questions:

(i) State second law of thermodynamics in terms of entropy.
Answer:
(0 Second law of thermodynamics in terms of entropy:
The second law of thermodynamics states that, “The total entropy of the system and its surroundings (universe) increases in a spontaneous process”.

(ii) Write Arrhenius equation and explain the terms involved in it.
Answer:
00 Arrhenius equation k = \(A e^{\frac{-E_0}{R T}}\)
where k is the rate constant Ea is the activation energy, R molar gas constant T temperature in kelvin and A is the pre-exponential factor.

(iii) Name two gases which deplete ozone layer.
Answer:
Two gases that deplete ozone layer are Nitric oxide (NO), chlorofluoro carbons (freons).

(iv) Write examples of addition polymers and condensation polymers.
Answer:
(a) Addition Polymers: Polythene, Teflon, Polyacrylonitrile, Polyvinylchloride (PVC), etc.
(b) Condensation Polymers: Terylene, Nylon-6,6, Bakelite, Novolac, etc.

(v) What are aromatic ketones?
Answer:
The compounds in which >C=O group is attached to either two aryl groups or one aryl and one alkyl group are called aromatic ketones.
eg.

(vi) Write reaction of P-toluenesulfonyl chloride with Diethylamine.
Answer:

(vii) Which nanomaterial is used for tyres of car to increase the life of tyres?
Answer:
Carbon black

(viii) Presence of disulphide link gives rise to which structure of protein?
Answer:
Tertiary structure of protein

Section B
Attempt any Eight of the following questions

Question 3.
The osmotic pressure of CaCl₂ and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate Van’t Hoff factor for CaCl₂.
Answer:
Given: Osmotic pressure of CaCl2 solution = 0.605 atm
Osmotic pressure of urea solution = 0.245 atm
To find: The value of Van’t Hoff factor
Formulae: π = MRT, π = iMRT
π = MRT
0.245 atm = MRT …….(i)
For CaCl₂ solution
π = iMRT
0.602 atm = iMRT …(ii)
For equations (i) and (ii)
\(\frac{0.605}{0.245}\) = \(\frac{iMRT}{MRT}\)
The value of Van’t Hoff factor is 2.47.

Question 4.
What are bidentate ligands? Give one example.
Answer:
Bidentate Ligands: The ligands which bind to central metal through two donor atoms are called bidentate ligands.
e.g.,
1. Ethylenediammine binds to the central metal atom through two nitrogen atoms

2. Similarly, Oxalate ligand C₂O₄^2- utilises electron pair on each of its negatively charged oxygen atoms on linking with central metal.

Question 5.
How nanotechnology plays an important role in water purification techniques?
Answer:
Nanotechnology plays an important role in water purification techniques. Some of its uses are:

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

Question 6.
What is a salt bridge?
Answer:

• It provides electrical contact between two solutions and thereby completes the electrical circuit.
• It prevents the mixing of two solutions.
• It maintains electrical neutrality in both the solutions by the transfer of ions.

Question 7.
What two uses of the alloys.
Answer:
The two uses of alloys are:

• Bronze, an alloy of copper and tin is tough, strong and corrosion resistant. It is used for making statues, medals and trophies.
• Cupra-nickel, an alloy of copper and nickel is used for making machinery parts of marine ships, boots. For example, marine condenser tubes.

Question 8.
Why haloarenes are less reactive than halo alkanes?
Answer:
(i) The low reactivity of aryl halides is due to resonance effect and sp² hybrid state of carbon to which halogen atom is attached.

(ii) In aryl halides, one of the lone pairs of electrons on- halogen atom is in conjugation with p-electrons of the ring. Due to resonance, the C-X bond acquires partial double bond character. Thus, the C-X bond in haloarenes is stronger and shorter than haloalkanes. Hence, it is difficult to break C-X bond in haloarenes. (e.g. C-Cl bond length in chlorobenzene is 169 pm as compared to C-Cl bond length in alkyl chloride which is 178 pm).

Question 9.
Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid.
Answer:
9. Given:
Percent dissociation = 5% Concentration (c) = 1 decimolar
To find: Dissociation constant of acid (k_a)
Formulae: used
(i) Percent dissociation = α × 100
(ii) K_a = α²c
Calculation: Using formula (i),

α = \(\frac{\text { Percent dissociation }}{100}\)
= \(\frac{5}{100}\) = 0.05
c = 1 decimolar = 0.1 M .
Using formula (ii),
K_a = (0.05)² × (0.1)
= 2.5 × 10^-4
The dissociation constant of acid is 2.5 × 10^-4.

Question 10.
What is entropy? Give its units.
Answer:
Entropy is a measure of molecular disorder or randomness. An entropy change of a system is equal to the amount of heat transferred (Q_rev) to it in a reversible manner divided by the temperature (T) in Kelvin at which the transfer takes place. Thus,
∆S = \(\frac{Q_{\mathrm{rev}}}{T}\)
Units of entropy are: JK^-1

Question 11.
Give two evidences for presence of formyl group in glucose.
Answer:
Two evidences in favour of the presence of formyl group in glucose are:
(i) Glucose gets oxidised to a six-carbon monocarboxylic acid called gluconic acid on reaction with bromine water which is a mild oxidising agent. Thus, the carbonyl group in glucose is in the form of formyl (-CHO).
(ii) Hemiacetal group of glucopyranose structure is a potential aldehyde (formyl) group. It imparts reducing properties to glucose. Thus, glucose gives positive Tollen’s test or Fehling test.

Question 12.
What is the standard enthalpy of combustion? Give an exampLe.
Answer:
The standard enthalpy of combustion of a substance is the standard enthalpy change accompanying a reaction in which one mole of the substance in its standard state is completely oxidised.
Consider the reaction,

In the above reaction, the standard enthalpy change of the oxidation reaction,-1300 Id is the standard enthalpy of combustion of C₂H₂(g).

Question 13.
Explain the effect of dilution of solution on conductivity.
Answer:
(i) The electrolytic conductivity is the electrical conductance of unit volume (1 cm³) of solution. It depends on the number of current-carrying ions present in unit volume of solution.
(ii) On dilution total number of ions increases as a result of an increased degree of dissociation.
(iii) An increase in the total number of ions is hot in the proportion of dilution. Therefore the number of ions per unit volume of solution decreases.
(iv) This results in a decrease of conductivity with a decrease in the concentration of the solution.

Question 14.
What is difference between a double salt and a complex? Give one example.
Answer:
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.t Mohr’s salt, FeSO₄(NH₄)₂SO₄.6H₂O is a double salt while K₄[Fe(CN)₆] is a complex.

Section C
Attempt any Eight of the following questions:

Question 15.
While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity, why?
Answer:
In boiling point elevation and freezing point depression, we deal with the systems whose temperature is not constant. We cannot express the concentration of the solution in molarity because it changes with temperature whereas molality is temperature independent. Hence, while considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity.

Question 16.
Why aldehydes are more reactive toward nucleophilic addition reactions than ketones?
Answer:
Reactivity of aldehydes and ketones is due to the polarity of carbonyl group which results in electrophilicity of carbon. In general aldehydes are more reactive than ketones towards the nucleophilic attack. This can be well explained in terms of both the electronics effect and steric effect.
(i) Influence of electronic effects:
(a) Alkyl groups have an electron-donating inductive effect (+1). A ketone has two electron-donating alkyl groups bonded to carbonyl carbon which are responsible for decreasing its positive polarity and electrophilicity.
(b) In contrast aldehydes have only one electron- donating groups bonded to the carbonyl carbon. This makes aldehydes more electrophilic than ketones.

(ii) Steric effects:
(a) Two bulky alkyl groups in ketone come in the way of the incoming nucleophile. This is called steric hindrance to nucleophilic attack.
(b) On the other hand, nucleophile can easily attack the carbonyl carbon in aldehyde because it has one alkyl group and is less crowded or sterically less hindered. Hence, aldehydes are more easily attacked by nucleophiles.

Question 17.
Why dissociation of HCN is suppressed by the addition of HCl?
Answer:
HCN and HCl both dissociate to produce H+ ions which are common to both.
HCN is a weak electrolyte. It dissociates to a little extent.
HCN(aq) → H⁺ + CN^–
HCl is a strong electrolyte. It undergoes complete dissociation.
HCNl(aq) → H⁺ + Cl^–
both HCN and HCl provide H⁺ ions.
The concentration of H⁺ ions in the solution increases due to the complete dissociation of HCl.
According to Le-Chatelier’s principle, the effect of the stress (the addition of H⁺ ions from HCl) applied to the ionisation equilibrium of HCN is reduced by shifting the equilibrium in the backward direction.
H⁺ ions combine with CN^– ions to produce unionised HCN. Thus, the dissociation of HCN is suppressed by the addition of HCL.

Question 18.
Write structural formulae for:
(i) 3-Methoxyhexane
(ii) Methyl vinyl ether
(iii) 1-Ethylcyclohexanol
Answer:

Maharashtra Board Class 12 Chemistry Previous Year Question Papers