CBSE Sample Papers For Class 12 Chemistry Mock Paper 1

Q1. Benzoylation of phenols is usually carried out in the presence of aq. NaOH since benzoyl chloride is not readily hydrolysed by alkalis. This reaction is called
Opt:

Schotten-Baumann reaction.

Fischer-Spier reaction.

Dean-Stark reaction.

Riemer-Tiemann reaction.
Ans:

Schotten-Baumann reaction.

Q2. Pick the alkyl halide for which the rate of S_N1 reaction is fastest.

Opt:

Tertiary halide

Secondary halide

Primary halide

All of them react at the same rate

Ans:

Tertiary halide

Q3. Which of the following are basic oxides?

Mn₂O₇, V₂O₃, V₂O₅, CrO, Cr₂O₃

Opt:

Mn₂O₇ and V₂O₅

V₂O₃ and CrO

CrO and Cr₂O₃

V₂O₅ and V₂O₃

Ans:

V₂O₃ and CrO

Q4. In the reaction,

$2H_2\ \ +\ \ \ O_2\ \ \ \longrightarrow\ \ 2H_2O$

2H2​ + O2​ ⟶ 2H2​O

Rate of formation of water is 4.6X10^-4 molL^-1s^-1, rate of disappearance of hydrogen will be

Opt:

2.6X10^-4 molL^-1s^-1.

3.6X10^-4 molL^-1s^-1.

4.6X10^-4 molL^-1s^-1.

5.6X10^-4 molL^-1s^-1.

Ans:

4.6X10^-4 molL^-1s^-1.

Q5.

1. 0.345 g of copper is deposited by current of 1.3 amperes in 0.5 hour. Its electrochemical equivalent is
a. 0.002348
b. 0.000344
c. 0.001167
d. 0.000147

Or

2. The amount of chlorine (in grams) can be produced by the electrolysis of molten NaCl with a current of 2 A for 17 minutes is
a. 1.7001 g
b. 0.6194 g
c. 0.7455 g
d. 2.1070 g

Ans:

d. 0.000147

Or

b. 0.7455 g

Q6. On passing a current of 1.0 ampere for 16 min and 5 sec through 1 litre solution of

$CuCl_2$

CuCl2​ , all copper of the solution was deposited at cathode. The strength of

$CuCl_2$

CuCl2​ solution was (Molar mass of Cu= 63.5; Faraday constant = 96,500 Cmol^–1)

Opt:

0.01 N

0.01 M

0.02 M

0.2 N

Ans:

0.01 N

Q7. Aniline when treated with conc. sulphuric acid gives

Opt:

p-methylbenzenesulphonic acid.

m-nitrobenzenesulphonic acid.

o-Nitroaniline.

p-aminobenzenesulphonic acid.
Ans:

p-aminobenzenesulphonic acid.

Q8. The oxidation state of Fe in the brown ring complex [Fe(H₂O)₅NO]SO₄ is

Opt:

+3

+2

+1

+4

Ans:

+2

Q9. Anisole on reaction with HI gives
Opt:

Phenyl iodide + Methyl iodide.

Phenol + Methyl iodide. Phenol + ethyl iodide.

Benzyl alcohol + Methyl iodide.
Ans:

Phenol + Methyl iodide.

Q10. The order of basicity of ethylamines is different from methylamines. It is
Opt:

primary>secondary>tertiary.

secondary>tertiary>primary.

tertiary>primary>secondary.

secondary>primary>tertiary.
Ans:

secondary>tertiary>primary.

Q11. The participation of carbonyl group in resonance stabilisation of carbanion formed due to acidic nature of α-hydrogens is explained by

Opt:

Low electronegativity of carbon atom.

High electronegativity of oxygen atom.

Hybridisation of carbon atom.

Tendency of carbonyl group for resonance.

Ans:

High electronegativity of oxygen atom.

Q12.

The rate of a certain reaction,

at different times are as follows:

Time (Second)	Rate (mol L-1 s-1)
0	2.8 10-2
10	2.78 10-2
20	2.81 10-2
30	2.79 10-2

A. zero order.
B. first order.
C. second order.
D. third order.

OR

For visually challenged​

The minimum amount of energy that colliding molecules must have in order that the collision between them may be effective is called

A. activation energy.
B. threshold energy.
C. free energy.
D. collision energy.

Ans:

A. zero order.

OR

B. threshold energy.

Q13.

In the complexes [Fe(H₂O)₆]³⁺ , [Fe(CN)₆]^3–, [Fe(C₂O₄)₃]^3– and [FeCl₆]^3–, more stability is shown by

Opt:

[Fe(H₂O)₆]³⁺

[Fe(CN)₆]^3-

[Fe(C₂O₄)₆]^3-

[FeCl₆]^3-

Ans:

[Fe(C₂O₄)₆]^3-

Q14. The compound, not required for the reductive amination of aldehydes and ketones is

Opt:

Ammones

Tin

Nickel

Magnesium

Ans:

Nickel

Q15. In the question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below the question.

Assertion (A): Phenols show +R effect.

Reason (R): The oxygen atom of alcohols contains lone pair of electrons.

Opt:

Both assertion (A) and reason (R) are true, and the reason (R) is the correct explanation of the assertion (A).​

​Both assertion (A) and reason (R) are true, but the reason (R) is not the correct explanation of the assertion (A).​

​Assertion (A) is true, but the reason (R) is false.​

​Assertion (A) is false, but the reason (R) is true.​

Ans:

Both assertion (A) and reason (R) are true, and the reason (R) is the correct explanation of the assertion (A).​

Q16. In the question, a statement of Assertion (A) followed by the statement of Reason (R) is given. Choose the correct option out of the choices given below the question.

Assertion (A): Enzymes increase the speed of a reaction.

Reason (R): Enzymes increase the activation energy.

(a) Both assertion and reason are true, and the reason is the correct explanation of assertion.

(b) Both assertion and reason are true, but the reason is not the correct explanation of assertion.

(c) Assertion is true, but the reason is false.

(d) Both assertion and reason are false.

Ans:

Answer: (C). Assertion is true, but the reason is false.

Explanation: Enzymes increase the speed of a reaction. This is because enzymes reduce the magnitude of the activation energy of a reaction.

Q17. In the question, a statement of Assertion (A) followed by the statement of Reason (R) is given. Choose the correct option out of the choices given below the question.

Assertion (A): Zinc has a low enthalpy of atomisation.

Reason (R): Zinc is not considered a transition element.

(a) Both assertion and reason are true, and the reason is the correct explanation of assertion.

(b) Both assertion and reason are true, but the reason is not the correct explanation of assertion.

(c) Assertion is true, but the reason is false.

(d) Both assertion and reason are false.

Ans:

Answer: (b) Both assertion and reason are true, but the reason is not the correct explanation of assertion.

Explanation: Zinc is a d-block element having a low enthalpy of atomisation. This is due to completely filled d orbitals in zinc. Thus, the d electrons are not involved in metallic bonding as in the case of other transition elements. Due to completely filled d-orbitals, zinc is not considered a transition element.

Q18. Assertion(A) Basicity of CH₃CH₂NH₂ (I), NH₃(II), and C₆H₅NH₂(III) is in order I > II > III

Reason(R) Electron donating group (such as alkyl group) increase the basicity of amines and electron-withdrawing groups (such as aryl group) decrease the basicity of amines

Opt:

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not correct explanation of A.

A is true but R is false.

A is false but R is true.

Ans:

Both A and R are true and R is the correct explanation of A.

Q19. (i) N₂O₅ (inCCl₄) → 2NO₂ + 1/2 O₂

This reaction is of first order and rate constant of reaction is 6.2 × 10^–4s^–1. What is the value of rate of reaction when [N₂O₅] = 1.75 mol L^–1.

(ii) What is the molecularity of the reaction?

Ans:

(i) Rate = k [N₂O₅]
Rate = k[N₂O₅]
= 6.2 × 10^–4 × 1.75
or = 10.93 molL^–1s^–1

(ii) Molecularity of the reaction is 1.

Q20. Explain mutarotation taking D-glucose as an example.

OR

Write the major classes in which the carbohydrates are divided depending upon whether these undergo hydrolysis, and if so, on the number of products formed.

Ans:

Mutarotation

–Specific rotation of

–D-glucose is +111º and that of β- D-glucose is +19.2º. When either

–D-glucose or β- D-glucose is dissolved in water and allowed to stand, they convert into other form and form equilibrium mixture having specific rotation +52.5º. This spontaneous change in optically active compound with time to an equilibrium value is called mutarotation.

OR

On the basis of hydrolysis products the carbohydrates are divided into following categories
(i) Monosaccharides-They are simplest carbohydrates and they do not hydrolyse. For example-Glucose.
(ii) Disaccharides-On hydrolysis they produce two molecules of monosaccharides-.For example-Sucrose on hydrolysis produces one molecule of glucose and one molecule of fructose.
(iii) Oligosaccharides-On hydrolysis they produce 3-10 molecules of monosaccharides. For example-Raffinose.
(vi) Polysaccharides-On hydrolysis they produce many molecules of monosaccharides. For example-Starch.

Q21. (a) Why are haloalkanes more reactive towards nucleophilic substitution reactions than halorenes
(b) Which one of the following two substances undergoes S_N1 reaction faster and why?

Or

How will you prepare DDT?
Ans:

(a) Haloarenes are resonance (Structure I to V) stabilized but haloalkanes are not. As a result energy of activation for displacement of halogen for haloalkanes is much lower than that from haloarenes. As a result, haloalkanes are more reactive than haloarenes towards nucleophilic substitution reactions.

(b)

undergoes S_N1 reaction faster. This is because of the fact

that S_N1 reaction takes place via carbocation formation and it forms 2⁰ carbocation, which is more stable than that of 1⁰ carbocation produced by later one.

Or

Preparation of DDT: – It is prepared by heating chlorobenzene with chloral in the presence of concentrated H₂SO₄.

Q22. Using the valance bond approach, predict the shape and magnetic behavior of [Cr(NH₃)₅Cl]²⁺ ion. (Atomic number of Cr is 24)
Ans:

The oxidation state of chromium in [Cr(NH₃)₅Cl]²⁺ ion is +3. The electronic configuration of Cr is [Ar] 3d⁵ 4s¹. Hence, we have

The shape of the complex is Octahedral. And the magnetic behaviour is paramegnetic.

Q23. When a non-volatile solute is added to pure water its vapour pressure decreases by 4 mm Hg. Calculate molality of solution. (Vapour pressure of pure water is 40mm Hg)

Opt:

Ans:

Q24. A reaction A Product is found to be third order reaction. What will happen to rate of reaction when concentration of A is doubled?
Ans:

r₁ = k [A]³ (Given)

Conc. of A is doubled.

∴ It becomes 2A.

∴ r₂ = k [2A]³ = 8[A]³ =8r₁

∴ Rate of reaction will increase by factor of 8.

Q25. Write the mechanism of hydration of ethene to yield ethanol.
Ans:

The mechanism of hydration of ethene to form ethanol involves three steps.

Step 1:

Protonation of ethene to form carbocation by electrophilic attack of H₃O⁺:

Step 2:

Nucleophilic attack of water on carbocation:

Step 3:

Deprotonation to form ethanol:

Q26. Draw the structures of optical isomers of
(i) [Co(en)₃]³⁺.
(ii) cis-[CoCl₂(en)₂]⁺.
(iii) [Co(en)Cl₂(NH₃)].
Ans:

(i) [Co(en)₃]³⁺

(ii) cis-[CoCl₂(en)₂]⁺

(iii) [Co(en)Cl₂(NH₃)]

Q27. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Ans:

Let x g of Na₂CO₃ be present in the mixture.

So, (1 – x) g of NaHCO₃ is present in the mixture.

Molar mass of Na₂CO₃ = (2 x 23 + 12 + 3 x 16)

= 106 g/mol

Molar mass of NaHCO₃ = (23 + 1 + 12 + 3 x 16)

84 g mol^–1

Moles of Na₂CO₃ in x g= x/106

And Moles of NaHCO₃ in (1 – x) g = (1 – x)/ 84

As the mixture contains equimolar amounts of two components, therefore,

x106=1−x84Or, 106-106x=84xOr, x=106190g=0.558gThus, moles of Na2CO3=0.558/106=00.00526And moles of NaHCO3=(1−0.558)/84=0.442/84=0.00526Reaction of Na2CO3 and NaHCO3 with HCl are as follows:Na2CO3+2HCl→2NaCl+H2O+CO2NaHCO3+HCl→NaCl+H2O+CO21 mole of Na2CO3 requires 2 moles of HCl 0.00526 moles of Na2CO3 requires =(0.00526×2)moles of HCl=0.01052 mole1 mole of NaHCO3 requires 1 mole of HCl 0.00526 moles of NaHCO3 requires 0.00526 moles of HCl=0.01052 mole1 mole of NaHCO3 requires 1 mole of HCl 0.00526 moles of NaHCO3 requires 0.00526 moles of HCl∴Total HCl required =(0.01052+0.00526) moles=0.01578moles 0.1 mole of 0.1 M HCl are present in 1000 mL of HCl∴ 0.01578 moles of HCl will be present in =(10000.1)×0.01578=157.8mL

Thus, 157.8 mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both.

Q28.

(a) Will Chlorobenzene be highly soluble in water or not? Explain with reason.
(b) (i) Pedict that 2-Bromopropanoic acid is optically active or not?
(ii) What are the differences between enantiomers and diastereomers?

OR

Write the equations for the preparation of 1-iodobutane from

1. 1-butanol
2. 1-chlorobutane
3. but-1-ene.

Ans:

(a) Chlorobenzene is slightly soluble in water. Although it is a polar molecule, but due the presence of large hydrocarbon part (benzene ring) hydrogen bonds formed between water and chlorobenzene are not as stronger as hydrogens bond present between water molecules. As a result, chlorobenzene is slightly soluble in water.

(b) (i) 2-Bromopropanoic acid is an optically active compound because it has one chiral centre and has non-superimposable mirror images.
(Where * indicate one chiral carbon)
2-Bromopropanoic acid

(ii) Differences between enantiomers and diastereomers:

OR

Q29. Case-Proteins are amino acid biomolecules with a high molecular mass. Except for glycine, all -amino acids have chiral carbon atoms, and majority of them are L-configured. Polypeptides are formed when a large number of α-amino acids combine. Proteins are peptides with a very large molecular mass (more than 10,000). The protein structure can be classified as primary, secondary, or tertiary, or quaternary. In an aqueous solution, amino acids exist as zwitter ions.

(a) Draw the structure of the amino acids alanine and aspartic acid

(b) Describe the secondary structure of proteins.

OR

(d) How do amino acids form zwitter ions. explain using an example.

Ans:

(a) The structures of amino acids alanine and aspartic acid are

(b) The secondary structure of protein refers to the shape in which a polypeptide chain exists. They are found to exist in two different types of structures

(i) α-helix, and

(ii) β-pleated sheet structure.

Secondary structures arise due to regular folding of polypeptide chain due to hydrogen bonding between and –NH– groups of the peptide bond. In α-Helix, a polypeptide chain forms hydrogen bonds by twisting into a right-handed screw (helix) with the –NH group of each amino acid residue hydrogen bonded to the C=O bond of an adjacent turn of the helix. In β-pleated sheet structure, all peptide chains are stretched out and then laid side by side which are held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery.

OR

(c) In aqueous solutions, amino acids exist as zwitter ions. A zwitter ion is formed by loss of proton from COOH group and gain of proton by NH₂ group. Thus, it has positive NH₃⁺ and negative COO^– group in the same molecule. For example, alanine exists as a zwitter ion

Q30. Calculate the magnetic moment of the following:

(a) [CrCl₃(py)₃]

(b) K[Mn(CN)₆]

Ans:

Q31. Give the chemical reactions for the following conversions.

(i) Butanal into butanoic acid
(ii) 3-nitrobromobenzene into 3-nitrobenzoic acid
(iii) Butyl alcohol into butanoic acid
(iv) Benzyl alcohol into phenylethanoic acid
(v) Cyclohexene into hexan-1,6-dioic acid

Ans:

(i) Butanal into butanoic acid : –

(ii) 3-nitrobromobenzene into 3-nitrobenzoic acid: –

(iii) Butyl alcohol into butanoic acid : –

(iv) Benzyl alcohol into phenyleth-1-oic acid : –

(v) Cyclohexene into hexan-1,6-dioic acid: –

Q32. Calculate the vapour pressure of a solution prepared by mixing 349.5 gm of component A (Mol. Mass 134) and 416.8 gm of component B (Mol. Mass 169) at 298 K. Vapour pressures of pure components A and B are 421mm Hg and 562mm Hg respectively. Also calculate mole fraction of A and B in vapour phase.

Or

(a) Define the term ‘osmosis’ and ‘osmotic pressure’. Describe how the molecular mass of a substance can be determined on the basis of the osmotic pressure measurement.
(b) 0.855 g of sugar (molar mass = 342 u) is dissolved in 250 cm³ of solution at 298 K. Determine the osmotic pressure of the solution (Given R = 0.083 L bar K^–1 mol^–1).
Ans:

(i) Butanal into butanoic acid : –

(ii) 3-nitrobromobenzene into 3-nitrobenzoic acid: –

(iii) Butyl alcohol into butanoic acid : –

(iv) Benzyl alcohol into phenyleth-1-oic acid : –

(v) Cyclohexene into hexan-1,6-dioic acid: –