Home > CBSE > 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:
Q2. Pick the alkyl halide for which the rate of SN1 reaction is fastest.
Opt:
Tertiary halide
Secondary halide
Primary halide
All of them react at the same rate
Ans:
Q3. Which of the following are basic oxides?
Mn2O7, V2O3, V2O5, CrO, Cr2O3
Mn2O7 and V2O5
V2O3 and CrO
CrO and Cr2O3
V2O5 and V2O3
Q4. In the reaction,
2H2 + O2 ⟶ 2H2O2H_2\ \ +\ \ \ O_2\ \ \ \longrightarrow\ \ 2H_2O
2H2 + O2 ⟶ 2H2O
Rate of formation of water is 4.6X10-4 molL-1s-1, rate of disappearance of hydrogen will be
2.6X10-4 molL-1s-1.
3.6X10-4 molL-1s-1.
4.6X10-4 molL-1s-1.
5.6X10-4 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
d. 0.000147
b. 0.7455 g
Q6. On passing a current of 1.0 ampere for 16 min and 5 sec through 1 litre solution of
CuCl2CuCl_2
CuCl2 , all copper of the solution was deposited at cathode. The strength of
CuCl2 solution was (Molar mass of Cu= 63.5; Faraday constant = 96,500 Cmol–1)
0.01 N
0.01 M
0.02 M
0.2 N
Q7. Aniline when treated with conc. sulphuric acid gives
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
+3
+2
+1
+4
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:
Q11. The participation of carbonyl group in resonance stabilisation of carbanion formed due to acidic nature of α-hydrogens is explained by
Low electronegativity of carbon atom.
High electronegativity of oxygen atom.
Hybridisation of carbon atom.
Tendency of carbonyl group for resonance.
Q12.
The rate of a certain reaction,
at different times are as follows:
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.
A. zero order.
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
[Fe(H2O)6]3+
[Fe(CN)6]3-
[Fe(C2O4)6]3-
[FeCl6]3-
Q14. The compound, not required for the reductive amination of aldehydes and ketones is
Ammones
Tin
Nickel
Magnesium
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.
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.
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.
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.
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
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.
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?
(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.
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.
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.
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 SN1 reaction faster and why?
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 SN1 reaction faster. This is because of the fact
that SN1 reaction takes place via carbocation formation and it forms 20 carbocation, which is more stable than that of 10 carbocation produced by later one.
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)
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?
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?
Write the equations for the preparation of 1-iodobutane from
(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:
2.
3.
They have identical physical properties.
They have specific rotation equal in magnitude but opposite in sign.
They have different physical properties.
They differ in magnitude of specific rotation.
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.
(d) How do amino acids form zwitter ions. explain using an example.
(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.
(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]
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
(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.
(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: