Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

  • There are 35 questions in this question paper with internal choices.
  • Section A consists of 18 multiple-choice questions carrying 1 mark each.
  • Section B consists of 7 very short answer questions carrying 2 marks each.
  • Section C consists of 5 short answer questions carrying 3 marks each.
  • Section D consists of 2 case-based questions carrying 4 marks each.
  • Section E consists of 3 long answer questions carrying 5 marks each.
  • All questions are compulsory.
  • Use of log tables and calculators are not allowed.

SECTION – A (18 Marks)
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mart There is no internal choice in this section.)

Question 1.
Which of the following option will be the limiting molar conductivity of CH3COOH if the limiting molar conductivity of CH3COONa is 91 Scm2mol-1? Limiting molar conductivity for individual ions are given in the following table: [1]

S.No. Ions Limiting molar conductivity/Scm2mol-1
1. H+ 349.6
2. Na+ 50.1
3. K+ 73.5
4. OH 199.1

(a) 350 Scm2mol-1
(b) 375.3 Scm2mol-1
(c) 390.5 Scm2mol-1
(d) 340.4 Scm2mol-1
Answer:
(c) 390.5 Scm2 mol-1

Explanation:
The limiting molar conductivity (Λm) for strong and weak electrolyte can be determined by using Kohlrausch’s law which states that “the limiting molar conductivity of an electrolyte can be represented as the sum of the indivudual contributions of the anion and cation of the electrolyte.”
Λ°CH3COONa = Λ°CH3COO + Na+
91 = Λ°CH3COO+ 50.1
⇒ ΛCH3COO = 40.9
⇒ ΛCH3COO = 40.9Scm2mol-1

Question 2.
Curdling of milk is an example of: [1]
(a) breaking of peptide linkage
(b) hydrolysis of lactose
(c) breaking of protein into amino acids
(d) denauration of proetin
Answer:
(d) denauration of proetin

Explanation:
The skin that forms on the curded milk is the phenomenon which occurs due to the denaturation of the proteins.

Related Theory
Denaturation is the process in which the proteins lose their native form on the application of the external factors such as strong acid, base, solvents (Both organic and Inorganic), radiation etc . in such cases proteins lose their native forms such as their quaternary, tertiary and secondary structures due to the breakage of the bonds and the unfolding in the molecules.

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 3.
When 1 mole of benzene is mixed with 1 mole of toluene The vapour will contain: (Given: vapour of benzene = 12.8 kPa and vapour pressure of toluene = 3.85 kPa).
(a) equal amount of benzene and toluene as it forms an ideal solution
(b) unequal amount of benzene and toluene as it forms a non ideal solution
(c) higher percentage of benzene
(d) higher percentage of toluene
Answer:
(c) higher percentage of benzene

Explanation:
Liquids which are more volatile exert more vapour pressure. The volatile liquids are those which have weak intermolecular force of attraction. The vapour pressure of benzene is more than the vapour pressure of toluene, thus benzene is more volatile in nature and it will have higher percentage of vapours thus the solution will have higher percentage of benzene.

Question 4.
Which of the following is the reason for zinc not exhibiting variable oxidation state? [1]
(a) inert pair effect
(b) completely filled 3d subshell
(c) completely filled 4s subshell
(d) common ion effect
Answer:
(b) completely filled 3d subshell

Explanation:
The electronic configuration of Zn is [Ar] 3d10 4s2, it has fully filled 3d subshell and it does not shows variable oxidation state as well since the fully filled shells are more stable and they do not react easily.

Question 5.
Propanamide on reaction with bromine in aqueous NaOH gives: [1]
(a) Propanamine
(b) Etanamine
(c) N-Methyl ethanamine
(d) Propanenitrile
Answer:
(b) Ethanamine

Explanation:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 1

Question 6.
The standard reduction potentials at 25°C for the follow half celL reactions are given against each: [1]
Zn2+(aq) + 2e -> Zn(s), E°cell = – 0.762 V
Cr3+(aq) + 3e —> Cr(s), E°cell = – 0.740 V
2H+ + 2e -> H2(g), E°cell = – 0.00 V
Fe3+ + e -> Fe2+, E°ceii = – 0.770 V
Which of the following is strong oxidizing agent?
(a) Fe2+
(b) Zn
(c) Cr
(d) H2
Answer:
(a) Fe2+

Explanation:
The value of standard reduction potential is directly proportional to strong reducing agent.

Question 7.
Which set of ions exhibit specific colours?
(Atomic number of Sc = 21, Ti = 22, V=23, Mn = 25, Fe = 26, Ni = 28 Cu = 29 and Zn =30)
(a) Sc3+, Ti4+, Mn3+
(b) Sc3+, Zn2+, Ni2+
(c) V3+, V2+, Fe3+
(d) Ti3+, Ti4+, Ni2+
Answer:
(c) V3+, V2+, Fe3+

Explanation:
These set of ions V3+, V2+, Fe3+ The electronic configuration of V3+ = [Ar] 3d3 4s2 The electronic configuration of V2+ = 1s2 2s2 2p6 3s2 3p6 3d3
And that of Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5.
Due to the incompletely filled d-subshells these ions undergo d-d transitions and thus shows colour due to the electronic transitions.

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 8.
Identify A, B, C and D:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 2
(a) A = C2H4, B = C2H5OH, C = C2H5NC, D = C2H5CN
(b) A= C2H5OH, B = C2H4, C = C2H5CN, D = C2H5NC
(c) A = C2H4, B = C2H5OH, C = C2H5CN, D = C2H5NC
(d) A = C2H5OH, B = C2H4, C = C2H5NC, D = C2H5CN
Answer:
(a) A = C2H4, B = C2H5OH, C = C2H5NC, D = C2H5CN

Explanation:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 3

Question 9.
The propionic acid when treated with aqueous sodium bicarbonate, liberates CO2. The ‘C’ of CO2 comes from: [1]
(a) methyl group
(b) carboxylic acid group
(c) methylene group
(d) bicarbonate
Answer:
(d) bicarbonate

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 4

Question 10.
Reagent used in Schiff’s base: [1]
(a) Pyridine
(b) SOCl2
(c) Raney Ni
(d) None of the above
Answer:
(c) Raney Ni

Explanation:
Schiff’s base on reduction with hydrogen in the presence of Raney nickel gives secondary alcohol

Question 11.
The alcohol which does not react with Lucas reagent is: [1]
(a) isobutyl alcohol
(b) n-butanol
(c) tert-butyl alcohol
(d) sec-butyl alcohol
Answer:
(b) n-butanol

Explanation:
In Lucas reagent test, turbiditys given immediately by tertiary alcohol and after 5 minutes turbidity’s given by secondary alcohol wlate primary alcohol does not give this test.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 5

Question 12.
Identify which is the allylic:
(a) H2C = CH – CH2– CH2OH
(b) CH3CH2OH
(c) OH-CH2CH2-OH
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 6
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 7

Explanation:
Allylic alcohol is an organic compound which has the structural formula CH2 = CHCH2OH. In other words, in these alcohols, the the-OH group is attached to sp2 hybridized carbon next to the carbon-carbon double bond, that is to an allylic carbon.

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 13.
Which among the following is a false statement? [1]
(a) Rate of zero order reaction is independent of initial concentration of reactant.
(b) Half life of a third order reaction is inversely proportional to square of initial concentration of the reactant.
(c) Molecularity of a reaction may be zero or fraction.
(d) For a first order reaction, =0.693k
Answer:
(c) Molecularity of a reaction may be zero or fraction.

Explanation:
Molecularity of reaction is defined as number of reactant molecules (or atoms or ions) taking part in an elementary reaction. Minimum vaLue of molecularity is one so cannot be fractional or zero.

Related Theory:
Molecularity can be explained by the following examples:
N2O5(g) → 2N02(g) + 21O2(g)

In this reaction, only one molecule is taking part in the reaction, therefore the rate law expression for this reaction is:
Rate = k[N2O5]
Hence, the reaction is unimolecular and first order. Similarly, a reaction can be bimolecular or trimolecular depending on the number or molecule taking part in the reaction.
Like O3(g) → 2O2(g) is a bi-molecular gaseouse
reaction tri-molecular gaseous reaction and 2NO + H2 → N2 + H2O2 is tri-molecular gaseous reaction.

Question 14.
For a reaction: H2 + Cl2 2HCl Rate = k
What is the order and molecularity of this reaction? [1]
(a) Zero-order reaction and molecularity is two.
(b) First-order reaction and molecularity is two.
(c) Second-order reaction and molecularity is one.
(d) Zero-order reaction and molecularity is one.
Answer:
(a) Zero-order reaction and molecularity is two.

Explanation:
Since Rate = k the reaction is not dependent on the concentration of any reactant hence the order is zero and two reactants are involved hence the molecularity is two.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false
(d) (A) is false but (R) is true

Question 15.
Assertion: Acylation of amines gives a monosubstituted product, whereas alkylation of amines gives polysubstituted Droduct.
Reason. Acyl group sterically hinders the approach of further acyl groups. [1]
Answer:
(c) (A) is true but (R) is false

Explanation:
Amines on acetylation give monosubstituted product, while on alkylation gives polysubstitution product as well. Thus, Assertion is correct statement but reason is wrong statement.

Question 16.
Assertion: The two strands are complementary to each other.
Reason: The hydrogen bonds are formed between specific pairs of bases. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
The two strands of DNA are complementary to each other the nucleotides held together on the each strand are complementary , since the nitrogenous bases A, T, G and C are bonded through hydrogen bonds with each other. Adenine and Guanine are purine bases present in DNA and cytosine and guanine are the pyrimidines bases present in DNA and RNA both, but the pyrimidine base uracil is present in RNA whereas the base thymine is present in DNA. Thus, assertion and reason both are correct statements and reason is correct explanation for assertion.

Related Theory
Adenine pairs with thymine with 2 hydrogen bonds. Guanine pairs with cytosine with 3 hydrogen bonds. This creates a difference in strength between the two sets of Watson and Crick bases. Guanine and cytosine bonded base pairs are stronger than thymine and adenine bonded base pairs in DNA.

Question 17.
Assertion: Propan-l-ol and propan-2-ol are distinguished byb iodoform test
Reason: Propan-l-ol gives positive iodoform test. [1]
Answer:
(c) (A) is true but (R) is false

Explanation:
Propan-l-ol and propan-2-ol are distinguished by iodoform test as propan-2-ol gives positive iodoform test

Question 18.
Assertion: Aquatic species are more comfortable in cold waters rather than in warm waters.
Reason: Different gases have different KH values at the same temperature. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
The amount of oxygen dissolved in the water decreases with rise in the temperature of the water. The amount of oxygen dissolved in water for per unit area is more for cold water as compared for the warm water. Hence aquatic species are more comfortable in cold waters rather than in warm waters. Thus, assertion and reason both are correct statements but reason is not correct explanation for assertion because the KH value of gases has nothing to do with these conditions needed for the survival of the aquatic species.

SECTION – B
(The following questions are very short answer type with internal choice in two questions and carry 2 marks each.)

Question 19.
Carry out the following conversions in not more than 2 steps: [2]
(A) Aniline to chlorobenzene
(B) 2-Bromopropane to 1- bromopropane
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 8

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 20.
A glucose solution which boils at 101 04°C at 1 atm. What will be relative lowering of vapour pressure of an aqueous solution of urea which is equimolal to given glucose solution? (Given: Kb for water is 0.52 K kg mol-1) [2]
Answer:
ΔTb = Kfm
ΔTb = 101.04-100= 1.04°C
or m = 1.04/0.52 = 2
Relative lowering of VP = x2
Relative lowering of VP = n2/n1 + n2
= 2/2 + 55.5 = 2/57.5 = 0.034 atm

Related Theory
The relative lowering in the vapour pressure is a coiligative property since-it depends on the number of the moles and can be calculated by using the different parameters such as molarity, mole fraction etc.

Question 21.
(A) Write the electronic configuration of iron ion in the following complex ion and predict its magnetic behaviour: [Fe(H2O)6]2+
(B) Write the IUPAC name of the coordination complex: [CoCl2(en)2]N03
OR
(A) Predict the geometry of [NiCN4]2+
(B) Calculate the spin only magnetic moment of [Cu(NH3)4]2+ ion. [2]
Answer:
(A) t2geg Paramagentic

The electronic configuration of the iron ion in the complex ion [Fe(H2O)6]2- can be written as t2g4eg2 and it will be paramagnetic in nature due to the presence of unpaired electrons in the ion.

(B) Dichloridobis(ethane-l,2-diamine) cobalt (III) nitrate
OR
(A) Square planar
(B) Cu2+ = 3d9, it means Cu2+ has 1 unpaired electron so \(\sqrt{13}\) = 1.73BM

The outer electronic configuration of Cu2+ is 3d9, that is it has 1 unpaired electron so magnetic moment can be calculated as
\(\sqrt{13}\) = 1.73 BM.

Related Theory:
Using the VSEPFi theory, the electron bond pairs and lone pairs on the centei atom the geometry of the molecule can be easily depicted whereas the shape of a molecule is determined by the location of the nuclei and its electrons.

Question 22.
For a reaction the rate law expression is represented as follows:
Rate = k [A][B]1/2
(A) Interpret whether the reaction is elementary or complex. Give reason to support your answer.
(B) Write the units of rate constant for this reaction if concentration of A and B is expressed in moles/L.
OR
The following results have been obtained during the kinetic studies of the reaction:
P + 2Q → R+ 2S
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 9
Determine the rate law expression for the reaction. [2]
Answer:
(A) Reaction is a complex reaction. Order of reaction is 1.5.

Molecularity cannot be 1.5. it has no meaning for this reaction. The reaction occurs in steps. so it is a complex reaction.
For the given rate = k [A][B]1/2
Since the order of the reaction is V, = 1.5 and the molecuLarity of the reaction cannot be equal to 1.5 thus such type of reactions are never elementary in nature they are complex in nature.

(B) Unit of k is mol1/2L1/2S-1

Related Theory:
Order and Molecularity are the two closely related terms in the context of the chemical reactions to know the change in the concentrations of the reactants and products with the change in time but they are different. The molecularity of the reaction is the number of molecules that come together to react in an elementary (single-step) reaction and is equal to the sum of stoichiometric coefficients of reactants in this elementary reaction whereas order of the reaction is the sum of the concentration powers given in the rate law.

OR

Let the rate law expression be Rate = k [P]x [Q]y from the table we know that,
Rate 1 = 3.0 × 10-4 = k (0.10) x (0.10)y
Rate 2 = 9.0 × 10-4 = k (0.30) x (0.30)y
Rate 3 = 3.0 × 10-4 = k (0.10) x (0.30)y
Rate 1/ Rate 3 = (1/3)y or 1 = (1/3)y
So y = 0
Rate 2/ Rate 3 = (3)x or 3 = (3)x
So x = 1
Rate = k[P]

Related Theory
The order of the reaction is given by the sum of the concentration powers given in the rate law; Let the hypothetical reaction be
r = k [A]x [B]y
The concentration powers of A = x
And the concentration power of B = y
Therefore the sum of the powers is = {(x) + (y)}

Question 23.
Name the reagents used in the following reactions:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 10
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 11

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 24.
The following haloalkanes are hydrolysed in presence of aq KOH.
(1) 1-Chlorobutane
(2) 2-Chloro-2-methylpropane
Which of the above is most likely to give (A) an inverted product (B) a racemic mixture: Justify your answer. [2]
Answer:
(A) Inverted product will be given by 1 chlorobutane as it undergoes SN2 reaction.
(B) Racemic mixture will be given by 2 chloro- 2-methylpropane as it undergoes SN1 reaction.

Question 25.
Zinc / silver oxide cell is used in hearing aids, instruments and other low power devices. The Following reaction occurs :
Zn(s) + Zn2+(aq) + 2e; E°Zn/Zn2+ = – 0.76
Ag2O + H2O + 2e → 2Ag + 2OH; E°Ag+/Ag = 0.344V
Calculate:
(A) Standard potential of the cell.
(B) Standard Gibbs Energy. [2]
Answer:
(A) In this reaction:
Zn is oxidized and Ag2O is reduced
cell = E°cathode – E°anode
= 0.344 – (-0.76) = 1.104V

(B) ΔG° = – nFE°cell
= – 2 × 96500 × 1.104
= -2.13 × 105 J mol-1

Related Theory
The value of Δg should always be negative for a reaction to be spontaneous. This is possible only when E°cell = E°cathode – E°anode is positive.

SECTION – C (15 Marks)
(The following questions are short answer type with internal choice in two questions and carry 3 marks each.)

Question 26.
Give reasons for the following:
(A) Transition elements act as catalysts
(B) It is difficult to obtain oxidation state greater than two for copper.
(C) CrO is basic but Cr2O3 is amphoteric.
OR
Observed and calculated values for the standard electrode potentials of elements from Ti to Zn in the first reactivity series are depicted in figure given below:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 12
Explain the following observations:
(A) The general trend towards less negative E° values across the series.
(B) The unique behaviour of copper.
(C) More negative E° values of Mn and Zn. [3]
Answer:
(A) Due to large surface area and ability to show variable oxidation states Transition elements act as cataLysts due to the Large surface area and ability to show variable oxidation states.

(B) Due to high value of third ionisation enthalpy. It is difficult to obtain oxidation state greater than two for copper because the value of third ionisation enthalpy is very high for a copper atom.

(C) Oxidation state of Cr in Cr2O3 is +3 and of CrO is +2. When oxidation number of a metal increases, ionic character decreases so CrO is basic while Cr2O3 is omphoteric. Cr0 is basic but Cr2O3 is amphoteric because oxidation state of Cr in Cr2O3 is +3 and of Cr in CrO is +2. When oxidation number of o metal increases, ionic character decreases so CrO is basic while Cr2O3 is omphoteric.

OR

(A) The general trend towards less negative EV values across the series is related to the general increase in the sum of the first and second ionisation entholpies.

(B) The high energy to transform Cu(s) to Cu2(aq) is not balanced by its hydration enthaLpy.

(C) The stability of the half-filled d sub-shell in Mn2+ and the completely filled d10 configuration in Zn2+ are reLated to their more negative E° values

Question 27.
Give any three tests to differentiate between aldehydes and ketones. [3]
Answer:
Three tests to differentiate between aldehydes and ketone are:
(i) Reaction with UAIH4: aldehydes gives primary alcohols and ketone gives secondary alcohols on reaction with LiAlH4.
(ii) Reaction with NaOH: aldehydes gives brown resinous mass and ketone shows no action with NaOH.
(iii) Schiff’s reagent test: aldehydes restores pink color andketones shows no action.

Question 28.
The graph below contains dashed lines representing the measured vapor pressure, and solid lines representing the ideal vapor pressure for a mixture of volatile liquids A and B.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 13
(A) What kind of deviation is depicted in the above graph?
(B) What would be the sign of enthalpy of mixing and volume for such a deviation ?
(C) How is Raoult’s Law treated as a special case of Henry’s law? [3]
Answer:
(A) Negative deviation occurs when the total vapour pressure is less than what it should be according to Raoult’s Law. Pa < Pa° XA and PB < P°B XB as the total vapour pressure (PA° XA + P°BXb) is less than what it should be with respect to Raoult’s Law.

(B) The enthalpy of mixing is negative that is, Amix H < 0 because more heat is released when new molecular interactions are formed.
The volume of mixing is negative that is, AmiX V < 0 as the volume decreases on the dissolution of components A and B.

(C) According to Raoult’s Law: PA = PA° XA
By Henry’s law P = KHX
In both the above laws the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. The only difference in the two expressions is the proportionality constant PA° (in Raoult’s law) and KH 0n Henry’s law). Therefore, Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to vapour pressure of the pure component

Caution:
In the case of solution of a gas in a liquid, the gaseous component is volatile component. Its solubility is governed by Henry law.

Question 29.
When treated with NaOH, an organic substance (A) having a distinctive odour transforms into two chemicals (B) and
(C). The oxidation of compound (B), which has the chemical formula C,H,0, results in compound (A). The sodium salt of an acid is compound (C). When (C) is heated with soda lime, an aromatic hydrocarbon results (D). Establish the structures of (A). (B). (C) and (D). [3]
Answer:
Since, compound (A) has characteristic odour, undergoes Cannizzaro reaction when treated with NaOH, and compound (B) having formula C7H8O gives back compound (A) on oxidation, compound (A) should be benzaldehyde. The sequence of reactions can be written as follows.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 14
Hence the structure or compounds (A), (B) (C) and (D are as follows
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 15

Question 30.
(A) Define threshold energy.
(B) What is the value of slope of graph log10k and 1/t indicates?
(C) What is the effect of positive catalyst in the reaction? [3]
Answer:
(A) Threshold energy is the minimal amount of energy that molecules colliding must have n order for a chemical reaction to take place.
(B) Slope = tan θ = Ea/2.303 R
(C) The function of a positive cutaiysi is co lowei down the “activation energy” The greater the decrease in the activation energy caused by the catalyst, higher will oe the reaction rate.

SECTION – D (8 Marks)
(The following questions are case-based questions. Each question has an internal choice and carries
4 (1 + 1 + 2) marks each. Read the passage carefully and answer the questions that follow.)

Question 31.
An efficient aerobic catalytic system for the transformation of alcohols into carbonyl compounds under mild conditions, copper- based catalyst has been discovered. This copper-based catalytic system utilizes oxygen or air as the ultimate, stoichiometric oxidant, producing water as the only by-product.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 16
A wide range of primary, secondary, allylic, and benzylic alcohols can be smoothly oxidized to the corresponding aldehydes or ketones in good to excellent yields. Air can be conveniently used instead of oxygen without affecting the efficiency of the process. However, the use of air requires slightly longer reaction times.
This process is not only economically viable and applicable to large-scale reactions, but it is also environmentally friendly.
(Reference: Ohkuma, T., Ooka, H., Ikariya, T., & Noyori, R. (1995). Preferential hydrogenation of aldehydes and ketones. Journal of the American Chemical Society, 117(41), 10417-10418.)
(A) Name any one reaction used to convert through the copper based catalyst. 1
(B) Write the reaction of oxidation of ethanol based on copper catalyst by ozonolysis.
(C) Benzyl alcohol on treatment with this copper-based catalyst gives a compound ‘A’ which on reaction with KOH gives compounds ‘B’ and ‘C’. Compound ‘B’ on oxidation with KMnO4. KOH gives compound ‘C’. Identity the compounds ‘A’, ‘B’ and ‘C’.
OR
An organic compound ‘X’ with molecular formula C3H80 on reaction with this copper based catalyst gives compound ‘Y’ which reduces Tollen’s reagent. ‘X’ on reaction with sodium metal gives ‘Z’. What is the product of reaction of ‘Z’ with l-chloro-2-methylpropane? [2]
Answer:
A propone

Explanation:
This convers or came out in the presence of the copper catalyst as well.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 17

(B) The oxidation of ethanoi based on the copper catalyst is carried out by the ozonolysis and the following product is obtained:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 18

(C) The sequence of the reactions is shown below:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 19
CH2 – C(CH3)2

Explanation:
The sequence of the reactions is shown below:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 20

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 32.
Amines are one of the most important classes of organic compounds which can be derived when we replace one or more hydrogen atoms of ammonia molecules with an alkyl group. An amine is generally a functional group with a nitrogen atom having a lone pair.
Compounds of nitrogen connected to a carbonyl group are called as amides; they have a structure R-CO-NR’R” and varies in properties with amines. Amines are organic compounds that contain nitrogen atoms with a lone pair. Basically, they are derived from ammonia (NH3) in which one or more hydrogen atom is replaced by an alkyl or aryl group, and so they are known as alkyl amines and aryl amines respectively. Anilines are the organic compounds in the class of group coming in organic chemistry which are also called as amino benzene or phenyl amine. These compounds are said to be toxic in nature and also known to be one of the classes of aromatic amines. These are used in a wide variety of industries and are known to possess all the characteristics of an aromatic compound. The aniline compounds are said to have the formula C6H5NH2 wherein the amino group is supposed to be attached to the Phenyl group.
(A) Why salts of benzene diazonium are water soluble. [1]
(B) Give one test to differentiate between cyanides and isocyanides. [1]
(C) Arrange the primary, secondary and tertiary amines in descending order on the basis of the following property:
(i) Boiling point
(ii) Basic strength
OR
Complete the following conversions.
(i) Nitromethane to dimethylamine
(ii) Methylamine to ethylamine. [2]
Answer:
(A) Because diazonium salts are polar by nature and water is a polar solvent, benzene diazonium salts are soluble in water. Consequently, the polar diazonium saLt dissolves in the water (like dissolve like)

(B) On reduction, cyanides give primary amines and isocyanides give secondary amines.

(C) (i) Primary amine > secondary amine > tertiary amine
(ii) secondary amine > primary amine > tertiary amine

OR
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 21

SECTION – E (15 Marks)
(The following questions are tong answer type and carry 5 marks each. Two questions have an internal choice.)

Question 33.
(A) How are conductance data used to determine cell constant?
(B) Differentiate between potential difference and the e.m.f. in three points.
(C) What flows in the internal circuit of a galvanic cell. [5]
Answer:
(A) Cell constant = \(\frac{\text { specific conductance }}{\text { observed conductance }}\)
(B)

Emf. Potential difference
1. It is difference between electrode potential of two electrodes when no current is flowing through circuit 1. It is difference of potential between electrode in a closed circuit.
2. It is the maximum voltage obtained from a cell. 2. It is less than maximum voltage obtained from a cell.
3. It is responsible for steady flow of current. 3. It is not responsible for steady flow of current.

(C) Ions flows is the internal circuit of a galvanic cell.

Question 34.
Three amino acids are given below:
Alanine CH3CH(COOH)(NH2)
Aspartic acid HOOC-CH2CH(COOH)(NH2)
Lysine H2N-(CH2)4-CH(COOH)(NH2).
(A) Make two tripeptides using these amino acids and mark the peptide linkage in both cases.
(B) Represent alanine in the zwitter ionic form.
(C) Which enzyme is used in conversion of starch to maltose?
OR
(A) State any four difference between globular and fibrous proteins.
(B) Identify the name used in the following conversions:
(i) Starch to glucose
(ii) Protein to amino acids
(iii) Urea to ammonia. [5]
Answer:
(A) Tripeptide 1 made up of three amino acids Alanine, Aspartic acid and lysine.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 22

(B) Alanine in the zwitter ionic form can be represented as follows:
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 23
(C) Diastase enzyme is used.
OR
(A)

Fibrous proteins  Globular proteins
1. These proteins possess thread Like structures.  1. These proteins possess folded spheroidal structures
2. The polypeptide chains are held together by stronger intermolecular hydrogen bonds.  2. The intermolecular hydrogen bonding is comparatively weaker.
3. These proteins are insoluble in water.  3. These proteins are soluble in water
4. They are stable to moderate changes in temperature and pH.  4. They are very sensitive to manges in temperature and pH.

(B) (i) Amylase
(ii) Trypsin
(iii) Urease

CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions

Question 35.
(A) State Kohlrausch law.
(B) Calculate the emf of the following cell at 298 K.
Al(s)|Al3+ (0.15M)||Cu2+(0.025M)|CU(S) (Given E°(Al3+/Al) = -1.66 V, E°(Cu2+/Cu) = 0.34V, log 0.15 = -0.8239, log 0.025 = -1.6020)
OR
(A) On the basis of E° values identify which amongst the following is the strongest oxidising agent:
Cl2(g) + 2 e > 2CI’ E° = +1.36 V,
MnO4 + 8H+ + 5e → Mn2+ + 4H2O, E° = +1.51 V
Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O, E° = +1.33 V
(B) The following figure, represents variation of (Λm) vs √c for an electrolyte. Here Am is the molar conductivity and c is the concentration of the electrolyte.
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 24
(i) Define molar conductivity.
(ii) Identify the nature of electrolyte on the basis of the above plot. Justify your answer.
(iii) Determine the value of Λ°m for the electrolyte.
(iv) Show how to calculate the value of Λ°m for the electrolyte using the above graph. [5]
Answer:
(A) A limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Kohlrausch law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
Λm = v+λ+ + vλ

v+ and v are the stoichiometric coefficients for the cation and anion in the electrolyte. λ°+ and λ° are the ionic conductance of individual ions (cations and anions).

(B) E°cell = E°cathode – E°anode
=0.34 – (-1.66) = 2.00 V
Ecell = E°cell – \(\frac{0.059}{n}\) log\(\frac{\mid A l^{3+} \hat{p}}{\mid C u^2+p^3}\)

Here n = 6
Ecell = 2 – \(\frac{0.059}{n}\) log\(\frac{10.15)^{{2}}}{[0.025]^3}\)
= 2 – 0.059/6 (2log 0.15 – 3 log 0.025)
= 2 – 0.059/6 (- 1.6478 + 4.8062)
= 2 – 0.0311
= 1.9689 V

In the given electrolytic reaction:
Al(s)|Al3+ (0.15 M) ||Cu2+ (0.025 M) | Cu(s)
Al is undergoing oxidation hence Aluminium is anode and copper electrode is cathode. Standard EMF of the given cell can be given as follows:
cell = E°cathode – E°anode = 0.34 – (-1.66) = 2.00 V

Net EMF of the given cell can be calculated as follows:
Ecell = E°cell – \(\frac{0.059}{n}\) log\(\frac{\mid A l^{3+} \hat{p}}{\mid C u^2+p^3}\)
Here n = 6
Ecell = 2 – \(-\frac{0.059}{6}\)log\(\frac{[0.15]^2}{[10.025]^3}\)
= 2 – 0.059/9 (- 1.6478 + 4.8062)
= 2 – 0.0311 = 1.9689V
Hence, EMF if the given cell is 1.9689 V.

OR

(A) MnO4
The given species can be arranged as follows according to their increasing oxidation potential-
Cr2O72- (+1.33 V) < Cl2(+ 1.36 V) < MnO (+ 1.51 V)
Hence, MnO4 is the strongest oxidising agent.

(B) (i) Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length.
(ii) Strong electrolyte, For strong electrolytes, Am increases slowly with dilution
(iii) Λm = Λm° – √ c
Therefore Λm° = 150 Scm2mol-1
CBSE Sample Papers for Class 12 Chemistry Set 6 with Solutions 25
Λm° = -Slope
= = – (149 – 147.8/ 0.010 – 0.022)
= 100 Scm2mol-1/(mol/L-1)1/2.