CBSE class 12 chemical sample paper 2023 and solutions (2023)

Class 12 chemical sample paper 2023

Chemical class 12 sample paper 2023: CBSE has published the chemical -testing paper 2023 of class 12 together with solution and marking scheme. Now the students have to attract themselves in order to start their revision for the chemical test, and the loosening of the chemistry sample paper in class 12 could be the best wayBeing to increase yourself and improve your trust for the actual examination. Look for the sample paper in class 12 chemistry 2023 for your chemical test according to the new CBSE test pattern? Then you are in the right place, as we have provided this hereCBSE class 12 chemical sample paperThis is issued by the CBSE on its official website according to the latest examination pattern.

Chemical sample paper class 122023

The Central Board of Secondary Education has also published the design and marking system for examining class 12 Chemistry Board 2023.In the exam, five sections are divided with 35 questions for 70 points.The complete examination pattern that was followed is as follows.

1st section A consists of 18 questions, each mark 1 [Multiple -Choice questions].

2. Section B consists of 7 questions, each of which bears 2 points [very short answer questions].

3rd section C consists of 5 questions, each of which bears 3 markings [short answer questions].

4. Section D consists of 2 questions, each of which has 4 markings [case -based questions].

5th section E consists of 2 questions that bear 5 points each [long answer questions].

Chemical class 12 sample paper 2023 PDF

The students who carry out their revision for the upcoming CBSE class 12 chemistry test must download the official sample paper PDF together with its solution from the following link. With this CBSE sample paper of the chemical class 12 it would be easier for the studentsTypes of questions to try the CBSE in the Board exam 2023 should be asked and analyze your experiment with the solution provided by CBSE.The link for the same is given below.

CBSE class 12 chemicals paper 2023- PDF download

CBSE class 12 chemical sample paper solution

CBSE class 12 chemical sample paper and solution 2023

Chemistry is one of the main subjects of class 12. Boards.It consists of a total of 100 points that continue to include 70 points for theory and 30 points for the practical test.In the following you will find the sample question from the official paper. Extra do you thoroughly to achieve good grades in chemistry, since chemistry is seen as one of the evaluation subjects in the board's exams.

Section A

The following questions are multiple-choice questions with a correct answer.

Question 1. The main product of the acid-catalyzed dehydration of 1-methylcyclohexanol is:

a.1-Methylcyclohexan

b.1-Methylcyclohexene

c.1-Cyclohexylmethanol

d.1-Methylcyclohexan

Solution:B.1-methylcycloHexen. After the rule of Saytzeff, d.H. The high -substituted alken is the main product. A dehydrate reaction takes place, an alken is formed due to the removal of a water molecule.

Question 2. Which of the following connections reacted to the SN1 reaction?

a.CH2 = CHCH2BR

b.C6H5CH2BR

c.C6H5CH (C6H5) BR

D.C6H5CH (CH3) BR

Solution:c.C6H5CH (C6H5) BR

C6H5CH (C6H5)+ carbocation is more stable.

Question 3. KMNO4 is colored because:

A.D-D transitions

B. Ligand transfer transfer to metal

C.unge paarte electrons in the D orbital of MN

D. Charging transfer from Metall to Ligand

Solution:B. Ligand of Ligand to Metall The MN -Atom in KMNO4 has an oxidation state of +7 with electron configuration [AR] 3D 04S0, since there are no unpaired electrons, D -D transitions are not possible. The molecule should therefore be colorless.His intensive purple through L → M (Ligand to Metal) charge transfer 2P (L) from O to 3D (m) of MN.

Question 5.The molar conductivity of CH3COOH with infinite dilution is 390 SCM²/mol. Using the diagram and the given information, the molar conductivity of CH3COOK is:

a.100 scm²/mol

b.115 scm²/mol

c.150 scm²/ mol

d.125 scm²/mol

Solution:b.115 scm² /mol λOCH3COOK = λO CH3COOH +λOKCL - λOHCL = 390 +150-425 = 115 scm² /mol

Question 6. For the reaction A +2b - AB2, the order W.R.T.Reaktant A is 2 and W.R.T.Reaktant B. How is the reaction speed changed if the concentration of A doubles and B halved?

A. is four times

B. takes four times

C. climbed twice

D. not change

Solution:A. is about 4 times the rate = [a] ², if [a] doubled, rate = [2a] ² = 4 [a] ² = 4 rate

Question 7. Arrange the following in the increasing order of your boiling points: A: Butanamin, B: N, N-Dimethylethanamin, C: N-Etthylethanaminamine

a.C <b <a

b.A <b <c

c.A <c <b

d.B <c <a

Solution:D.B <c <a.im primary amin, the intermolecular association is maximum due to the H bond, while it is minimal in the tertiary.

Question 8. The Kfste of [Cocl6] 3- IS 18000 cm-1 The CFSE for [Cocl4]- will be:

a.18000 cm-1

b.8000 cm-1

c.2000 cm-1

d.16000 cm-1

Solution:b.8000 cm-1 ∆t = (4/9) x 18000cm-1 = 8000 cm-1

Question 9. What would be the main product of the following reaction?

C6H5 -CH2 -OC6H5 + HBR = A + B

a.A = C6H5CH2OH, B = C6H6

b.A = C6H5CH2OH, B = C6H5BR

c.A = C6H5CH3, B = C6H5BR

d.A = C6H5CH2BR, B = C6H5OH

Solution:d.A. = C6H5CH2BR, B = C6H5OH, C6H5CH2OC6H5 H+ C6H5CH2OC6H5

Question 10. Which of the following statements is not correct for Amine?

A.The alkylamine are more fundamental than ammonia solution.

B.Der PKB value of ethylamine is lower than benzylamine.

C.CH3NH2 in response with laughing disease releases NO2 gas.

D.hinsberg's reagentzs reacts to sulfonamids with secondary amines.

Solution:C.CH3NH2 In response with laughing disease, NO2 Gas false statement releases. The development of nitrogen gas takes place.

Question 11. Which of the following tests/ reactions are given by both aldehydes and ketones?

a.Fehings Test

b.Great’s test

c.2,4 DNP -Test

D.Cannizaro -Anaktion

Solution:C.4 DNP test Fehlings, Tollens and Cannizzaro's reactions are only shown by alcohol.

Question12. The equation of Arrhenius can be shown graphically as follows:

The (i) intercept and (ii) slope of the graphic are:

a.(I) ln a (ii) Luft / r

B. (i) and (ii) air

c. (I) LN or (II) - air / r

d. (i) and (ii) -ea

Solution:12. C. (I) LN A (Ii) - air / r

Frage13.13. The number of ions that are formed when a molecule of FESO4 is dissolved.(NH4) 2S4.6H2O in water is:

a.3

b.4

c.5

d.6

Solution:13. C.5 1FE2 +, 2 SO4 2 and 2 NH4 + ions

Question14.The oxidation from Toluol to benzaldehyde by chromyl chloride is mentioned

A.etard reaction

b.Riemer-Tiemann Reaction

C.Stephen's reaction

D.Cannizzaro's reaction

Solution:A. Etard's reaction

Frage15.Below are two statements called "Association (A) and Reason (R).

Assertion (a):An ether is more fleeting than alcohol with a comparable molecular flash mass.

Basic (R):Ether are polar nature.

Select the most suitable answer from the options below:

A.Sowohl A and R are true and r is the correct explanation of A

B.Sowohl A and R are true, but R is not the right explanation of A.

C.A is true, but R is wrong.

D.A is wrong, but r is true.

Solution:B both a and R are true, but R is not the correct explanation of A and R and R are two different statements about ether.The right reason for this is that hydrogen binding does not exist between ether molecules

Question16. The following are given two statements called "Association (A) and Reason (R).

Assertion (a):It is found that proteins have two different types of secondary structures, namely the alpha-helix and beta-pace sheet structure.

Basic (R):The secondary structure of proteins is stabilized by hydrogen bond.

Select the most suitable answer from the options below:

A.Sowohl A and R are true and r is the correct explanation of A

B.Sowohl A and R are true, but R is not the right explanation of A.

C.A is true, but R is wrong.

D.A is wrong, but r is true.

Solution:B both a and r are true, but r is not the correct explanation of A.

Question17. The following are given two statements called "Association (A) and Reason (R).

Claim:Magnetic moment values of actinids are less than the theoretically predicted values.

Reason:Actinide elements are very paramagnetic.

Select the most suitable answer from the options below:

A.Sowohl A and R are true and r is the correct explanation of A

B.Sowohl A and R are true, but R is not the right explanation of A.

C.A is true, but R is wrong.

D.A is wrong, but r is true.

Solution:B both a and r are true, but R is not the correct explanation of A. The magnetic moment is lower, since the 5f electrons of actinids are less effective, which leads to deleting orbital contributions.Due to the presence of unpaid, they are strongly paramagnetic electrons

Question18. The following are given two statements called "Association (A) and Reason (R).

Assertion (a):Tertiary amines are more fundamental than the corresponding secondary and primary amine in a gaseous state.

Basic (R):Tertiary amines have three alkyl groups that cause the +I effect.

Select the most suitable answer from the options below:

A.Sowohl A and R are true and r is the correct explanation of A

B.Sowohl A and R are true, but R is not the right explanation of A.

C.A is true, but R is wrong.

D.A is wrong, but r is true.

Solution:Both a and r are true and r is the correct explanation of A.

Section B

This section contains 7 questions with internal selection in two questions. The following questions are very short response types and each bear 2 markings.

Question 19. A first order reaction takes 69.3 min for 50%. How is the time required for 80% of the reaction to be completed? (Given: Log 5 = 0.6990, log 8 = 0,9030, log 2 = 0.3010)

Solution:19.HALF-LIFE T1 /2 = 0.693 /K

K = 0.693/69.3 = 1/100 = 0.01 min - - 1

For the first order reaction

𝑘 = 2,303/ 𝑡 𝑙𝑜𝑔 [𝑅𝑜]/ [𝑅]

𝑡 = 2.303 /0.01 𝑙𝑜𝑔 100/20

𝑡 = 230,3 log 5 (log 5 = 0,6990)

t = 160,9 min

Question 20.Conent for the following:

A.Ses gives 5 oh groups in glucose

B. glucose is a reducing sugar

Solution:A.The acetylation of glucose with acetic acid anhydride results in glucosententa acetate that confirms the presence of five OH groups.

OR

Question 20. What happens when D - glucose is treated with the following reagents

a.Bromwasser

B.Murter

Question 21. Enter a reason for the following:

A. During the electrophilic substitution reaction of haloarens, the para-substituted derivative is the main product.

B.Das formed during SN 1 reaction is a racemic mix.

Solution:A.in the Ortho position is a higher steric obstacle, so Para isomer is usually predominant and is preserved in the main amount.

B. SN 1 -mechanism is hybridized with the formed medium carbocation of SP2 and planar in nature. This enables the attack of nucleophils from both sides of the level, which leads to a racemic mix.

OR

Question 21. Enter a reason for the following:

A.nine the appropriate alcohol and the suitable reagent from which 2-chlorine 2-methylpropan can be made.

B. Welches from chlormethane and fluorethane have a higher dipol moment and why?

Solution:A.tter butyl alcohol or 2-methyl-propane-2-ol using Lucas-Ragenz, a mixture of conc.hcl and ZNCL2 The reaction follows SN 1 path.

B.Chlorethan has a higher dipol moment. Beyond the smaller fluorus size, the dipole moment of flouromethane is comparatively lower.

Question22.The Formula CO (NH3) 5co3Cl could represent a carbonate or chloride. Write the structures and names of possible isomers.

Solution:[CO (NH3) 5Co3] CL and [CO (NH3) 5Cl] CO3

Pentaamminecarbonatocobalt (iii) Chloride

Pentaamminechlorocobalt (III) Carbonate

Question 23. Corrosion is an electrochemical phenomenon. The oxygen in moist air reacts as follows:

O2 (G) + 2h2o (L) + 4E - → 4oh - (AQ).

Write the possible reactions for the corrosion of zinc, which occur at anode, cathode and overall reaction to form a white layer of zinc hydroxide.

Solution:Anode: Zn (s) → Zn2 + (aq) + 2 e--

Kathode: O2 (G) + 2h2o (L) + 4E - → 4oh- (AQ).

In total: 2 ZN (s) + O2 (g) + 2H2O (L) → 2 ZN2 + (AQ) + 4OH- (AQ)

2 Zn (s) + O2 (g) + 2H2O (l) → 2 Zn (OH) 2 (PPT)

Question24. Explain how and why the reaction speed for a certain reaction is influenced if

A. A catalyst is added

B. the temperature in which the reaction took place is reduced

Solution:The reaction speed increases. The catalytic converter reduces the activation energy of the reaction, which is why the reaction becomes faster. (1/2+1/2) B.which leads to a reduction in the reaction speed.

Question 25. Write the reaction and IUPAC names of the product that was formed when 2-methylpropanal (isobutyraldhyde) is treated with ethyl magnesium bromide, followed by hydrolysis.

Solution:

Section C

This section contains 5 questions with internal selection in two questions. The following questions are short response types and each bear 3 markings.

Question26. Write the equations for the following reaction:

A.Salicylic acid is treated with acetic acid anhydride in the presence of Conc.h2so4

B.tert butyl chloride is treated with sodium method.

C.Phenol is treated in the presence of Naoh with chloroform

Solution:(i) Aspirin is formed

(iii) o-hydroxybetzaldehyde is formed

Question27. Explain the following in relation to the paramagnetic complex [mn) 6] 3-

A.art of hybridization

B. magnetic moment

C.art of the complex - inner, outer orbital complex

Solution:[Von (be) i] p = [r] 3D54S2 Mn3+ = [R] 3D4 mn (Grend STAT)

Type of hybridization - D2SP3

Magnetic moment - √n (N+2) = √ (2 (2+2)) = 2.87 BM (n = No. No. of unpaired electrons)

Type of complex - inner orbital

Question 28. Answer the following questions:

A.State Henry's law and explain why the tanks used by divers with a helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen) explain.

B. long, argon exercises a partial pressure of 6 bar. You calculate the solubility of Argongas in water. (In view of the Henry's law, constant for Argon, which was dissolved in water, KH = 40 kbar)

Solution:A.Henry's law: The partial pressure of the gas in the steam phase (P) is proportional to the mola content of the gas (x) in the solution. The pressure under water is high so that the solubility of gases in the blood increases.The surface comes, the pressure decreases, as well as the solubility, which causes nitrogen bubbles in the blood to avoid this situation and maintain the same sub -pressure of the nitrogen under water, the dilution is diluted.

B.P = KH x molan break from Argon in water x = p/ k = 6/40 x 103 = 1.5 x 10-4

Question 29. Enter reasons for 3 of the following observations:

A.Anilin is acetylated before the nitration reaction.

B. PKB from Anilin is lower than the M-Nitroanilin.

C. The primary amin when treating benzene sulfonyl chloride forms a product that is soluble in NAOH.Secondary Amin, however, offers a product that is insoluble in Naoh.

D.Anilin does not react with methyl chloride in the presence of an anhydrous ALCL3 catalyst.

Solution:(Any 3)

A.Anilin is acetylated before the nitration reaction to avoid the formation of Terry oxidation products and protect the Amino group so that P -Nitro derivative can be preserved as the main product.

B. PKB from Anilin is lower than the M-Nitro-Anilin.The basic strength of anilin is more than M -Nitroanilin. The PKB value is the reverse to the basic strength. The presence of an electron withdrawal group reduces the basic strength.

C. The existence of the presence of acid hydrogen in the N-alkylbene scenesulphonamide, which is formed by the treatment of primary amines. (1)

D.anilin does not react with methyl chloride in the presence of an ALCL3 catalyst, since Anilin is a base and alcl3 Lewis -acid is that leads to the formation of salt.

Question 30identify the main product that is formed when 2-cycloHexylchlorethandergo is a dehydrohalogenation reaction.

B. WHARUM Haloalkanes react to nucleophilic substitution reactions as haloarene and vinyl -containing halogenids?

solution: A.The main product that was formed when 2-cycloHexyl chlorethanethan experiences a dehydrohalogenation reaction, 1 cyclohexyethen.

B.haloalkanes are more reactive than haloarene and vinylic halogenids, since there is a partial double binding character-C-X bond in haloarens and vinylhalogenidides.

OR

A.nine the possible alken, the 1-chlor 1 methylcycloHexan in your reaction with HCL, write the associated reactions.

B.Allchloride is easier than N-propylchloride hydrolyzed.

Solution:B.Allchloride shows a high reactivity, since the carbocation formed in the first step is stabilized by resonance, while with N-propylchloride there is no such stabilization of the carbocation.

Section d

The following questions are case -based questions. Jede Question has an internal selection and bears 4 (1+1+2) markings. Read the passage carefully and answer the following questions.

Question 31. Strengthening the foundation:

Chargaff formulates his "rules".Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, the DNA was first identified by the Swiss chemist Friedrich Miescher at the end of the 1860s, then in the decades after the discoveryFrom Mieschermithe.

Chargaff, an Austrian biochemist, as his first step in this DNA research, set out whether there were differences in the DNA between different types. After the development of a new paper chromatography method for separating and identifying small amounts of organic material,

Chargaff reached two main consequences: (i) The nucleotide composition of the DNA varies between the species. (II) Almost all DNA, regardless of which organism or tissue type it comes, reserves certain properties, even if its composition varies.a) The amount of thymine (t) similar, and the amount of guanine (g) approaches the amount of cytosine (c).

In other words, the total amount of purins (A + G) and the total amount of pyrimidines (C + T) are usually almost the same. This conclusion is known today as the "Chargaff's rule".These have either a striking DNA or RNA as a genetic material.

Answer the following questions:

A. A DNA segment has 100 adenin and 150 cytosin bases. How many nucleotides are there in this DNA segment?

B.an two places was found a hair and blood sample. Scientists claim that rehearsals belong to the same way. How did the scientists come to this conclusion?

C. The sample of a virus has been tested and it was found that it contains 20% adenin, 20% thymine, 20% guanin and the rest of the cytosine.-Single -Helix (C) -NA? What do you conclude from this data?

Solution:A.A = 100 SO T = 100 C = 150 SO G = 150 total nucleotide = 100+100+150+150 = 500

B.sie examined the nucleotide composition of DNA.Tes was the same and came to the conclusion that the rehearsals belong to the same way.

C.A = T = 20%, but G is not the same as the double helix is excluded. The base pairs are ATGC and not Augc, so it is not RNA, the virus is a single helix -dna virus.

OR

Question 31.How can the rule of Chargaff be used to conclude that the genetic material of an organism is a double helix or single helix?

Solution:According to the rule of Chargaff, all double helix DNA has the same amount A and T and C.

So she is dared.Henna examines the melting point of different salt solutions. You produce a salt solution with 10 ml of water with a known mass of salt. She puts the salt solution in a freezer and lets it freeze.Frozen salt solution melts. They repeat every experiment.

Assuming the melting point of pure water as a 0OC, you answer the following questions:

A. A temperature in the second result set does not fit the pattern. What is the temperature of your answer.

B.Why did Henna collect two results?

C.Anpell from NACL, if Henna had used glucose, what would the melting point of the solution be with 0.6 g glucose in it?

Solution:The melting point of the ice cream is the freezing point of the water. In this case we can use the depression in the property of the freezer point

A.3.Readers for 0.5 g It has to take an increase in the depression of the freezing point and thus that freezing also decreases, so that the melting point decreases when the amount of salt is increased, but the trend is not followed in this case.

B.Two sentences help to avoid errors in data acquisition and to give more objective data.

c.Δtf (Glucose) = 1 x kf x 0,6 x 1000 /180 x 10

Δtf (NaCl) = 2 x kf x 0,6 x 1000 /58,5 x 10

3,8 = 2 x kf x 0,6 x 1000 /58,5 x 10

Equation 1 by 2 parts

Δtf (Glucose) / 3,8 = 58,5 / 2x 180

ΔTF (glucose = 0.62 freezing point or melting point = - 0.62 ° C

OR

So she is dared.What is the predicted melting point if 1.2 g salt is added to 10 ml of water? Justify your answer.

Solution:Depression at freezing is directly proportional to molality (mass of the dissolved substance if the amount of solvent remains the same.)

0.3 g depression is 1.9 ° C.

0.6 g depression is 3.8 ° C.

1.2 g depression is 3.8 x2 = 7.6 ° C

Section e

The following questions are long response types and each bear 5 points. Two questions have an internal choice.

So he returned.A.Why does the cell voltage of a mercury cell remain constant during your 10 lifespan?

B. Write the reaction that occurs on the anode and the cathode and the products of the electrolysis of AQ KCL.

C. What is the pH value of the HCL solution if the hydrogen gas electrode shows a potential of -0.59 V at standard temperature and printing?

Solution:The cell potential remains constant during its lifespan, since the overall reaction does not include an ION in a solution, the concentration of which can change over the course of its lifespan.

B.KCl (aq) à K + (aq) + cl- (aq) h2o (aq) h2o (l) + e-½CL2 (Aq) + E- (1/2) Net-Jare: kcl (aq) + h2o (al) + oh- (aq) + ).

C., the potential of the hydrogen gas electrode = –0.59 V

Electrodenreaktion: h + + E– → 0,5 h2

Apply equation, E (H+ /H2) = EO (H+ /H2) -0.059 Log [H2] 1/2 N [H+] (1) EO (H+ /H2) = 0 V E (H+ /H2) = -0.59V N = 1 [H2] = 1 bar –0.59 = 0 - 0.059 ( - log [H+]) (1/2) -0.59 = –0.059ph

∴ pH = 10

OR

A.The molar conductivity of the substance "A" is 5.9 × 103 s/m and "B" 1 x 10-16 S/M. What is the most likely copper metal and why?

B. How is the amount of electricity in coulombs necessary to produce 4.8 g mg made of melted MGCL2? How much approx.

Mg = 24 u, nuclear mass of approx. = 40 u).

C. What is the standard change of free energy for the following reaction at room temperature? Is the reaction spontaneous?

Sn (s) + 2cu2 + (AQ) à SN2 + (AQ) + 2c + (s)

Solution:a. "A" is copper, metals are ladder and therefore have a high value of the conductivity.

B.MG2 + + 2E mg 1 mol magnesium ions wins two electron moles or 2F to form 1 mol mg

24 g mg requires 2 f electricity

4.8 g mg requires 2 x4.8/24 = 0.4 f = 0.4 x96500 = 38600C

Ca2 + + 2e- → Ca

2 F electricity is required to generate 1 mol = 40 g approx.

0.4 f electricity generates 8 g ca (1)

c.F = 96500c, n = 2,

Sn2 + (aq) + 2e - → sn (s) –0,14v

Cu2+ (aq)+ e- → cu+ (aq) 0.15 v

E ° ZELLE = E ° Cathode - And ° Anode

= 0,15-(-0,14) = 0,29 V (1)

δg ° = -nfecell

= -2 x96500x 0,29 = 55970 j/mol

Faraji 4.A hydrocarbon (a) with molecular formula C5H10 on ozonolysis results in two products (b) and (c).b) has a silver mirror in the reaction with great reagent.

A. identify (a), (b) and (c).

B. Write the reaction of B with Tollens Reagentz

C. Write the equation for the iodoform test for C.

D. Write down the equation for the Aldol condensation reaction of B and C.C.

OR

An organic compound (a) with molecular formula C2CL3O2H is obtained when (b) reacts with red P and CL2.

A. identify A and B

B. Write the reaction for the formation of A from B. What's this reaction?

C. Enter a method with which organic compound B can be made from its corresponding acid chloride.

D. What will the acidic connection (a) or (b) be? Why that?

E. Write down the reaction to establish methane from the connection (B).

Question 35. Answer the following:

A.Why are all copper halogens known except for this copper iodide?

B. Is the EO value (V3+/V2+) comparatively low for vanadium?

C.Whl shouldn't HCL be used for potassium perpetrators?

D. Clarify the observation at the end of each period that the nuclear radius of D block elements increases slightly.

E. How does the pH value affect dichromat solution?

Solution:.a.cu2+ oxidized iodidion on iodine.

For example, low value for V is related to the stability of V2+ (half -filled T2G).

C.Permanganate titles in the presence of hydrochloric acid are unsatisfactory because hydrochloric acid is oxidized to chlorine.

D. The DABILATE is full of ten electrons and protects the electrons available in the higher S orbital to a larger extent, which leads to an increase in size.

E.The chromate and dichromats depend on the pH value of the solution in an aqueous solution. Eerhow of the pH value (in a basic solution) of dichromations A color change of orange is observed when dichromations change to chromations