CBSE Sample Papers for Class 12 Chemistry Paper 1

CBSE Sample Papers for Class 12 Chemistry Paper 1 are part of CBSE Sample Papers for Class 12 Chemistry. Here we have given CBSE Sample Papers for Class 12 Chemistry Paper 1.

CBSE Sample Papers for Class 12 Chemistry Paper 1

Board	CBSE
Class	XII
Subject	Chemistry
Sample Paper Set	Paper 1
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme, as prescribed by the CBSE, is given here. Paper 1 of Solved CBSE Sample Paper for Class 12 Chemistry is given below with free PDF download solutions.

Time Allowed : 3 Hours
Max. Marks: 70

General Instructions

• All questions are compulsory.
• Question number 1 to 5 are very short answer questions and carry 1 mark each.
• Question number 6 to 10 are short answer questions and carry 2 marks each.
• Question number 11 to 22 are also short answer questions and carry 3 marks each.
• Question number 23 is a value based question and carry 4 marks.
• Question number 24 to 26 are long answer questions and carry 5 marks each.
• Use log table, if necessary. Use of calculators is not allowed.

Questions

Question 1.
Write the reason for appearance of charge on particles of dispersed phase.

Question 2.
Alkoxide ions are stronger base than hydroxyl ion explain.

Question 3.
Write IUPAC name for

Question 4.
Name the crystal defect due to which density of solid increases.

Question 5.
Discuss the role of coordination compounds by giving an example for metallurgy of metals.

Question 6.
State Raoult’s law for the solution containing volatile components.

Question 7.
Complete the following chemical equations:

1. PH₃ + HgCl₂ →
2. XeF₂ + H₂O →

Question 8.
What is actinoid contraction, how does it differs from lanthanoid contraction?

Question 9.
(i) Find limiting molar conductivity for aluminium sulphate if ionic conductivity for Al³⁺ (λ_0Al³⁺) = 80S cm² mol^-1 and for sulphate ion 40S cm^-2 mol^-1.
(ii) Write a relation between molar conductivity, limiting molar conductivity and degree of dissociation.

OR
Define primary batteries and write cell reaction for dry cell.

Question 10.

1. Flalogen have maximum negative electron gain enthalpy in the respective periods of periodic table. Why?
2. Fluorine exhibits only-1 oxidation state unlike other halogens. Explain.

Question 11.
Write the basic principle involved is the following processes:

1. Vapour phase refining
2. Zone refining
3. Liquation

Question 12.
Explain the following observations:

1. Soap solution at higher concentration appears hazzy.
2. Purification of water caused by alum.
3. Platinum as colloidal sol acts as better catalyst.

Question 13.
An element having molar mass 60 g/mol forms a cubic emit cell of edge length 300 pm and density 7.41 g/cm³. What is the rank of this unit cell?

Question 14.
The vapour pressure of pure liquid A and B are 450 and 750 mm Hg respectively at 350 K. Find out the composition of the solution if the total pressure is 600 mm Hg. Also find the composition of these components in the vapour phase.

Question 15.
Represent the cell corresponding the following reaction:
Mg(s) + 2Ag⁺ (0.0001 M) → Mg²⁺ (0.130 M) + 2 Ag (s)
Calculate the emf of cell if E_cell° = 3.17 V.

Question 16.
How will you synthesize the following?
(i) 1-phenyl ethanol from a suitable alkene.
(ii) Cyclohexyl methanol using an alkyl halide by S_N2 reaction.
(iii) Pentan-l-ol using a suitable alkyl halide.

Question 17.
(i) Draw the geometrical isomers of the complex [Ni(NH₃)₂(CN)₂]
(ii) Draw crystal field splitting of d-orbital for tetrahedral complex.
(iii) Write the hybridisation type and magnetic behaviour of the complex [Cu(NH₃)₄]²⁺

Question 18.
Explain the following observations:

1. Size of Zr and Hf are comparable.
2. CuCl a white compound turns blue in aqueous solution.
3. The transition metals form interstitial compounds.

Question 19.
Give reasons for the following:

1. Ozone is a powerful oxidising agent
2. SO₂ acts as bleaching agent
3. ICl is better halogenating agent than I₂.

Question 20.
How will you convert the following:
(i) Aniline to benzene
(ii) p-Xylene to benzene
(iii) Ethanoic acid to methanamine

OR
(a) Write short notes on
(i) Cannizzaro’s reaction
(ii) Kolbe’s reaction
(b) Write the mechanism of dehydration of ethanol.

Question 21.
(a) Write the reaction involved when D-glucose is treated with
(i) HI
(ii) Ethanoic anhydride
(b) Define a hormone.

Question 22.
Answer the following questions:

1. Which class of drug is used for depression?
2. What are biodegradable detergents?
3. Aspirin can also be used to prevent heart attack. Explain.

Question 23.
Polymer has wide range of application. Raman a friend of Madan, has a fatal accident and there is need of surgery. Madan suggested doctor to use self dissolving threads instead of normal thread. On the basis of above passage answer following questions.

1. Name the polymer, which is used as suture
2. What is biodegradable polymer? Give an example.
3. Write values associated with advice of Madan.

Question 24.
(a) The time required for 25% completion of a first order reaction at 310 K is double to that of 10% completion of a reaction at 300 K. Find energy of activation for this reaction.
(b)

1. Define specific rate constant
2. Order of reaction

OR
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:
CH₃COOCH₃ + H₂O → CH₃COOH + CH₃OH

t/s	0	30	60
[CH3COOCH3] mol L-1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction as the concentration of water remains constant.
(ii) Calculate average rate of reaction between time interval of 0 to 60 second.

Question 25.

1. Define the following:
   • Racemic mixture
   • Meso compound
2. • Alkyl halides though polar are immiscible with water.
   • Grignard reagent should be prepared under anhydrous condition.
   • The dipole moment of chlorobenzene is greater than that of cyclohexyl chloride.

OR
(i) Complete the following reaction:

(ii) Distinguish between following pairs of compounds

1. Chlorobenzene and benzyl chloride
2. Ethanol and methanol

Question 26.
An organic compound A(C₉H₁₀O) forms orange red precipitate ‘B’ on reaction with 2,4-DNP. ‘A’ also gives yellow precipitate ‘C and colourless compound ‘D’ on treatment with NaOI. ‘A’ does not reduces Tollens’ reagent, Fehling’s solution and Br2 water. On drastic oxidation with chromic acid it gives carboxylic acid ‘E’ of molecular formula C₇H₆O₂. Deduce the structure of the organic compound A to E.

OR
An organic compound ‘A’ on treatment with ethyl alcohol gives carboxylic acid ‘B’ and compound ‘C’. Hydrolysis of ‘C’ under acidic condition gives ‘B’ and ‘D’. Oxidation of ‘D’ with KMn04 also gives ‘B’. ‘B’ on heating with Ca(OH)₂ gives ‘E’ with molecular formula (C₃H₆O). ‘E’ does not give Tollens’ test or reduce Fehling solution but forms 2, 4-DNP derivative. Identify A to E.

Answers

Answer 1.
The ions of opposite charge got adsorbed on surface of precipitate during peptization leads to appearance of charge on dispersed phase.

Answer 2.
Due to +I-effect of alkyl group present in alkoxide increases negative charge on O and thus makes it stronger base than hydroxyl ion.

Answer 3.
2-Bromopropanoyl chloride.

Answer 4.
Interstitial defect

Answer 5.
Coordination compounds used during concentration of ore, extraction of metal and purification of extracted metal in one metal or other e.g.
Concentration of Al₂O₃ by leaching with NaOH
Al₂O₃ (S) + NaOH (aq) → Na[Al(OH)₄] (aq)
Soluble complex impurities remains as such.

Answer 6.
Raoult’s law: It states that in solution of volatile liquid, vapour pressure of a component is directly proportional to its mole fraction in the solution.
P_A ∝ x_A or P_A = P⁰_Ax_A and P_B ∝ x_B or P_B = P⁰_Bx_B
P_Total = P⁰_Ax_A + P⁰_Bx_B (where x_A and x_B are mole fraction of A and B respectively)

Answer 7.

1. 2PH₃ + 3 HgCl₂ → Hg₃P₂ + 6HCl
2. 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

Answer 8.
The contraction in size from Ac (actinium) to Lawrencium (Lr) due to filling of electrons in 5f orbital is known as actinoid contraction. Element to element actinoid contraction is more than lanthanoid contraction due to poorer shielding of 5f orbital as compared to 4f orbital.

Answer 9.

where Δ^C_m = molar conductivity at concentration ‘C’
Δ⁰_m = Limiting molar conductivity.

OR
Primary batteries are non rechargeable batteries, in such cell the chemicals present gives energy by chemical reaction and such reactions can not be reversed
Reactions in dry cell are:

Answer 10.

1. Reason for maximum negative electron gain enthalpy for halogen
   • smallest size among a period
   • only le^– less for acquiring noble gas configuration
   • minimum bond dissociation enthapy in respective period.
2. Fluorine is the most electronegative element in the periodic table thus, no element can oxidise it to show positive oxidation states. Hence -1 is the only possible oxidation state for it.

Answer 11.

1. Vapour phase refining: Metal should form volatile complex which should be decomposed on further heating.
2. Zone refining: The impurity are more soluble in molten metal as compared to solid state of metal.
3. Liquation: Metal should have lower melting point as compare to the impurities.

Answer 12.

1. At lower concentration soap solution behaves as true electrolyte but at higher concentration colloidal solution forms and hence solution turns hazzy.
2. The silt or impurities present in water get coagulated by electrolytes e.g. alum, K₂SO₄. Al₂(SO₄)₃. (H₂O) large number of ions cause coagulation and purifies water.
3. In colloidal state its surface area becomes much larger and hence it acts as better catalyst.

Answer 13.
It is given that
Molar mass (M) = 60 g/mol
300 pm = 3 × 10^-8 cm
Density (d) = 7.41 g/cm³
Rank of solid (z) = ?

Answer 14.
From question
P⁰_A = 450 mm Hg
P⁰_B = 750 mm Hg
P = 600 mm Hg
Let mole fraction of A is x_A then x_B = 1 – x_A
From Raoult’s law

Answer 15.
Cell representation

Answer 16.

Answer 17.

Answer 18.

1. Due to lanthanoid contraction size of Zr and Hf is comparable.
2. Cu⁺ get disproportionate in aqueous solution hence the solution turns blue because of Cu²⁺ ion.
   
3. Transition metals have suitable size of void to accommodate smaller atoms like H, B, C, N etc and there is strong metal bond as a result these can not skip from voids.

Answer 19.

1. O₃ get dissociated to give nascent oxygen ‘O’ hence acts as powerful oxidising agent.
   O₃ → O₂ + ‘O’
2. In presence of moisture, SO₂ produce nascent hydrogen and thus acts as reducing agent as well as bleaching agent.
3. ICl an interhalogen compound has its bond dissociation energy less than I₂ so it acts as a better halogenating agent.

Answer 20.

OR
(a) (i) Cannizzaro’s Reaction: The aldehyde having no alpha hydrogen get oxidised and reduced in presence of concentrated alkali. This disproportionation reaction is known as cannizaro’s reaction
2HCHO \(\underrightarrow { NaOH(Conc.) }\) HCOONa + CH₃OH
(ii) Phenol on reaction with C02 and NaOH followed by acidic hydrolysis given salicylic acid.

Answer 21.

(b) Hormone: A biochemical messanger which gives message to a body organ to carry out certain function.

Answer 22.

1. Tranquilizer
2. The detergent which can be decomposed in smaller particles by microorganism are known as biodegradable detergent. The detergents having long unbranched alkyl chain are biodegradable detergent.
3. Aspirin dilutes blood and relax muscles and hence also used to prevent heart attack.

Answer 23.

1. Nylon-2-nylon-6
2. Such polymer get decomposed by microorganism e.g. PHBV, biological polymers like starch, rubber etc.
3. Knowledge of chemistry, concern for friend, critical thinking.

Answer 24.
(a) If k₃₀₀ is the rate constant for the reaction at 300 K and time required for 10% completion is t then for completion of 25% of reaction at 310 K the time required will be 2t and rate constant is k₃₀₀.
Since for first order reaction

(b)

1. Specific rate constant: It is equal to rate of reaction when concentration of all reactants is considered as unity.
2. Order of reaction: The sum of the exponents of the concentration of the reactants in the rate law expression is known as order of the reaction. If rate law equation for a reaction.
   aA + bB → cC + dD is
   rate = k[A]^x [B]^y then
   order of reaction is x + y.

OR
(i) For the hydrolysis of methyl acetate to be pseudo first order reaction, the reaction should be first order with respect to ester when water is constant. The rate constant k for pseudo first order reaction is

It can be concluded that since value of rate constant k₁ and k₂ are same, hence the given reaction is pseudo first order reaction in which [H₂O] is constant.

(ii) Average rate of reaction

Negative sign represents that concentration of reactant i.e., ester decreases with time.

Answer 25.

1. • Racemic mixture: A mixture containing two enantiomers in equal proportion will have zero optical rotation and this mixture is called racemic mixture.
   • Meso compound: The compound having two or more assymetric centres (chiral carbon) and optical rotation of one part of compound is counter balanced by other part and hence net optical rotation becomes zero.
2. • Immiscibility in water of alkyl halide is due to presence of hydrophobic alkyl group.
   • Grignard reagent get hydrolysed to give alkane and hence it is prepared under anhydrous condition.
     
   • Phenyl group in chlorobenzene is electron withdrawing and charge density of halogen also shifts on benzene ring due to resonance hence chlorobenzene has less dipole moment than that of cyclohexyl chloride.

OR

(ii) 1.

Reagent	Chlorobenzene	Benzyl chloride
NaOH + AgNO3	No. ppt. will form	White ppt. will form

2.

Reagent	Methanol	Ethanol
NaOH + I2, heat	No ppt. will form	Yellow ppt of CHI3 will

Answer 26.

OR

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