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Câu hỏi mô hình công cộng thương mại lần thứ 11 Tiếng Anh Trung bình P. KARTHIK   K. G. Trường trung học cơ sở S Matric, Tiruppur. Tải xuống

 

Câu hỏi một điểm Thương mại lần thứ 11 Tiếng Anh Trung bình P. KARTHIK   K. G. Trường trung học cơ sở S Matric, Tiruppur. Tải xuống

Đề thi công khai lần thứ 11 tháng 3 năm 2023 và đáp án

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 in presence of peroxide give  as major product.
Reason (R). Addition of HBr on alkene proceed by carbocation intermediate

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

Assertion (A). All isotopes of a given element show the same type of chemical behaviour
Lý do (R). Tính chất hóa học của nguyên tử được điều khiển bởi số lượng electron trong nguyên tử

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

Assertion (A). Both 44g CO2 and 16g CH4 have same number of carbon atoms
Reason (R). Both contain 1 g atom of carbon which contains 6. 023  {tex}\times{/tex}  1023 carbon atoms.

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

Class 11 Chemistry Sample Paper Section B

Calculate the molar solubility of Ni(OH)2 in 0. 10M NaOH. The ionic product of Ni(OH)2 is 2. 0 {tex}\times{/tex}  10-15.

What do you understand by exothermic reaction and endothermic reaction? Give one example of each type

Calculate. Number of gram atoms in 1. 4 grams of nitrogen. (Atomic mass, N = 14)

Draw the structure of the following compounds all showing C and H atoms

1. 2-methyl -3-isopropyl heptane
2. Dicyclopropyl methane

OR

Write equations for the preparation of

1. HC {tex}\equiv{/tex} CD and
2. DC {tex}\equiv{/tex} CD

What is the mass (me) of an electron?

Class 11 Chemistry Sample Paper Section C

Explain with the help of suitable example polar covalent bond

Answer the following

1. Two liters of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 liters. How much heat is absorbed and how much work is done in the expansion?
2. Identify the state functions and path functions out of the following
   Enthalpy, entropy, heat, temperature, work, free energy
3. If enthalpy of fusion and enthalpy of vaporisation of sodium metals are 2. 6 and 98. 2 kJ mol-1 respectively, what is the enthalpy of sublimation of sodium?

Calculate the standard Gibbs energy change for the formation of propane at 298 K
3C (graphite) + 4H2(g)  {tex}\to{/tex}  C3H8(g)
{tex}\Delta_fH^o{/tex}  for propane, C3H8(g) is – 103. 8 kJ mol-1
Given,  {tex}s_m^{\circ}{/tex}  C3H8(g) = 270. 2 JK-1mol-1
{tex}s_m^{\circ} C{/tex}  (graphite) = 5. 70 JK-1mol-1
and  {tex}s_m^{\circ}{/tex} H2(g) = 130. 7JK-1mol-1

In the reactions given below, identify the species undergoing oxidation and reduction

1. H2S(g) + CI2(g)  {tex}\rightarrow{/tex}  2HCl(g) + S(s)
2. 3Fe3O4(s) + 8 AI(s)  {tex}\rightarrow{/tex} 9 Fe(s) + 4Al2O3(s)
3. 2Na (s) + H2(g) {tex}\rightarrow{/tex}  2NaH(s)

Calculate the energy required for the process
He+ (g)  {tex}\longrightarrow{/tex}  He2+ (g) + e–
The ionization energy for the H atom in the ground state is 2. 18 {tex}\times{/tex}  10-18 J atom.

An element ‘X’ with atomic number 112 has been recently predicted. Its electronic configuration is . [Rn] 5f146d107s2 . Predict

1. its group
2. block in which this element would be placed
3. IUPAC name and symbol

Commercially available concentrated hydrochloric acid(HCl) contains 38% HCI by mass

1. What is the molarity (M) of the solution (density of solution = 1. 19 g mL-1)
2. What volume required of concentrated HCI is required to make 1. 0 L of an 0. 10 M HCI?

Class 11 Chemistry Sample Paper Section D

Read the text carefully and answer the questions
IUPAC (International Union of Pure and Applied Chemistry) system of nomenclature. Common names are useful and in many cases indispensable, particularly when the alternative systematic names are lengthy and complicated. A systematic name of an organic compound is generally derived by identifying the parent hydrocarbon and the functional group(s) attached to it. By using prefixes and suffixes, the parent name can be modified to obtain the actual name. In a branched-chain compound, small chains of carbon atoms are attached at one or more carbon atoms of the parent chain. The small carbon chains (branches) are called alkyl groups. An alkyl group is derived from a saturated hydrocarbon by removing a hydrogen atom from carbon. Abbreviations are used for some alkyl groups. For example, methyl is abbreviated as Me, ethyl as Et, propyl as Pr and butyl as Bu

1. Draw the structure of 3-Ethyl-4,4-dimethylheptane

2. OR

   Why CH4 after becoming-CH3 called a methyl group?

3. How is the numbering in branched chain hydrocarbon done?
4. Derive the structure of 2-Chlorohexane

Read the text carefully and answer the questions
When anions and cations approach each other, the valence shell of anions are pulled towards the cation nucleus and thus, the shape of the anion is deformed. The phenomenon of deformation of anion by a cation is known as polarization and the ability of the cation to polarize the anion is called as polarizing power of cation. Due to polarization, sharing of electrons occurs between two ions to some extent and the bond shows some covalent character
The magnitude of polarization depends upon a number of factors

1. Out of AlCl3 and AlI3 which halides show maximum polarization?
2. Out of AlCl3 and CaCl2 which one is more covalent in nature?
3. The non-aqueous solvent like ether is added to the mixture of LiCl, NaCl and KCl. Which will be extracted into the ether?

4. OR

   Out of CaF2 and CaI2 which one has a minimum melting point?

To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to create similar papers with their own name and logo

Class 11 Chemistry Sample Paper Section E

Attempt any five of the following

1. What happens to equilibrium constant when temperature increases for a reaction?
2. Explain why alkynes are less reactive than alkenes towards addition of Br2
3. What is a Lindlars’ catalyst?
4. Why do alkynes not show geometrical isomerism?
5. Which conformation of ethane is more stable?
6. State Le chatelier’s principle
7. How is alkene produced by vicinal dihalide?

Write a relation between {tex}\triangle G{/tex} and Q and define the meaning of each term and answer the following.

1. Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K
2. Explain the effect of an increase in pressure in terms of reaction quotient Q for the reaction
   CO(g) + 3H2(g)  {tex} \rightleftharpoons {/tex}  CH4(g) + H2O(g)

OR

Ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as
CH3COOH(l) + C5H5OH(l)  {tex}\rightleftharpoons{/tex}  CH3COOC2H5(l) + H2O(l)

1. Write the concentration ratio (reaction quotient), Qc, for this reaction (note. water is not in excess and is not a solvent in this reaction)
2. At 293 K, if one starts with 1. 00 mol of acetic acid and 0. 18 mol of ethanol, there is 0. 171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant
3. Starting with 0. 5 mol of ethanol and 1. 0 mol of acetic acid and maintaining it at 293 K, 0. 214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

Answer the following

1. Give the IUPAC name of the following compound
   {tex}{\text{C}}{{\text{H}}_3} – \mathop {\mathop {{\text{CH}}}\limits_. }\limits_{{\text{Br}}} – \mathop {\mathop {\text{C}}\limits_{. } }\limits_{\text{O}} – \mathop {\mathop {{\text{CH}}}\limits_. }\limits_{{\text{C}}{{\text{H}}_{\text{3}}}} – {\text{C}}{{\text{H}}_3}{/tex}
2. 0. 12 g of an organic compound containing phosphorous gave 0. 22 g of Mg2P2O7 by usual analysis. Calculate the percentage of phosphorous in the compound

OR

1. Which of the two structures CH3COOH and CH3COO– is more stabilized by resonance? Explain
2. Will CCl4 give a white precipitate of AgCl on heating it with AgNO3?

Class 11 – Chemistry Sample Paper Solution (2023-24)

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1. Class 11 Chemistry Sample Paper Solution Section A

2. (b) 4. 7g
   Explanation. Mass = Density  {tex}\times{/tex}  Volume
   = 3. 12 g mL-1  {tex}\times{/tex} 1. 5 mL = 4. 68 g = 4. 7 g
   significant figures as that of the least given  number. Therefore, correct answer is 4. 7 g
3. (d) charge on the electrons
   Explanation. The oil drop experiment was performed by Robert A. Millikan and Harvey Fletcher in 1909 to measure the elementary electric charge (the charge of the electron). The experiment entailed observing tiny electrically charged droplets of oil located between two parallel metal surfaces, forming the plates of a capacitor
4. (b) may be positive or negative
   Explanation. Standard molar enthalpy of formation of a compound from its elements can be +ve or –ve
   For example.   {tex}C+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H{/tex}  = 393. 5 kJ mol-1
   {tex}N_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow N_{2} O(g);{/tex}   {tex}\Delta_{r} H{/tex}  = +92 k J mol-1 
5. (d) calcium
   Explanation. calcium
6. (c) -778 kJ
   Explanation. Heat of reaction, {tex}\Delta_r{/tex} H = {tex}{\textstyle\sum_{}}\triangle_rH_{products}-{\textstyle\sum_{}}{\textstyle{\scriptstyle\triangle}_r}{\textstyle{\scriptstyle H}_{reactants}}{\textstyle\;}{\textstyle\;}{/tex}
   {tex}\Rightarrow{/tex}   {tex}\Delta_{r} H{/tex}  = [ {tex}\Delta_{f} H{/tex} (N2O) + 3 {tex}\Delta_{f} H{/tex} (CO2)] – [ {tex}\Delta_{f} H{/tex} (N2O4) + 3 {tex}\Delta_{f} H{/tex} (CO)] {tex}\Rightarrow{/tex}   {tex}\Delta_{r} H{/tex}  = [81 + {3  {tex}\times{/tex}  (-393)}] – 99. 7 + {3  {tex}\times{/tex}  (-110)}] kJ
   {tex}\Rightarrow{/tex}   {tex}\Delta_{r} H{/tex}  = -777. 7 kJ  {tex}\approx{/tex}  -778 kJ
   To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to create similar papers with their own name and logo
7. (d) {tex}\lambda{/tex} = {tex}\frac{h}{m v}{/tex}
   Explanation. Louis de-Broglie proposed that matter,  like light , has a dual character. It exhibits wave as well as  particle nature. The wavelength of the wave associated with a particle of mass m moving with velocity v is given by
   {tex}\lambda{/tex} = {tex}\frac{h}{m v}{/tex}
8. (a) NH4NO3
   Explanation. NH4+NO3–. In NH4+
   N is -3 and
   NO3– , N is +5
9. (a) {tex}{(C{H_3})_3}\mathop C\limits^ \oplus {/tex}
   Explanation. {tex}{(C{H_3})_3}\mathop C\limits^ \oplus {/tex}  i. e. the tertiary carbocation is most stable.  Alkyl groups directly attached to the positively charged carbon stabilise the carbocations due to inductive and hyperconjugation effects.
10. (b) alkaline KMnO4
    Explanation. Baeyer’s reagent is a dilute alkaline KMnO4 solution. It is violet in colour when this solution reacts with the double bond the colour diaper and the whole solution become colourless
11. (b) Both AI2O3 and As2O3
    Explanation. A12O3 and As2O3 are amphoteric in nature. Amphoteric oxides behave as acidic with bases and basic with acids
12. (a) Both Constant volume, qv or Constant pressure, qp
    Explanation. In calorimetry, vessel called calorimeter is immersed in a known volume of a liquid. Measurements are made under two different conditions
    1. At constant volume, qv
    2. At constant pressure, qp
13. (b) Sodium acetate
    Explanation. This is an example of Kolbe’s electrolysis method. The reaction is
    {tex} { 2CH_3COONa + 2H_2O } \overset{_{electrolysis}}\longrightarrow{ CH_3. CH_3 + 2NaOH + H_2 + 2CO_2}{/tex}
    The step-wise redox reactions occuring in the electrolytic cell are  depicted as under
    
    
    
    
    
    
    2H–  {tex}\rightarrow{/tex}  H2
14. (d) A is false but R is true
    Explanation. The resonance structure is hypothetical and individually does not represents any real molecules.  According to the resonance theory, the actual structure of benzene cannot be adequately represented by any of these structures
15. (b) Both A and R are true but R is not the correct explanation of A
    Explanation. Both A and R are true but R is not the correct explanation of A
16. (a) Both A and R are true and R is the correct explanation of A
    Explanation. All isotopes have the same atomic number means the same number of electrons and they bear similar chemical properties
17. (a) Both A and R are true and R is the correct explanation of A
    Explanation. 44g of CO2 = 1 mole {tex}\equiv{/tex} 1 g atom of C
    16g CH4 = 1 mole  {tex}\equiv{/tex}  1 g atom of C
    1 g atom of C = 12 g C
    12 g C contains 6. 023  {tex}\times{/tex}  1023 carbon atoms.

18. Class 11 Chemistry Sample Paper Solution Section B

19. According to the question, ionic product of Ni(OH)2 is 2. 0 × 10-15
    Reaction
    Ni(OH)2  {tex}\rightleftharpoons{/tex}  Ni2+ + 2OH–
    NaOH  {tex}\rightarrow{/tex}  Na+ + OH–
    Here, [OH–]total = (2S + 0. 10)
    So, Ksp = [Ni2+] [OH–]2
    {tex}\Rightarrow{/tex} 2. 0  {tex}\times{/tex}  10-15 = (S) (0. 10 +2S)2
    As Ksp is small, 2S << 0. 10,
    {tex}\Rightarrow{/tex}  2. 0  {tex}\times{/tex}  10-15 = S(0. 10)2
    {tex}\therefore{/tex}  [Ni2+] = S = 2. 0  {tex}\times{/tex}  10-13 M.
20. Exothermic reaction .  The reaction accompanied by release of energy in the form of light or heat is called exothermic reaction.   {tex}\Delta H{/tex}  is negative for exothermic reaction.
    Example. C+ O2  → CO2 ;  {tex}\Delta H{/tex} = -178. 3 kJmol-1.
    Endothermic reactions . The reaction accompanied by absorption of energy is called endothermic reaction.   {tex}\Delta H{/tex}  is positive for endothermic reaction.
    Example. 2NH3 → N2 +3H2. {tex}\Delta H{/tex}  = +91. 8kJmol-1.
21. Since, 1 gram atom N
    = atomic mass of nitrogen (N) in grams
    =  14g
    {tex}\therefore{/tex}   1. 4 g of N
    [ {tex}\frac{1}{14} \times 1. 4{/tex}   ]gram atoms
    = 0. 1 gram atoms
    1.  The struture of 2-methyl -3-isopropyl heptane is
    2.  The struture of Dicyclopropyl methane is
       

22. OR

    1. {tex}\mathrm{HC}\equiv\mathrm{CH} \stackrel{\mathrm{NaNH}_{2}}{\longrightarrow} \mathrm{HC} \equiv \mathrm{C}^{-} \mathrm{Na}^{+} \stackrel{\mathrm{D}_{2} \mathrm{O}}{\longrightarrow} \mathrm{HC}\equiv\mathrm{CD}{/tex}
       Here, Strong base NaNH 2   deprotonated the alkynes to give so-called acetylide ions.
    2. {tex}\mathrm{HC} \equiv \mathrm{CH} \stackrel{2 \mathrm{Na}}{\longrightarrow} \mathrm{Na}^{+} \mathrm{C}^{-}\equiv\mathrm{C}^{-} \mathrm{Na}^{+} \stackrel{\mathrm{D}_{2} \mathrm{O}}{\longrightarrow} \mathrm{DC} \equiv \mathrm{CD}{/tex}
       Here, sodium metal is one-electron reducing agent
23. Mass of an electron (me) =  {tex}\frac{Charge\;on\;electron}{Charge\;to\;Mass\;ratio}{/tex}  =  {tex}\frac { e } { \frac em }{/tex}   {tex}= \frac { 1. 602 \times 10 ^ { – 19 } C } { 1. 76 \times 10 ^ { 8 } C g ^ { – 1 } }{/tex}{tex}= 9. 10 \times 10 ^ { – 28 } g{/tex}   {tex}= 9. 1 \times 10 ^ { – 31 } k g{/tex}
    Therefore,  the mass of an electron is {tex}9. 1 \times 10 ^ { – 31 } k g{/tex} .
    Mass of an electron = {tex}\frac{1 }{1837}{/tex} th of the mass of a hydrogen atom.

24. Class 11 Chemistry Sample Paper Solution Section C

25. When two atoms with different electro-negativity are linked to each other by a covalent bond, the shared pair of electrons will not in the center because of the difference in electro-negativity. For example, in hydrogen fluoride
    molecule, fluorine has greater electronegativity than hydrogen. Thus the shared pair of electrons is displaced more towards fluorine atom, and acquire a partial negative charge ( {tex}\delta^-{/tex} ). At the same time, the hydrogen atom will have a partial positive charge ( {tex}\delta^+{/tex} ). Such a covalent bond is known as polar covalent bond or simply polar bond.
    lt is represented as
    {tex}\delta^+{/tex} {tex}\delta^-{/tex}
    H – F
    (2,1) (4,0)
26. Answer
    1. We have q = – w = pex (10 – 2) = 0(8) = 0 No work is done; no heat is absorbed
    2. Path function. heat, work
       State function.  enthalpy, entropy, temperature, free energy
    3. According to the question, enthalpy of fusion and enthalpy of vaporisation of sodium metal is 2. 6 kJ mol-1 and 98. 2 kJ mol-1
       We know that, enthalpy of sublimation =  {tex}\Delta _ { \mathrm { sub } } H ^ { \circ } = \Delta _ { \mathrm { fus } } H ^ { \circ } + \Delta _ { \mathrm { vap } } H ^ { \circ }{/tex}
       = 2. 6 + 98. 2
       = 100. 8 kJ mol-1
27. 3C(graphite) + 4H2(g) {tex}\to{/tex} C3H8(g)
    {tex}\Delta_rS{/tex} = {tex}\Sigma S_m^{\circ}{/tex} (products) – {tex}S_m^{\circ}{/tex} (reactants)
    {tex}\Delta_rS{/tex} = {tex}S_m^{\circ}{/tex} [C3H8(g) – 3 {tex}S_m^{\circ}{/tex} [C (graphite)] + 4 {tex}S_m^{\circ}{/tex} m[H2]] = (270. 2 – 3 {tex}\times{/tex} 5. 70 – 4 {tex}\times{/tex} 130. 7) JK-1mol-1
    = (270. 2 – 17. 10 – 522. 80) JK-1mol-1
    = (270. 2 – 539. 90) = – 269. 7 JK-1mol-1
    {tex}\Delta_rH^o{/tex} = {tex}\Delta_fH^o{/tex} products – {tex}\Delta_fH^o{/tex} reactants
    {tex}\Delta_rH^o{/tex} = {tex}\Delta_fH^o{/tex} (C3H8) – 4 {tex}\Delta_fH^o{/tex} [H2(g)] – 3 {tex}\Delta_fH^o{/tex} [C (graphite)] {tex}\Delta_rH^o{/tex} = – 103. 8 kJ mol-1; {tex}\Delta_rG^o{/tex} = {tex}\Delta_rH^o{/tex} – {tex}T\Delta_rS^o{/tex}
    {tex}=\left(-103. 8-\frac{298 \times(-269. 7)}{1000}\right) \mathrm{kJ} \mathrm{mol}^{-1}{/tex}
    = (-103. 80 + 80. 370) kJmol-1
    = – 23. 43 kJ mol-1
    1. 
       Thus, here, H2S undergoes oxidation because oxidation number of Sulphur atom get increased from -2 to 0 while Cl2 undergoes reduction because oxidation number of Cl atom get decreased from 0 to -1
    2. 
       Thus, here Fe3 O 4 undergoes reduction because oxidation number of Fe atom get decreased while AI atom undergoes oxidation because Oxidation number of Al atom get increased from 0 to +3.
    3. 2 Na(s) + H2(g)  {tex}\rightarrow{/tex}  2 NaH(s)
       Since Na is more electro-positive than H so, here Na atom get oxidised by losing electron while H-atom get reduced by accepting electron
28. The expression for the ionization energy atom
    En =  {tex}\frac{2. 18 \times 10^{-18} \times Z^{2}}{\mathrm{n}^{2}}{/tex}  J atom-1
    For H atom (Z = 1) En = 2. 18 {tex}\times{/tex} 10-18  {tex}\times{/tex} (1)2 J atom-1 (given)
    For He+ ion (Z = 2) En = 2. 18  {tex}\times{/tex} 10-18 (2)2 = 8. 72  {tex}\times{/tex}  10-18 J atom-1
    (one electron species)
29. The configuration of the element is. {tex}[\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^{2}{/tex}
    1. It belongs to the 12th group
    2. It belongs to d block
    3. IUPAC name is. Ununbium; Symbol. Uub
    1. Let assume the total mass of the solution is 100g
       38 % HCI by mass means 38 g of HCI is present in 100 g of solution
       The volume of solution (V) =  {tex}\frac { \text { mass } } { \text { density } } = \frac { 100 } { 1. 19 } = 84. 03 \mathrm { mL }{/tex}    (Density of solution = 1. 19 g/mL)
       Number of moles of HCl (nB) =  {tex}\frac{38}{36. 5}{/tex}  = 1. 04
       Molarity =  {tex}\frac{n_B \times 1000}{V (\ in \ mL)}{/tex} =   {tex}\frac{1. 04 \times 1000}{84. 03 \ mL}{/tex} = 12. 38 M
    2. From the molarity equation,  {tex}\begin{array} { l } { M _ { 1 } V _ { 1 } = M _ { 2 } V _ { 2 } } \\ { \text { acid } _ { 1 } \quad \text { acid } _ { 2 } } \end{array}{/tex}
       {tex}12. 38 \mathrm { M } \times V _ { 1 } = 0. 10 \mathrm { M } \times 1. 0 \mathrm { L }{/tex}
       {tex}\therefore{/tex}   {tex}V _ { 1 } = \frac { 0. 1 \times 1. 0 } { 12. 38 } = 0. 00808 \mathrm { L } = 8. 08 \mathrm { cm } ^ { 3 }{/tex}

30. Class 11 Chemistry Sample Paper Solution Section D

31. 2. OR

       CH4 after becoming-CH3 called a methyl group because an alkyl group is named by substituting ‘yl’ for ‘ane’ in the corresponding alkane

    3. The numbering is done in such a way that the branched carbon atoms get the lowest possible numbers
    4. ‘Hexane’ indicates the presence of 6 carbon atoms in the chain. The functional group chloro is present at carbon 2. Hence, the structure of the compound is CH2CH2CH2CH2CH(Cl)CH3

32. 1. AlI3  halides show maximum polarization. The most covalent halide is AlI3
       Since lesser, the electronegativity difference, the more covalent is the aluminum halide
    2. AlCl3 is more covalent in nature
    3. LiCl will be extracted into the ether

    4. OR

       CaI2 has a minimum melting point

33. Class 11 Chemistry Sample Paper Solution Section E

34. Attempt any five of the following
    1. Equilibrium constants gets changed if we change the temperature of the system
       • For Exothermic reactions, if the temperature is increased, the equilibrium will shift to favour the reaction which will decrease the temperature and the exothermic reaction is favoured
       • For Endothermic reactions, if the temperature is increased, the equilibrium will shift to favour the reaction which will reduce the temperature and the endothermic reaction is favoured
    2. The triple bonds of alkynes, because of its high electron density, are easily attacked by electrophiles, but less reactive than alkenes due to the compact C-C electron cloud. The three-membered ring bromonium ion formed from the alkyne (A) has a full double bond causing it to be more stained and less stable than the one from the alkene (B),
       
       
       Also, the carbon’s of (A) that are part of the bromonium ion has more s-character than (B), further making (A) less stable than (B)
    3. Lindlar’s catalyst. Partially deactivated palladised charcoal is known as Lindlar’s catalyst
       Uses.  Alkynes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes
    4. Alkynes have a linear structure.  Alkynes have triple bond. So, rotation is not possible.  Hence, alkynes cannot show geometrical isomerism
    5. Staggered conformation of ethane is more stable
       Structure
       
    6. Le chatelier’s principle. If a system at equilibrium is subjected to change in the temperature, pressure or concentration of the reactants or the products that govern the equilibrium, then the equilibrium shifts in that direction in which this change is reduced or nullified
    7. Alkene are produced from Vicinal dihalide by the process of dehalogenations. Vicinal dihalide on treatment with Zn metal lose a molecule of ZnX2 to from an alkene
       CH2Br-CH2Br + Zn → CH2=CH2 + ZnBr2

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    1. We know that,  {tex}\Delta G = \Delta G ^ { \circ } + R T \operatorname { ln } Q{/tex}
       Where,
       {tex} \Delta G ^ { \circ } {/tex}  = Change in free energy as the reaction proceeds
       {tex}\Delta G{/tex} = Standard free energy change
       Q = Reaction quotient
       T = Absolute temperature
       Also,  {tex} \Delta G ^ { \circ } = -RT\ lnK{/tex}
       {tex}\Rightarrow \Delta G = – R T \ l n K + R T l n Q \\\therefore \Delta G= R T \ l n \frac { Q } { K }{/tex}
       If Q < K,  {tex}\Delta G{/tex} will be negative. So, the reaction proceeds in the forward direction.
       If Q = K,  {tex}\Delta G{/tex}  = 0, reaction will be at equilibrium.
    2. Reaction
       CO(g) + 3H2(g)  {tex} \rightleftharpoons {/tex}  CH4(g) + H2O(g)
       On increasing the pressure equilibrium will shift in forward direction, it means Q < K

35. OR

    1. The concentration ratio (concentration quotient) Qc for the reaction is
       {tex}{Q_c} = \frac{{[C{H_3}COO{C_2}{H_5}(l)][{H_2}O(l)]}}{{[C{H_3}COOH(l)][{C_2}{H_2}OH(l)]}}{/tex}
    2. {tex}\begin{array}{*{20}{c}} {}&{C{H_3}COOH(l) + }&{{C_2}{H_5}OH(l) \rightleftharpoons }&{C{H_3}COO{C_2}{H_2}(l) + } \\ {Initial\;molar\;conc. }&{1. 0\;mol}&{0. 18\;mol}&0 \\ {Molar\;conc. \;at}&{(1 – 0. 171)}&{(0. 18 – 0. 171)}&{0. 171\;mol} \\ {equilibrium\;po\operatorname{int} }&{ = 0. 829\;mol}&{ = 0. 009}&{} \end{array}\begin{array}{*{20}{c}} {{H_2}O(l)} \\ 0 \\ {0. 171\;mol} \\ {} \end{array}{/tex}
       Applying Law of Chemical equilibrium
       {tex}{K_c} = \frac{{[C{H_3}COO{C_2}{H_3}](l)[{H_2}O(l)]}}{{[C{H_3}COOH](l)[{C_2}{H_3}OH(l)]}}{/tex}
       {tex} = \frac{{(0. 171\;mol) \times (0. 171\;mol)}}{{(0. 829\;mol)(0. 009\;mol)}} = 3. 92{/tex}
       Therefore, the equilibrium constant is 3. 92
    3. {tex}\begin{array}{*{20}{c}} {}&{C{H_3}COOH(l) + }&{{C_2}{H_5}OH(l) \rightleftharpoons }&{C{H_3}COO{C_2}{H_5}(l)} \\ {Initial\;molar\;conc. }&{1. 0\;mol}&{0. 5\;mol}&{0. 214} \\ {Molar\;conc. \;at}&{1. 0 – 0. 214}&{0. 5 – 0. 214}&{} \\ {equilibrium}&{ = 0. 786}&{ = 0. 286\;mol}&{} \end{array}\begin{array}{*{20}{c}} { + {H_2}O(l)} \\ {0. 214mol} \\ {} \\ {} \end{array}{/tex}
       {tex}{Q_c} = \frac{{[C{H_3}COO{C_2}{H_5}(l)][{H_2}O(l)]}}{{[C{H_3}COOH(l)][{C_2}{H_2}OH(l)]}}{/tex}
       {tex} = \frac{{(0. 214mol) \times (0. 214\;mol)}}{{(0. 286\;mol)(0. 786\;mol)}} = 0. 204{/tex}
       Since Qc value 0. 204 is less than Kc, value 3. 92 this means that the equilibrium has not been reached. The reactants are still taking part in the reaction to form the products
36. Answer the following
    1. Structure
       {tex}\mathop {{\text{C}}{{\text{H}}_3}}\limits^1 – \mathop {\mathop {\mathop {{\text{CH}}}\limits^2 }\limits_. }\limits_{{\text{Br}}} – \mathop {\mathop {\mathop {\text{C}}\limits_{. } }\limits_{\text{O}} }\limits^3 – \mathop {\mathop {\mathop {{\text{CH}}}\limits_. }\limits_{{\text{C}}{{\text{H}}_{\text{3}}}} }\limits^4 – \mathop {{\text{C}}{{\text{H}}_3}}\limits^5 {/tex}
       IUPAC name. 2-bromo-4-methyl-pentan-3-one
    2. % of P = {tex}\frac{62}{222} \times \frac{\text { weight of } \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \text { formed } \times 100}{\text { weight of organic compound }}{/tex}
       =  {tex}\frac{62}{222} \times \frac{0. 22}{0. 12} \times{/tex} 100
       = 51. 20%

    OR

    1. Resonating structures of CH3COOH are
       
        
       Resonating structures of CH3COO– are
       
       
       Structure (III) and (IV) are of equal energy and also contribute equally towards resonance hybrid of compound CH3COO–
       Structure II is less stable than Structure I because in structure II separation of +ve and-ve charges exists. Therefore, the contribution of structure (II) is less than that of structure (I) towards resonance hybrid of compound CH3COOH
       Therefore, CH3COO– is more stable than CH3COOH
    2. Carbon tetrachloride contains chlorine but it is bonded to carbon by covalent bond. Hence, it is not in ionic form
       So, it does not combine with AgNO3 solution
       Therefore, CCl4 does not give white precipitate with silver nitrate solution
       CCl4 + AgNO3 →No reaction

Marking Scheme for the Class 11 exam

SubjectBoard MarksPractical or internal MarksEnglish80 Marks20 MarksHindi80 Marks20 MarksMathematics80 Marks20 MarksChemistry70 Marks30 MarksPhysics70 Marks30 MarksBiology70 Marks30 MarksComputer Science70 Marks30 MarksInformatics Practices70 Marks30 MarksChemistry80 Marks20 MarksBusiness Studies80 Marks30 MarksEconomics80 Marks20 MarksHistory80 Marks20 MarksPolitical Science80 Marks20 MarksGeography70 Marks30 MarksSociology80 Marks20 MarksPhysical Education70 Marks30 MarksHome Science70 Marks30 Marks

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