K1505200340usolution 6-10 Chemistry Sqp 11th.pdf

Time : 3 Hours Maximum Marks : 70

CHEMISTRY CBSE Sample Question Papers

Solution to Question Paper

6

Self Assessment _______________________________________________

1. Halogens. 1 2. CaC2 + 2H2O → Ca(OH)2 + CH ≡ CH 1 3. Acetanilide, glycol. 1 4. Sodium thiosulphate helps in fixing the photographs thus making them stable and insensitive to light. 1 2

5.

1

3 4

6. Given : 7. (a) (b)

5 6

M = 3 mol L–1 Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5g Mass of 1 L solution = 1000 × 1.25 = 1250 g Since density = 1.25 gmL–1 Mass of water in solution = 1250 – 175.5 = 1074.5 g no. of moles of solute 3 mole Molality = = = 2.79 M mass of solvent in kg 1.0745 kg

pH of human muscle fluid = 6.83 pH = – log[H+] or ; 6.83 = – log[H+] log H+ = – 6.83 or ; [H+] = Antilog (–6.83) = 1.48 × 10–7 M Human stomach fluid pH = 1.2 pH = – log[H+] or ; –log [H+] = 1.2 [H+] = anti log 2.8 = 6.3 × 10–2 M = 0.063 M HOOC COOH H COOH

8.

C

C

H





1

4-Ethyl-3-propylhex-1-ene

C H

cis

1

1 1+1

C

HOOC

2

H trans



OR FeCl3 and SnCl4 1+1 9. Quantities that are dependent upon the state of the system only, are called state functions. They do not depend upon how that state has been achieved. Q + W is equivalent to DU i.e., change in internal energy.

2

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

DU will depend only upon the initial and final states. It is independent of the way in which the change has been brought about. 2 3 × ( atomic mass of calcium ) 10. Mass percent of Ca = × 100 molecular mass of Ca3 ( PO 4 )2 120u × 100 = 38.71% = 310u 2 × ( atomic mass of phosphorus ) Mass percent of P = × 100 molecular mass of Ca3 ( PO 4 )2

2 × 31u = × 100 = 20% 310u 8 × ( atomic mass of oxygen ) Mass percent of O = × 100 molecular mass of Ca3 ( PO 4 )2 8 × 16u = × 100 = 41.29% 310u

2

11. (a) H+ = 1s0 (b) Na+ = 1s2, 2s2, 2p6 (c) O2– = 1s2, 2s2, 2p6 1 + 1 + 1 12. In anhydrous AlCl3, aluminium atom is linked with three chlorine atoms by covalent bond. However, when aluminium chloride is dissolved in water, It undergoes hydration as Al2Cl6 + Water  2[Al(H2O)6]3+ + 6Cl– + Energy Hydration of Al2Cl6 is an exothermic reaction and the released energy is responsible for the removal of three electrons from Al to form Al3+, Thus AlCl3. 6H2O is ionic in nature. 3 13. (a) Na2O2 + 2H2O → 2NaOH + H2O2 1 (b) 4KO2 + 2H2O → 4KOH + 3O2 ½ or 2KO2 + 2H2O → 2KOH + H2O2 + O2 ½ (c) Na2O + CO2 → Na2CO3. 1 14. (i) Mists : These are particles of spray liquids and condensation of their vapours in air. Examples : Portions of herbicides and insecticides that are not used up, travel through the air to form mists. 1 (ii) Smoke : These are small soot particles produced due to combustion of fossil fuels and organic matter. Examples : Oil, tobacco and carbon smoke are typical examples. 1 (iii) Fumes : These include condensed vapours of metallurgical acid and alkali fumes. 1 15. (a) Let the O.N. of O be x. Since Ca is an alkaline earth metal, therefore, its O.N. is + 2. Thus, +2 x Ca O2 + 2 + 2(x) = O or x = –1 Thus,the oxidation number of O in CaO2 = –1. 1 (b) In NaBH4, H is present as hydride ion. Therefore, its O.N. is – 1. Thus, +1 x –1 Na B H4 1(+ 1) + x + 4(–1) = 0 or x = + 3 1 Thus, the O.N. of B in NaBH4 = + 3 (c) +1 x –2 H2 S2 O7 2(+1) + 2(x) + 7(–2) = 0 or x = 6 1 Thus, the O.N. of S in H2S2O7 = + 6 16. For CO2, M = 44 g mol–1 = 0.044 kg mol–1 ; T = 700 K (i) Root mean square speed,

C =

3RT = M

3 × (8.314 JK –1 mol –1 )(700K) (0.44 kgmol –1 )

1

Solution

3

= 39680.4545 =199.199 ms–1 8RT = (ii) Average speed, C = pM = 1925.32 (iii) Most probable speed,

C* =

2RT = M

8(8.314 JK –1 mol –1 )(700 K) 3.14(0.004 kgmol –1 )

2(8.314 JK –1 mol –1 ) (700 K) 0.044 kgmol –1

=

3706878.98 1 1



= 264536.4 ms –1 = 514.33 ms–1 17. (a) Orbitals which possess equal energies are called degenerate orbitals. For example 2px, 2py and 2pz orbitals which are oriented along X, Y and Z axis are degenerate. Similarly five d-orbitals dxy, dyz, dzx, dx2–y2 and dz2 are degenerate orbitals. 1 (b) The angular momentum of an electron in a given stationary state can be expressed as h mvr = n where n =1, 2, 3, ............ (Bohr's equation) 2p or

2pr = n

h mv

n But = l (de Broglie's equation) 2 mv From the two equations, we have 2pr = nl or Circumference of Bohr orbit = nl Thus, we can say that circumference of Bohr's orbit is an integral multiple of de Broglie wavelength. 18. (i) PCl5(g)  PCl5(g) + Cl2(g) At equlibrium 0.5 × 10–1 mol x mol L–1 x mol L–1 2 x KC = = 8.3 × 10–3 (Given) 0.5 × 10 –1 or, or

x2 = (8.3 × 10–3) (0.5 × 10–1) = 4.15 × 10–4 x = 4.15 × 10 –4 = 2.04 × 10–2 M = 0.02 M

Hence, [PCl3]eq = [Cl2]eq = 0.02 M (ii) KP = KC (RT) Dng = 8.3 × 10–3 × 0.082 × 473 = 0.3216 atm

2

1

393K 19. (i) 2(CaSO4.2H2O) → (CaSO4)2. H2O + 3H2O 1 Gypsum Plaster of Paris (ii) CaCO3  CaO + CO2, DH = 179.9 kJ 1 Quicklime (iii) CaO + H2O → Ca(OH)2 1 Slaked lime OR (i) Ionization enthalpies of alkaline earth metals are higher than those of alkali metals. This is because of smaller size of alkaline earth metal corresponding to alkali metal of the same period. 1 (ii) The oxides of alkali and alkaline earth metal dissolve in water to form their respective hydroxides. These oxides are strong bases. However, the oxides of alkali metals are more basic than those of alkaline earth metals. This is because the ionization enthalpy of alkali metals is lower. The electropositive character of alkali metals is higher than that of corresponding alkaline earth metal so that M–OH bond in alkali metals can ionize more easily. 1 MOH → M+ + OH–

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

4

(iii) Because of small size and high charge, the lattice enthalpies of alkaline earth metals are much more than those of alkali metals and therefore, the solubility of alkaline earth metal hydroxides is less than that of alkali metals. 1 CH3 20. (i)

+ CH3Cl

Al2Cl6

Benzene Br (ii)

CH2CH3

Toluene

+ 2 Na + H3CCH2Br

dry ether



Ethylbenzene

The above reaction is known as Wurtz - fitting reaction CH3

(iii) 3CH3—C  CH

873 K

1

Catalyst

CH3

CH3



1

+ 2NaBr



1

+ HCl

Mesitylene

Propyne



21. (i) Test for Nitrogen : The sodium fusion extract is boiled with iron (II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen.

xH 2 O 3[Fe(CN)6]4– + 4Fe3+  → Fe4[Fe(CN)6]3.xH2O 1 Prussian blue (ii) Test for Sulphur : The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur. 1

S2– + Pb2+ → PbS





Black

(iii) Test for halogens : The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine. 1 X– + Ag+ → AgX (X represents Cl, Br or I)



22. Important conditions required for the linear combination of atomic orbitals to form molecular orbitals are : (a) The combining atomic orbitals should have comparable energies. (b) The combining atomic orbitals must have proper orientation e.g., same symmetry, so that they are able to overlap to a considerable extent. 3 (c) The extent of overlapping should be large. Greater the overlap, greater will be the electron density between the nuclei. Be2+ < Mg2+ <

23. (i)

(39 pm) –

(ii) (a) Br or Br

(65 pm)

Na+ < Cl–

(95 pm)

½+½+½+½

(181 pm)

1

(iii) Discrimination will be removed from our society, all the people would have same status. It will be an ideal situation but then nobody will try to work of lower levels. 24. Distillation means the conversion of a liquid into vapours, which on condensation by cooling gives a pure liquid. This method is used for purification of liquids which boil without dicomposition

Solution

5

and contain non-volatile impurities. This method can also be used for separating liquids having sufficient differences in their boiling points. 1 Distillation under reduced pressure also involves conversion of a liquid into vapours by heating followed by condensation of the vapours thus produced by cooling under reduced pressure. In this method, the pressure acting on the system is not atmospheric pressure but is reduced with the help of a vacuum pump. This method is used to purify liquids which decompose below their normal boling points. 2 Steam distillation is similar to distillation under reduced pressure even though there is no reduction in the total pressure acting on the solution. It is used for the separation and purification of a liquid which is appreciably volatile when the sum of the vapour pressure of the organic liquid (P1) and that of water (P2) becomes equal to atmospheric pressure i.e., (P) = P1 + P2 2 OR The process of displacement of σ-electrons along the saturated carbon chain due to the presence of a polar covalent bond at one end of the chain is called inductive effect and is denoted as I-effect. This is a permanent effect and is generally represented by an arrow with is head in the middle of the covalent bond pointing in the direction of displacement of electron as shown below : +''' C

+'' C

+' C

+ C

– C

3 However, it may be noted that this effect decreases sharply as we move away from the atoms involved in the initial polar bond and becomes negligible from the fourth atom onwards. The phenomenon of movement of electrons from one atom to another in a multiple bond at the demand of attacking reagent is called electromeric effect and is denoted as E-effect. The electromeric shift of electrons occurs only at the moment of reaction. Like the inductive effect, the electromeric effect is also classified as : (a) The given order can be explained on the basis of -I effect O O O Cl Cl Cl C C O H CH C O H Cl CH2 C O H Cl Cl 1 More is the number of halogen atoms higher is the -I effect and hence is more polar resulting in higher acidic character. (b) The give order can be explained by +I effect O CH3 O O Cl CH3 C C O H CH C O H CH3CH2 C O H Cl CH3 1 As the number of alkyl groups increases, the +I effect increases, resulting in decrease in acidic strength. 25. Standard free energy of formation (Df G°) for the reaction 1 ½N2(g) + O2(g) → NO(g) is positive (+86.7 kJ mol–1), 2 Hence, the reaction is non-spontaneous under the standard condition. OR (i) For mole of an ideal gas,

5

 DU  CV =    DT 

or DU = CVDT For an isothermal process, T is constant so that DT = 0

2

6

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

∴ DU = 0 DH = D(PV) (ii) For an ideal gas, PV = RT ∴ DH = DU + D(RT) or = DU + RDT Since T is constant, DT = 0 DH = 0 3 26. (a) The first element of a group often shows similarities to the second element of the neighbouring group on the right. This type of behaviour is known as diagonal relationship. Group I Group II Group III Group IV Li

Be

B

C

Na

Mg

Al

Si

(b) Plaster of Paris is prepared by heating gypsum CaSO4. 2H2O at 393 K.

3

393K

2[CaSO4.2H2O] → [CaSO4]2.H2O + 3H2O

Gypsum Plaster of Paris Its chief property due to which it is widely used is that when it is mixed with one-third of its weight of water it sets with expansion into a hard mass of interlocking crystals of gypsum within 5-15 minutes. The setting is due to hydration of plaster of paris into gypsum. OR (a) B atom in BCl3 has only six electrons in the valence shell and therefore is an electron deficient molecule. It can readily accept a pair of electrons donated by water and hence undergoes hydrolysis to form boric acid and HCl. 2 BCl3 + 3H3O → H3BO3 + 3HCl In contrast, C atom in CCl4 has 8 electrons in its valence shell. It does not have vacant d-orbitals to extend its octet. Therefore, it is an electron precise molecule and hence neither accepts nor donates a pair of electrons. Therefore, it does not accept a pair of electrons from H2O molecule and Hence CCl4 does not undergo hydrolysis in water. 2 (b) Boric acid is not a protic acid because it does not ionize in water to give a proton. But it acts as a Lewis acid by accepting electrons from hydroxide ion : H H + (HO)3B + O (HO)3B– O [B(OH)4]– + H+ H H

Or B(OH)3(aq) + 2H2O → [B(OH)4]–(aq) + H3O+(aq) 3 – Thus, boric acid acts as Lewis acid because it reacts with OH (Lewis base) ions rather than acting as a Bronsted acid.

qqq

Time : 3 Hours Maximum Marks : 70

CHEMISTRY CBSE Sample Question Papers

Solution to Question Paper

7

Self Assessment _______________________________________________ 1. The relation may be written as : E2 – E1 = hn or DE = hn 1 Where h = Planck's constant (6.626 × 10–34 Js) and n = frequency of a photon emitted or absorbed. 2. Fluorine is more electronegative than iodine and, therefore, there is more charge separation in HF than HI. Hence, HF is more polar than HI. 1 3. Change in O.N. : Mn : +4 to 0; Al : 0 to 2; O : –2 to 2. MnO2 is reduced and Al is oxidised. ½+½ 4. The atom or group of atoms present in a molecule which determines its chemical properties is called as functional group. 1 + 5. pH = – log [H ] = – log [10–7] = 7 ( logx2 = 2log x) ( log10 = 1) 1 6. The lowest region of atmosphere in which the human beings along with other organisms live is called troposphere. 1 It extends upto the height of approximately 10 km from sea level. 1 7. (i) CH3—CH—CH2—CH2–CH2—CH3

Br 2- Bromohexane (ii) CH2—CH2—CH2—CH2–CH2—CH3 Br





1- Bromohexane



1+1

OR For preparation of alkanes containing odd number of carbon atoms, a mixture of two alkyl halides has to be used. Since, two alkyl halides can react in three different ways, therefore, a mixture of these alkanes instead of the desired alkane would be formed. e.g. CH3CH2CH2—Br + 2Na + Br—CH2CH2CH3

1-Bromopropane

1-Bromopropane

Dry ether 

CH3CH2CH2CH2CH2CH3 + 2NaBr Hexane



1

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

8

CH3CH2CH2—Br + 2Na + Br—CH2CH2CH2CH3 1-Bromopropane

Dry ether 

CH3CH2CH2CH2CH2CH2CH3 + 2NaBr

1-Bromobutane

Heptane

8.

1  1 n = 1.09677 × 107  2 – 2 2 4

or

 1 1  –1 n = 1.09677 × 107  –  m  16 4 

1

 –1 m 

or

 3  –1 n = 1.09677 × 107  –  m  16 

or

n = 2.0564 × 106 m–1 (Ignoring the negative sign)

2

9. At normal pressure the boiling point of glycerol is 563 K but it decomposes on this temperature. Therefore simple distillation cannot be used. Under a reduced pressure of 12 mm Hg, glycerol can be distilled at 453 K without decomposition. 2 10.



pV =

m Nu2

1

Where

N = Number of molecules of gas present,



V = Volume of the container,



P = Pressure exerted by the gas,



m = mass of each molecule at the temperature T



u = Root mean square velocity

1

11. Carbon dioxide (CO2) is slightly acidic in nature. Hence, KOH which is a strong alkali which dissolves CO2 (evolved from the compound containing carbon) and reacts with it to form K2CO3. Increase in the mass of potassium carbonate will give the weight of CO2 evolved.

2KOH(aq) + CO2(g) → K2CO3 + H2O(l).

12.

N2(g) + 3H2(g) → 2NH3(g)



28 g



34 g

28 g of nitrogen reacts with = 6 g of hydrogen 2 × 103 g of nitrogen react with =

6 × 2 × 103 g of hydrogen 28

= 0.4286 × 103 g of hydrogen

(i)

6g

2

28 g of nitrogen produce = 34 g of NH3 2 × 103 g of nitrogen produce =



34 × 2 × 103 g NH3 28

= 2.4286 × 103 g NH3

1 1

(ii) Yes, hydrogen will remain unreacted.

(iii) (1 × 103 – 0.4286 × 103)g or 0.5714 × 103 g of hydrogen

1

13. (i) BeH2 < CaH2 < TiH2

1

(ii) LiH < NaH < CsH



1

(iii) F–F < H–H < D–D



1

14. (i) BeCl2

Cl : Be : Cl,.



The central atom has only two bond pairs and one lone pair, i.e.,it is of the type AB2. hence, shape is linear. ½

Solution

9

(ii) BCl3 =

The central atom has only 3 bond paris and no lone pair, i.e., it is of the type AB3. Hence, shape is trigonal planar. ½

..

Cl

B

.. Cl



.

Cl .

Cl Cl Sl Cl (iii) SiCl4 = Cl



Bond pairs = 4, lone pairs = 0, it is of the type AB4. Shape = Tetrahedral

..

½

Cl

Si

..Cl

Cl

..

..

Cl

Tetrahedral F F (iv) AsF5 = F As F F



Bond paris = 5, lone pair = 0, i.e., it is of they type AB5 shape = Trigonal bipyramidal. .. F

½

F ..

.. F As F

F ..

Trigonal bipyramidal (v) H2 S = H S H, Bond pairs = 2, lone pairs = 2, i.e., it is of the type AB2 L2. Shape = Bent/V shaped. .. S .. H

½

H Bent

(vi) PH3 = H P H, H Bond pairs = 3, lone pair = 1, i.e., it is of the type AB3L. Shape = Trigonal pyramidal.

½

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

10

.. P

H H

H

Trigonal pyramidal

15. When total volume of container is given, we can use nRT P = V n =

For SO3 :

19.5 = 0.243 mol R = 0.0821 L bar mol–1 K–1 80.1

T = 20 + 273 = 293 K V = 5L 0.243 × 0.821 × 293 Pso3 = 5



= 1.17 bar 1 Similarly for He : n = = 0.250 mol 4 PHe =



0.250 × 0.0821 × 293 5

3

= 1.20 bar HgCl2(aq) + 2KI(aq) → HgI2(s) + KCl(aq)

16. (i)

There is no change in the oxidation number of any element.



Thus, it is not a redox reaction. D → 2Fe(s) + 3CO2(g) Fe2O3(s) + 2CO(g) 

(ii)

Oxidation number of Fe changes from +3 to 0. Thus, Fe2O3 is reduced or it acts as an oxidising agent. Oxidation number of C changes from = +2 to +4. Thus, CO is oxidised or it acts as a reducing agent. It is a redox reaction. 1 PCl2(l) + 2H2O(l) → 3HCl(aq) + H3PO3(aq). 1

(iii)

1

It is not a redox reaction because there is no change in the oxidation number of P and Cl.

17. (i) 4BF3 + 3LiAlH4 → 2B2H6 + 3LiF + 3AlF3 1

Z

X

(ii) B2H6 + 6H2O → 2H3BO3 + 6H2

X



1



1

Y D

→ B2O3 + 3H2O (iii) B2H6 + 3O2 



X OR



Two alloys of aluminium containing magnesium :

(i) Magnalium : Al = 95%, Mg = 5% (ii) Duralumin : Al = 95%, Mg = 0.5%, Mn = 0.5% Cu = 4% Addition of magnesium to aluminium : (a) makes alloy harder and stronger. (b) increases durability. (c) makes the alloy malleable. 18.

2

½ ½

CaSO4  Ca2+(aq) + SO24 – (aq)

If s is the solubility of CaSO4 in moles L–1, then

½

Solution

11 Ksp = [Ca2+] [ SO24 – ] = s2



S =



K SP =

½

9.1 × 10 –6 = 3.02 × 10–3 mol L–1 –3

½

–1

= 3.02 × 10 × 136 gL = 0.411 gL–1 Thus, for dissolving 0.411g water required = 1 L 1 ∴ For dissolving 1 g of CuSO4, water required = L = 2.43 L 0.411

½ 2 ½

19. (a) Compounds having the same molecular formula but different functional groups is called functional isomerism. For example, 1 CH3CH2OH CH3—O—CH3 Ethanol Methoxymethane (b) When a covalent bond adjoining two atoms A and B breaks in such a way that both the electrons of the covalent bond (i.e., shared pair) are taken away by one of the bonded atoms, the mode of bond cleavage is known as heterolytic cleavage. 1

A : B → A+ + : B– or A : B → A:– + B+

(c) Higher alkanes on heating to higher temperature decompose into lower alkanes, alkanes etc., Such a decomposition reaction into smaller fragments by the application of heat is called as pyrolysis or cracking. 1 C6H12 + H2 C6H14

773 K

C4H8 + C2H6 C3H6 + C2H4 + CH4

20. (a) DU = wad (wall is adiabatic) as q = 0 ; DU = q + w or DU = 0 + wad (b) DU = – q (thermally conducting walls) as w = 0 ; q = – q ; DU = – q + O, DU = – q (c) DU = q – w (closed system) as w = – w ; q = + q ∴ DU = – q + w, DU = – q 21. Condensed structure

Bond line structure

1+1+1

Functional group

(a) (CH3)2C—CH2C(CH3)3

Nil O

O

OH 2

(b) CH2(COOH)C(OH)(COOH)2

1

HO

—OH and —COOH

3

HO

O O

H

(c) OHC(CH2)4CHO

—CHO H O



3

22. Smog is a combination of water droplets of fog, smoke and fumes discharged into air from factories, homes and automobiles. These harmful fumes sometimes do not rise high into the atmosphere; instead a layer of warm air is formed over the smog and covers it like a lid. Smog is a major problem of the metropolitan area. Harmful effects of smog : (a) It reduces visibility, thereby affects air traffic and the movement of Vehicular traffic :

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

12

(b) In London in the year 1952, because of smog the city remained covered with the brown coloured photochemical smog for five days. As a result, 4000 people died and many suffered from bronchitis and heart problems. 3 23. (i) Electron gain enthalpy

Electronegativity

(a)

The enthalpy change accompanying the process of addition of electron to a neutral gaseous atom to convert it into a negative ion, is called electron gain enthalpy.

A qualitative measure of ability of an atom in a chemical compound to attract shared electrons to itself is called electron negativity.

(b)

It is the property of isolated atoms.

It is the property of a bonded atom.

(c)

It can be experimentally measured.

It cannot be experimentally measured. It is only a relative number.

2 (ii) The electronegativity of nitrogen will not be 3 in all its compounds. It depends upon the other atoms attached to it. Electronegativity also depends on the state of hybridization and the oxidation state of the element. 1 (ii) Our thoughts get largely affected by our society just like the electron negativity varies depending upon the element to which it is bound. 1 24. (a) Carbon monoxide is very toxic because it combines with haemoglobin in the blood preventing its function as an oxygen carrier. it is particularly dangerous because it is odourless and therefore, is not easily detected. 2 (b) (i) Inert pair effect : It is defined as the tendency of s-electrons to remain together or the reluctance of s-electrons to participate in a reaction, because the energy required to unpair the ns2 electrons is not compensated by energy released in forming the additional bonds. 1 (ii) Allotropy : The phenomenon of existence of a chemical element in two or more forms differing in physical properties but having same chemical nature is known as allotropy. This phenomenon is due to the difference either in the number of atoms in the molecules e.g., in O2 and O3 or arrangement of atoms in the molecules in graphite. 1 (iii) Catenation : The tendency of formation of long open or closed chains by the combination of same atoms is known as catenation, It is maximum in carbon and decreases down the group e.g., C >> Si > Ge = Sn >> Pb. 1 OR (a) Aluminium shows a uniform oxidation of + 3. Like aluminium, Tl also shows + 3 oxidation state in some of its compounds like TiCl3, Tl2O3, etc. Like Al, Tl also forms octahedral complexes : [AlF6]3– and [TlF6]3– Group 1 metals shown an oxidation state of + 1. Like group 1 metals, Tl due to inert pair effect also shows +1 oxidation state in some of its compounds such as Tl2O, TlCl, TlClO4 etc. Like group 1 oxides, Tl2O is strongly basic. 2 (b) Since metal X reacts with NaOH to first give a white ppt. (A) which dissolves in excess of NaOH to give a soluble complex (B), therefore, metal (X) must be Al; ppt (A) must be Al(OH)3 and complex (B) Must be sodium tetrahydroxoaluminate (lll). 2Al + 3NaOH → Al(OH)3 + 3Na+ (X) Aluminium hydroxide (ppt) Al(OH)3 + NaOH → Na+ [Al(OH)4]– A B Sod. Tetra hydroxoaluminate (lll) Since (A), i.e., Al (OH)3 reacts with dil. HCl to give (C), therefore, (C) must be AlCl3. Al(OH)3 + 3HCl → AlCl3 + 3H2O (A) (C) 3 Since (A) on heating gives (D) which is used to extract metal (i.e., electrolysis of Al2O3 gives Al metal), therefore, (D) must be alumina (Al2O3).

Solution

13 D → Al2O3 + 2H2O 2Al(OH)3  (A) (D)

25. (i) Irreversible expansion takes place when external pressure (Pext) remains constant Wirrev = –Pext (V2 – V1) = –Pext DV 1 (ii) Reversible expansion takes place when internal pressure is infinitesimally greater than external pressure (Pint ≈ Pext) at every stage. Thus, external pressure has to be adjusted through out. V2 Wrev = –nRT ln V 1 1 (iii) The expansion is irreversible, further, as Pext = 0, therefore, W = –Pext DV = 0 × (DV) = 0 1 (iv) When gas is compressed, work is done on the gas. For isothermal reversible compression, P2 1 W = + nRT ln P1 (v) For the adiabatic expansion, temperature falls, i.e., T2 < T1 1 W = CV(T2 – T1), i.e., W is negative OR (a) –DG° = RT In K. Thus, if DG° is less than zero i.e., it is negative, then In K will be positive and hence K will be greater. 1 (b) Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction. 2 (c) DG = DH – TDS. At low temperature, TDS is small. Hence, DH dominates. At high temperature, TDS is large, i.e., DS dominates the value of DG. 2 26. Low of constant composition or definite proportion states : "A chemical compound is always found to be made of the same elements combined together in the same fixed proportion by weigh." 1 Examples : CO2 may be prepared in the laboratory as follows : Heat → CaO + CO2 (↑) (i) CaCO3  ½ Heat → CO2 (↑) (ii) C + O2  (iii) CaCO3 + 2HCl → CaCl2 + CO2 (↑) + H2O Heat → Na2CO3 + CO2 (↑) + H2O (iv) 2NaHCO3 

½ ½

½ In all the above examples, CO2 is made up of the same elements, i.e., carbon (e) and oxygen (O) combined together in the same fixed proportion by weight 12 : 32 or 3 : 8 by weight. OR (a)

Molarity =

No. of moles of solute (HCl) , Volume of the solution in litres

Mass of solute 38 No. of moles of HCl = Mol. mass of HCl = 36.5 Volume of the solution : 3.8 g of HCl is present in 100 g of the solution 100 g Mass Volume of 100 g of the acid = = = 84.05 cm3 1.19 g cm –3 Density



Molarity =

38/36.5 = 12.38 M 84.05/1000

(b) We know that for monobasic acid M1V1 = M2V2 12.38 M × V1 = 10M × 1000 mL 10 × 1000 V1 = = 807.7 mL 12.38

1

1 1

1 1

qqq

Time : 3 Hours Maximum Marks : 70

CHEMISTRY CBSE Sample Question Papers

Solution to Question Paper

8

Self Assessment _______________________________________________ 1. R = 8.314 Joule kg–1 K–1. 1 2. The given radiations in increasing order of frequency cosmic rays <X-rays
1 1 1

6. Element

Percentage

Atomic maas

Relative number of atoms

simple ratio of atoms

K

31.84

39

31.84 = 0.816 39

0.816 =1 0.814

Cl

28.98

35.5

28.98 = 0.814 35.5

0.814 =1 0.814

O

39.18

16

39.18 = 2.449 16

2.449 =3 0.814

Empirical formula = KClO3 2 7. Since x-axis, is being considered as the internuclear axis, and y-axis is perpendicular to x-axis, therefore 2py and 2py atomic orbitals can overlap only sidewise. 2 Snice overlapping of p-orbitals sidewise results in the formation of pi (p) bond, therefore 2py and 2py will not form a sigma bond. 8. As we move down the group atomic size increases. 1 , hence it is expected that on moving down the group I.E. should We know I.E. ∝ Atomic size decrease. 1 The deviation is due to weak screening or shielding effect of d-orbital due to which atomic size decreases and Ionisation energy increases. 1

Solution

15

OR (a) Gain of electron : When a neutral atom gains one electron to form an anion, its radius increases. ½ The reason being that the number of electrons in the anion increases while its nuclear charge remains the same as the parent atom. ½ (b) Loss of electrons : When a neutral atom loses one electron to form a cation, its atomic radius decreases. ½ Reason being that the number of electrons in the cation decreases while its nuclear charge remains the same as the parent atom. ½ 9. 2 SO2(g) + O2(g)  2SO3(g) [SO3 ]2 (1.9)2 KC = = = 12.229 mol/L 2 [SO2 ]2 [O2 ] (0.60)2 (0.82) 10. The process of producing ‘syn-gas’ from coal is called ‘coal gasification’. The mixture of CO and H2 is called syngas. 1270K C(s) + H2O(g) → CO(g) + H2(g) 1 The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst. 673K

→ CO2(g) + H2(g) CO(g) + H2O(g)  Catalyst This reaction is called as water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite. 1 23 11. (i) 1 mole of Argon (Ar) contain 6.023 × 10 atoms 52 moles of Ar contains 6.023 × 1023 × 52 atoms = 3.13 × 1025 atoms 1 (ii) 4 u of Helium (He) = 1 atom of He 1 52 u of He = × 52 = 13 atoms 1 4 (iii)

4 g of He contains = 6.023 × 1023 atoms 6.023 × 1023 × 52 52 g of He contains = = 78.299 × 1023 atoms 4

12. (i) Two important uses of caustic soda : (a) It is used in paper, soap, rayon and textile industries. (b) It is used in refining of petroleum. (ii) Two important uses of sodium carbonate : (a) It is used in manufacturing of glass, borax and caustic soda. (b) It is used in water softening, laundering and cleaning. (iii) Two important uses of limestone : (a) It is used as building material and road aggregate. (b) It is used in the manufacturing of quiklime. 13 D O2 ( g )  13. (i) C4H10(g) + → 4CO2(g) + 5H2O(g) 2 Butane 15 D O2 ( g )  (ii) C5H10(g) + → 5CO2(g) + 5H2O(g) 2 Pentene 17 D O2 ( g )  (iii) C6H10(g) = → 6CO2(g) = 5H2O(g) 2 Hexyne

1

1

1

1 1

1

1

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

16 14.

Cl H2S = S

SiCl4 =

O

Si

2–

CO3 =

H H



2–

C

Cl Cl Cl O O 2– 2 – 15. The average O.N. of S in S 2 O3 is + 2 while in S 4 O6 it is + 2.5. The O.N. of S in

1+1+1 is +6. Since SO2– 4

Br2 is a stronger oxidation agent than I2 , it oxidizes S of S 2 O23 – to a higher oxidation state of +6 and

2– hence forms SO2– 4 ion. I2, however, being a weaker oxidizing agent oxidizes S of S 2 O3 ion to a

lower oxidation of +2.5 in S 4 O26 – ion. It is because of this reason that thiosulphate reacts differently with Br2 and I2. 16. T1 = 27 + 273 = 300 K T2 = 477 + 273 = 750 K By Charle’s Law, V1 V = 2 T1 T2 V1 300 2 = or V2 = 2.5 V = or V2 750 5

Volume of air expelled = 2.5 V–V = 1.5 V 1.5 3 Fraction of air expelled = = = 0·6. 2.5 5

3 1

1

1

17. (a) 5-chloro, 3-ethyl pent-2en-1-ol. 1 (b) O2N—CH2—CHO– is expected to be more stable than CH3—CH2O–. In O2N—CH2—CH2O–, due to the –I effect exerted by –NO2 group, there occurs a dispersal of the negative charge, wheras in CH3—CH2O– due to +I effect exerted by CH3, there occurs an intensification of the negative charge. Dispersal of the charge leads to more stability of the ion. 2 H H O 18. (i) CH3CH = CH2 + O3

CH3—CH O

CH2 O

95%H 2 SO 4 (ii) CH3CH2OH CH2=CH2 + H2O 433 – 443K 1 Ethanol Ethene

Zn/H2O – ZnO

CH3—C = O + H—C = O Ethanol Methanol

1

OH

(iii)

+ Zn

+ ZnO

1

Benzene 19. Pesticides : (i) They are basically synthetic toxic chemicals with ecological repercussion. (ii) Earlier DDT was used to control insect borne diseases in crops later, as insect resistance of DDT was increased other organic toxins such as Aldrin and Dieldrin were used as pesticides. 1½ Herbicides : (i) These are sodium chlorate and sodium arsenite. (ii) Most are toxic to mammals but are not as persistent as organo chlorides. 1½ (iii) These chemicals decompose in few months. (iv) Some herbicides cause birth defects.

Solution

17

20. (a) Dipole moment may be defined as a product of charge and distance between two opposite charges. 1 r –q +q  = qr

(b) Because of electronegativity difference between B and Cl. B-Cl has a dipole moment. However, the overall dipole moment of a compound depends on its geometry. BCl3 is planar molecule in which the three B-Cl bonds are situated at an angle or 120°. 2 Therefore, the resultant of two B-Cl bond gets cancelled by equal and opposite dipole moment of the third B-Cl bond. Hence, B-Cl3 molecule has in zero dipole moment. Cl Cl

B

120° Cl

Resultant

=0

OR The important features of Maxwell's distribution of velocities are : (i) The fraction of molecules with very low or high velocities is very small. 1 (ii) The fraction of molecules processing higher velocities goes increasing till it reaches the peak and further it starts decreasing. 1 (iii) The maximum fraction of molecules possesses a velocity, corresponding to the peak in the curve. This velocity is referred to as most probable velocity. 1 21. (a) For n, the values of l are 0, 1, ......... (n – 1) ∴ For n = 5 ; the values of l are 0, 1, 2, 3, 4 For l = 4, g subshell can exist. Hence, lowest value of n = 5 1 (b) When n = 3, l = 2 (Since it is in 3d subshell) ml = – 2, – 1, 0, + 1, + 2 1 (c) Pauli Exclusion Principle 1 10 10 × 68 N 22. Normality of 10 volumes of H2O2 = = 1 5.6 22.4 × 17 Applying normality equation N1V1 = N2V2 (H2O2) (KMnO4) 10 × V1 = 2 × 200 1 5.6 V1 = 224 cm3 1 23. (a) The following assumptions are made in the derivation of the gas equation : (i) there is no force of attraction between the molecules of the gas. (ii) Actual volume of the gas mdecules is negligible compared to the volume of the gas. These conditions are not always satisfied, hence, PV is not alwyas equal to KT. 2 (b) Equation for a real gas for n mole is  an2   p + 2  (V – nb) = nRT where, a and b are van der Waal's constants. 1½ V  

(c) Jyoti is helping, caring and intellectually smart. ½ 24. (a) Saytzeff rule : Whenever two alkanes are theoretically possible during a dehydro halogenation reaction, it is always the more highly substituted alkene which predominates. Br CH3—CH2—CH—CH3 2-Bromobutane





alc. KOH

2

CH3—CH = CH—CH3 + CH3—CH2—CH = CH2 But-2-ene (80%) (More highly substituted alkane : more stable)

But-1-ene (20%) (Less highly substituted alkane : less stable)



OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

18 CH3

CH3

(b)

CH2 = C—CH2—CH3 2-Methyl but-1-ene

CH3

CH3—C = CH—CH3

CH3—CH—CH = CH2

2-Methyl but-2-ene

2-Methyl but-1-ene

3

OR (a) (i) Chlorobenzene rel="nofollow"> p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. (ii) Toluene > Para—CH3—C6H4—NO2 > Para—O2N—C6H4—NO2 CH3 NO2

NO2

1+1

NO2

(b) Nitration of benzene is an electrophilic substitution reaction NO2 + HNO3



H2SO4

+ H2O

NO2+ (nitronium) ion is attacking electrophile. In toluene, one H of the ring is substituted by – CH3 group which is an electron releasing group and makes the electrons more readily available to electrophile, whereas in m-nitrobenzene, the Two –NO2 group which are electron withdrawing in nature makes the availability of electrons of ring difficult to the electrophile. 3



That is why nitration of toluene easier than others.

25. (a) Hydrolysis of alkyl trichlorosilanes gives cross-linked silicones. R R Cl—Si—Cl + 3H2O

– 3HCl

Cl R

OH R



O.....

.....O—Si—O—Si—O—Si—R

R n OH—Si—OH

1

HO—Si—OH

Polymerization – (n – 1)H2O

O

O

O

1

.....O—Si—O—Si—O—Si—R

OH



R

R

O.....

(b) CO2 is produced during combustion reactions. It is released into the atmosphere. However, if the concentration of CO2 in the atmosphere increases beyond a certain limit due to excessive combustion, some CO2 will always remain unutilized, The excess CO2 absorbs heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is redirected back towards earth's surface which heat up the atmosphere. As a result, temperature of the earth and other bodies on the earth increases. This is called as green house effect. But when this greenhouse effect causes a large rise in temperature of the Earth, it is called global warming. It has serious consequences. 3 OR 2

(a) BF3 is a planar molecule in which B is sp hybridized.It has an empty 2p-orbital. F-atom has three lone pairs of electrons in the 2p-orbitals. Because of similar sizes, pp – pp back bonding occurs in which a lone pair of electrons is transferred from F to B. As a result of this back bonding, B–F bond acquires some double bond character. In contrast, in [BF4]– ion, B is sp hybridized and hence does not have an empty p-orbital available to accept the electons donated by the F atom.

Solution

19

–

Consequently, in [BF4] , B–F is a purely single bond. Since double bonds are shorter than single bonds, therefore, the B–F bond length in BF3 is shorter (130 pm) than B–F bond length (143 pm) in [BF4]–. 2 F B

F

F

(b) Structure of diborane : In diborane, the four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B–H bonds are regular bonds while the two bridge (B—H—B) bonds are different and known as banana bonds (3-centre-2-electron bridge bonds). H H

B B

120°

97° B

H

119 pm

134 pm H



H

1½ Structure of boric acid : Boric acid has a layer structure in which H3BO3 units are joined by hydrogen bonds. Carbon H bond B

H Hydrogen bond

O

O

H

H

O

O

H

B

1½ O

O H

H

B O O H

H

H

O

B

H O

O H

26. (i) The free energy change of a reaction is given DG = DH – TDS For a reaction to be spontaneous, DG should be –ve 1 Such reactions are therefore, called entropy driven. This can happen in either of the following Two ways : (a) DS should be so large that even it T is low, TDS should be greater than DH. 1 (b) If DS is small, T should be so large that TDS > DH. 1 (ii) In an ideal gas, there are no intermolecular forces of attraction. Hence, no energy is required to overcome these forces. Moreover, when a gas expands against vacuum, work done is zero (because Pext = 0). Hence, internal energy of the system does not change, i.e., there is no absorption or evolution of heat. 2 OR (i) (a) Exothermic reactions are generally thermodynamically spontaneous because even if it is accompanied by decrease randomness (e.g., in the condensation of a gas or solidification of a liquid), the heat released is absorbed by the surroundings so that the entropy of the surroundings increases to such an extent that DStotal is positive. 1

20

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

(b) The molecules in the vapour state have greater freedom of movement and hence greater randomness than those in the liquid state. Hence, entropy increases in going from liquid to vapour state. 2 (ii) A substance has a perfectly ordered arrangement of its constituent particles only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formations is the heat change involved in the formation of one mole of the substance from its element. An element formed from itself means no heat change, i.e., DH0 = 0. 2

qqq

Time : 3 Hours Maximum Marks : 70

CHEMISTRY CBSE Sample Question Papers

Solution to Question Paper

9

Self Assessment _______________________________________________

1. Cr+ ⇒1s2, 2s2, 2p6, 3s2, 3p6, 3d5. 1 2. Graphite. 1 3. This is due to inert pair effect. The s electrons do not participate in bonding as we move downward in the group. 1 5 4. O.N. of S in Na2 S2 O3 is +2 and in Na2S4O6 is +2.5/ + 1 2 5. Average atomic mass is the average of atomic masses of all the isotopes of an element. 200 g of H 2 6. Moles of H2 = = 100 moles of H2 2 g mol –1 From the balanced equation it is clear that 1 mol of NH3 is produced when



3 mole of H2 is consumed. 2

3 × 100 mol of H2 consumed 2 = 150 mole of NH3 17g Mass of NH3 produced = 150 mole of NH3 × = 1132.2 g mol NH 3 =

1

2

7. Reactions affected will be those in which (np = nr) gaseous. By applying Le Chatelier's principle, we can predict the direction. Increase of pressure will shift the equilibrium to the side in which the number of mole is less. (a) np = 2, nr = 1, i.e., np > nr, reaction will go in backward direction. 1 (b) np = 3, nr = 3, i.e., np = nr, reaction will not be affected by pressure. 1 60g 8. Moles of Al, n = = 2.22 mol 27g mol –1 Molar heat capacity Cm = 24.0 J mol–1 K–1 DT = (55 – 35)°C = 20 °C = 20 K Now Q = Cm × n × DT 2 = 24.0 J mol–1 K–1 × 2.22 mol × 20 K = 1065.6 J = 1.067 kJ 9. Hydrogen is the first element in the periodic table. It has 1s1 configuration. ½ It shows resemblances with alkali metals of Group-I as it has ns1 configuration. It has tendency like them to form H+ ion. ½

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

22

Like Halogens group 17, it has tendency to gain one electron and acquire inert configuration. As it shows resemblance with group I and group 17, its position is anomalous. 10. CH3OCH2CH2CH3. Its IUPAC name is 1-Methoxy propane. CH3

½ ½

— CH3O—CH—CH3 Its IUPAC name is 2-Methoxypropane. 1+1 OR Electrophiles are electron loving reagents. They are electron deficient and so, act as Lewis acid. e.g. : H+, H3O+. Nucleophiles are nucleus loving reagents. They are electron rich and so, act as Lewis base. e.g., : Cl–, Br–. 2

(ii) 235 (iii) 94 Be 3 92 U 12. Anhydrous barium peroxide is not used because barium sulphate forms a protective layer around unreacted barium peroxide which prevents its further participation in the chemical reaction. BaCO3 + H2SO4 → BaSO4 + H2O2 1 – – (a) Cl2 + H2O2 + 2OH → 2Cl + 2H2O + O2 1 (b) Ag2O + H2O2 → 2Ag + H2O + O2 1 13. (a) H+2 = 1, O2– = 17 11. (i)

35 17 Cl

(b) (i) 3py (ii) 1s (c) Heisenberg uncertainty principle : "It is impossible to measure simultaneously both the position and momentum of a microscopic particle with absolute certainity. 1+1+1 14. (a) Given : Mass of iron, m = 10 g Dt = (500 – 25)°C = 475°C Specific heat capacity of iron, C = 0.45 J(°C)–1 g–1 Energy needed = m × c × Dt = 10g × 0.45 J(°C–1) g–1 × 475°C = 2.1 × 103J 1½ (b) Given : Specific heat of gold = 0.13J J(°C–1) g–1 × 475°C Energy needed = m × c × Dt 2.1 × 103 J = m × 0.13 J(°C–1)g–1 × 475°C m =

2.1 × 103

1½ 0.13(°C –1 )g –1 × 475°C = 34.008 g 15. (i) Lithium is the strongest reducing agent but least reactive amongst all the alkali metals. It produces monoxide only as it does not react with excess of air. 1 1 2Li + O2 → Li2O 2 (ii) Sodium carbonate in aqueous solution gets hydrolysed to give alkaline HC + O3– and OH– ions. C O23 – + H2O → HC + O3– + OH– 1 (iii) Sodium being powerful reducing agent cannot be isolated by ordinary reducing agent like C Or CO. Hence, it is prepared by electrolytic reducing method. 1 16. (a) An alkali metal – I (b) An alkaline earth mental – II (c) a non-metal – III 1+1+1 17. (i) Two postulates of kinetic theory of gases are not valid at all conditions of temperature and pressure. These are : (a) Gases consist of large number of molecules which are so small and so a part that the actual volume of the molecules is negligible. This assumption is not valid at low temperature and high pressure. 1

Solution

23

(b) There is no force of attraction between the molecules. This assumption is not valid at low temperature and high pressure. 1 That is why gases deviate from ideal behaviour. (ii) (a) Density decreases ½ (b) Viscosity decreases ½

18. QC =

[NH 3 ]2

[N 2 ][H 2 ]3

=

(0.5)2 = 0.0104 (3.0)(2.0)3

1

As Qc ≠ Kc, reaction is not in equilibrium. 1 As Qc < Kc, reaction will proceed in forward direction 1 19. (i) Laboratory method. diethyl ether (a) 4BF3 + 3Li AlH4  → 2B2H6 + 3LiF + 3AlF3 (b) 2NaBH4 + I2 → B2H6 + 2NaI + H2 (ii) Industrial method :

1

450 K 2BF3 + 6NaH → B2H6 + 6NaH 20. Given : T = 2.0 × 10–10 s 1 1 Frequency = = = 5 × 10–11 s–1 T 2.0 × 1010

1



1 1

We have the relation c = υλ l =

Wage length,

Substituting the values of c and v we have 3 × 108 ms –1 = = 6.0 × 10–2 m 5 × 109 s –1 1 v = Wave number, l v =



1 = 16.66 m–1 6.0 × 10 –2 m

3

21. The upper stratosphere consists of considerable amount of ozone (O3) which protects us from the harmful UV radiations (λ = 255 mm) coming from the Sun. The main reason for depletion is CFCs. When released in the atmosphere, CFCs mix with the normal atmospheric gases and eventually reach the stratosphere. In stratosphere, they get broken down by powerful UV radiations, releasing chlorine free radical. hn

.

CF2Cl2(g) → Cl (g) + CF2Cl(g) The chlorine free radical (Cl) then reacts with stratospheric ozone to form chlorine monoxide radicals (ClO) and molecular O2.

.

.

Cl + O3(g) → Cl O(g) + O2(g) Reaction of ClO with atomic oxygen

.

.

Cl O + O(g) → Cl (g) + O2(g) The chlorine radicals are continuously regenerated and cause the break down of ozone. Thus, CFCs are transposing agents for continuously generating chlorine radicals into the stratosphere and damaging ozone layer thus causing its depletion. 3 22. The necessary condition for any system to be aromatic are : (i) The molecule should be planar. (ii) It should be cyclic with alternate single and double bonds and the cyclic p cloud should sorround all the carbon atoms of the ring.

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

24

(iii) It should contain (4n + 2)p electrons where n = 0, 1, 2, 3, ..........etc. A molecule which does not satisfy any one or more of the above conditions is said to be nonaromatic. 3 OR S. No.

Inductive effect

Resonance effect

1.

Use σ-electrons.

Use p-electrons or lone pair of electrons.

2.

Move upto 3-carbon atoms

All along the length of the conjugated system.

3.

Slightly displaced electrons.

Complete transfer of electrons take place.

3 –1 –1 23. (i) (a) F(–328 kJ mol ) has more negative electron gain enthalpy than O (–141 kJ mol ). (b) Cl (–349 kJ mol–1) has more negative electron gain enthalpy than F (–328 kJ mol–1). 1 – (ii) When an electron is added to oxygen atom to form O ion energy is released. But when another electron is added to O– to form O2– energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged O– and the new electron being added. The second electron gain enthaply of O is positive. Thus, first electron gain enthalpy : O(g) + e–(g) → O–(g); DHeg = 141 kJ mol–1 1 (energy is released) Second electron gain enthalpy : O–(g) + e–(g) → O2–(g); DHeg = +780 kJ mol–1 1 (energy is released) (iii) Sameer is freindly and intellectually smart. 1 24. (a) While preparing higher alkanes by Wurtz's reaction only these alkanes are preferably prepared which contain even number of carbon atoms, e.g., ethane, butane, hexane containing respectively 2, 4, 6 carbon atoms. Dry ether

→ CH3—CH3 + 2NaBr CH3Br + 2Na + BrCH3  Ethane 1½





Dry ether

→ C4H10 + 2NaBr C2H5Br + 2Na + BrC2H5  Butane



When we prepare propane or pentane, there are chances of the formation of side products, e.g., by starting with 1-bromopropane and 1-bromobutane besides the main product is heptane hexane and octane are formed as side products. CH3CH2CH2Br + BrCH2CH2CH2CH3

Dry ether

C7H16

+ C6H14 + C8H18

Heptane Side product

(b) Sodium acetate is used, for the preparation of methane. CaO → Na2CO3 + CH4 CH3COONa + NaOH 



1½

(c) Since alkene A on ozonolysis gives a mixture of ethanol and 3-pentanone, its structure is O— O

C2H5

—

CH3 CH C—CH2—CH3

Zn/H2O

CH3CHO ethanal +

O O

—

3 - ethylpent - 2 - ene

— —

—

—

CH3—CH = C—CH2—CH3 + O3

CH3—CH2—C—CH2—CH3



pentan - 3 - one



2

Solution

25

OR (a) Since acetylene is very weak acid Ka = 10–25 therefore, it does not react with NaOH or KOH. 1 (b) In alkane, butane and 2-methyl propane are isomeric while in alkene pent-2-ene and 2-methylbut 1-ene are isomeric. 1 In phenol, benzene ring has alternate single and double bonds while cyclohexanol is an alicyclic compound. OH OH 1 Cyclohexanol

Phenol

(c) Hex-2-ene is CH3—CH2—CH2—CH=CH—CH3. Its cis and trans forms are given below : H3C CH3CH2CH3 H3C H C

C

C

H

H (cis-form)

C

H

CH2CH2CH3 (trans-form)

The cis isomer will have higher melting point due to more polar nature leading to stronger intermolecular dipole-dipole interactions thus requiring more heat energy to separate them whereas trans from being non-polar (or weakly polar) have weak induced dipole interactions and so have lower melting point. 2 25. (i) Reaction for the enthalpy of formation of NH3(g) is : 1 3 N2(g) + H2(g) → NH3(g) 2 2

∴ DfH° = – 92.4/2 = – 46.2 kJ mol–1 (ii) The solubility of Sr(OH)2 at 298 K = 19·23 g L– 1 Sr(OH)2 →Sr2 + + 2OH– Molar concentration of dissolved 19.23 Sr(OH)2 = = 0·1581 M 121.6 ∴ Sr(OH)2 Sr2+ + 2OH–

∴ [Sr2+] = [Sr(OH)2]diss. = 0·1581 M [OH–] = 2 × 0·1581 = 0.3162 M Kw 1.0 × 10 –14 ∴ [H+] = = 0.3162 [OH – ] –14 = 3·1625 × 10 M ∴ pH = – log [H+] = – log 3·1625 × 10–14 = – (– 14) – log 3·1625 = 14 – log 3·1625 = 13·50. OR (i) For an isolated system with DU = 0, the spontaneous change will occur if DS > 0. [Salt] (ii) pH = pKa + log [Acid]

4 = – log (1·8 × 10– 4) + log

[Formate] [Formic Acid]

2

1

1

1 1 1

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

26

4 = 3·74 + log

log



[Formate] [Formic Acid]

[Formate] = 4 – 3·74 = 0·26 [Formic Acid] Formate = 1·8 Formic Acid



1

The buffer capacity of this solution would be maximum near the pKa of the acid. \ For maximum buffer capacity pH = pKa = – log Ka = log 1.8 × 10–4 = 3.74. 26. (i) CH2=C=O ; sp2, sp2 (ii) (CH)3—CH = CH2 : sp3, sp2, sp2 (Iii) (CH3)2 C = O : sp3, sp3, sp2 (iv) CH2=CHCN : sp2, sp2, sp (v) C6H6 : sp2 OR KMnO2

5

CH3—CH—CH—CH3 + MnO2 + KOH

—

298–303 K

—

(a) CH3CH = CHCH3 + H2O + [O]

1

OH OH Butane–1, 2–diol

CH2OH

—

KMnO2

(b) CH2 = CH2 + H2O + [O]

298–303 K



CH2OH

—

CH3

(c)

+ CH3Cl

Al2Cl6



+ HCl Toluene

(d)

—

CH2CH3

—

Br + 2Na + H3CCH2Br

dry ether

ethyl benzene

—

CH3

+ H3C—CH—CH3

Grignard reagent

—

(e)

CH

Br

—

—

MgBr

CH3

—



ether

Isopropyl benzene

5

qqq

Time : 3 Hours Maximum Marks : 70

CHEMISTRY CBSE Sample Question Papers

10

Solution to Question Paper

Self Assessment _______________________________________________ 1. Newton metre–1 (Nm–1).



1

2. No change occurs on dilution.

1

2

2

6

2

6

3

3. 1s , 2s , 2p , 3s , 3p , 3d



1

3



4. (i) 5.608

(ii) 1.790 × 10



5. Henry's law : The mole fraction of gas in the solution is propotional to the partial pressure of the gas over the solution. 1

1

6. Anaemia is caused by lead contamination in drinking water while methemoglobinaemia is caused by nitrate contamination in drinking water. 2

7. (i) Heisenberg's uncertainty principle states that it is not possible to measure simultaneously both the position and momentum (or velocity) of a microscopic particle, with absolute accuracy. 1

(ii) 4f.

1

8. No the structures (1) and (2) cannot be taken as canonical forms of H3PO3 because the positions of atoms have been changed. Two structures to represent resonance should have same positions of atoms. 1 9. In MnO4– , Mn is in the highest oxidation state i.e., + 7. Therefore, it does not undergo disproportionation. MnO2–4 undergoes disproportionation as under : 2 3MnO2–4 + 4H+ → 2 MnO4 + MnO2 + 2H2O



10. (i) Benzene to p-nitrobromobenzene Br + Br2



Fe

Br HNO3/H2SO4

1 Bromobenzene

(ii) Benzene to m-nitrobromobezene

NO2 p-nitrobromobenzene

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

28

NO2 Br2/Fe

H2SO4

+ HNO3

NO2

1 Br





m-nitrobromobenzene

Nitrobenzene

OR O

O

+ CH3—N

(i)

–

+

CH3—N O

1

–

O



NH2

NH2

+ NH 2

+ NH 2 –

(ii)

+ NH 2 –

1

– 11. When elements are arranged from top to bottom in increasing order of their standard reduction electrode potential, a series called electrochemical series is obtained. For example, reduction potential of Zn : Zn2+/Zn = – 0.76 V and reduction potential of Cu : Cu2+/Cu = + 0.34 V. 1 The characteristics of electrochemical series are : (i) Tendency for reduction to occur increases. (ii) Tendency to gain electrons increases. (iii) Power as oxidising agent increases. (iv) A metal placed above in electrochemical series can displace the metal kept below from its salt solution. 2 12. Mass of dioxygen = 70.6 g 70.6 No. of moles of dioxygen = = 2.21 32



Mass of neon = 167.5 g 167.5 No. of moles of neon = = 8.375 20 Total no. of moles = 2.21 + 8.375 = 10.585 2.21 Mole fraction of dioxygen = = 0.208 10.585 Mole fraction of neon =

8.375 = 0.791 10.585

Partial pressure of dioxygen = 0.208 × 25 = 5.82 Partial pressure of neon = 0.791 × 25 = 19.775 13. (i) (a) By the action of steam on coke at high temperature

3

1270 k C(s) + H2O(g) → CO(g) + H2(g)

Here, C(s) is a reducing agent (b) By the action of water on sodium : 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Here, Na(s) is a reducing agent. (ii) When metal reacts with acid, it also evoles dihydrogen. Cu + H2SO4 → CuSO4 + H2

1

1

Solution

29

(iii) Reaction of steam on hydrocarbons at high temperature.

1270 k → CO(g) + 3H2(g) (a) CH4(g)+ H2O(g)  catalyst



1270 k → 3CO(g) + 7H2(g) (b) C3H8 + 3H2O(g)  catalyst

14. (i)

KC =

(ii)

K' =

1

[PCl 3 ( g )][Cl 2 (g )] [PCl 5 (g)]

1

1 1 = = 120.48 KC 8.3 × 10 –3

1

(iii) No effect. 1 15. The dipole moment of NH3 (1.46 D) is much higher than that of NF3. This is due to different directions of the bond moments of N—H and N—F bonds. In the first case, N is more electronegative but in the second case F is more electronegative. This in NH3 the dipole moments of N—H bonds are in the same direction as that of the lone pair but in NH3, the dipole moments of N—F oppose that of the lone pair (as shown in the diagram below). 2

Resultant 3N—H bond

N H



H

H

Resultant 3N—F bond

N F

F

F

Molecular orbital formula of O22 – = σ1s2, σ*1s2, σ2s2, σ*2s2 σ2 p z2

1 (p2px)2 = (p2py)2, (p*2px)2 = (p*2py)2 16. (a) Halogens act as good oxidising agents on account of their high value of reduction potential, they have more tendency to attract electrons. 1 (b) Electron gain enthalpy of noble gases is almost zero on account of their fully filled configuration, they have least tendency to attract electrons. 1 (c) Na and Mg+ have same number of electrons but removal of electron from Mg+ requires more energy because Mg+ have small size in comparison to Na, hence more energy is required. 1 1 17. (a) Small size ∝ Hydration energy ∝ mobility. 1 ionic (b) Lithium on account of its small size, shows anamolous behaviour. 1 (c) BaO is soluble but BaSO4 is insoluble on account of lattice energy parameter. 1 18. (i) The given data is in accordance with the law of multiple proportions, which states : When two elements combine to from two or more than two compounds, the mass of one of the two elements which combines with the fixed mass of the other, bears a simple ratio to one another. In the said question, if we fix the weight of dinitrogen at 14 g, then the weights of dioxygen which combines with the fixed weight (= 14 g) of dinitrogen will be 32, 64, 32, 8 which are in the simple whole number ratio of 2 : 4 : 2 : 5. ½ 6 15 (ii) (a) 1 kg = 10 mm = 10 pm ½ (b) 1 mg = 10–6 kg = 106 ng. ½

(c) 1 mL = 1 mL +

L = 10– 3 L 1000 ml

½

= 10–3 dm3. 19. (i) SI unit would be D Pa(m 3 )2 (K)2 = mol

1½

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

30

Nm–2 (m3 )2 (K)2 or Nm4K2 mol–1 mol

or 1 L atm ×

(ii)

10 –3 m 3 101.325Pa × = 101.325 Pa m3 1L atm

We know that Pa =

N  N  , therefore we can write 101.325 Pa m3 = 101.325  2  m3 m  m2

= 101.325 Nm = 101.325 J 1 20. Formal charge = V – S – L 2

1½

1

Where V → Valence electrons on free atom L → Total no of unshared valence electrons S → Total no of shared electrons O 2 O1



O3



½

(1) O1 atom Formal charge = 6 – 1 (4) – 4 = 0 ½ 2 (2) O2 atom Formal charge = 6 – 1 (6) – 2 = + 1 ½ 2 (3) O3 atom Formal charge = 6 – 1 (2) – 6 = – 1 ½ 2 21. Classical smog

S. No.

Photochemical smog

1.

This type of smog was first observed in London in 1952.

This type of smog was first observed in LosAngels in 1950.

2.

Occurs in cool, humid climate.

Occurs in warm, dry and sunny climate.

3.

It is a mixture of smoke, fog and sulphur dioxide.

It is formed due to photochemical reaction taking place when air contains NO2 and hydrocarbons.





3

22. (a) (CH3)3CCH2CH(CH3)2



1

—

OH



—

—

—O —C—C—C— O— OH OH

—

OH

—O C—

O OH

(b) HOOC.CH2C(OH)COOH.CH2COOH

—

OH

— —

—

Functional groups : = COOH, OH

— —



O

1

Solution

31

O

(c) OHC(CH2)4CHO Functional group : CHO.

1 O

OR Chromatography is the technique of separating the components of a mixture in which seperation is achieved by the differential movement of individual components through a stationary phase under the influence of a mobile phase. The stationary phase can be either a solid or tightly bond liquid on a solid support while the mobile phase can be either liquid or a gas. There are majorly two types of adsorption chromatography : (i) Column chromatography (ii) Thin layer chromatography Solvent front y x Starting line

Chromatogram 3 23. (i) d-block elements have close packed structure in which d-block elements are linked with powerful metallic bonds. As a consequence, d-block elements are hard solid with high density. 2 (ii) Transition metals have higher ionisation enthalpies in comparison to alkaline earth metals. Therefore, transition metals are less electropositive than alkaline earth metals. 1 (iii) Raman's father has scientific attitude and he knows better that how curosity can be satisfied by knowledge. 1 + 24. (i) (a) CH3 is most stable. The replacement of H by Br increases positive charge on carbon atom because Br is more electronegative than H and consequently the species becomes less stable. 1 – (b) CCl3 is most stable because Cl is more electronegative than hydrogen. On replacing hydrogen by chlorine, negative charge on C is reduced and the species becomes stable. 1 (ii) S. No.

Inductive effect

Resonance effect

1.

This involves displacement of s - The resonance involves displacement of electrons. p-electrons or lone pair electrons.

2.

It operates in saturated compounds.

3.

This effect moves upto three carbon This effect moves all along the length of atoms and becomes negligible from the conjugated system. the fourth carbon atom onwards.

It operates only conjugated systems.

in

unsaturated

3 OR Benzenoid compound, H2C CH2 Nonbenzenoid compound (a) 1+1 (b) The compounds having same molecular formula but different arrangement of carbon chain (Skeleton) with the molecule are called chain isomers or skeletal isomers and the phenomenon is termed as chain isomerism. e.g., Butane, 2 methyl propane.

OSWAAL CBSE Sample Question Papers, Chemistry, Class-XI

32

CH3—CH2—CH2—CH3 Butane



CH3—CH—CH3 CH3 2-methyl propane

The compounds which have the same molecular formula but differ in the position of the function groups, carbon-carbon multiple bond or substituent group are called position isomers and this phenomenon is termed as position isomerism. e.g.: Propan-1-ol, Propon-2-ol. 3 OH CH3—CH2—CH2OH

CH3—CH—CH3

Propan-1-ol

Propan-2-ol

The compounds having same molecular formula but different functional groups in the molecule are called functional group isomers and this phenomenon is termed as functional group isomerism. e.g. : Ethanol, Methoxymethane. CH3—CH2OH CH3—O—CH3 Ethanol Methoxymethane – 25. (i) F is a powerful ligand in comparison to Cl–, therefore, it can easily form co-ordination compound with SiF4. 1 – 2– SiF4 + 2F → [SiF6] (ii) Since diamond exists as a three dimensional network solid, it is the hardest covalent substance known with high density and melting point. 1

(iii) Boric acid is a weak monobasic acid (Ka = 1.0 × 10–9). It may be noted that boric acid does not act as a protonic acid (i.e., proton donor) but behaves as a Lewis acid by accepting a pair of electrons from OH– ion. 1 – + B(OH)3 + 2H—O—H → [B(OH)4] + H3O (iv) Boron tri-fluoride is an electron deficient species, therefore, it behaves like a Lewis acid. 1 (v) Carbon dioxide is a linear molecule while silicon dioxide is covalent three dimensional molecule therefore SiO2 is a solid while CO2 is a gas. 1 OR (i) Lattice energy : It is energy required to break one mole of the solid salt into its ions.

1

(ii) Since boric acid release only proton in aqueous solution, therefore it is monobasic in nature. 1 H—OH + B(OH)3 → [B(OH)4]– + H+ + p = 9.25 (iii) In heavier elements electrons from s-orbitals do not take part in bond formation. Such a pair of electrons is called inert pair and this effect is called as inert pair effect. 1 (iv) Since Pb4+ is less stable due to inert pair effect, therefore PbCl4 is a good oxidising agent. Pb (v)

4+

–

+ 2e → Pb H

1

2+

B H—N

N—H Borazine, B3N3H6

H—B

B—H N



H

7 26. We are given C2H6(g) + O2(g) → 2CO2(g) + 3H2O(l), DH° = – 372 kcal 2 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l), DH° = – 530 kcal C(s) → C(g), DH° = 172 kcal



1 ...(i) ...(ii) ...(iii)

Solution

33

°

H2(g) → 2H(g), DH = 104 kcal 1 H2(g) + O2(g) → H2O(l), DH° = –68 kcal 2

...(iv) 1 ...(v)

C(g) + O2(g) → CO2(g), DH° = – 94 kcal ...(vi) –1 –1 Suppose the bond energy of C—C bond = x kcal mole and the of C—H bond = y kcal mole H H C2H6(g), i.e.,



2C(g) + 6H(g),

H = x + 6y

C3H8(g), i.e.,



(vii) 1

H H H H H

and for

H—C—C—H

H—C—C—C—H

3C(g) + 8H(g), H = 2x + 8y

...(viii) 1

H H H

To get Eqn. (vii), operate eqn. (i) + 2 × Eqn. (iii) + 3 × Eqn. (iv) – 3Eqn. (v) – 2 × Eqn (vi)

1

This gives DH = 676 kcal To get eqn (viii), operate eqn (ii) + eqn (iii) + 4 × eqn (iv) – 4 × eqn (v) – 3 × eqn (vi) This gives DH = 956 kcal Thus,

x + 6y = 676

2x + 8y = 956 On solving these equations, we get x = 82, y = 99 Hence, C—C bond energy = 82 kcal mol–1 and C—H bond energy

1

–1

= 99 kcal mol

OR 1560 Energy left unutilized = = kJ = 780 kJ 2

(i)

For losing 44 kJ of energy water to be evaporated = 1 mole = 18 g 18 ∴ For losing 780 kJ of energy, water to be evaporated = × 780 g = 319 g 44 (ii) Electrolysis of water takes palce as : 2H2O(l) → 2H2(g) + O2(g) 2 Thus, 2 moles of H2O, i.e., 2 × 18 = 36 g of H2O on electrolysis produce 2 moles of H2 gas and one mole of O2 gas, i.e., total 3 mole of the gases. 3 ∴ 100 g of water will produce gases = × 100 = 8.33 moles 36

Volume occupied by 8.33 moles of gases at 25°C and 1 atm pressure is given by



V =

(8.33 mole)(0.0821 L atm K – mol –1 )(298 K) nRT = 1 atm P

= 203.8 L Taking the volume of liquid water as negligible, DV = 203.8 L ∴ W = –Pext DV = – 1 atm × 203.8L = – 203.8 L atm = – 203.8 × 101.3 J = – 20.6 kJ

3

qqq