AITS_CHEM_(FT_1)_US

November 4, 2017 | Author: venkitheboss | Category: Colloid, Particle, Chemical Bond, Solution, Chemical Compounds
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test series of narayana and chaitanya...

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NARAYANA IIT ACADEMY Color : GREEN

(Andhra Pradesh, Kota and Delhi) Max. Marks : 150 Time: 2hrs.

AITS (Full Test) – CHEMISTRY

Note : (1) Each question answered correctly is given +3 marks (2) Each question answered incorrectly is given  1 mark (3) 1  V questions are comprehensive containing 4 questions each

1.

A liquid which is immiscible in water was steam distilled at 95.2 0C at a total pressure of 0.983 atmosphere what is the mass of the liquid is present per gram of water in the distillate. Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84atm. (a) 1 gm (b) 1.27 gm (c) 0.787 gm (d) 13.43 gm O

2.

+ CH2 = CH - CN

dil. HO

-

(A)

(i) LiAlH4 (ii) H

+

(B)

(i) C2H5Br(excess) (ii) OH,

(C) (an unsaturated compound)

Identify (C) in the given reaction above N(C2H5)2

(a)

(b) N(C2H5)2

Et Et CH2

(c)

3.

Compound C2H2O3(A) on a reaction with 50% NaOH produces (B) and (C) .Compound (C) can be prepared by heating sodium formate at 3600C. What is the molecular formula of (B): (a)

HC-COOH || O

COONa

| (c) CH OH 2

4.

(d) None of these

COONa

(b) |

COONa

COONa

(d) |

CHO

What is the ratio of number of atoms per unit cell for molybdenum cubic crystal to the palladium cubic crystal? Use the data given below for Mo and Pd. Mo Pd a/pm (edge length) 314 pm 388 pm -3 /g cm 10.3 12.16 Molar mass 95.9 g/mol 106.4 g/mol NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 2

(a) 1 (c) 1/2

(b) 2 (d) 3/2

5.

What is the pH of a solution containing Na 2CO3 and NaHCO3. If 50 mL of the solution requires 20 mL of 0.1N HCl solution in presence of phenolphthalein indicator for the first equivalence point. A further 25 mL of same HCl solution is required when methyl orange is added after the first equivalence point the pH of the log 2 = 0.3010. The pKa 1 and pKa2 of H2CO3 is 6.38 and 10.28 respectively. (a) 8.98 (b) 10.89 (c) 3.18 (d) 5.18

6.

A solution was prepared by dissolving 0.1 gm CrCl 3. 6H2O in sufficient water and volume made to 20mL. 2mL of this solution when treated with excess AgNO3 solution gave a precipitate. The precipitate is filtered, dried and weighed to be 0.01076 gm. What is the formula of the complex?(At. Masses of Cr = 52, Ag = 108) (a) [Cr(H2O)6]Cl3 (b) [Cr(H2O)5Cl]Cl2 . H2O (c) [Cr(H2O)4Cl]Cl .2H2O (d) [Cr(H2O)3Cl3].3H2O

7.

Identify the end product in the given reaction: OH

Ph

(i) P + Br2

O O

(a)

B

Ph

O Ph

A

(ii)OH

H+

Ph

O

(b)

O

Ph

O O

(c)

Ph

O O

Ph

(d)

Ph

COOH

O

8.

An orange solid (A) on heating produces a gas (B), which on reaction with Mg(s) followed by hydrolysis gives another gas (C). (C) form brown precipitate with K2HgI4 in presence of KOH. What could be (A), (a) NH4Cl (b) (NH4)2Cr2O7 (c) (NH4)2CO3 (d) (NH4)2 C2O4

9.

Which of the following does not exist as dimmer in benzene: (a) AlCl3 (b) HCOOH (c) BF3 (d) All of them

10.

In which compound all the sulphur atoms are the centre of tetrahedron: (a) H2S2O5 (b) H2S2O7 (c) H2S2O6 (d) Na2S4O6

11.

How many number of moles can be oxidized of FeS2 by 1 mole of potassium permagnate in acidic medium so as to convert it into Fe3+ and SO2. (a) 1.2 (b) 5 (c) 0.454 (d) 11 NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 3

12.

What is the end product of the following sequence of reaction: (NH4)2S

NO 2

A

(i) NaNO2+HCl, 50C

(ii)

NO 2 O 2N

OH

O 2N

(a)

NH2

N

OH

H 2N

(c)

N

N

OH

(d)

N 2Cl

O

Arrange the following in their increasing order of stability. -

-

-

CH2

C:

CH

(I)

(II)

(a) II, I, III, IV (c) IV, III, I, II 14.

N

(b)

NH2

13.

B

(III)

.. C

CH3

(IV)

(b) III, IV, I, II (d) IV, III, II, I

Which of the following is the correct order of the acidity: O H +

N H

+

H

N H

(I) (a) (III) > (I) > (II) (c) (II) > (I) > (III)

(II)

H

N

+

H

(III) (b) (I) > (II) > (III) (d) (III) > (II) > (I)

15.

Which of the following statement is correct: (a) when a molecule can best be represented as a series of resonance forms each of these forms always contributes to the same degree in the hybrid. (b) A node is a region of high electron density between the two atoms in a covalent bond. (c) For a carbon atom, the 2s and 2p orbitals are equal in energy (d) When filling two or more orbitals of the same energy with electrons, the single electron is filled into different orbital rather than pair up in the same orbital

16.

Identify the incorrect statement amongst the following (a) Liquid nitrogen does not sticks to the polls of magnate whereas liquid oxygen does. (b) NO2 is brown in colour at room temperature but colourless below 00C. (c) Bond order of CN and N2 are equal but CN is more reactive (d) NO+ is having lower bond strength then NO.

17.

The heat of neutralization of HF, HNO2, HCN and HOCl are –p, -q, -r, and –s respectively. Where p > q > s > r. The correct order of pKa values of the given acids are NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 4

(a) HF > HNO2 > HOCl > HCN (c) HCN > HOCl > HNO2 > HF 18.

(b) HCN > HNO2 > HOCl > HF (d) HOCl > HCN > HNO2 > HF

Match the following ions/molecules structure ICl (a) (i) octahedral 4 (b) (c) (d)

ClF+ 4

+

I3

SO3

Shape (p) sea-saw

(ii) tetrahedral

(q) square pyramidal

(iii) triangle bipyramidal (iv) trigonal planer

(r) bent shape (s) trigonal planer

(a) a – (iii) – p, b – (iv) – r, c – (ii) – s , d - (i)- q (b) a – (iii) – p, b – (i) – q, c – (iv) – r , d - (ii)- s (c) a – (ii) – q, b – (iii) – p, c – (i) – s , d - (iv)- r (d) a – (i) – q, b – (iii) – p, c – (ii) – r , d - (iv)- s 19.

The plots of compressibility factor versus pressure is shown for the CH4, CO2 and NO2. Which of the gas will occupy least volume, the Vander Waal constant b is almost same for the given gases. C H

4

C O

2

N O

Z

2

Id ea l G as

P (a tm )

(a) CH4 (c) NO2 20.

(b) CO2 (d) all occupy the same volume

Which of the following statements are true (i) Sucrose does not reduce fehling solution although both the constituent monosaccharides glucose and fructose reduces. (ii) Glucose and fructose are epimers (iii) Ketoses exists both in pyranose forms and as furanose derivative (iv) Monosaccharides are insoluble in water but soluble in ether. (a) (ii), (iii), (iv) (b) (i), (iii), (iv) (c) (i), (ii), (iii) (d) (i), (ii), (iv) CH2COOH

21. CH2COOH

(i) CaO, (ii) NH2OH

A

P2O5 H+

B; Identify the compound B.

CH2CONH2

(a)

(b)

NH

CH2CONH2

(c)

N

CH2CN

H

(d) CH2CN

O

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 5

H

H N

22.

O

O

N

O

O

H

H

N

N

O

CHO

CHO

O

(II)

(I)

(IV)

(III)

order of basic strength for the given compounds : (a) (I) > (IV) > (III) > (II) (b) (IV) > (I) > (III) > (II) (c) (II) > (III) > (I) > (IV) (d) (IV) > (II) > (III) > (I) 23.

Electrode potential data given below Cl2 + 2H2O  2ClO + 4H+ + 2e; E0 = -1.61 volt ClO + 2H2O Cl O3 + 4H+ + 4e; E0 = -0.50 volt Based on these data which is the spontaneous reaction (a) Cl2 ClO + ClO3(b) ClO Cl2 + Cl O3ClO3(c) ClO3- Cl2 + ClO (d) ClO + Cl2

24.

A dianion [x] decolourises the dichromate solution in acidified medium. Acidifying the sodium carbonate extract with acetic acid and adding calcium chloride, a white precipitate is obtained. The precipitate dissolves in dilute H2SO4 form [y] which on heating produce two gases. The [x] is (a) SO32(b) CO32(c) C2O 2-4 (d) S2O42-

25.

Correct order of basicity is: CH3 N

CH3

CH3

CH3 N CH3

CH3

CH3

II

I (a) IV > III > I > II (c) III > IV > I > II

N

CH3

CH3 N

CH3

CH3

O 2N

III (b) I > II > IV > III (d) I > II > III > IV

IV

26.

What will be the surface area of catalyst which absorbs 224 cm3 of X2 (g) at S.T.P. per gram in order to form monomolecular layer of X2(g). The effective surface area occupied by a gas molecule on the surface of catalyst is 1.67  10-16 cm2 (a) 2.24  105 cm2 (b) 106 cm2 (c) 1.67  105 cm2 (d) 6  105 cm2

27.

A reactant X(g) decomposes in the two way simultaneously 3Z (g)  K2  X

K1

2Y(g)

The decomposition follows first order kinetics and the rate constant K1 and K2 are 1.8  10-4 and 3.0  10-4 sec-1 respectively. What is the ratio of the concentration of Y and Z at 100 sec, when 1 M of X is initially taken NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 6

(a) 3 : 5 (c) 5 : 3 28.

(b) 2 : 5 (d) 5 : 2

The following questions consist of two statement, one labelled ASSERTION (A) and the another labelled REASON (R) select the correct answers to these questions Assertion(A) : Thiosulphuric acid, H2S2O3 cannot be prepared by adding acid to a thiosulphate. Reason(R) : The free acid decomposes in water into a mixture of S, H2S, H2Sn, SO2 and H2SO4 (a) both A and R are true and R is the correct explaination of A. (b) both A and R are true but R is not the correct explaination of A (c) A is true but R is false (d) both A and R are false CH3

29.

NaF H3C S H3C

Br

CH3

(X) ;

CHCl

O

CH3

NaF H 2O

(X) + (Y)

(Y) ;

H

+

(Z)

Compound (X) and (Z) are respectively (a)

CH3 F

CH3

and

(b)

and

OCHMe2 CH3

(c)

CH3

OCHMe2

(d)

CH2 and

CH3

and

The relative rate of extraction of H and D is 2.6 : 1. The percentage yield of the product A and B in the following reaction is: C 2 H 2 D 4 + Cl2   uv C2 HD 4Cl + C 2 H 2 D3Cl (A)

(a) 27.8%, 72.2% (c) 56.52%, 43.48% P-1.

CH3

OCHMe2

OCHMe 2

30.

CH3

(B)

(b) 72.2%, 27.8% (d) 43.48%, 56.52%

Glycerol is a trihydric alcohol when it is treated with potassium hydrogen sulphate it form a compound (A). Compound (A) gives the following reaction: ( A)    (ii )(Hi )2OO3/ Zn ( B) (C ); (C ) 

( i ) Conc .OH  ( ii ) H 

( D) [C4 H 4O4 ]

( B )  NH (E) 3  Substance (E) can be used in medicinal purpose.

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 7

31.

Compound (D) in the given sequence of reaction is O

O (b)

CHCOOH (a)

CHCOOH HO

O HO

O

CH

(c)

HC

(d)

O CH2

O

HO

O

HC

O H H O

32.

When compound (B) is treated with 50% NaOH, then it gives two compound which on reaction on reaction in an acidified medium terms (a) CH3COOH (b) HCOOCH3 (c) CH3COOCH3 (d) CH3COOC2H5

33.

Compound (C) is (a) Glycolic acid (c) Oxalic acid

34.

(b) Glycerol (d) Glyoxal

Which of the following statements is true for (C): (I) It reduces Tollen’s reagent (II) It reduces Fehling solution (III) It gives Canizzarro reaction with conc. Base. (a) I only (b) I, II (c) I, III (d) III only H3C

35.

Compound (A)

CH2

+

Product ; CH2

The product of the reaction is : (a)

H3C (b) H3C H3C

CHO H3C (c)

(d)

CHO

P-2.

CHO

CHO

Colloid is heterogeneous system is which are substance is dispersed as very fine particles in another substance called dispersion medium A solution is a homogeneous system. The essential difference between a solution and a colloid is one of particle size. In a solution, the particles are ions or small molecules in a colloid, the dispersed phase may consist of particles of a single macromolecule (such as protein or synthetic polymer) or an aggregate of many atoms, ions or molecules. Colloidal particles are larger than simple molecules but small enough to remain suspended. Depending upon mlether the dispersed phase and dispersion medium are solids, liquids and gases, eight types of colloidal systems are possible. Dispersed phase Dispersion medium Types of colloid. NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 8

Solid Solid Solid Liquid Liquid Liquid Gas Gas

Solid Liquid Gas Solid Liquid Gas Solid Liquid

Solid solution Solution Aerosol Gel Emulsion Aerosol Solid solution Foam

Depending upon the type of the particles of the dispersed phase, colloids are classified as multimolecular and associated colloids: 36.

Which of the following does not result formation of solution. Fe(OH)  3 3HCl (A) FeCl3  3H 2 O (B) SO2  2H 2S (C) As 2S3  2H 2S (D) None of these

37.

 3S

2H 2 O

As2S3

3H 2O

Hair cream is an example of: (a) Gel (c) Solution

(b) Emulsion (d) Aerosol

38.

Which of the following is not an example of aerosol: (a) Fog (b) Smoke (c) Froth (d) cloud

39.

Depending upon the nature of interaction between the dispersed phase and the dispersion medium colloidal sols are lyophillic and lyophobic. Which of the following is not result lyophobic colloidal solution on mixing with suitable liquid (a) Gelatin (b) Starch (c) Rubber (d) all of them

40.

Match the followings: (A) Sol (B) Gel (C) Foam (D) Emulsion (a) (A)–(I) , (B) –(II), (C)-(II), (D)–(IV) (c) (A)–(II) , (B) –(IV), (C)-(III), (D)–(I)

P-3.

(I) Milk (II) Whipped cream (III) Cheese (IV) Paints (b) (A)–(III) , (B) –(II), (C)-(I), (D)–(IV) (d) (A)–(IV) , (B) –(III), (C)-(II), (D)–(I)

The rate constant for most of the reactions increases with the increase of temperature. It has been found that a plot of log10K against 1/T gives a straight line (as shown in fig) represented Ea by the equation log10K = log10 A  2.303RT

lg10 K (1/T)

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 9

The equation is known as Arrhenius equation and also be written as K  Ae  Ea / RT For an equilibrium reaction A(g) ‡ˆ ˆ† ˆˆ B(g) The rate constant of forward reaction increases by 130% when its temperature is raised from 300 K to 310 K. The ratio of activation energy of forward reaction to backward reaction is 1 : 3. [Log2.3 = 0.361, antilog (-0.723) = 0.189] 41.

By increasing the temperature, the effect on equilibrium constant : (a) It decreased by 130% (b) It decreases by 19% (c) It decreases by 81% (d) It does not change

42.

The activation energy of the forward reaction is (a) 7.92KJ (b) 64.41 KJ (c) 16.2 KJ (d) 6.44 KJ

43.

For the given reaction, the temperature coefficient of the forward reaction is (a) 1. 3 (b) 0.77 (c) 0.43 (d) 2.3

44.

What fraction of molecules are in the activated state in forward reaction [antilog (0.79) = 6.61] (a) 6.16  10-12 % (b) 6.16  10-10 % (c) 6.16  10-8 % (d) 6.16  10-6 %

45.

For the reactions (i) X Y ; Ea = 40 KJ mol-1 and (ii) X Y ; Ea = 80 KJ mol-1 0 for the same 10 C rise in temperature which of the following statement is correct (a) Both the reaction’s rate increases to an equal extent. (b) Increase in rate of reaction (i) is more than that of (ii) (c) Increase in rate of reaction (ii) is more than that of (i) (d) The rate will not be affected

P-4.

An alkene can be converted to alcohol by the following three methods: (a) acid catalysed hydration (b) oxymercuration – demercuration reaction (c) Hydroboration – oxidation the reactions are as shown below: +

(a)

alkene

H

H /H2O

OH

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 10

OH

(b)

OH

(i) Hg(OAc) 2 / H2O

(ii) NaBH4 HgOAc

H

(i) B2H6.THF

(c)

(ii) H2O 2/OH H

H

B

OH

In first method formation of alcohol takes place via formation of carbocation so there is a possibility of rearrangement where as in oxymeration –demercuration and hydroborationoxidation formation of alcohol takes place without any rearrangement of structure because the formation of carbocation does not takes place. An alkene (X) C10H12 from three isomeric alcohols by the above three methods as X  (a) A ; X  (b) B ; X  (c) C ; 0 0 0 (2 alc.)

(3 ) Alcohol

(1 alc.)

Alkene (X) with hot KMnO4 gives C7H6O2 which gives effervcence with NaHCO3 46.

What is the structure of (X) : CH3

CH3

(a) H3C

C

(b)

CH2

C

CH

CH3

CH3

(c) H3C 47.

CH2

CH

CH2

(d)

CH

CH

CH2

The end product for the given sequence of reactions: (X)

(i) B 2H6.THF

(C)

(ii) H2O2/OH CH3

(a)

C

CH2CH3

H

+

C 6H 6

Product

(b)

CH2

CH

CH2CH3

CH3

(d) H3C

CH3

(c)

48.

CH2

HC

CH2

C

CH3

CH2

When 1-ethyl-6-methyl cyclohexene is treated with mercuric acetate in presence of methyl alcohol followed sodium borohydride gives-

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AITS_CHEM(FT_01) - 11

OCH3

(a)

H3C

(b)

CH3 C2 H 5

C2 H5

H3CO

(c) H3C

C2 H5

(d)

C2 H5

OCH3 C 2H 5

OCH3

49.

HO H3C

CH2 CH3

(i) Hg(OAc) 2 / H2O

Product

(ii) NaBH4

Identify the product in the given reaction: HO

(a)

OH CH3

H3C

O

H3C

CH3

(d)

H3C

50.

CH3 O

CH3

(c)

OH

(b)

H3C

O

Identify the final product of the given reaction: CH3 CH2

(i) B2H6.THF

H

+

(ii) H2O 2/OH

(i) B2H6.THF (ii) H2O 2/OH CH3

CH3

(a)

CH3

(b)

OH

OH

CH3

(c)

Product

CH3

OH

CH3

(d)

OH

CH3

***

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AITS_CHEM(FT_01) - 12

NARAYANA IIT ACADEMY Color : GREEN

(Andhra Pradesh, Kota and Delhi)

AITS – CHEMISTRY ANSWERS 1

2

3

4

5

6

7

8

9

10

B

D

C

C

B

B

A

B

C

B

11

12

13

14

15

16

17

18

19

20

C

B

C

A

D

D

C

D

C

C

21

22

23

24

25

26

27

28

29

30

C

B

B

C

B

B

B

A

B

C

31

32

33

34

35

36

37

38

39

40

B

B

D

C

B

D

B

C

D

D

41

42

43

44

45

46

47

48

49

50

C

B

D

B

C

D

A

C

C

D

Hints and Solution 1.

PT = 0.983 atm 0 Pwa0 ter  0.84atm ; Pliquid  0.143atm



( wt)liquid



Pliq0 uid  M liquid 0 water

( wt) water P ( wt)liquid 1.27  ( wt) water 1

 M water



0.143  134.3 19.205  0.84  18 15.12

CH2 CH2 CN

.. + CH2 = CH - CN

2.

H2N

O

O

OH

LiAlH4 CH2 H

Et N

CH2 = CH2 +

Et

CH2

+

H

+ Et

N

CH2

OH

(i) C2H5Br (excess)

N

(ii) OH

(C)

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 13

4.

For M0 a 3 N A M N Mo 1a13 N A M 1   2 32 = N Pd M1  a2 N A 2

N

For Mo Pd

N=2 N=4

BCC FCC

lattice lattice

1 Meq Na2CO3 = 20  0.1 = 2 2

5.

Meq Na2CO3 = 4 1 Meq Na2CO3 + Meq NaHCO3 = 25  0.1 = 2.5 2

Meq NaHCO3 = 0.5 2 =0.04 50 0.5  Molarity  NaHCO3 = =0.01 50 |CO32- | pH=pK 2 +log |HCO3- | 0.04 = 10.28 + log = 10.28 + 0.6020 = 10.88 0.01

 Molarity  Na CO 2

3

=

Let formula of CrCl3.6H2O  [Cr(H2O)nClx] Cl3-x .(6 – n)H2O [Cr(H2O)nClx]Cl3-x (6 – n)H2O + (3-x)AgNO3  (3 –x) AgCl(ppt.) wt. of complex compound for 2 mL is 0.01 gm

6.

 3  x  

0.01 0.01076   143.5  266.5

solving x = 1  Formula is [Cr(H2O)5Cl]Cl2. H2O O O

7. Ph

9.

Ph

(ii) OH

OH

HC

COOH

H+

C Ph

OH

C

O C

O

Ph

O

AlCl3 and HCOOH exist as dimmer in benzene.

10. HO

11.

(i) P + Br2

O

O

S

S O

O

FeS2 + (n = 11) 

nKMnO4 nFeS2



O

(H2S2O7) OH

KMnO4   H (n = 5)



Fe3+ + SO2 + Mn2+

110 5

 nFeS = 0.454 mole 2

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 14 (NH4)2S

NO 2

12.

NH2

O2 N

O2 N

HCl + NaNO2

+

N2 Cl

0

5C O2N

-HCl

N

N

OH

OH

NO 2

19.

The order of ‘a’ value NO2 > CO2 > CH4 (as per graph). If the gases are having same value of b but different values of a, then the gas having larger value of a will occupy lesser volume. This is due to larger force of attraction hence the lesser distance between its molecules. CH2 COOH

21.

CH2

CaO,

O

CH2 COOH

H2NOH

N - OH

CH2

P 2O 5 , H

+

NH O

24.

C2O42

+ Cr2O72  Cr 3 + CO2 H2C2O4   H2O + CO + CO2

26.

No. of molecules  N =

224  6.023  1023 = 6.023  1021 22400

Area/molecule = 1.67  10-16 cm2 total surface area covered by X2 (g) / gram of the catalyst = 1.67  10-16  6.023  1021 = 10.06  105 cm2 27.

3Z (g)  K2  X

K1

2Y(g)

| y |  K1 2  1.8  104 2    | z | 3K 2 3  3 104 5

NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

AITS_CHEM(FT_01) - 15

CH3

29

NaF H3C S H3C

Br

CH3

NaF H 2O

CHCl

CH3

O (X) CH3

CH3

CH3

+

CH

OH

CH CH3

CH3

[via E2 machanism] [F- is solvated ]

(Y)

OH

H

CH3

+

OCHMe2

CH3

30.

35.

CH2

CH2

H3C

+

CHO A

P-3.

(Z)

Product A 2.6× 2 5.2 = = Product B 4×1 4 5.2  100  56.52%  %A= 9.2

CHO

H3C

CH2

K 310 =2.3 K 300 K 310 E af  1 1 =    K 300 2.303R  300 310 log(2.3)  2.303R  300  310  Eaf = = 64.41 kJ 10 E af 1 = As   H  r×n =E af -E ar  2E af E ar 3 The reaction is exothermic so by increasing the temperature the equilibrium constant decreases  K  2 -2E af  1 1 log      K  1 2.303R  300 310 K K log 2 = -2  log2.3  2  0.189 K1 K1 log

K2  0.19 K1  equilibrium constant decreases. And it decreases by 81%. CH3

50.

CH2

CH3

(i) B2H6THF (ii) H2O 2/OH

H

+

OH OH

CH3 C2 H 5

CH3 C 2 H5

(i) B2H6.THF (ii) H 2O 2/OH

*** NARAYANA IIT ACADEMY South Centre: 47 B, Kalu Sarai, N. Delhi-16. North West Centre: 2/40, Adj. Central Mkt., W. Punjabi Bagh, N. Delhi-26.

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