iit jam_ chemistry_free solved expected paper.pdf

7/17/2019 IIT JAM_ CHEMISTRY_FREE SOLVED EXPECTED PAPER.pdf

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C SIR NET, GATE, IIT-JAM, UGC NET , TIFR, IISc , JEST, JNU, BHU , ISM , IBPS, CSAT, SLET, NIMCET, CTET

Phone: 0744-2429714 Mobile: 9001297111, 9829567114, 9001297243Website: www.vpml!sses.om "-M!il: vpml!sses#$!hoo.om %in&o#vpml!sses.om

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For IIT-JAM, JNU, GATE, NET, NIMCET and Other Entrance Exa!

Pattern of questions : MCQs + Fill in the blanks + Subjective

PART-A:- Attempt ALL the objective questions(Questions 1-10). Each of these questions carriestwo marks. 0.25 negative mark for each wrong answer.

PART-C:- Attempt ALL questions(Questions 21-30). Each of these questions carriesfive marks

PART-B:- Attempt ALL questions(Questions 11-20). Each of these questions carriesthree marks

IIT-JAM - CHEMISTRY

1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429714

Web Site www.vpmclasses.com [email protected]

Total marks : 100

Duration of test : 3 Hours

MOCK TEST PAPER(Acco ding to new pattern)r





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GUIDELINES FOR JAM ASPIRANTS:

1. CHANGE OF SYLLABUS :

(i) After 8 years , IIT-JAM syllabus of Chemistry has been changed.

(ii) The following new topics have been added from JAM 13 onwards.

Allenes, Biphenyls, cyclohexanes Acid-Base , Redox reactions , Organometallic

Reagents,Solid state,Bioinorganic Chemistry, NMR, IR spectroscopy, Adsorption

etc.

(iii) The JAM aspirants should specially stress upon these above mentioned topics &

should practice the problems related to these topics.

2. CHANGED EXAM PAPER PATTERN:

(i) IITs have changed the exam paper pattern also & have Introduced Fill in the blank

type of questions for the first time .

(ii) The given mock test paper reflects the paper pattern and the toughness level of JAM

papers. The questions of new syllabus have also been incorporated in this pattern.

(iii) Students can start preparing for IIT-JAM from B.Sc. First year itself, so that in the

course of three years duration, their preparation level reaches to the level of JAM

paper. This helps in raising the percentage in university exams also.





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PART-A (1-10)

1. The relat ionship betw een the dissociation energy of N2 and N2+ is

(A) N2 = N2+

(B) Unpredictable

(C) N2 > N2+

(D) N2+ > N2

2. The correc t absolute conf iguration ass igned for compound (I) and (II) respectively is –

(A) R, R

(B) R, S

(C) S, S

(D) S, R

3. The correc t order of acidic strength is –

(A) SiO2 < CO2 < N2O5 < SO3

(B) SiO2 < N2O5 < CO2 < SO3

(C) CO2 < SiO2 < N2O5 < SO3

(D) SO3 < CO2 < N2O5 < SiO2

4. Which of the following has maximum number of lone pairs on central atom?

(A) ClO3 –

(B) XeF4

(C) SF4

(D) I3 –





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5. Which of the following is an arachno borane –

(A) [B6H6]2–

(B) [B5H9]

(C) [B2H6]

(D) [B6H12]

6.

Which of the following is incorrect for above reaction?

(A) Product is optically active

(B) It is an oxidation reaction

(C) Overall reaction is stereospecific syn addition

(D) B2H6 adds on alkene by forming a cyclic transition state.

7. The Miller indices of crystal plane w hich cuts through the crys tal axis at 2a, –3b, –3c is ,

(A) 2 3 3

(B) 3 2 6

(C) 6 2 3

(D) 3 2 2

8. IR spectra of benzaldehyde and acetophenone will show distinguishing peaks in the region

(A) 3000 - 3600 cm –1

(B) 1200- 1300 cm –1

(C) 2700-2800 and 2860-2975 cm –1

(D) 1600 and 1740 cm –1

2 6

2 2

1. B H / T.H.F.

2. H O / OHΘ →





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9. Which is not correctly matched?

(A) XeO3: trigonal bipyramidal

(B) CIF3 : T shaped

(C) XeOF4 : square pyramidal

(D) XeF2 : l inear shape

10. If standard reduction potentials of Ni2+ /Ni and Au3+ /Au couples are –0.25 V and 1.50 V

respectively, then the EMF of the cell

Ni/Ni2+ (0.01 M) || Au3+ (0.1 M) Au w ill be

(A) unpredictable

(B) = EMFº

(C) < EMFº

(D) > EMFº

PART-B (11-20)

11. Compound (X) __________ on reduction w ith LiAlH4 g ives a hybride (Y) containing 21.72%

hydrogen along w ith other products. The compound (Y) reacts w ith air explosively resulting

in boron trioxide. The structure of Y is ___________.

12. ______ and _____ meta lloenzymes are responsible for the removal of hydrogen peroxide.

.

13. D-Glucose and D-Galactose are _______ w hile α-D and β -D Glucose are ______.

14. The structure of an alcohol C5H11OH is _________. It gives a ketone on oxidation. When

the alcohol is dehydrated and the resulting alkene is oxidised, a mixture of a ketone and an

acid results.

15. At certain temperature the r.m.s. speed of molecules of Nitrogen gas is U.When the

temperature is doubled, the molecules dissociate into individual atoms. The new r.m.s.

speed of atoms is _________.





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16. Diff erence betw een ∆H and ∆U for the combustion of benzene at 300 K is ________.

17. The molar conductance at infinite dilution of HCl, NaCl and NaZ (Sodium Crotonate) are 425

× 10 –4, 125 × 10 –4 and 80 × 10 –4 Sm2 mol –1, respectively. The specific conductance of

0.001 M aqueous solut ion of crotonic acid (HZ) is 3.8 × 10 –3 Sm –1. The degree of

dissociation is _____.

18. ------------- are responsible for electron transf er in plants and bacteria.

19. In Plastocyanin Cu(I) and Cu(II) are respectively arranged as------------ respectively

20. 10 moles of an ideal gas expand isothermally and reversibly f rom a pressure of 10 atm to 2

atm at 300 K. _______ is the largest mass w hich can be lifted through a height of 1 metre in

this expansion

PART-C (21-30)

21. When 16.8 g of w hite solid, X w ere heated, 4.4 g of acid gas A, that turned l ime-w ater milky

was driven off together w ith 1.8 g of a gas B w hich condensed to a colourless liquid. The

solid that remained, Y, dissolved in w ater to give an alkaline solution, w hich w ith excess of

barium chloride solution gave a w hite precipitate Z. The precipitate eff ervesced with acid

giving of carbon dioxide. Identify A, B and Y and w rite dow n the equation for the thermal

decomposition of X.

22. Acetylene + CH3MgBr → X + CH4

X + CO2 → YH+

→ Z(C3H2O2)

Z 2 2 4 4H O, H SO , HgSO →W(C

3

H

4

O

3

)

W + KMnO4 → CH2(COOH)2

Identify X, Y, Z , W and w rite complete reaction equation.

23. Bromobenzene + Mgether

→ E(C6H5MgBr)





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E + ethylene oxideH+

→ F(C8H10O)

F + PBr3 → G(C8H9Br)

G + NaCN → H(C9H9N)

H + H2SO4 2H O

∆ → I(C9H10O2)

I + SOCl2 → J(C9H9OCl)

J + anhydrous HF catalyst → K(C9H8O)

K + H2 catalyst → L(C9H10O)

L + H2SO4,Warm

→ M(C9H8)

Identify E, F, G, H, I, J, K, L, M and write complete reaction equation.

24. The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution, containing 7.0 g of the

salt per 100 g of w ater at 100ºC is 70%. If the vapour-pressure of water at 100ºC is 760 mm,

calculate the vapour-pressure of the solution.

25. Heat of combustion of ethylene (C2H4) is – 337 kcal mol –1 under standard state at 298 K.

Assuming 70.0% eff iciency, how many kg of w ater at 20ºC can be conver ted into steam at

100ºC burning 1.00 m3 of C2H4 gas measured at STP? Specific heat of water is 1 kcal kg –1

K –1 and its latent heat of vaporization is 540kcal kg –1.

26. Identify the intermediate products X, Y, Z and W in the follow ing sequence of reactions.





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+ (X) (Y) (Z)

(W) – co2

27. The vapour-pressure of ethanol and methanol are 44.5 and 88.7 mm Hg respectively. An

ideal solution is formed at the same te mperature by the mixing 60 g of ethanol w ith 40 g of

methanol. Calculate the total vapour-pressure of the solution and the mole-fraction of

methanol in the vapour.

28. The rate law for the reac tion 2NO(g) + O2(g) → 2NO2(g) is

– [O ]2t

∆∆

= k[NO]2[O2]

The proposed mechanism is

NO(g) + O2(g) → NO3(g)

NO(g) + NO3(g) → 2NO2(g)

(a) The reaction mechanism does not include a collision among three molecules. Since the

reaction order is three, shouldn’t the mechanism include such a collision ?

(b) Which step in the proposed mechanism is the slow est step ?

(c) What is the molecularity of each step ?

29. (a) Cv for uranium metal is 3.04 JK –1 mol –1 at 20K. Calculate the absolute entropy of the

metal in JK –1 mol –1 at 20K.(b) Calculate the uncertainty in the velocity of an electron if uncertainty in its position is 1Aº.

EtO NaΘ ⊕

→Zn/H+

→2, (–H O)∆

→

2OH, H O

hydrolysis

Θ

→





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30. If ν is the volume of a gas adsorbed on the surf ace of a solid, the p lot of p/ ν versus p w here

p is the gas pressure in the Langmuir adsorption isotherm, gives a straight line. What w ill be

the slope of this line?

ANSWER KEY

HINT & SOLUTION

1.(C) According to MOT the bond order of N2 is 3 w hereas that of N2+ is 2.5. So the bond

dissociation energy of N2

w ill be greater than that of N2

+. Bond dissociation energy is

directly proportional to bond order.

(B.O αααα BDE)

2. (B)

H

CH3

NH2

(R)

Priority SequenceNH > COOH > CH > H

2 3

1

2

3

H

CH SH2

NH2

(S)

Priority SequenceNH > >

2CH SH COOH

2

3

1

2

3. (A) These are the anhydrides of following acids:

H2SO4, HNO3, H2CO3 , H2 SiO3.

Question 1 2 3 4 5 6 7 8 9 10 Solution C B A D D A C A D





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The acid strength order of these acid is H2SO4, > HNO3>H2CO3 >H2 SiO3. The oxides

also follow the same order.

4. (D) I3 – — sp3d Hybr idization w ith 3 lone pairs

Ι : Ι

O –

: :

5. (D) Arachno boranes have the general formula - BnHn+6

w ith (n+ 3) electron pairs B6H12 is arachno borane.

6. (A) 2 6

2 2

1 . B H

2.H O /OHΘ →

7.(D) a b c

2 –3 –3

1

2 –

1

3 –

1

3

3 –2 –2

∴ Miller indices are ( 322) .

8.(C) C6

H5

CHO and C CH3

O

will diff er in the absorption peaks due to aldehydic C – H

(2700-2800 cm –1

) and sp3 C – H stretch (2860-2975 cm

–1).

9. (A) XeO3 has trigonal pyradimal shape .





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10.(D) EMF = EMFº + log = EMFº + log = EMFº + log 104

= EMFº + × 4. Hence EMF > EMFº.

11. BCl3,

Sol. Y is 4BCl3 (X) + 3LiAlH4 2B2H6 (Y) + 3AlCl3 + 3LiCl

% of hydrogen in B2H6 =6

27.62 × 100 = 21.72

B2H6 + 3O2 (air) explosion → B2O3 + 3H2O

Structure of B2H6

12. Catalase and peroxidase.

Sol. Catalase and peroxidase are tw o heme enzymes that catalyse reactions of hydrogen

peroxide.

Catalases prevent the accumulation of poisonous H2O2 and other peroxy compounds

resulting from the various oxidase reactions in the biological system. Catalase catalyses

disproportionation of H2O2 according to overall reaction :

2H2O2 Catalase

⇀↽ 2H2O + O2

0.059

6

3 2

2 3

[Au ]

[Ni ]

+

+

0.059

6

2

3

(0.1)

(0.01)0.059

6

0.059

6

→





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The process occurs in two steps :

PFeIII + H2O2 ⇌ Compound I + H2O

Compound I + H2O2 ⇌ PFeIII + H2O2 + O2

Peroxidase reacts by mechanism similar to catalase, but the reaction catalysed is the

oxidation of a w ide var iety of organic and inorganic substrate by H2O2

H2O2 + AH2 Peroxidase

→ 2H2O + A

or H2O2 + Substrate H2 (Reduced form) Peroxidase

→ 2H2O + Substrate

(Oxidised form)

13. C-4 epimers, Anomers

Sol.

14.

Sol.





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15. 2u

Sol. u1 =3RT

M, T2 = 2T; M2 =

M

2

u2 =3R(2T) 3RT

2M 2 M

= = 2u

16. 3.74 kJ

Sol. Combustion of Benzene is represented by

C6H6(l) +15

2 O2(g) → 6CO2(g) + H2O(l)

∆n = 6 –15

2 = –

3

2

∆H = ∆ U –∆nRT

or ∆H – ∆ U = – ∆nRT = – –3

2

(8.314)(300) × 10 –3 = 3.74 kJ

17. 0.10

Sol. Crotonic acid is formed as follows :HCl + NaZ → NaCl + HZ (crotonic acid)

where Z stands for crotonate ion.

Crotonic acid is a w eak organic acid and HCl, NaZ an d NaCl are s trong electrolytes.

Hence, using Kohlraush’s law of independent migration of ions,

Dehydration →





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o o o o

mHZ mHCl mNaZ mNaCl – Λ = Λ + Λ Λ = (425 + 80 – 125) × 10 –4 S m2 mol –1 = 380 × 10 –4 S m2

mol –1

Also, at the given concentration of crotonic acid,

Λm = –3 –1

3 –3

3.8 10 S m

c 0.001 10 mol m

κ ×=

× = 38.0 × 10 –4 S m2 mol –1

α = –4 2 –1

m

o –4 2 –1m

38.0 10 S m mol

380 10 S m mol

Λ ×=

Λ × = 0.10

18.. Ferredoxins;

Sol. Ferredoxins are group of non-haem iron-sulphur proteins which are responsible for electron

transfer in plants and bacteria.

19. Flattened tetrahedron and octahedrally;

Sol. n plastocyanin Cu(I) and Cu( II) are respectively arranged as f lattened tetrahedron and as

octahedron respectively .Cu(I) forms distorted tetrahedron, while Cu(II)adopts octrahedral

arrangement

20. 409.28 kg

Sol. Work done in this case is given by

w = –nRT ln( P1 /P

2) = –10 mol(8.314 J K –1 mol –1)(300 K) ln (10atm/2atm) = –40.15 × 103 J

Let M be the mass w hich can be lifted through a height of 1 m.

Work done in lif ting the mass = Mgh = M × 9.81 m s –2 × 1 m

∴ M × 9.81 × 1 m2 s –2 = 40.15 × 103 J

∴ M = 409.28 kg

21. (X) is NaHCO3 (molecular w t. = 84)

Reactions involved are given below :

3 2 3 2 2( A) (B)( X) (Y)

2NaHCO Na CO (s) CO (g) H O(g)∆

→ + +

2 × 84 = 168 g 106 g 44 g 18 g





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16.8 g≈ 10.6 g 4.4 g 1.8 g

2 2 3 2lim e water white p pt.

CO Ca(OH) CaCO H O+ → ↓ +

H2O(g) is condensed to liquid w ater

2 3 2 3(Y ) (Z)

3 2 2 2(Z )

Na CO BaCl BaCO NaCl

BaCO 2HCl BaCl H O CO

+ → +

+ → + +

22. H – C≡ C – H + CH3MgBr HC ≡ CMgBr + CH4

H – C CMgBr O C O

(X)

≡ + = = →

H⊕

→

=

O

HO – C – C CH≡

(Z) 2 2 4 4H O, H SO , HgSO →

=

O

H – C – CH – C – OH2 =

O

(W) 4KMnO

→

CH2

COOH

COOHMalonic acid

23.

Br

+ MgEther

→

MgBr

(E)

+

-

CH – CH2 2

O

C6H5 – CH2 – CH2 – OMgBr 3H O+

→

6 5 2 2C H – CH – CH – OH

(F)3

PBr → 6 5 2 2C H – CH – CH – Br

(G)

NaCN → 6 5 2 2C H – CH – CH – CN

(H)

2 4

2

H SO

H O, →

∆ I

6 5 2 2C H – CH – CH – COOH

( )

24. 2 –

3 2 3Ca(NO ) Ca 2NO+

+⇌

1 0 0 Before dissociation

1 – α α 2α After dissociation

∴ Total moles at equilibr ium = (1 + 2α) = 1 + 2 × 0.7 ( α = 0.7)

= 2.4

For Ca(NO3)2 : n or

exp

m

m = 1 + 2α

O = C – OMgBr

C CH≡

(Y)





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∴ mexp = norm 164

1 2 0.7 2.4=

+ × = 68.33

Also at 100ºC, P0 = 760 mm, w = 7g

W = 100 g

Raoult’s Law

∵ 0 s

s

P – P wM

P Wm=

Now, o s

s

P – P 7 18P 68.33 100

×=×

= 0.0184 ors

PºP

– 1 = 0.0184

∴ ps =760

1.0184 = 746.26 m

25. 1.00 m3 at STP = 1000 L =1000

22.4 mol

Enthalpy change due to burning of 1.00 m3 ethylene = – 337 ×1000

22.4 kcal

Due to 70% eff iciency, useful heat is = 337 ×1000

22.4

×70

100

kcal

H2O(l) is converted into H2O(g) in tw o stages

H2O(l, 20ºC) → H2O(l, 100ºC), ∆H1 = 80 kcal kg –1 (r ise in temperature = 80ºC)

H2O(l, 100ºC) → H2O(g, 100ºC), ∆H2 = 540 kcal kg –1

∆H(Total) = 620 kcal kg –1

thus, w ater converted into steam =heat evolved 337 1000 70 1

heat required 22.4 100 620

×= × × = 16.98 kg

26. (X) = ; (Y) = ;





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(Z) = ; (W) =

27. Total vapour-pressure of solution:

Pm = º º º º

eth eth me th meth e th e th me th meth

eth meth

P M.F. P M.F. P x P x

P P

× + × = × + ×

Given that:

º

ethP = 44.5 mm Hg, p0

meth = 88.7 mm Hg

w t. of ethanol = 60 g, w t. of methanol = 40 g

∴ neth

=60

40 = 1.304 ( mol. w t. of ethanol = 40)

nmeth

=40

32 = 1.25 ( mol. w t. of methanol = 32)

xeth

= eth

eth meth

n 1.304 1.304

n n 1.304 1.25 2.554= =

+ + = 0.51

xmeth = meth

meth eth

n 1.25n n 2.554

=+

= 0.49

∴ Pm

= 44.5 × 0.51 + 88.7 × 0.49 = 22.67 + 43.46 = 66.13 mm Hg

mole-f rac tion of methanol in vapour = meth

m

P 43.46

P 66.13= = 0.66 m m Hg

28. (a) No, A collision among three molecules is not likely. The [NO]2

term results because the

slow est step in the mechanism depends on [NO] and some other concentration ter m that is

directly related to [NO][O2].

(b) The slow est step is the second step in the proposed reaction mechanis m. Assuming

that the first step is fast and has reached a state of equilibrium,

( ) ( ) ( )

( ) ( ) ( )

1

2

K

2g g 3 g

3g g 2 gK

NO O NO

NO NO 2NO

+

+ →

⇀↽





'(()ess: 1-*-8, +hee l! *how(h!)$ o!(, ++, '/W', ', ''+', 324005

then the concentration of NO3(g) is related to the concentrations of NO(g) and O2(g) by

the equilibrium relation K1 = [NO3]/[NO] [O2] or [NO3] = K1[NO][O2]. The rate law for the

reaction, as derived f rom the slow est step, is ∆[O2]/ ∆t = k2[NO][NO3] = K2 [NO]

K1[NO][O2] = k1k2[NO]2[O2] = k[NO2]2[O2].

(c) Both steps are biomolecular.

29. At temperatures (0< T < 20 K) , Cp = Cv = aT3

[Debye’s T3

law ]

( )33 –1 –1 –5 –4 –1a C / T 3.04JK mol / 20K 38.03 10 JK mol ν= = = ×

Hence , ( )3 –5 –1 –4 3pC aT 38.3 10 Jmol K T= = ×

From Eq.14, dS = (Cp/T)dT = 38.03×10 –5

J mol –1

K –4

T2 dT

Or ( )3 –5 –4 –1

20 0S – S 38.03 10 JK mol 20K / 3= ×

Or [ ] –1 –1

20 0S 1.01JK mol S 0= ∴ =

30. A/C to langmuir isother m

p

P

K

1 Kθ =

+

P

1 11

K= +

θ

Let θ =mono

ν ν

Where νmono is the volume corresponding to complete

coverage

mono

p

11

K

ν= +

ν

Multiplying throughout bymono

p

ν





'(()ess: 1-*-8, +hee l! *how(h!)$ o!(, ++, '/W', ', ''+', 324005

mono mono

p p 1

K= +

ν ν ν

y = mx + C

Slope =mono

1

ν

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