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http://www.chemistrycrest.com/ Page 1

PAPER-1Maximum Marks: 80

Question paper format and Marking scheme:

1. In Section I ( Total Marks: 21), for each question you will be awarded 3 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,

minus one (1) mark will be awarded.

2. In Section II (Total Marks: 16), for each question you will be awarded 4 marks if you darken ALL the

bubble(s) corresponding to the correct answer(s) ONLY and zero marks other wise. There are no negativemarks in this section.

3. In Section III (Total Marks: 15), for each question you will be awarded 3 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,

minus one (1) mark will be awarded.

4. In Section IV (Total Marks: 28), for each question you will be awarded 4 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in thissection.

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SECTION I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of

which ONLY ONE is correct.

1. Resonance energy of C6H6 using Kekule formula for C6H6 from the following data:

(1) 0H for C6H6 = 358.5 kJ mol1

(2) Heat of atomisation of C = 716.8 kJ mol1

(3) Bond energy of C__H, C-C, C = C, and H__H are 490, 340, 620 and 436.9 kJ mol1

A) 140 kJ B) 150 kJ C) 160 kJ D) 170 kJ

Sol. (B)

6 6

2 6 6

tan

exp

6 ( ) 3 ( ) ; 358.5

[6 716.8 3 436.9] [3 340 3 620 6 490]

208.5

358.5 ( 208.5) 150

cal reac ts pro ucts

cal

cal

cal

ForC H

C s H g C H H kJ

H BE B

H x x x x x

H kJ

RE H H

kJ

+ =

=

= + + +

=

=

= =

d

2. Which of the following oxidation reactions incorrectly matched?

(A)( )Cr VI

CH -CH OH CH CHO3 2 3+H O

3

(B)

( ) ( )CrO , H SO

3 2 4CH - CH C C CH CH CH CH C C C CH3 2 3 3 2 3propanone,2 2

OH O

(C)

(D)

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Sol. (A)

In aqueous medium, over oxidation takes place to give acid rather than aldehyde.

( )

O O|| ||Cr VI

CH -CH -OH CH - C -H CH - C -OH3 2 3 3+H

3. Chromyl chloride test is employed for the detection of chloride ions. A similar test can be

employed for(A) Br (B) I (C) Both (D) None of these

Sol. (D)K2Cr2O7 will oxidise Br

& I

into Br2 and I2 which give colourless solutions with NaOH

4. D fructose is a ketose, but it reduces Tollens reagent. It can be explained on the basis of

(A) Mutarotation (B) Reimer Tiemann reaction

(C) Amadori reararangement (D) Lobry de Bruyn Van Ekenstein rearrangement

Sol. (D)Glucose withdil. NaOH solution gives a mixture of D-Glucose,D-Fructose and D-Mannose .This

mixture is obtained even when the starting material is either fructose or mannose.This is called Lobry-

debruyn- van-Ekenstein rearrangement. Due to this,though fructose is a ketose, it can reduce tollens test.

5. Chlorine evolved by the reaction of 45.31 g of pyrolusite (impure) and excess of HCl is found to

combine completely with the hydrogen produced by the reaction of 10 g of magnesium and excess

of dilute hydrochloric acid. Hence, the percentage purity pyrolusite sample isA) 80.56% B) 60.56% C) 40.56% D) 90.56%

Sol. (C)

2 22Mg HCl MgCl H+ +

24g 73g 2g

2 2 2 24MnO HCl MnCl Cl H O+ + +

87g

2 22H Cl HCl+

2g 71g

2g H obtained from 24 g of Mg will combine completely with

71 g of Cl2 produces from 87 g of pure MnO2(Pyrolusite)

So, when 10 g of Mg are used, mass ofpure MnO2 needed =10 87

36.2524

x=

Therefore, % purity =36.25

100 80.04%45.31

x =

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6. An optically pure compound X gave an [ ]25 0

D 20.0= . A mixture of (X) and its enantiomer (Y) gave

[ ]25 0

D

10.0= + . Select the True / False statements regarding this mixture.

(1) It has 50% dextrorotatory isomer

(2) It has 25% laevorotatory isomer

(3) It has 50% optical purity

(A) TTF (B) FFT (C) FTT (D) TFT

Sol. (C)

% optical purity of mixture =[ ]

[ ]obs

pure

100

10100 50%

20= =

So 50d l =

100d l+ =

2 150d =

75%d = and l = 25%

7. Match the following

ListI List II

(Crystal System) (axial distances)

1) Trigonal P) cba

2) Monoclinic Q) cba = 3) Rhombic R) a = b = c

4) Hexagonal

ListIII List IV

(axial angles) (unit cells possible)

a)o

90==i) P, F, E, B

b)o

90===ii) Primitive

c)oo 120,90 ==

iii) Primitive

d)

oo 120,90 ===iv) P, E

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A) 1 - P, a, iii; 2 - R, b, iv; 3 - P, d, i; 4 - Q, c, iii

B) 1 - R, b, i; 2 - P, c, ii; 3 - P, a, iii; 4 - Q, d, iv

C) 1 - Q, a, iii; 2 - P, b, iv; 3 - R, d, ii; 4 - P, c, i

D) 1 - R, a, ii; 2 - P, c, iv; 3 - P, b, i; 4- Q, d, iii

Sol. (D)

Crystal systemAxial

distancesAxial angles

Unit cells

possible

Cubic a b c= = 090 = = = PBF

Tetragonal a b c= 090 = = = PB

Orthorhombic (or)

Rhombica b c 090 = = = PBFE

Monoclinic a b c 090 = = , 0120 PE

Triclinic a b c 090 P

Hexagonal a b c= 0 090 , 120 = = = P

Rhombohedral (or)Triogonal

a b c= = 090 = = P

P =Primitive, B = Body centered, F = Face centred, E = end centred

SECTION II (Total Marks : 16)(Multiple Correct Answers Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of

which ONE OR MORE may be correct.

8.

Lucas test is used to distinguish 1

0, 2

0and 3

0alcohols. Correct options regarding this are

2.

2ROH H Oconc White turbidity

anhydrous ZnClHCl RCl+ +

A) ROH behaves as a Lewis baseB) Greater the value of pKa (alcohol), greater is the reactivity with conc. HCl and thus faster the formation

of white turbidity.

C) Alcohol which readily reacts with Na metal, will give turbidity readily

D) Alcohol which gives red colour in Victor Meyer test, will give turbidity at slower rate than those giving

blue or white colour in Victor Meyer test.

Sol. (A,B,D)In victor Meyer test, 10, 20 and 30 alcohols give red , blue and white colours respectively

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9. The values of E0

of some reactions are givenbelow.

I2

+ 2e 2I

E0

= 0.54 volts

Sn+4

+ 2e Sn2+

E0 = 0.152 volts

Cl2 + 2e 2Cl

E0 = 1.36 volts

Fe+3

+ e Fe

2+E0 = 0.76 volts

Ce+4

+ e Ce

3+E0 = 1.6 volts

Hence

(A) Fe3+

oxidizes Ce+3

(B) Ce4+ can oxidize Fe2+

(C) Sn2+ will reduce Fe3+ to Fe2+

(D) Cl2 will be liberated from KCl by passing I2.

Sol. (B,C)Greater the SRP, greater is the tendency to undergo reduction, greater is the tendency to act as oxidant

and vice cersa

B) As SRP is more for Ce+4/Ce

+3, it undergoes reduction by oxidizing Fe

+2

C)As SRP is less for Sn

+4/Sn

+2, it undergoes oxidation by reducing Fe

+3

10. Which of the following reactions generate tertiary alcohol function

(A)

CH

3CH

3

O CH O3

CH OH3

H O

2

(B)

CH3

CH3

CH O2

Ag +

CH2

Cl

(C)

CH3

Br

H O2

+Ag

(D)

CH3

OCH MgBr

3

CH3

H3

O+

Sol. (A,B,C)

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(A)

CH3

CH3

OCH O

3

H+

CH C CH OCH3 2 3

CH3

OH

(B)

+Ag

CH2

ClCH

3

CH3

CH3

CH3

CH2

+

(C)

CH3

Br

+Ag

+

CH3

CH3 H O2

+-H

+

alkylic and 3carbocation

0

CH3

OH

(D)

CH3

OCH MgBr

3

H C3

CH C C OCH3 3

CH3

+ O

HH

H

11. Which of the following on hydrolysis produces ammonia(A) CaNCN (B)Borazine

(C) Li3N (D) NCl3

Sol. (A,B,C,D)

2 3 3CaNCN 3H O 2NH CaCO+ +

3 3 6 2 3 3 3 2B N H 9H O 3H BO 3NH 3H+ + +

3 3Li N 3H2O 3LiOH NH+ +

3 2 3NCl 3H O NH 3HOCl+ +

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SECTION-III (Total Marls : 15)(Paragraph Type)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions andbased on the other paragraph 3 multiple choice questions have to be answered. Each of these questions

has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.

Paragraph for Question Nos. 12 to 13

Estimation of Carbon and Hydrogen in an organic compound can be carried out by

liebigs method and nitrogen by either Dumas method or by Kjeldahls method. Although

Kjeldahls method is easy to carry out in the laboratory, it is not of general applicability.

12. 0.50g of an organic compound was Kjeldahlised and the NH3 evolved was absorbed in 50ml of 0.5

M H2SO4. The residual acid required 60ml of 0.5 M NaOH. The percentage of nitrogen in theorganic compound is

A) 14 B) 28 C) 56 D) 42

Sol. (C)Moles of H2SO4 reacting with

Ammonia = (moles of H2SO4 absorbed ) (moles of NaOH solution required/2)

= ( )3

3 60 0.5 1050 0.5 102

=20x0.5x10-3

% of N can be calculated using the formula,

1.4volume of acid Normality of acid

wt of organic compd taken

1.4 20 (2 0.5)56

0.5

x x x=

13. 0.46g of an organic compound containing carbon, hydrogen and oxygen was heated strongly in a

stream of N2 gas. The gaseous mixture thus obtained was passed first over heated coke at 1373 K

and then through a warm solution of iodine pentoxide when 1.27 g of I2 was liberated. The

percentage of oxygen present in the organic compound is

A) 87.4 B) 86.96 C) 47.38 D) 38.47

Sol. (B)

% of O can be calculated using the following formulae

2wt of I formed 1005 162 127 wt of org.compd

2

wt of CO formed 10016

44 wt of org.compd 5x16 1.27

%'O' x 86.96%2x127 0.46

= =

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Para. Q. Nos (14 - 16)

Addition of a non-volatile solute to a solvent lowers its vapour pressure. Therefore, the vapour

pressure of a solution (i.e., V.P. of solvent in a solution) is lower than that of pure solvent, at the

same temperature. A higher temperature is needed to raise the vapour pressure upto one

atmosphere pressure, when boiling point is attained. However increase in b.pt. is small. For

example 0.1 molal aqueous sucrose solution boils at 100.05oC. Sea water, an aqueous solution,

which is rich in Na+ and Cl ions, freezes about 1OC lower than frozen water. At the freezing

point of a pure solvent, the rates at which two molecules stick together to form the solid and leave

it to return to liquid state are equal when solute is present. Fewer solvent molecules are in contact

with surface of solid.

However, the rate at which the solvent molecules leave the surface of solid remains unchanged.

That is why temperature is lowered to restore the equilibrium. The freezing point depression in a

dilute solution is proportional to molality of the solute.

14. The freezing point of benzene solution was 5.4oC. The osmotic pressure of same solution at 10oC

is (boiling point of benzene = 5.5oC).Assume solution to be dilute.

Kffor C6H6 is 4.9K molality 1].

A) 0.274 atm B) 0.474 atm C) 0.674 atm D) 0.874 atm

Sol. (B)0

f

f

5.5 5.4 0.1 C

0.1 k .m

0.1

m 4.9

CST

M m

0.1x0.0821x2830.474atm

4.9

= =

=

=

=

=

= =

15. The amount of ice separated out on cooling a solution containing 50 g ethylene glycol in 200 g

water to 9.3oC is [Kffor H2O = 1.86 K molality1]

A) 38.71g B) 61.29g C) 138.71g D) 161.29g

Sol. (A)Ethylene glycol is CH2OH- CH2OH

.

0 ( 9.3) 1.86

50 10009.3 1.86

62

161.29

Ice seperated=200 -161.29 = 38.71

f fT k m

xm

xx

xW

W

=

=

=

=

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16. 2 g of benzoic acid dissolved in 25 g of C6H6 shows a depression in f.pt. equal to 1.62 K. Kf for

C6H6 is 4.9 K molality1. The percentage association of acid if it forms doubls molecules in

solution is

A) 0.8% B) 99.2% C) 90.2% D) 9.8%

Sol. (B)

6 5 6 5 22 ( )C H COOH C H COOH

1 /2

i = 1- + /2 =1+ /2. .

2 10001.62 (1 ).(4.9) .

2 122 25

(1 ) 0.5042

( ) 0.4952

0.9916 99.2%

f fT i k m

x

x

=

=

=

=

= =

SECTION-IV (Total Marks : 28)(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a single digit integer, ranging

from 0 to 9. The bubble corresponding to the correct is to be darkened in the ORS.

17. Co-ordination number of Cr in CrCl3.5H

2O is six. The volume of 0.1 N AgNO

3needed to ppt. the

chlorine in outer sphere in 200 ml of 0.01 M solution of the complex is/are indicated .

(1) 140 ml (2) 40 ml (3) 80 ml (4) 20 ml

Sum of the correct options is

Sol. (6)

Case : 1

[ ]2 5 2Cr(H O) Cl Cl

0.01x2000.002

1000

0.002x2 0.1xV(lit)

V(lit) 0.0440ml

=

=

==

Case : 2

[ ]2 4 2 2Cr(H O) Cl Cl.H O

0.01x2000.002

1000

0.002x1 0.1xV(lit)

V(lit) 0.0220ml

=

=

==

18. The number of products obtained when propane is subjected to vapour-phase nitration is

Sol. (4)

|CH CH CH NO , CH CH CH

3 2 2 2 3 3

NO2

CH NO , CH CH NO3 2 3 2 2

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19. A chemical reaction takes place in three steps having rate constant k1, k2, k3 respectively. If

E ,E

a a1 2

and E

a3

are 50, 40 and 30 respectively in kJ and the overall rate constant, 1 3

2

k kk=

k

.

The overall energy of activation is x x 10 kJ. x is

Sol. (4)Arrhenius equation,

/. Ea RTK A e=

1/// 32; . ; .

1 1 2 2 3 3

Ea RTEa RTEa RTK A e K A e K A e

= = =

As/2

1 31 3 1 3

2 2

E E Ela RTa aK K A AK e

K A

+ = =

50 30 40 40

1 3 2

E E E E KJa a a a

= + = + =

20. Green vitriolO300 C anhydrous salt high

tempL M N + + . If oxidation state of the central atom in

L,M,N are X,Y,Z respectively (X

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Let the activity due to impurity be a cpm.

And that due to Na is (1000 a) cpm.

After 30 hrs a would be reduced to 101

a( )2 cpm

and (1000 a) would be reduced to1

(1000 a)4

cpm

total activity after 30 hrs would be

101a( )2

+1

(1000 a)4

= 200 (given)

Solving, we get

250 1

a4

= 200

a = 200Hence 20% activity was due to impurity.

22. A hydrogen like species (atomic number Z) is present in a higher excited state of quantum

numbern. this excited atom can make a transition to the first excited state by successive emission

of two photons of energies 10.20 eV and 17.0 eV respectively. Alternatively, the atom from the

same excited state can make transition to the second excited state by successive emission of two

photons of energy 4.25eV and 5.95 eV respectively. Value of Z is

Sol. (3)

22

2

1 117 13.6

32eV Z

=

2 17 36

13.6 5Z

=

9=

3Z =

23. There are 100 persons sitting at equal distance in a row XY . N2O (laughing gas) is released from

the side X and tear gas (molar mass = 176) from side Y at the same moment and at the same

temperature. The person who will have a tendency to laugh and weep simultaneously is at z

from side Y. Unit place of z is

Sol. (3)

Let 1 be N2O and 2 be tear gas

5

4

3

2

1

1st

excited

(2nd excited

4.25 eV

5.95 eV10.20 eV

17 eV

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2

2 1

176

100 44

M

M

x

x

=

=

1d

d

d is distance travelled, M is mol wt

From X = 67

From Y = 33

jms-5 paper -1

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