jms-3 paper -1 sol

8/2/2019 Jms-3 Paper -1 Sol

1/15

JMS 3 PAPER 1

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.


2/15

JMS 3 PAPER 1


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. A current is passed through 2 voltameters connected in series. The first voltameter contains XSO4

(aq.) and second has Y2SO4. The relative atomic masses of X and Y are in the ratio of 2 : 1. The

ratio of the mass of X liberated to that of Y liberated is

(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) None of the above

Sol. (A)

G.At.wt c tm

Valency 96500=

212

1 1

1

x

x x

yy

y

A

m V

Am

V

= = =

2. STATEMENT 1 : Sulphide ores are concentrated by froth floatation.

STATEMENT 2 : Pine oil act as collector in froth floatation.

(A) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.

(B) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for

Statement 1.

(C) Statement 1 is True, Statement 2 is False.

(D) Statement 1 is False, Statement 2 is True.

Sol. (B)

Froth floatation is used for removing gangue ( impurities ) from sulphide ores.

In this process, a suspension of the powdered ore is made with water and to this collectorsand froth

stabilisers are added.

Collectors (e. g., pine oils, fatty acids, xanthates) enhance non-wettability of the mineral particles

Froth stabilizers (e. g., cresols, aniline) stabilise the froth.


3/15

JMS 3 PAPER 1


3. Which of the following Corey-house synthesis will give poor yield ?

(A)

CuLi

2

CH I3

+

CH3

(B)

( )CH CH CuLi+CH3 2 32 C Br CH CH C CH

3 2 3

CH3

CH3

CH3

CH3

(C)

( )CH CuLi+CH Br3 32 CH CH3 3 (D)

( )CH CH CH CuLi+CH CH Br3 2 3 22 CH CH CH CH CH3 3 2 3 H C

3

CH

3 Sol : (B)

Corey-House synthesis proceeds through nucleophilic substitution bimolecular (SN2) mechanism.

So, it gives best result with primary halide.Alkyl halide used in B is tertiary . So , it gives poor yield .

4. Among the following, the one formed by condensation polymerization is

(A) Teflon (B) Polystyrene(C) PVC (D) Dacron

Sol. (D)

Dacron ot Terylene is a condensation polymer.

Teflon, Polystyrene and PVC are formed by addition polymerizationCondensation polymers are formed by repeated condensation reaction between two different bi-

functional or tri-functional monomeric units. In these polymerisation reactions, the eliminationof small molecules such as water, alcohol, hydrogen chloride, etc. take place.

Addition polymers are formed by the repeated addition of monomer molecules possessing double or

triple bonds,

Monomer of Teflon is Tetra fluoro ethylene, that of polystyrene is Ethenyl benzene or vinyl benzene

That of PVC is vinyl chloride or Chloroethene and that of Dacron are Hexamethylene diamine and

adipic acid


4/15

JMS 3 PAPER 1


5. The end product of the following reaction is

A OH

OHC

OHC

NH2

B) OH

OHC

C

Cl

N

C

OH

NH2

Cl

CHO

D OH

NC

ClOHC

Sol. (D)Reimer-tiemann reaction as well as carbene addition

6. ColumnI Column II

a) 3 CuO + 2NH3

p) PH3

b) PH4I + NaOH q) Metal

c) Ba(N3)2 r) H3PO4

d) 2H3PO2 s) Nitrogen

t) H2O

A) a-s,t ; b-p,s ; c-q,s,t ; d-p,rB) a-q,t ; b-q,t ; c-p,q,s ; d-t,r

C) a-q,s,t ; b-p,q,t ; c-p,s ; d-p,q,r

D) a-q,s,t ; b-p,t ; c-q,s ; d-p,r

Sol. (D)

3 2 23 2 3 3CuO NH Cu N H O+ + +

4 3 2PH I KOH KI PH H O+ + +

3 2 2( ) 3 Ba N Ba N +

3 2 3 4 32 H PO H PO PH

+


5/15

JMS 3 PAPER 1


7. Arrange following Amines for rate of reaction with CHCl3 + KOH (Hoffmann isocynide test)

(P)

NH2

CH3 (Q)

NH2

NO2 (R)

NH2

(S)

NH2

(A) S > P > R > Q (B) Q > R > P > S (C) S > P > Q > R (D) S > Q > R > P

Sol. (A)

Kb of amines Rate of reaction. (See the mechanism of carbyl amine reaction)Greater the kb , lesser the P

kb , greater is the basic nature

Pkb values of given compounds

S (3.36)

P (8.88)

R (9.42)

Q (12.98)

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 ofwhich ONE OR MORE may be correct.

8. Which of the following form only one oxime on reaction with NH2OH solution.

(A) I (B) II (C) III (D) IV

Sol. (B , C)I and IV give two oximes as the carbonyl group has different gro ups attached to it

C

H

=O+H NOH2

C = N

H

O - H

H

C = N

O - H


6/15

JMS 3 PAPER 1


II and III give only one oxime as the carbonyl group has only one or same type of groups attached to it

+ H NOH2

= N

O - H

9. In the structure given below

H

H

C

H

C = C C C H

H

H

(A) one carbon is in zero oxidation state (B) one c is in 4 state

(C) Two carbons are in 2 state (D) Two carbons are in 1 state

Sol. (A,C,D)

Individual oxidation state = Actual valency No of e-belonging to it after bonding

e-belong to more electronegative atom after bonding

If difference in electronegativity is 0 , both the bonded atoms get equal share of bonded e-

10. From the following diagram, the potentials Eo1and Eo2 are

A) 1.496 V B)1.392 V C)1.425 V D) 1.532 V


7/15

JMS 3 PAPER 1


Sol. (A,C)

For E20

:

0

3 2 1

0

2 2 2

0

2 3

0

3 2

2 4 4 ; G

12 ; G

2

1G

2

3 6 6 ; G

BrO H O e BrO OH

BrO H O e Br OH

Br e Br

BrO H O e Br OH

+ + +

+ + +

+ ;

+ + +

0 0 0 0

1 2 3

0

0

0

G G G G

FE F(1.47)] F(1.6)] F(1.07)]

6 5.88 1.6 1.07

1.425

E

E V

= + +

6 = [4 + [1 + [1

= + +

=

For E10

:

0

3 2 1

2 2

0

3 2 2

2 4 4 ; G

12

21

3 5 6 ; G

2

BrO H O e BrO OH

BrO H O e Br OH

BrO H O e Br OH

+ + +

+ + +

+ + +

0 0 0

1 2

0

0

G G G

FE F(1.47)] F(1.6)]

1.496E V

= +

5 = [4 + [1

=

11. Which of the following reactions give product having two-D atoms at trans position?

(A)

CH3

CH3

Br

BrD

D Zn

alcohol

(B) NaC H C C CH

2 5 3 liq.ND3

(C)

CH COOD3

(D)D

2C H C C CH2 5 3 Lindlar catalyst


8/15

JMS 3 PAPER 1


Sol. (A , B)

I )

Dehalogenation (E2)

Trans Anti - Meso (TAM)

CH3

CH3

Br

BrD

D E2

anti-eliminationC C=

CH3

D

DD

(meso)(trans)

II)

Birch reductionAnti addition

III)

Syn addition

DD

H H .

IV)

Lindlars catalyst Syn addition

CH3

D

C C=

C H2 5

D


9/15

JMS 3 PAPER 1


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

In qualitative analysis, III group includes Fe3+

, Al3+

and Cr3+

. III group reagent is NH4OH in

presence of NH4Cl. NH4Cl is added to suppress the ionisation of NH4OH so that only the III group

radicals are precipitated as their insoluble hydroxide and not the IV and V group radicals as the

solubility products of their hydroxides are much higher. Before adding group reagent to the filterate

of II group, one drop of concentrated HNO3

(oxidising agent) is added and the contents are boiled

for 2-3 minutes and then cooled because the solubility product of Fe(OH)2 is very high as compared

to Fe(OH)3.

When a light bluish green crystalline compound containing cations of IIIrd group and zero group is

analysed, it responds to following tests /reactions.

(a) Its aqeous solution gives a reddish brown precipitate with alkaline solution of potassium

tetraiodomercurate (II)

(b) Its aqueous solution after boiling with conc. HNO3 gives a reddish brown ppt. with sodium

hydroxide.

(c) Solution of reddish brown ppt. in dil. HCl gives blood red colouration with ammonium

sulphocyanide and prussian blue with potassium ferrocyanide.

(d) Its very dilute solution gives white ppt. with (CH3COO)2 Pb solution and white ppt. is only soluble in

hot ammonium acetate.

12. Identify the correct statement :

A) NH4Cl is added along with NH4OH so that only III group cations can be precipitated as their

hydroxides

B) In place of NH4Cl, (NH4)2 SO4 can not be used, as barium ( V group radical) will also be precipitated

as BaSO4along with Al+3 , Fe+3 & Cr+3

C) Aqueous solution of ammonium sulphate also produces white ppt. with BaCl2

solution which is

insoluble in conc. HCl.D) All of these

Sol. (D)

A) It is because , Ksp of hydroxides of Al+3 , Fe+3 & Cr+3 are low and NH4Cl suppresses the ionization ofNH4OH

B) 24 2 4 4 4( ) ( ) 2 NH SO Ba BaSO white NH + ++ +

C) 2 24 4 ( )SO Ba BaSO white ++


10/15

JMS 3 PAPER 1


13 The reddish brown precipitate formed in (a) is ofA) Mercury amido iodide

B) Mercury iodide

C) Oxydimercuric ammonium iodideD) Mercury amido iodide and mercury

Sol. (C)

2 4 3 2 22 3 7 2 0K HgI NH KOH NH Hg O Hg I KI H + + + +

2 NH Hg O Hg I , Oxydimercuric ammonium iodide, Iodide of Millons base(Brown ppt)

Paragraph for Question Nos. 14 to 16

One of the most important applications of chemical reactions is in the production of energy in the

form of useful work. For example in combustion, the heat generated is used to create steam forproduction of mechanical work, or electrical work from a dry cell or storage battery. The quantity

G is called free energy because G represents the maximum quantity of energy released in a

process occuring at constant temperature and pressure that is free-or-available to perform useful

work. G is a driving force in a chemical change that can be used to perform useful work.

14 . The e.m.f. of a certain cell varies as E = 0.4108 + 0.0032 T, where E is in volts and T is in Kelvin

temperature. The H of the following cell reaction is nearly

( ) C25atZnClCeCl2ZnCeCl20

23S4 ++

(A) -79 kJ (B) -59 kJ

(C) -49 kJ (D) -19 KJ

Sol. (A)

;G

G H T S G H T T P

= =

( )

GG H T

T P

nFEG H T

TP

= +

= +

( )

( )

nFEnFE H T

TP

EnFE H nFT

TP

= +

=

0.4108 0.0032

H EE T

nF T P

E T

= +

= +


11/15

JMS 3 PAPER 1


0.0032

0.4108 0.9536 1.3644

E

T P

E

=

= + =

1.3644 300 0.00322 96500

H= +

78.049H J =

15. The enthalpy change of a reaction at 270C is 15KJ/mole. The reaction is feasible if the entropy

change is(A) 15 J mole-1 K-1 (B) - 50 J mole-1 K-1

(C) greater than 50 JK

-1

moles

-1

(D) lesser than 50 J K

-1

mole

-1

Sol. (C)

; 15 300G H T S G KJ S = =

If S is > 50 J K-1 1mole , then G is ve; So process is feasible

16 The change in free energy observed when four moles of N2 and one mole of O2 (assumed to

behave as ideal gases) are mixed at 298K and constant pressure is nearly

A) - 6.2 KJ B) 6.2 KJC) - 3.1 KJ D) 3.1 KJ

Sol. (A)

2.303 logS R n X mix i i

=

( ) 18.314 2.3 4 log 0.8 log 0.2 J K= +

120.8JK=

;G H T Smix mix mix

=

for ideal solution, 0Hmix

=

298 20.8Gmix

=

= - 6198.4 - 6200 J = - 6.2 KJ


12/15

JMS 3 PAPER 1


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. The plot of log (V V) versus t (where V is the volume of nitrogen collected u nderconstant temperature and pressure conditions) for the decomposition of C6H5N2Cl is

given at 50C with an amount of C6H5N2Cl equivalent to 58.3 cc N2.

5 10 15 20 25 30

1.0

1.75

time(min.)

log(V V)

Rate constant for the reaction in hr1 in a single significant digit is

Sol. (4)

2 1

2 1

y yslope

x x

=

1.75 1.0

25 0slope

=

= 0.03

k = 2.303 (slope) min1

= 2.303 ( 0.03) min1

= 0.06909 min1

= 4.14 hr1

4 hr1

18. Total number of DBE present in the product obtained when cyclohexanone reacts with

2,4-DNP is

Sol. (8)

The product is


13/15

JMS 3 PAPER 1


C H N O12 14 4 4

The general formula to calculate DBE of a compound is C aHbNcOd

Then DBE = 12

b ca +

12, 14, 14, 4a b c d = = = =

14 4 1012 1 13 13 5

2 2DBE

= + = =

DBE = 8

19. Amongst the following the total number of compounds which show positive test towards tollens

reagent are

Glucose , Fructose , Formaldehyde , Acetaldehyde , Formic acid , Acetic acid

Sol. (5)

Glucose , Fructose , Formaldehyde , Acetaldehyde , Formic acid give positive test towards tollens test.

20. The structure of B3N3H6 is as follows

N

B

B

N

N

B

H

H

H

H

H

H

How many derivatives of B3N3H4X2 can be derived from the basic structure, by the replacement of

two hydrogen atoms?

Sol. (4)

Replacing any two hydrogen atoms of B3N3H6 , 4 derivatives can be obtained

21. The number of total carbon atoms in the little amount of ester formed during the Kolbes

electrolysis of sodium propanoate is

Sol. (5)

Kolbes electrolysis occurs via free radical mechanism

Free radicals

O

CH CH C O3 2

and CH CH3 2

combine to give ester as the side product.


14/15

JMS 3 PAPER 1


22 No of species that contain N N = structural unit in their structures in the following

52 2 3 2 2 2 4 2 4 2 2 2 2, , , , , , ,N N O HN N H N H N O N O H N O

Sol. (4)

2 3 2 2 2 2 2, , ,N O HN N H H N O contain N N = bonds.

Hydrazoic acid

Nitrous oxide

Diazene

Hyponitrous acid

23. 0.5 mole of Borax and 0.2 mole of Boric acid are mixed in 250 ml of water. pH of the solution

formed is x. Now the solution is diluted to 1 lit. pH of diluted solution is y. The value of

xy

y+ is

( Ka of boric acid is 2.5 105

) (consider borax remains normal)

Sol : (6)

pH

of an acidic buffer, Boric acid and its salt with strong base,NaOH i.e., borax is

[salt]log

[acid]ak pH p= +

0.5(5 log 2.5) log

0.2

5

pH

x

= +

= =


15/15

JMS 3 PAPER 1


On dilution, pH of buffer does not change

So, y = 5

55 6

5

xy

y+ = + =

jms-3 paper -1 sol

Documents