jms i paper 2 solutions

8/2/2019 Jms i Paper 2 Solutions

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JMS I PAPER 2

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

PAPER-2

Maximum Marks: 80

Question paper format and Marking scheme:

1. In Section I (Total Marks: 24), 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 negative

marks in this section.

3. In Section III (Total Marks: 24), 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 this

section.

4. In Section IV (Total Marks: 16), for each question you will be awarded 2 marks for each row in which

you have darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marksotherwise. Thus each question in this section carries a maximum of 8 Marks. There are no negative marksin this section.


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JMS I PAPER 2


SECTION I (Total Marks : 24)

(Single Correct Answer Type)

This Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of

which ONLY ONE is correct.

1. 1 ml of certain type of activated charcoal has a surface of 1000 m2. Calculate the

volume of ammonia at STP) which is adsorbed on 10 ml of this active charcoal if the cross

section of single ammonia molecule is 9 (A0)

2.

(A) 4.13 L (B) 3.14 L (C) 29 10 L (D) 29 10 L

Sol. (A)

Total area available on charcoal = 10 ml 1000 m2/ml = 10000 m

2

Area of each NH3 molecule =20 29 10 m

No. of NH3 molecules adsorbed = 2010000

9 10

Vol. of NH3 at S.T.P =nRT

P

20 23

10000 1 0.0821 273

9 10 6.023 10 1

=

= 4.13 L

2.

( )

( )

( ) ( )

( )

P O 1 CH MgBr 1 I +Ca OH22 5 3 2C NH A B C,

2 + 2 2 H O3

O

C is

(A)

C H3

C

O

(B)

C OH

O

(C)

C

O

(D)

C CH CH2 3

O

Sol. (C)


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JMS I PAPER 2


C

O

Ca

C

C

O

O

( )Take2 moles C OH+CHI

3

O

O

O

( )I +Ca OH 22Haloform

reaction

3. Bleaching powder reacts with iodide ion according to the following unbalanced equation.

OCl-+ I

-+ H

+ I2 + Cl

-+ H2O

A 0.6 g sample of bleaching powder requires 35.24 ml of 0.1084 M Na2S2O3 to titrate the

liberated iodine.The percentage of available Cl in the sample is

(A) 12.59 (B) 11.29 (C) 22.58 (D) 45.16

Sol. (C)

(3.55).( )( )%of available chlorine

Wt of bleaching powder

X Y=

( where X = Molarity of hypo, Y = Volume of hypo in ml)

(3.55).(0.1084)(35.24)%of available chlorine

0.6=

= 22.6


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JMS I PAPER 2


4. A solid is made up of elements X, Y and Z. it has BCC arrangement of atoms with X

occupying body centre and Y and Z are at alternate corners. If atoms from two corners

along a body diagonal are removed, the formula of compound is

(A) 3 8 8X Y Z (B) 2 3 3X Y Z (C) 8 3 3X Y Z (D) 3 3XY Z

Sol. (C)

X body centre = 1

Y, Z alternate corners = 4When atoms from two corners along a body diagonal are removed,

(i.e. one atoms of Y and Z each),

Then

X = 1,1

38

Y= ,1

38

Z=

8 3 3X Y Z

5. In the given reaction

Product is

(A) 1 mole of HCOOH and 1 mole of HCHO

(B) 2 mole of HCHO and 1 mole of HCHO(C) 2 mole of HCHO and 1 mole of HCOOH

(D) 2 mole of HCHO and 4 mole of HCOOH

Sol. (D)

Acetal hydrolysis


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JMS I PAPER 2


6. Match the following

List I List - II

(Refining process) (Metal Used)

a) Distillation P) Ge

b) Monds process Q) Tic) Zone refining R) Ni

d) Van Arkel method S) Hg

A) a R ; b - P ; c Q ; d - S

B) a - Q ; b R ; c S ; d - P

C) a S ; b R ; c P ; d - Q

D) a R ; b S ; c Q ; d P

Sol. (C)

Distillation

method is useful for low boiling metals like zinc and mercury.Mond Process is useful for the refining of nickel. In this process, nickel is heated in astream of carbon monoxide forming a volatile complex, nickel tetracarbonyl:

Ni + 4CO Ni(CO)4

This carbonyl is subjected to higher temperature upon which it is decomposedgiving the pure metal:

Ni(CO) Ni + 4CO

Zone refining is very useful for producing semiconductor and other metals of very high

purity, e.g., germanium, silicon, boron, gallium and indium.


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JMS I PAPER 2


Van Arkel Method is used for Refining of metals like Zirconium or Titanium: This method is

very useful for removing all the oxygen and nitrogen present in the form of impurity in

certain metals like Zr and Ti. The crude metal is heated in an evacuated vessel with iodine. The

metal iodide being more covalent, volatilisesZr + 2I2 ZrI4

The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K. The

pure metal is thus deposited on the filament.

ZrI4 Zr + 2I2

7. , A is

(A)

NO2

Me (B)

NO2

Me (C)

Me

NO2

(D)

NO2

Sol. (D)F.C. reaction is not applicable for strong deactivating ring like Ph-NO2

8. Statement-I: Nuclide 30Al13

is less stable than 4020

Ca

Because

Statement-II: Nuclides having odd number of protons and odd number of neutrons are

generally unstable.

(A) If both Statement-I and Statement-II are true and Statement-II is a correctexplanation ofStatement-I.

(B) If both Statement-I and Statement-II are true and Statement-II is not a correct

explanation ofStatement-I.(C) IfStatement-I is true but Statement-II is false.

(D) IfStatement-I is false but Statement-II is true.

Sol. (A)

4020

Ca has even no. of protons and even no. of neutrons

3013

Al has odd no. of protons and odd no. of neutrons


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JMS I PAPER 2


Nuclides having even number of protons and even number of neutrons are

most stable due to max spin pairing of protons among protons and neutrons among

neutrons

Nuclides having odd number of protons and odd number of neutrons areleast stable due to min spin pairing of protons among protons and neutrons among

neutrons.

Stability order of nuclides is

Even protons - even neutrons > even protons odd neutrons >

odd protons even neutrons > odd protons odd neutrons

SECTION II (Total Marks : 16)(Multiple Correct Answer(s) 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.

9. Which of the following is/are Isolobal

A) CH3+ B) H3O

+ C) CH3 D) NH3

Sol. (B,C, D)

Two fragments are isolobal if the number, symmetry properties, approximate energy, shape of

the frontier orbitals and no of e- in them are similar.

10.

Which of the following regents can perform this conversion successfully?

(A) H2, Raney Ni,

(B) Mg, THF, H2O(C) NaBH4, C2H5OH, H+, H2O (D) HCHO, OH

Sol. (C, D)

H2, Raney Ni, can reduce the complete system

Mg, THF, H2O is not effective

NaBH4 can reduce aldehydes,ketones and acid halides.

HCHO, OH gives crossed Cannizaro reaction.


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JMS I PAPER 2


11. Which of the following combination/s is/are incorrect

(Polymer) (Monomer Unit)

A) Nylon 6, 6 Terepthalic acid and ethylene glycolB) Bunas Buta1, 3diene & Acrylo nitrile

C) Nylon 6 Caprolactam

D) Dacron Hexamethylene diamine & Adipic acid

Sol. (A,B, D)

Dacron :

Nylon 6 :

Nylon 6, 6 :

n H2N-(CH2)6-NH2 + n HOOC(CH2)4 COOH

( - NH-(CH2)6 - NHCO - (CH2)4 - CO - )n

Bunas :

[ ]2 2n CH CH CH CH = = + [ ]6 5 2n C H CH CH =

Buta -1,3-diene Styrene


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JMS I PAPER 2


12. In which of the following compound electrophilic aromatic substitution takes place in

phenyl ring present on right hand side

(A)

C O

O

(B)

(C) (D)

CH2

O

C

Sol. (A, B, C)

Benzene rings activated due to lone pair of electrons on attaching atom. Therefore, electrophilicaromatic substitution takes place in phenyl ring present on right hand side

SECTION-III (Total Marks : 24)

(Integer Answer Type)

This section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0

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

13. A substance is analyzed by paper chromatographs, giving the chromatogram shown

Rf value of the substance represented by the spot at 10.0 cm Z x 10-1

. Z is

Sol. (8)

10 2 80.8

12 2 10R

f

= = =

14. The standard oxidation potential of2/ Ni Ni + electrode is 0.2116 V. If this is combined

with hydrogen electrode in acid solution, at what pH of the solution, emf will be zero at

250C.

Sol. (4)


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JMS I PAPER 2


2+Ni

=1 M and 1

2

PH

= atm.

Std oxidation potential of Ni electrode = 0.2116Std oxidation potential of H electrode = 0.0V Ni is oxidized, H+ is reduced.

2

2

2

2

2

2 2

2

Ni Ni e

H e H

Ni H Ni H

+

+

+ +

+

+

+ +

Nernst equation,[ ]

[ ]0 0.059

logtan

cell cell

productsE E

n reac ts=

( )0.059 1

0 0.0 0.2116 log22

H

=

+

20.0590 .2116 log

2o H+ = +

log 4H+ =

pH = 4

15. Unit place of EAN of each Mn in Mn2CO10 is

Sol. (6)

EAN of each Mn = 25 (electrons from each Mn) + 10 (electrons from 5 CO ligands)

+ 1 ( electron from Mn- Mn bond) = 36


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JMS I PAPER 2


16. A gaseous mixture of Helium and Hydrogen contain 20% H2. The number of diffusionstages required if H2 is to be enriched from 20% to 80% by means of gaseous

diffusion are

Sol. (8)Separation factor

( )( )

( )2

2

42

===

Her

Hrf

( )

( )

( )

( )( )n

initialn

fHeN

HN

HeN

HN

=

22

number of diffusion stages requried.

( )n280

20

20

80=

( ) ( ) 2/4 22216 nn == 84

2== n

n

17. If methyl bromide and ethyl bromide are mixed in equal proportions,and the mixture istreated with sodium in presence of dry ether, the no of possible organic products are

Sol. (5)

A case of Wurtz reaction (See Daily Crackers for detailed mechanism)


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JMS I PAPER 2


18. What is the min pH required to prevent the precipitation of ZnS in a solution that is 0.01M

ZnCl2 and saturated with 0.1M H2S.

21 2010 , 10

1 2K K K

Sp a a

= =

Ans. (1)

22 2

ZnCl H S ZnS HCl+ +

2 2K Zn S

Sp+ =

21102 19100.01

S

= =

12

Ka H S H HS+ +

22K

a HS H S +

+

22

.1 2

2

H S

K Ka a

H S

+

=

( )2 191020100.1

H+

=

0.1H+ =

1pH =


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JMS I PAPER 2


SECTION-IV (Total Marks : 16)

(Matrix-Match Type)

This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and

five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching

with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B

matches with the statements given q and r, then for the particular question, against statement B, darken the

bubbles corresponding to q and r in the ORS.

19. Column I Column-II

(Oxide of Nitrogen) (Characteristic)

A) N2

O P) Zn + dil. HNO3

B) N2O3 Q) Planar

C) NO2 R) Presence of NN linkage

D) N2O5 S) Acidic

T) Cu + Conc. HNO3

Sol. A P,R B Q,R,S C S,T; D Q , S

A)

Linear, colourless gas, neutral

4Zn + 10HNO3(dilute) 4 Zn (NO3)2 + 5H2O + N2O

B)

blue solid,acidic, planar,has N-N linkage


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JMS I PAPER 2


C)

Angular, Brown gas , acidic

Cu + 4HNO3(conc.) Cu(NO3)2 + 2NO2 + 2H2O

D)

Colourless solid, planar, acidic

20. Match the reactions in column I with appropriate types of steps/reactive intermediate

involved in these reactions as given in column II.

Column I Column-II(Process (or) reaction)

(A)

O

Zn-Hg

(P) A hydrocarbon is formed

(B)

(Q) An amine (or) a derivative of amine is

Formed


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JMS I PAPER 2


(C)

Ph

H

H C3

COOH

( )( )1 SOCl2

2 NaN3

( )3 /2

H O

(R) Rearrangement takes place

(D)

(S) Substitution takes place

Sol. A P, B Q , C Q,R,S; D P , S

A is Clemmensen reduction : Reduction of ketones (or) aldehydes

to alkanes using zinc amalgam and hydrochloric acid

Product is

B is cope elimination : Pyrolytic elimination of amine oxide is cope eliminmation.It is a syn

elimination and gives hofmannproduct .

Product is

C is a substitution reaction involving Curtius rearrangementProduct is

Ph

H

H C3

NH2


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JMS I PAPER 2


D is Corey-House synthesis. It is a substitution reaction.(Gilmans reagent) R2CuLi + R'-X R-R' + RCu + LiX

Product is

CH3

jms i paper 2 solutions

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