Jms-2 Paper -2 - Solutions

8/2/2019 Jms-2 Paper -2 - Solutions

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JMS 2 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 thebubble 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: 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 marks

in this section.


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JMS 2 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 ofwhich ONLY ONE is correct.

1. Choose the incorrect statement.

(A) C6H6 is more reactive than C6D6 towards nitration

(B) C6D6 is more reactive than C6T6 towards sulphonation

(C) C6H6 is more reactive than C6D6 towards sulphonation

(D) C6D6 and C6H6 have equal reactivity towards nitration

Sol. (A)

Order of reactivity towards nitration is6 6 6 6 6 6

C H C D C T = =

Order of reactivity towards Sulphonation is6 6 6 6 6 6

C H C D C T > >

It is because sulphonation shows isotopic effect i.e., rate of the reaction slows down when H atoms are

replaced by heavier isotopes like Deuterium or Tritium, which inturn is because, the loss of proton from

carbocation is slow i.e., rate determining stepn in the mechanism of sulphonation

2. Octane, C8H18 a constituent of gasoline burns according to the equation.

C8H18 (l) +1

122

O2 (g) 8 CO2(g) + 9 H2O(l)

0.257 g of C8H18 was burnt in the presence of excess O2 in a bomb calorimeter. The heat capacity

of the calorimeter was 1.726 x 103 J/oC. The temperature of the calorimeter and 1.2 x 103 g of

water rose from 21.22oC to 23.05oC. The heat of combustion per mole of octane is

A) - 6266.9 KJ mole1

B) - 5476.9 KJ mole1

C) - 5269.9 KJ mole1

D) - 4572.9 KJ mole1

Sol. (B)

Q = m s t

2 1200 4.184 (23.05 21.22) 9188H OQ x x J = = 1726 (23.05 21.22) 3159

BombcalorimeterQ x J= =

2reaction H O BombQ Q Q= +

= 3159 + 9188 = 12347J

Heat of combustion of 1 mole of Octane

12347 114

0.257

5476879 /

5476.9 /

x

J mol

KJ mol

=

=

=


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3.

Which of the following set of reagents is the most appropriate to perform the above conversion

(A)HIO4; OH ; ZnHg/conc HCl

(B) Cold, alkaline KMnO4; Pb(OAC)4, OH; Li/Liq. NH3

(C) O3, (CH3)2S; OH; Li/Liq NH3

(D) KMnO4, H+; OH; ; NH2 NH2, OH

, 150oC

Sol. (D)

4. The height of a HCP unit cell is 5.7150A. Volume of the unit cell in 0

3

A ?

A) 91 B) 182 C) 273 D) 82.5

Sol. (B)

. The height of the HCP unit cell =2

4 .r = 5.7153

05.715 3

r = = 1.75 A

2 4

The base area of the unit cell =2

6 3 r The volume of the unit cell = base area x height

= 3 0 324 2 r 181.8 A=

5. Identify A in the following scheme.

)OHC(A 147+ ,H

B )HC( 127

alcohol2oCOOHCH,Zn/O 33

altanhepoxo6


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(A)

3CH

OH(B)

3CH

3CHHO

(C)

3CH

OH(D) 3

CH

3CH

OH

Sol. (C)

6. Calculate the cell emf in mV for

pt, H2 (1 atm) / HCl (0.01 M) / AgCl (s) / Ag (s) at 298 K

if0

fG values at 250C are -109.56 KJ mole

-1for AgCl and -130.79 KJ mole

-1for

( ) . H Cl aq+ + (A) 456 (B) 654 (C) 546 (D) 564

Sol. (A)


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JMS 2 PAPER - 2


+H + 2 AgCl 2 Ag + 2 H + 2 Cl2

0 0

cell f

G G = (products) - 0f

G (reactants)

( ) ( )0 2 130.79 2 109.56cellG =

= -42.46 KJ mole-10 o

cell cellG nFE =

042460 2 96500cell

E = 0 0.22cell

E V=

( )4

8

0.0590.22 log 0.01

2

0.0590.22 log(10 )

20.059

0.22 ( 8)2

0.22 0.059( 4)

0.22 0.236

0.456

456

cellE

V

mv

=

=

=

=

= +

=

=

= 456 mV

7. Match the following.

Column-I Column-II

(a) H2S2O2 (P) SOS linkage

(b) H2S2O3 (Q) Diprotic

(c) H2S2O4 (R) S = S linkage

(d) H2S2O7 (S) S S linkage

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

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

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

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

Sol. (B)

Thio sulphurous acid (H2S2O2)


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Thio sulphuric acid (H2S2O3)

Dithionous acid (H2S2O4)

Pyrosulphuric acid : (H2S2O7)

8. For the reaction, A Products

=

a

1

xa

1

t

1k

A graph between (ax)1

and time t is of the type

(A)

p

t

1)xa(

(B)

p

t

1)xa(

(C)

p

t

1)xa(

(D)

p

t

1)xa(

Sol. (C)

1 1 1K

t a x a

=

1 1Kt

a x a

= +

y mx C = + So, the given graph must be the one with positive slope (with slope = K and intercept = 1/a)


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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. When a mxiture of NaCl and K2Cr2O7 is heated with conc. H2SO4.

A) A deep red vapour is evolved

B) The vapour when passed into NaOH solution gives a yellow solution of Na2CrO4

C) Deep red vapour is of chromyl chloride

D) Chlorine gas is evolved

Sol. (A,B,C)

4 6 2 6 32 2 7 2 4 2 2 4 2

K Cr O NaCl H SO CrO Cl NaHSO H O+ + + +

4 2 22 2 2 4 2

CrO Cl NaOH Na CrO NaCl H O+ + +

Deep red vapour yellow solutionof chromyl chloride

10. 3 2 24NH +5O 4NO+6H O ; if reaction starts with 3mole of NH3 and X mole of O2 (obtained by

complete decomposition of 5 lit H2O2 solution of 11.2 volume strength), then choose the correct

option(A) Oxygen is limiting reagent

(B) 2 mole of NO is formed after complete reaction(C) Normality of H2O2 solution is 2

(D) % strength of H2O2 solution is 3.4

Sol. (A,B,C,D)

4 5 4 63 2 2

NH O NO H O+ +

Initial 3 X = 2.5 0

Reacted 2 2.5 0

Left or 1 0 2

Formed

5L, 11.2 volume strength H2O2 = 56 lit of O2 at S.T.P. i.e, 2.5 moles

11.22

5.6N = =

% strength =11.2

3.4%3.3

= =


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JMS 2 PAPER - 2


11. Which of the following polymersdo not have ester linkage

(A) Nylon (B) Bakelite (C) Teflon (D) Terylene

Sol : (A,B,C)

(a) Nylon has amide linkage

(b) Bakelite is phenolformaldehyde resin(c) Teflon's monomer is tetrafloroethylene.

(d) Terylene has COO group, with monomer units Ethylene glycol and terephthalic acid.

12. Phosphine is obtained, whenA) Red P is heated with NaOH B) White P is heated with NaOH

C) 3 2Ca P reacts with water D) Phosphorus trioxide is treated with hot water

Sol : (B,C,D)

In the laboratory, phosphine is prepared by heating white phosphorus with concentrated NaOH

solution in an inert atmosphere of CO2

2 3 2 2P 3NaOH 3H O PH 3NaH PO+ + + Phosphine is prepared by the reaction of calcium phosphide with water or dilute HCl.

( )3 2 2 32Ca P 6H O 3Ca OH 2PH+ + 3 2 2 3

Ca P 6HCl 3CaCl 2PH+ +

P4O6 reacts with hot H2O violently and forms PH3.4 6 2 3 4 3

6 3P O H O H PO PH + +

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. heat2 6 3 23B H 6NH 3X 2Y 12H+ + .In the anionic part of the intermediate X, the covalency of

the central atom is

Sol : (4)

( ) [ ] heat2 6 3 2 3 4 3 3 6 223B H 6NH 3 BH NH BH 2B N H 12H+

+ +

14. 3 atoms of hydrogen are excited from ground state to n = 4, then maximum number of different

radiations,when all H atoms come at ground state finally is

Sol. (5)


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Different possible transitions for the three H atoms during de excitation are

15. The number of organic products obtained when but-1-ene reacts with Cl2 in CH3OH medium

containing LiBr is

Sol. (3)

16. Air contains O2 and N2 in the ratio of 1 : 4. The ratio of their solubilities in terms of mole

fractions at atmospheric pressure and room temperature is Z x 10-1

. Z is

(Given, Henrys constant for O2 =73.3 10 torr, for N2 =

76.6 10 torr)

Sol. (5)( ) .P K A H AA =

( )

PA

AK

H A

=

( )( )

7

7

22 2

2 22

0.2 6.6 10

0.8 3.3 10

1

2

K NO POH

PN K ON H

x x

x x

=

=

=

( % of O2 and N2 are taken as 20 and 80 respectively)

17. n-Butylethanoate when heated to high temperature produces a hydrocarbon. The number of

carbon atoms in the hydrocarbon is

Sol. (4)It is pyrolytic elimination of ester (Ei). It gives syn elimination and Hofmann product .


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18. The approximate change in pH when a 0.1M solution of CH3COOH in water at 250C is diluted to

a final concentration of 0.01 M is Z x 10-1

. Z is

( )51.85 10 , log1.36 0.1335, log4.3 0.6335aK= = =

Sol. (5)

. H K C

a

+ =

Case-I : 5 31.85 10 0.1 1.36 10H + = =

M

3 log1.36 3 0.1335 2.8665pH = = =

Case-II: 5 41.85 10 0.01 4.3 10H + = =

M

4 log 4.3 4 0.6335 3.3665pH = = =

Change in pH = 3.3665 2.8665 = 0.5

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. Match the following.Column-I Column-II

(A) -decay (P) Results in increase of n/pratio

(B) -decay (Q) Results in decrease of atomicnumber by 1

(C) Positron emission (R) Accompanied by emission of

a neutrino

(D) K-electron capture (S) Results in production of X-

rays

(T) Does not affect mass number

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


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(A) decay increases n/p ratio to2

2

n

p

Eg:238 234

4292 90

U Th He +

(B) -decay does not affect mass number

14 14 06 7 1

C N e +

(C) Positron emission results in increase in n/p ratio

Decrease in atomic number by 1

Accompanied by emission of anti neutrino

Does not affect mass number.

( )13 13 07 6 1 N C e + +

+

( )p n + + +

(D) K-electron capture results in increase in n/p ratio

Decrease in atomic number by 1

Results in production of X rays or neutrino

Does not affect mass no

K shell

p e n

+ +

7 0 7 04 1 3 0

Be e Li v+ +

133 0 13356 1 55

Ba e Cs X rays+ +


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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

(A) (P) Product formation through

intramolecular reaction

(B) O||

Ph C H OH

(Q) Reaction has carbanion

intermediate

(C)

3 2 5

O||

CH C O C H 2 5C H O

2 5C H OH

(R) Product is keto ester

(D) O||C

2 5O C H

O

2 5O C H

2 5C H O Na +

2 5C H OH

(S) Aldol product is not formed

in given condition

(T) Product is aldol

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

A-P, Q, T :

B-S :


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C-Q, R, S :

D-P,Q,R,S :

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Jms-2 Paper -2 - Solutions