SAFE HANDS & IIT -i

SAFE HANDS & IIT-ian’s PACE JEE Adv Paper-II CHEMISTRY SOLUTIONS

SECTION –I (SINGLE CORRECT)

*This section contains 05 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.(+3/-1)

1. 5 gm of a sample containing two metal M and M, was boiled with dil. H2SO4 and diluted to 500 ml. 20 ml of this solution required 1 ml 0.04 M KMnO4 in acidic medium. In the 2nd experiment 50 ml of the same solution was treated with excess KI and I2 liberated required 5 ml of 0.02 M S2O3

2–

solution. Calculate the percentage of two metals in that sample?The mineral is of MMS2 type and

given that initially M and M both have (+2) oxidation state in the sample and other stable states

are (+3) for M and (+1) for M. [Also given atomic weight of M = 64 and M = 57] (A)6.4%, 1.14% (B)6.4 % , 1.9% (C)1.14%, 5.2% (D)4.5%, 2.2%

Ist Case: In the sample M is present into its lower oxidation state but M is present into its

highest oxidation state. So in KMnO4 solution only M will be oxidised from (+2) to (+3) and M will remain intact.

gm eq. of MnO4– = gm eq. of M

Let in the sample M is x gm and M is y gm

In 20 ml solution eq. of M2+ in the sample is 1500

20

64

x

[n-factor is 1 for M2+ M3+]

1000

504.01

25

1

64

x

x = 1000

2564504.0 = 0.32 gm

percentage of metal M in the sample is 1005

32.0 = 6.4%

2nd Case:Now in presence of KI solution only M will be reduced from M(2+) to M(1+). So

no. of gm eq. of M reacted.

1500

50

57

y (here again n-factor is 1)

= gm eq. of I2 liberated = gm eq. of S2O3

2– required

1000

102.05

10

1

57

y

y = 100

5702.05 = 0.057 gm

percentage metal M in the sample is 1005

057.0 = 1.14%

2. C6H5–CO–CH2–CH2 –CH2–COOH C6H5–CO–CH2–CH2–CH2–CD2OH. This conversion is done by (1) NaBH4 / H3O

+ followed by LiAlD4 / H2O (2) LiAlD4 / H2O followed by NaBH4 / H3O

+ (3) (CH2OH)2 followed by LiAlD4 / H3O

+ (4) DMgBr / H3O

+

(3)


OH OHPh-C-CH

2-CH

2-CH

2-COOHPh-C-CH

2-CH

2-CH

2-COOH

OOO

LiAlD4/H

3O+

Ph-C-CH2-CH

2-CH

2-CD

2OH

O

3. Calculate the pH at which Mg(OH)2 begins to precipitate from solution containing 0.1 M Mg+2 ions.

Ksp for Mg(OH)2 is 1.0 x 10–11 (1) 4 (2) 9 (3) 5 (4) 8 (2)

22 1110Mg OH ,

112

101010 5

0.1

OHOH P

PH=9

4. An organic compound upon hydrolysis produces two compounds one product gave silver mirror test, other product reacts with Hinsberg reagent to produce an alkali insoluble product.The organic compound is

(1)

3 2 3

||

O

CH CH C NHCH

(2)

H C

O3CH

N

3CH

(3) 3NHC

2CH3CH

O

(4) C 2CH 3CH

O

H NH

(2)

H-C-N

OCH

3

CH3

H3O+

H-COOH + CH3-N-H

Ph-SO2Cl

CH3 - N - S - Ph

Silver mirror test

NaOH

O

O

CH3 CH

3

5. The Gibb’s Energy for the decomposition of

2 3Al O at 5000C is as follows

2 3 22 4

; 965 /3 3

rAl O Al O G KJ mole

The potential difference needed for electrolytic reduction of 2 3Al O at 5000C is at least

(1) 2.5V (2) 3V (3) 4.5V (4) 5.0V (1)

3965 102.5

4 96500

GG nFE E

nF

The potential difference needed for the reduction = 2.5V


SECTION –II (INTEGER TYPE)

*This section contains 05 questions. The answer to each question is a NUMERICAL VALUE. (+4/0). For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. – 0.33, 384.45, -6.00, -12.40, 5.00 etc.) 6. Among the complex ions, [Co(NH2–CH2–CH2–NH2)2Cl2]+, [CrCl2(C2O4)2]3–, [Fe(H2O)4(OH)2]+, [Fe(NH3)2(CN)4]–, [Co(NH2–CH2–CH2–NH2)2 (NH3)Cl]2+ and [Co(NH3)4(H2O)Cl ]2+, the number of complex ion(s) that show(s) cis-trans isomerism is Ans. 6.00 All the complexes given show cis-trans isomerism

7. The number of hydroxyl group(s) in Q is

Ans.4.00


8. Among the following, the number of aromatic compound (s) is?

Ans. 5.00 9. A metal crystallizes in two cubic phases, face centered cubic (fcc) and body centered cubic bcc whose unit cell length are 3.5 and 3.0A° respectively. Calculate the ratio of density of fcc and bcc. Ans. 1.25

Solution: = 3

A

m

aN

Mn

For face center cubic (fcc) n = 4, a = 3.5 Å

(fcc) = 3

A

m

)5.3(N

M4

--------------- (I)

For bcc lattice N = 2, a = 3.0 Å

bcc = 3

A

m

)0.3(N

m2

---------------- (ii)

From equation (i) equation (ii)

5.35.35.32

3334

)5.3(

3

2

43

3

)bcc(

)fcc(

= 1.259

10. In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio Which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of fuel. A burner which has been adjusted for methane as fuel (with x litre/hour of CH4 and 6 x litre/hour of O2) is to be readjusted for butane C4H10. In order to get same calorific output, what should be the rate of supply of butane and oxygen. Assume that losses due to incomplete combustion etc., are the same for both fuels and that gases behave ideally. Heats of combustion: CH4 = 809kJ/mol; C4H10 = 2878 kJ/mol. Ans. 5.48

Solution: CH4 + 2O2 CO2 + 2H2O; H = –809 kJ mol–1

Initial volume/hr x 6x x 6x (in litre) Let the temperature be T and assume volume of 1 mole of a gas is V litre at this

condition

V litre or 1 mole CH4 gives energy on combustion = 809 kJ

x litre of CH4 gives energy on combustion = V

)x(809kJ

2878 kJ energy is obtained by 1 mole or V litre C4H10

V

)x(809 kJ energy is obtained by =

2878V

V)x(809

litre C4H10


= 0.281 (x) litre C4H10

Thus butane supplied for same calorific output = 0.281 x litre

C4H10 + 2O2

13 4CO2 + 5H2O; H = –2878 kJ / mole.

Volume of O2 required = 3 Volume of O2 for combustion of C4H10

= 3 2

13 volume of C4H10 = 3

2

13 0.281 (x)

= 5.48(x) litre O2

SECTION –III

*This section contains SIX questions of matching type.(+3/-1) * This section contains TWO tables (each having 3 columns and 4 rows) *Based on each table, there are THREE questions *Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct *For each question, darken the bubble corresponding to the correct option in the ORS. * For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases 11. (A) 12. (B) 13. (C) 14. (A) 15. (B) 16. (A) Direction: Read the assertion and reason carefully to mark the correct option out of the options given below : 1. Both assertion and reason are correct and reason explains assertion correctly. 2. Both assertion and reason are correct but reason does not explain assertion correctly. 3. Assertion is correct and reason is wrong. 4. Assertion is wrong and reason is correct. 5. Both assertion and reason are wrong.

(1)Assertion :The enthalpy of physisorption is greater than chemisorption. Reason : Molecules of adsorbate and adsorbent are held by van der Waals forces in physisorptio and by chemical bonds in chemisorptions Ans. 4

Hint: Assertions is false but reason is true.

The enthalpy of chemisorption is of the order of 1200 kJmol while for physical adsorption it is of the

order of 120 kJmol .

(2)Assertion:Wolframite impurities are separated from cassiterite by electromagnetic separation. Reason :Cassiterite being magnetic is attracted by the magnet and forms a separate heap. Ans. 3 Wolframite being magnetic is attracted by the magnetic roller and forms a heap under it

(3)Assertion : Heterolytic fission involves the breaking of a covalent bond in such a way that both the electrons of the shared pair are carried away by one of the atoms.

Reason : Heterolytic fission occurs readily in polar covalent bonds. Ans. 2 Heterolytic fission occurs when the two atoms differ considerably in their electronegativities and

shared pair of electrons is carried by more electronegative atom


(4) Assertion : F atom has a less negative electron affinity than Cl atom Reason : Additional electrons are repelled more effectively by 3p electrons in Cl atom than by 2p

electrons in F atom Ans. 3. The lower value of electron affinity of F is due to electron-electron repulsion in 2-p orbitals of F-atom

is stronger.

Q. NO. ANS Q. NO. ANS Q. NO. ANS Q. NO. ANS

41 A 46 5.00 51 C 57 A

42 A 47 9.00 52 B 58 D

43 A 48 7.00 53 A 59 A

44 B 49 8.00 54 A 60 A

45 C 50 0.00 55 C

56 D

MOCK TEST-01 (ADVANCE)

PAPER-2 (MATHS) ANSWER KEY

SECTION I SECTION II SECTION III SECTION IV

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