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IIT JEE 2011 Paper 1 (Code 4)

Part - I: Chemistry

SECTION - I (Total Marks: 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has 4 choices (A), (B),

(C) and (D) out of which ONLY ONE is correct.

1. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density

1.15 g/mL. The molarity of the solution is

(A) 1.78 M (B) 2.00 M

(C) 2.05 M (D) 2.22 M

Sol. (C)

Mass of solution = 1120 g

So, volume of solution =Mass 1120

973.9mlDensity 1.15

wt. of solute 1000Molarity

mol. wt. of solute volume of solution120 1000

Molarity 2.05M60 973.9

2. AgNO3 (aq.) was added to an aqueous KCl solution gradually and the conductivity of

the solution was measured. The plot of conductance ( ) versus the volume of AgNO3is

(A) (P) (B) (Q)

(C) (R) (D) (S)

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Sol. (D)

Ag+ and K+ have nearly same ionic mobility so first part of the curve does not show

increase in conductivity. After the equivalence point i.e. in second part of the curve,

conductivity increase because of increase in concentration of Ag+ and NO3- ions.

3 3AgNO + KCl AgCl + KNO

3. Among the following compounds, the most acidic is

(A) p-nitrophenol (B) p-hydroxybenzoic acid

(C) o-hydroxybenzoic acid (D) p-toluic acid

Sol. (C)

o-hydroxybenzoic acid is most acidic because of stabilisation of its conjugate base by

intramolecular hydrogen bonding and ortho effect.

4. The major product of the following reaction is

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Sol. (A)

5. Extra pure N2 can be obtained by heating

3 4 3

4 2 2 7 3 2

(A) NH with CuO (B) NH NO

(C) (NH ) Cr O (D) Ba(N )

Sol. (D)

Extra pure N2 can be obtained by heating Ba(N3)2 since there is no gaseous by

product formed in the reaction whereas other reactants involve formation of

gaseous by products.

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3 2 2Ba(N ) Ba(s) + 3N

6. Geometrical shapes of the complexes formed by the reaction of Ni2+

with Cl-

, CN-

and

H2O, respectively are:

(A) Octahedral, tetrahedral and square planar

(B) Tetrahedral, square planar and octahedral

(C) Square planar, tetrahedral and octahedral

(D) Octahedral, square planar and octahedral

Sol. (B)

+2 - -24

+2 - 24

+2 +22 2 6

Ni + 4Cl [NiCl ] (tetrahedral)

Ni + 4CN Ni(CN) (square planar due to strong field ligand)

Ni + H O Ni(H O) (octahedral)

7. Bombardment of aluminium by - particle leads to its artificial disintegration in twoways, (i) and (ii) as shown. Products X, Y and Z respectively are:

(A) Proton, neutron, positron (B) Neutron, positron, proton

(C) Proton, positron, neutron (D) Positron, proton, neutron

Sol. (A)

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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. Amongst the given options, the compound(s) in which all the atoms are in one plane

in all the possible conformations (if any), is (are)

Sol. (B, C)

A. Only two conformers cis- and trans- will have all atoms in same plane. Two

hydrogens will lie in different plane due to rotation along C2- C3 bond.

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B. All atoms lie in one plane hence it is planar.

C. All atoms lie in one plane hence it is planar.

D. It has terminal hydrogens lying perpendicular to each other and hence don't lie in

same plane.

9. According to kinetic theory of gases

(A) Collisions are always elastic

(B) Heavier molecules transfer more momentum to the wall of the container

(C) Only a small number of molecules have very high velocity

(D) Between collisions the molecules move in straight lines with constant velocities

Sol. (A, B, C)

According to Kinetic theory of gases

(i) Collisions are always elastic

(ii) Momentum transferred on wall by one collision along x component = 2 mvx

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(iv) During each collision, the speed and the direction of motion of molecules

undergo a change.

10. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is

(are)

(A) Adsorption is always exothermic

(B) Physisorption may transform into chemisorption at high temperature

(C) Physisorption increases with increasing temperature but chemisorption decreases

with increasing temperature

(D) Chemisorption is more exothermic that physisorption, however it is very slow

due to higher energy of activation

Sol. (B, D)

A. Adsorption is not always exothermic; chemisorption of H2 on glass is an

endothermic process.

B. At high temperature, sufficient activation energy for chemical adsorption is

provided so physisorption may transform into chemisorption.

C. Physisorption occurs at low temperatures and increases with increasing

temperature but chemisorption occurs at high temperatures and decreases with

increasing temperature.

D. Chemisorption is more exothermic than physisorption. Enthalpy of adsorption for

chemisorptions is larger than that for physical adsorption.

11. Extraction of metal from the ore cassiterite involves

(A) Carbon reduction of an oxide ore (B) Self reduction of a sulphide ore

(C) Removal of copper impurity (D) Removal of iron impurity

Sol. (A, C, D)

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Cassiterite ore contains 0.5-10% of meal as SnO2, the rest being impurities of Fe,

Cu.

o1200-1300 C2SnO + C Sn CO

SECTION - III (Total Marks: 15)

(Paragraph Type)

This section contains 2 paragraphs. Based upon one of the paragraphs, 2 multiple

choice questions and based upon the second paragraph, 3 multiple choicequestions have to be answered. Each of these questions has four choices (A), (B),

(C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions Nos. 12 and 13

An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only

organic product through the following sequence of reactions, in which Q is an

intermediate organic compound.

12. The structure of compound P is

Sol. (D)

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13. The structure of the compound Q is

Sol. (B)

Paragraph for Questions Nos. 14 to 16

When a metal rod M is dipped into an aqueous colourless concentrated solution of

compound N, the solution turns light blue. Addition of aqueous NaCl to the blue

solution gives a white precipitate O. Addition of aqueous NH3 dissolves O and gives

an intense blue solution.

14. The metal rod M is

(A) Fe (B) Cu

(C) Ni (D) Co

Sol. (B)

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3 3 2Cu + AgNO Cu(NO ) + Ag

(M) (N) Blue solution

The metal rod dipped in the solution is of Copper.

15. The compound N is

3 3 2

3 3 3 2

(A) AgNO (B) Zn(NO )

(C) Al(NO ) (D) Pb(NO )

Sol. (A)

N is silver nitrate solution in which the metal rod is dipped.

+3 3 2Cu + AgNO Cu(NO ) + Ag

(M) (N) Blue solution

16. The final solution contains

2+ 2- 3+ 2+3 4 4 3 4 3 4

+ 2+ + 2+3 2 3 4 3 2 3 6

(A) [Pb(NH ) ] and [CoCl ] (B) [Al(NH ) ] and [Cu(NH ) ]

(C) [Ag(NH ) ] and [Cu(NH ) ] (D) [Ag(NH ) ] and [Ni(NH ) ]

Sol. (C)

+

White ppt (O)

+3 3 2

2+ 23 3 4

Ag + NaCl AgCl

AgCl + 2NH [Ag(NH ) ] Cl

Cu 4NH [Cu(NH ) ]

(Bluesolution)

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 answer

is to be darkened in the ORS.

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17. The work function ( ) of some metals is listed below. The number of metals which

will show photoelectric effect when light of 300 nm wavelength falls on the metal is

Sol. (4)

Energy of incident photon =34 8

9 19hc 6.62 10 3 10

4.14 eV300 10 1.6 10

Kinetic Energy; E = h - work function

Energy of incident photon should be higher than work function to slow photoelectric

effect. Li, Na, K and Mg have work function less than 4.14 eV so they'll show

photoelectric effect.

18. To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1

mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm. at 0C)

are introduced. Considering the ideal gas behavior, the total volume (in litre) of the

gases at 0C is close to

Sol. (7)

Partial pressure of He = 1 - 0.68 = 0.32 atm.

If V is the total volume of gases at 0oC, then

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23. The maximum number of electrons that can have principal quantum number, n = 3,

and spin quantum number, s1

m , is2

Sol. (9)

Number of orbitals for n = 3 is n2 = (3)2 = 9

So, total no. of electrons in n = 3 is = 2n2 =