AIEEE 2010PAPER

8/2/2019 AIEEE 2010PAPER

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MAHESH JANMANCHI AIEEE 2010

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Maximum Marks: 144

Question paper format and Marking scheme:

1. This question paper has 30 questions.

2. Q. No 4 to 9 and 13 to 30 consist of4 marks each and Q. No 1 to 3 and 10 to 12 consist of8 marks.

3. each (one fourth) of total marks allotted to each question will be deducted for indicating incorrectresponse.

4. No deduction from the total score will be made if no response is indicated for an item in the answer

sheet.


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1. The standard enthalpy of formation of NH3 is 46.0 kJ mol1

. If the enthalpy of formation of H2

from its atoms is 436 kJ mol1

and that of N2 is 712 kJ mol1

, the average bond enthalpy of NH

bond in NH3 is

(1) -1964kJmol (2) -1352kJmol+ (3) -11056 kJmol+ (4) -11102kJmol

Sol. (2)

Enthalpy of formation of NH3 = - 46 kJ/mole1

2 2 33 2 2 46

fN H NH H kJ mol + =

Bond dissociation is endothermic (+ve) and bond formation is exothermic( - ve)

Assuming x is the bond energy of N-H bond ((kJ mol-1

), then

( )712 3 436 6

46 2

x +

=

x = 352 kJ/mol

2. The time for half life period of a certain reaction A products is 1 hour. When the initial

concentration of the reactant A, is 2.0 mol L1

, how much time does it take for its concentration

to come from 0.50 to 0.25 mol L1

if it is a zero order reaction ?

(1) 4 h (2) 0.5 h (3) 0.25 h (4) 1 h

Sol. (3)For a zero order reaction 0

[ ] [ ]k =

t

tA A (1)

Since [A0] = 2M, t1/2 = 1 hr;

2k = 1

2x1=

when t = t1/2 , [At] = [A0]/2

[ ]0

12

Ak=

2t

(2)

sub in equation (1)

0.5 0.250.25hr

1t

= =


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3. A solution containing 2.675 g of COCl3. 6 NH3 (molar mass = 267.5 g mol1) is passed through a

cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give

4.78 g of AgCl (molar mass = 143.5 g mol1

). The formula of the complex is

(At. Mass of Ag = 108 u)

(1) ( )3 36Co NH Cl (2) ( )2 3 4CoCl NH Cl

(3) ( )3 3 3CoCl NH (4) ( )3 25CoCl NH Cl

Sol. (1)

AgNO3CoCl .6NH x Cl x AgCl

3 3

( ) ( )3 3n AgCl = X x n CoCl .6NH

4.78 2.675= X = 3

143.5 267.5X

The complex is ( )3 36Co NH Cl

4. Consider the reaction:

( ) ( ) ( ) ( ) ( )2 2 2 2Cl aq H S aq S s H aq Cl aq+ + + +

The rate equation for this reaction is rate = [ ][ ]2 2k Cl H S

Which of these mechanisms is/are consistent with this rate equation?

(A) ( )2 2Cl H S H Cl Cl HS slow+ +

+ + + +

( )Cl HS H Cl S fast + +

+ + + (B) ( )2H S H HS fast equilibrium

+ +

( )2 2Cl HS Cl H S fast equilibrium ++ + +

(1) B only (B) Both A and B (C) Neither A nor b (D) A only

Sol. (4)Rate equation is to be derived from slow or rate determining step.

(* Since both the reactants are involved in RDS)

from mechanism (A), Rate = k[Cl2] [H2S]

5. If 104

dm3

of water is introduced into a 1.0 dm3

flask to 300 K, how many moles of water are inthe vapour phase when equilibrium is established?

(Given : Vapour pressure of H2O at 300 K is 3170 pa; R = 8.314 J K-1

mol-1

)

(1) 35.56 10 mol (B) 21.53 10 mol (C) 24.46 10 mol (D) 31.27 10 mol

Sol. (4)Volume occupied by water molecules in vapour phase is ( 1- 10 -4)dm3. i.e., approximately 1dm3(1L)

PVn =

RT

-53

-1 -1

3170 10 atm 1L= 1.27 10 mol

0.0821Latm k mol 300K

=


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6. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a

molecular mass of 44 u. The alkene is

(1) propene (2) 1butene (3) 2butene (4) ethene

Sol. (3)But-2-ene is a symmetrical alkene. So, it gives two moles of the same aldehyde.

O3CH CH=CH CH 2.CH CHO

3 3 3

2Zn/H O Molar mass of CH CHO

3is 44 u.

7. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous

solution, the change in freezing point of water ( ( )fT , when 0.01 mol of sodium sulphate isdissolved in 1 kg of water, is (kf= 1.86 K kg mol

1)

(1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K

Sol. (2)

Vant Hoff factor (i) for Na2SO4 = 3 (assuming complete ionization)2

2 4 42Na SO Na SO+ +

. .f fT i k m =

0.013 1.86 0.0558

1K= =

8. From amongst the following alcohols the one that would react fastest with conc. HCl and

anhydrous ZnCl2, is(1) 2-Butanol (2) 2Methylpropan2ol

(3) 2Methylpropanol (4) 1Butanol

Sol. (2)

3 alcohols react fastest with Lucas reagent i.e., an. ZnCl2/conc.HCl due to formation of 3 carbocation

2methyl propan2ol is the only 3 alcohol.

9. In the chemical reactions,

NaNO2

HCl 278KA B

HBF4

NH2

the compounds A and B respectively are

(1) nitrobenzene and fluorobenzene (2) phenol and benzene

(3) benzene diazonium chloride and fluorobenzene (4) nitrobenzene and chlorobenzene

Sol. (3)


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This is Schiemann reaction (also called the Balz-Schiemann reaction). It is a chemical reaction inwhich anilines are transformed to arylfluorides through diazonium fluoroborates .

10. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahls method

and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acidrequired 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen

in the compound is

(1) 59.0 (2) 47.4 (3) 23.7 (4) 29.5

Sol. (3)Moles of HCl reacting with

Ammonia = (moles of HCl absorbed ) (moles of NaOH solution required)

= ( ) ( )3 320 0.1 10 15 0.1 10 = moles of NH3 evolved

= moles of nitrogen in organic compound

wt. of nitrogen in org. compound = 30.5 10 14

= 37 10 g

% wt =3

3

7 10100 23.7%

29.5 10x

=

11. The energy required to break one mole of ClCl bonds in Cl2 is 242 kJ mol1. The longest

wavelength of light capable of breaking a single Cl Cl bond is

( )8 -1 23 -1Ac = 310 ms and N = 6.0210 mol (1) 594 mm (2) 640 mm (3)700 mm (4) 494 mm

Sol. (4)

Energy required for 1 Cl2 molecule =3

23

242 10

6.02 10x

Joule.

The energy is contained in photon of wavelength ' ' .34 8

3

23

6.625 10 3 10

242 10

6.02 10

hcE

=

=

-6= 0.494 x 10 m = 494 nm


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12. Ionisation energy of He+

is 19.6 x 1018

J atom1

. The energy of the first stationary state (n = 1) of

Li2+

is

(1) 16 14.4110 Jatom (2) 17 1-4.4110 Jatom

(3) 15 1-2.210 Jatom (4) 17 18.8210 Jatom

Sol. (2)

IE 2

2

2

2

2

3

He

Li

IE

IE

+

+

=

2

18 2

2

19.6 10 2

3Li

x

IE +

=

2

18 919.6 104Li

IE x x+=

Therefore 218 919.6 10

4LiE x x+

= (IE ans E are same in magnitude but opposite in sign)

=17 1-4.4110 Jatom

13. Consider the following bromides:

The correct order of 1N

S reactivity is

(1) B > C > A (2) B > A > C (3) C > B > A (4) A > B > C

Sol (1)

Sol. 1NS reaction proceeds via carbocation intermediate. As the formation of carbocation is rate

determining step,the most stable one forms the product faster.Hence reactivity order for A, B, C depends on stability of carbocation formed.

B forms 20allyl carbocation ,which is the most stable among given.

The next stable carbocation is formed by C. It forms a 20

carbocation.

The least stable carbocation is formed by A. It forms a 10

carbocation.


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14. Which one of the following has an optical isomer?

(1) ( ) ( )2+

Zn en NH3 2

(2) ( ) 3+Co en 3

(3) ( ) ( )3+

Co H O en2 4

(4) ( )2+

Zn en2

Sol. (2)

15. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two

liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of

the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane

= 100 g mol1

an dof octane = 114 g mol1

).

(1) 72.0 kPa (2) 36.1 kPa (3) 96.2 kPa (4) 144.5 kPa

Sol. (1)Mole fraction of Heptane =

25/100

25 35

100 114

0.250.557

0.45

+

=

=

HeptaneX 0.45=

Mole fraction of octane = 1 - 0.45 = 0.55

Total pressure = 0i iX P

( ) ( )105 0.45 45 0.55 kPa= +

= 72.0 kPa


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16. The main product of the following reaction is ( ) ( )conc.H SO

2 4C H CH CH OH CH CH6 5 2 3 2

?

(1)

CC

H( )CH CH3 2

H C5 6 H

(2)

CC

CH3

C H CH6 5 2

HCH

3

(3)

CC

( )CH CH3 2C H

6 5

HH (4)

C CH2

H C CH CH5 6 2 2

H C3

Sol. (1)


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17. Three reactions involving2 4

H PO

are given below :

(i)3 4 2 3 2 4

H PO H O H O H PO+ + +

(ii)2

2 4 2 2 4 3H PO H O H PO H O ++ +

(iii)2

2 4 3 4H PO OH H PO O + +

In which of the above does2 4

H PO act as an acid?

(1) (ii) only (2) (i) and (ii) (3) (iii) only (4) (i) only

Sol (1)

Proton donor is an acid acc to Bronsted Lowry theory.

( )3 4 2 3 2 4

i H PO H O H O H PO

conjugate baseacid

+ + +

( ) 22

2 4 2 4 3ii H PO H O H PO H O

acid conjugate base

++ +

( ) 22 4 3 4

iii H PO OH H PO O

base conjugate acid

+ +

So, only in reaction (ii)

2 4

H PO is acting as acid.

18. In aqueous solution the ionization constants for carbonic acid are-7 -11

1 2K = 4.2 10 and K = 4.8 10

Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

(1)The concentration of 23CO

is 0.034 M

(2)The concentration of 23CO

is greater than that of3HCO

(3)The concentration of H+

and3HCO are approximately equal

(4)The concentration of H+

is double that of 23CO

Sol. (3)

2 3 3A H CO H +HCO+ 71K 4.2 10

=

2

3 3B HCO H +CO + 112K 4.8 10

=

As K2


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19. The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the

cation is 110 pm, the radius of the anion is

(1) 288 pm (2) 398 pm (3) 618 pm (4) 144 pm

Sol. (4)In an FCC lattice,

( )+2 r + r = edge length

( )2 110 + r = 508

= 144pmr

20. The correct order of increasing basicity of the given conjugate bases (R = CH3) is

(1) RCOO< HC =C < R < NH

2

(2) R < HC C < RCOO < NH

2

(3) R COO < NH HC C < R2

< (4) RCOO < HC C< NH R2

<

Sol. (4)Correct order of increasing basic strength is

( ) ( ) ( ) ( )RCOO < CH C < NH R

2

<

Order of basicity can be explained on the basis of acidity of the acids of the given conjugate base.

Stronger the acid, weaker is the conjugate base and vice versa.

As RCOOH is the strongest acid among given, RCOO is the weakest base.

Acetylene is also acidic due to sp carbon. It is a weak base but stronger than RCOO -.

sp3

carbon is less electronegative than sp3

nitrogen, R-

is less acidic or more basic than NH2-

21. The correct sequence which shows decreasing order of the ionic radii of the elements is

(1) 3+ 2+ + 2Al > Mg > Na > F > O (2) + 2+ 3+ 2Na Mg Al > O F > > >

(3) + 2+ 2 3+Na F Mg O Al > > > > (4) 2 + 2+ 3+O F Na Mg Al > > > >

Sol. (4)

For isoelectronic species ( all have 10 e-), higher the

Z

eratio, smaller the ionic radius

2Z 8for O 0.8

e 10

= =

9F 0.9

10

= =

11Na 1.1

10

+ = =

2 12Mg 1.210

+ = =

3 13Al 1.310

+ = =


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(Or)

In case of isoelectronic species, as the magnitude of negative charge increases,

size increases and as the magnitude of positive charge increases, size decreases

22. Solubility product of silver bromide is 5.0 1013

. The quantity of potassium bromide (molar

mass taken as 120 g of mol1

) to be added to 1 litre of 0.05 M solution of silver nitrate to start the

precipitation of AgBr is

(1) 101.2 10 g (2) 91.2 10 g (3) 56.2 10 g (4) 85.0 10 g

Sol. (2)

Ag Br AgBr+ +

Precipitation just starts when ionic product just exceeds solubility product ( Q >Ksp)

spK Ag Br + =

13115 10 10

0.05

SPK

BrAg

+

= = =

i.e., precipitation just starts when 1011

moles of KBr is added to 1L of AgNO3 solution.

No. of moles of KBr to be added = 1110

= 1110 120 91.2 10 g

23. The Gibbs energy for the decomposition of Al2O3 at 500C is as follows :

1

2 3 2

2 4

, 9663 3 rAl O Al O G kJ mol

+ = +

The potential difference needed for electrolytic reduction of2 3Al O at 500

0C is at least

(1)4.5 V (2) 3.0 V (3) 2.5 V (4) 5.0 V

Sol. (3)

The ionic reactions are

Cathode : 32

2 44

3 3Al e Al+ +

Anode : 23 2

24

3O O e +

So, the no of electrons transferred = 4

G nFE =

GE

nF

=

3966 10

4 96500E

=

= - 2.5 V

The potential difference needed for the reduction = 2.5 V


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24. At 25C, the solubility product of Mg(OH)2 is 1.0 1011

. At which pH, will Mg2+

ions start

precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+

ions?

(1) 9 (2) 10 (3) 11 (4) 8

Sol. (2)

( )22

2Mg OH Mg OH+ +

22

spK Mg OH + =

4

210

spKOH

Mg

+ = =

4OHP = and pH = 10

At pH

= 10, Mg2+

starts precipitating.

25. Percentage of free space in cubic close packed structure and in body centred packed structure are

respectively

(1) 30% and 26% (2) 26% and 32% (3) 32% and 48% (4) 48% and 26%

Sol. (2)

Packing efficiency of cubic close packing = 74%

Therefore % free space = 26%Packing efficiency of body centred packing = 68%.

Therefore % free space = 32%

26. Out of the following, the alkene that exhibits optical isomerism is(1) 3methyl2pentene (2) 4methyl1pentene

(3) 3methyl1pentene (4) 2methyl2pentene

Sol. (3)

2H C = HC 2 5C H

H

3CH

Only 3-methyl pent-1-ene has a chiral carbon

27. Biuret test is not given by(1) carbohydrates (2) polypeptides (3) urea (4) proteins

Sol. (1)

It is a test characteristic of amide linkage ( - CONH) . Urea also has amide linkage like proteins.


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28. The correct order of0

2 /E

M M+with negative sign for the four successive elements

Cr, Mn, Fe and Co is

(1)Mn > Cr > Fe > Co (2) Cr > Fe > Mn > Co

(3) Fe > Mn > Cr > Co (4) Cr > Mn > Fe > Co

Sol. (1)

E0

values for Mn ,Cr , Fe and Co are -1.18V, -0.91V , -0.44 V and -0.28V respectively

29. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is

(1) Teflon (2) nylon 6,6 (3) polystyrene (4) natural rubber

Sol. (2)

Nylon 6, 6 is a polymer of adipic acid and hexamethylene diamine. It contains amide linkages.

So, it will have very strong intermolecular hydrogen bonding between NH and - C=O groups of

two polyamide chains

C ( )CH2 4 ( )C NH CH NH2 6 O O

n

30. For a particular reversible reaction at temperature T,

H and

S were found to be both +ve. IfTe is the temperature at equilibrium, the reaction would be spontaneous when

(1) T Te

> (2) T Te

> (3) Te

is 5 times T (4) T Te

=

Sol. (2)

G H T S =

At equilibrium 0G =

for a reaction to be spontaneous G should be negative.

As both H and S are +ve , for ve G T Te

>

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