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Test - 13 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2011

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TEST - 13 (Paper - II)

ANSWERS

CHEMISTRY MATHEMATICS PHYSICS

39. (A)

40. (A)

41. (A)

42. (D)

43. (A)

44. (A)

45. (8)

46. (1)

47. (3)

48. (4)

49. (9)

50. (B)

51. (B)

52. (C)

53. (C)

54. (D)

55. (A)

56. A (t)

B (s)

C (q)

D (p)

57. A (q, t)

B (r, t)

C (s, t)

D (p)

20. (B)

21. (C)

22. (A)

23. (B)

24. (B)

25. (B)

26. (1)

27. (9)

28. (0)

29. (5)

30. (3)

31. (B)

32. (A)

33. (D)

34. (D)

35. (C)

36. (B)

37. A (s, t)

B (q)

C (p, q, r, s)

D (q, r, s, t)

38. A (t)

B (p, q, r, s, t)

C (p)

D (q, r, s, t)

1. (A)

2. (B)

3. (C)

4. (B)

5. (D)

6. (A)

7. (5)

8. (3)

9. (8)

10. (7)

11. (4)

12. (C)

13. (B)

14. (D)

15. (A)

16. (B)

17. (D)

18. A (p, q, t)

B (p, s, t)C (p, q, r, t)

D (r, t)

19. A (q, r)

B (p, s)

C (q, r)

D (p, s, t)

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All India Aakash Test Series for IIT-JEE 2011 Test - 13 (Paper - II) (Answers & Hints)

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PART - I (CHEMISTRY)

ANSWERS & HINTS

1. Answer (A)

G H T S =

cell cellP

nFE H T.nF ET

=

cell cell

P

HE T E

nF T

= +

For Ecell

versus T,

cellP

Slope ET

= =

Temperature coefficient

Intercept = HnF

2. Answer (B)

CH CNO + N OH3 2 CH CNO + H O CH CNO3 2 2 3 2

H H

(A)

O NOH(Nitrous acid)

(Nitrolic acid)(P)

NONa

(Red solution)

NaOH

RCNO + N = O2 RCNO + H O2 2

(B)

R

(Pseudo-nitrole)

H HO

R

N=O

Gives blue colour with aq. NaOH

RCNO2 No reaction

(C)

R (Due to absence of H in C)

RHNO2

3. Answer (C)

2

213Cl 3 mole

71= =

NH3

= 8 mole and 1 mole

8NH3

+ 3Cl2

N

2+ 6NH

4Cl

NH3

+ 3Cl2NCl3 + 3HCl

N2

evolved = 28 g

NH4Cl formed = 321 g

NCl3

formed = 120.5 g

HCl formed = 109.5 g

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4. Answer (B)

Aldol condensation and cross-aldol condensation products are formed. ( unsaturated carbonyl compounds)

5. Answer (D)

rms(I)

3R(t 273)v

M

+= , mp(I)

2R(t 273)v

M

+=

rms(II)

3R 546v

2M

= , mp(II)

2R 546v

2M

=

rms(III)

3R(2t 546)v

(M / 2)

+= , mp(II)

2R(2t 546)v

(M / 2)

+=

6. Answer (A)

(I)

H

Me

H

BrH

H

Anti arrangement(Favourable)

Syn arrangement(Not favourable)

(II) Major product is H

OH

Me

H

(III) Walden inversion

7. Answer (5)

Brown ring complex = [Fe(H2O)

5NO].SO

4

8. Answer (3)

Adiabatic slope(Poisson's ratio)

Isothermal slope=

2 2He O CO

5 7 4 28 1x y z 3

3 5 3 9 9 = = = = +

9. Answer (8)

4 alcohols and 3 ethers

Butan-2-ol has both d- and l- form

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10. Answer (7)

Phenolphthalein end point:

NaOH + HCl NaCl + H2O

NaCl + HCl No reaction

NaHCO3

+ HCl No reaction

Na2CO

3+ HCl NaHCO3 + NaCl

x = 1 + 4 = 5 moles

Methyl orange end-point:

NaHCO3

+ HCl NaCl + H2CO3

y = 3 + 4 = 7 moles

xy 7 57

5 5

= =

11. Answer (4)

Anti fluorite structure Na2O

Na+ at all tetrahedral voids. O 2in CCP.

1st C.N. of O2at a corner = 8 = x

(All 8 Na+ in tetrahedral voids in eight cubes sharing a corner)

2nd C.N. of O2at a corner = 12 = y

(12 O2, each three O2in a cube at adjacent face centres sharing a corner)

y x = 12 8 = 4

12. Answer (C)

13. Answer (B)

14. Answer (D)

Solution of Q. 12 to Q.14

NiCl2

+ 2DMG H C C = N3

H C C = N3Ni

N = C CH3

N = C CH3

O -----------H O

O H -----------O

(Cherry red ppt.)

Ni2+ (3d8) Square planar complex due to strong ligand field

dsp2

No unpaired electrons. So, = 0 B.M.

Two chelate rings are present, each are five membered

15. Answer (A)

The order of reactivity can be explained in terms of

(i) Leaving group ability of X

(ii) Electron donating resonance effect ofX

Order of leaving group of X:

Cl > R Cl > R O > R NH() ()

O

O()

()

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16. Answer (B)

RC

O

OHRCOH RCOH RCO

OH

R O 18..

(+)

H

..

H

OH

..

O18

RH +

:OH

18OR

(+)

H

HRCO +

18H O2

OH

..

(+)

H

RCO R18

O

R

17. Answer (D)

COOH

COOH

P O2 5O

O

O

Dry NH3NH

O

O

0C

Br , NaOH2NBr

O

O

(Succinicacid)

(Succinicanhydride)

(Succinimide) (N-bromosuccinimideor NBS)

18. Answer A(p, q, t), B(p, s, t), C(p, q, r, t), D(r, t)

A.Vapour

pressure

Temperature (T)

Tb

SolutionSolvent

B.Stopping

potential

(V

)S

Wavelength ( )

0s

0

hc hcV

e e=

C.

7

Vol. of base (ml)

(titration of weak acidand weak base)

pH

D.

Rate ofbackwardreaction

Time (t)

19. Answer A(q, r), B(p, s), C(q, r), D(p, s, t)

A. All have planar shape.

All have 2 lone pairs.

B. ( 1)2

= +l

hl l

Total nodes = n 1

C. Sideways destructive overlapping between porbitals (in one plane)

2 nodes+

+

D. Single electron species.

2

2

zE

n

zVn

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20. Answer (B)

After simplifying k= 1, n= 2

21. Answer (C)

P1

O

P2

z1

z2

By rotation it can be easily seen z2

= iz1

(i)

and if we interchange z1

and z2

then we have z1

= iz2

(ii)

Now, clearly |z1

z2| = |z

1+ z

2| for (i) and (ii)

1 2 1 2 0z z z z + = for (i) and (ii)

2

1

Re 0z

z

=

22. Answer (A)

2

161024[ln ] 6[ ] 2011

1

x xx ex

+ = ++

So, for a solution to exist 216

1

x

x+must be an integer.

Now, 22

1 11

x

x

+

So,2

2

1

x

x+may take values 1, 0, 1 as integral values.

But in all cases R.H.S. is ODD while LHS is divisible by 2.

Hence the equation has no solution.

23. Answer (B)

As the letter boxes are identical so we will be only concerned with the number of letters.

The distribution can be as follows

5

4

3

3

2

2

1

0

1

2

1

2

1

1

0

0

0

1

1

1

1

0

0

0

0

0

1

1

0

0

0

0

0

0

1

54

53

5 23 1

5 32 2

52

1 1

5

10

1( . ) 10

2

1( ) 15

2!

10

1 1

C

C

C C

C C

C

=

=

=

=

=

=

=

Ways of distribution Number of ways

So, total number of ways = 1 + 5 + 10 + 10 + 15 + 10 + 1 = 52

PART - II (MATHEMATICS)

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24. Answer (B)

62

, 1

( )i ji ji j

P x x=

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