for


Time: 2 Hours Marks: 150
INSTRUCTIONS
(i) This paper contains 50 objective questions under two sections.
(ii) Section

I contains 35 objective questions with four options of which only one choice is correct. For each question you will be awarded 3 marks if you have marked the right answer. For any unattempted question you will be awarded zero mark. In all other cases you will be awarded 1 mark .
(iii) Section

II contains 3 comprehension passages followed by objective questions with four options of which only one choice is correct. Read the passage carefully before answering the questions and mark the most appropriate option. For each question you will be awarded 3 marks if you have marked the right answer. For any unattempted question you will be awarded zero mark. In all other cases you will be awarded -1 mark .
(iv) Fill the circle of the correct alternative(s) with HB pencil on the answer sheet given.
In case you wish to change an answer, erase the old answer completely using a good eraser.
(v) Use of calculator is not allowed.
(vi) Extra paper for rough work will not be provided.
(vii) Useful Data:
Avogadro’s constant,
N
AV
= 6.023

10 23
Gas constant, R = 8.314 J. mol

1 .K

1 = 0.0821 L.atm.mol

1 .K

1 = 2 cal mol

1 K

1
Rydberg’s constant,
R
H
= 109678 cm

1
Planck’s constant = 6.626 
10

34 J
 s
; Mass of electron = 9.1

10

31 kg
Atomic masses :
H = 1, C = 12, N = 14, O = 16, Na = 23, K = 39, Cu = 63.5
Name of the candidate
: ………………………………………………
: ……………………………………………… Enrollment Number
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST

1.
Identify the compound (X) in the following reaction.
C

C – CH
3
2.
(a)
COOH
     ) 
(X)
(b)
C

C – CH
3
OH
OH
OH OH
C = C – CH
3
(c) (d)
C

C – CH
3
O
O
CH = CH – CH
2
COOH
 
(A) . The compound (A) may be
Me
(a)
Me
OH
(b)
CH = CH – CH
2
COOH
F
Me
F
CH
2
–CHCH
2
COOH
(c) (d)
F
CH
2
CH
2
CH
2
COOH
Me Me

Space for rough work

PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
3.
The correct order of bond angles of the X

N

X bond (where X is the surrounding atom or group of atoms)

(a) NH
4
> N(CH
3
)
3
> NH
3
> NF
3
(c) NH
3
> NF
3
> NH

4
> N(CH
3
)
3
(b) NH

4
< N(CH
3
)
3
< NH
3
< NF
3
(d) NH
3
> NF
3
> N(CH
3
)
3
> NH

4
4.
The time required for electroplating of 39 g of chromium from a Cr(III) solution, using an average current of 30 ampere with 75% efficiency, is
(a) 1.68 hours
(c) 5.4 hours
(b) 2.68 hours
(d) 10.8 hours
5.
Which of the following compound responds to iodoform reaction?
(a) C
2
H
5
– C – CH
O
2
Cl (b) CH
3
– C – CH
2
– COOH
O
(c) CH
3
– C – CH
2
– C – OC
2
H
5
O O
(d) CH
3
– CH – CH
3
Br
6.
CH
3
–CH
2
– CH – CH
3
F
  H   C 
5
 
?
The product of the above reaction is
(a) CH
3
CH
2
CH = CH
2
(c) CH
3
CH
2
– CH – CH
3
(b) CH
3
CH = CHCH
3
(d) None of the above
7.
Which one of the following is not correct systematic name of corresponding compound?
(a) NaMn(CO)
5
: Sodium pentacarbonylmangnate(I)
(b) SnCl
4
(Et
2
NH)
2
: Tetrachlorobis(diethylamine)tin(IV)
(c) Ni(CO)
2
(PPh
3
)
2
: Dicarbonylbis(triphenylphosphine)nickel(0)
(d) Na
2
[Fe(CN)
5
NO]: Sodium pentacyanonitrosylferrate(II)
8.
If
 o
is the threshold wavelength for photoelectric emission,

the wavelength of light falling on the surface of a metal and m is the mass of the electron, then the velocity of ejected electron is given by
(a)
(c)


2 h m
(
 o
OC
2
H
5
 
)


1 / 2



2 hc m


 o

 o
 




1 / 2
(b)
(d)





2 hc m
(
 o
2 h m


1
 o
 

) 
1 / 2

1






1 / 2
9. Which of the following reaction involves inversion of configuration in the product?
(a)
H
3
C
CH

CH
2
OH      
(b) CH
3
CH
2
Br
 
H
3
C
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
(c)
CH
3

CH

OK   
(d)
CH
3

CH

OTs   
C
2
H
5
10. A real gas at a very high pressure occupies
C
2
H
5
(a) more volume than that of an ideal gas under identical conditions.
(b) less volume than that of an ideal gas under identical conditions.
(c) same volume than that of an ideal gas under identical conditions.
(d) can’t predict
11. The correct order of acidity for the compounds
O O O O
F
3
C
O
(A)
O
CF
3
H
3
C
(B)
O
CH
3
O
H
3
CO OCH
3
F
3
C CF
3
(C)
CO
2
CH
3 would be
(a) (C) > (B) > (A) > (D)
(c) (A) < (B) < (C) < (D)
(D)
(b) (C) < (B) < (A) < (D)
(d) (C) < (B) < (D) < (A)
12. If the pressure of hydrogen gas is increased from 1 atm to 100 atm, keeping the hydrogen ion concentration and temperature constant, the voltage of hydrogen half cell will change by
(a) 0.059 V
(c) 0.118 V
(b) 0.59 V
(d) 0.0295 V
13. For the reaction,
M x +
+

MnO

MO
3

+ Mn
2+
+ ½O
2 if one mole of MnO oxidizes 1.67 moles of M x +
4
to MO

3
, then the value of x in the metal ion is
(a) 5
(c) 2
(b) 3
(d) 1
14. The vapour pressure of the distillate obtained on condensation of the vapour formed above a solution containing 2 mole of a liquid A ( P
A o
= 100 mm of Hg) and 3 mole of a liquid B
( P
B o
= 150 mm of Hg) at same temperature would be
(a) 130 mm of Hg
(c) 140 mm of Hg
(b) 135 mm of Hg
(d) 145 mm of Hg
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
15.
 –hydroxy acids can be prepared by which of the following reaction/sequence of reactions?
(a) Aldol condensation
(c) Intramolecular Cannizaro reaction
(b) Claisen condensation
(d) Perkin reaction
16. Inorganic benzene is
(a) B
3
N
3
(c) B
6
H
6
H
6
(b) B
(d) B
2
3
H
O
6
3
H
6
17. The increasing order of molar conductivity of the following compounds
(I) K[Co(NH
3
)
2
(NO
2
)
4
] (II) [Cr(NH
3
)
3
(NO
2
)
3
]
(III) [Cr(NH
3
)
5
(NO
2
)]
3
[Co(NO
2
)
6
]
2 is
(IV) Mg[Cr(NH
3
)(NO
2
)
5
]
(a) (II) < (I) = (IV) < (III)
(c) (II) < (I) < (IV) < (III)
(b) (III) > (IV) = (I) > (II)
(d) (III) > (IV) > (I) > (II)
18.
The cleavage of an unsymmetrical ether Ph—O—CH
2
CH
2
— —NO
2 by HI produces
(a) PhCH
2
OH (b) PhCH
2
CH
2
OH
(c) PhI (d) NO
2
— —CH
2
CH
2
I
19. At pH = pK in
+ 1 (where K in
= dissociation constant of the indicator), the ratio [In

]/ [HIn] in the solution is
10
1

. For this pH, the percentage dissociation of the indicator is
(a) 91
(c) 10
(b) 90
(d) none of these
20.
The solution of a compound (X) reacts with AgNO
3
solution to form white precipitate of
(Y) which dissolves in NH
4
OH to give a complex (Z). When (Z) is treated with dil. HNO
3
,
(Y) reappears. The compound (X) could be
(a) NaCl
(c) NaBr
(b) CH
3
(d) NaI
Cl
21.
When magnesium turning is added to an aqueous solution of aluminium sulphate, the gas which evolves with effervescence is
(a) O
2
(c) SO
3
(b) SO
2
(d) H
2
22. Which of the following species is iso-structural with XeF
4
?
(a) TeF

5
(b) I

3
(c)
4

BrF (d) XeO
3
23. n

Butyraldehyde can be synthesized from n
 propyl magnesium bromide and
(a) ethyl ortho formate (b) acetyl chloride
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
(c) n
 butyl acetate (d) acetamide
24. An inorganic salt (A) on reaction with H
2
O gives a solution (B) and a gas (C). Gas (C) on passing into ammonical AgNO
3
gives white precipitate of (D). CO
2
gas turns solution (B) milky. The Compound (A) would be
(a) BaSO
4
(c) CaC
2
(b) Na
2
SO
4
(d) CaCO
3
25.
The heat of combustion of yellow phosphorous and red phosphorus are

9.91 kJ/mole and

8.78 kJ/mole respectively, then the heat of transition of yellow phosphorus to red phosphorus is
(a)

18.69 kJ
(c) +18.69 kJ
(b) +1.13 kJ
(d)

1.13 kJ
26. The differential rate law for the hypothetical reaction, aA(g)

Product, taking place at temperature T may be written as

1 a d [ A ] dt

K c
[ A ] n
or

1 a dP dt
A
The value of K p
would be equal to
(a) K c
( RT ) n
 a
(c) K c
( RT )
1
 n

K p
( P
A
) n
(b)
(d)
K c
( RT ) a
 n
K c
( RT ) n

1
27.
At 40°C, in a 5 litre flask LiCl.3NH
3
(s) starts to decompose to give LiCl.NH
3
(s) and
NH
3
(g). After sufficient time the pressure of NH
3
was observed to be 3 atm. Now more
LiCl.3NH
3
(s) is added at the same temperature, but no increase of NH
3
is observed even after a long time, then
(a) K p
= 9
(c) K p
= 3
(b) K
(d) K p p
< 9
> 4
28. 3 mg of AgCl (molecular weight = 144) is put in 2 litre of H
2
O. The solubility of AgCl in 5

10

6
M BaCl
2
is 10

5
M. The amount of solid in water is
(a) 1 mg (b) 0.12 mg
(d) 2 mg (c) 2.88 mg
29.
Aluminium is not present in the mineral
(a) Feldspar
(c) Cryolite
(b) Fluorspar
(d) Mica
30. According to MOT, if the bond
 order of a molecule is coming out to be fractional, this means that
(a) the molecule is not stable.
(b) all the bonds in the molecule are odd electron bond.
(c) the molecule is stabilized by resonance.
(d) both (a) and (c)

Space for rough work

PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
The questions below (31 to 35) consist of an ‘Assertion’ in Column 1 and the ‘Reason’ in Column 2. Use the following key to choose the appropriate answer.
(A) If both assertion and reason are CORRECT and reason is the CORRECT explanation of the assertion.
(B) If both assertion and reason are CORRECT but reason is NOT the CORRECT explanation of the assertion.
(C) If assertion is CORRECT but reason is INCORRECT.
(D) If assertion is INCORRECT but reason is CORRECT.
Assertion (Column

1) Reason (Column

2)
31. Liquid ammonia is used for refrigeration. It vapourizes quickly.
32. Benzyl bromide when kept in acetone water produces benzyl alcohol.
33. BaCO
3
is more soluble in HNO
3
than in plain water.
Questions
36.
What is the volume of single oil molecule?
(a) 1.56

10

20
cm
3
The reaction follows S
N
2 mechanism.
Carbonate is a weak base and reacts with the
H
+
from the strong acid, causing the barium salt to dissociate.
34. The micelle formed by sodium stearate in water has

COO

groups at the surface.
35. The O

O bond length in H
2
O
2
is shorter than that of O
2
F
2
.
Surface tension of water is reduced by the addition of stearate.
H
2
O
2
is an ionic compound.

C OMPREHENSION

1
Renowned statesman Benzamin Franklin was also an inventor and a scientist. He is more known for his demonstration that lightning is electricity. Very less number of people know he also conducted an experiment for the simple estimation of molecular size and
Avogadro’s number through oil spread on water.
He fetched out one table spoon (5 ml) of oil of density 0.95 g cm

3
on the water of a large pond. He saw it spread itself with surprising swiftness upon the surface. Such a small quantity of oil produced an instant calm over an area of about 2

10
7
cm
2
making it as smooth as a looking glass.
The calculation may go like this:
Avogadro’s number is the number of molecules in a mole. So, if we can estimate both the number of molecules and the number of moles in Franklin’s teaspoon of oil, we can calculate Avogadro’s number.
In calculation he assumed that the oil molecules are tiny cubes that pack closely together and formed a layer only one molecule thick and the molar mass to be 240 g mol

1
.
(b) 2.56

10

17
cm
3
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
(c) 5.62

10

19
cm
3
(d) 6.23

10

18
cm
3
37.
What is the total number of molecules of oil?
(a) 1.23

10 19
(c) 3.21

10 20
(b) 2.25

10 18
(d) 7.23

10 17
38.
Number of moles of oil spread out in the oil film are?
(a) 9.18

10

2
(c) 8.91

10

1
(b) 1.98
(d) 9.56


10
10


2
3
39.
What value we will get for Avogadro’s number from given information
(a) 6.023

10 23
(c) 1.62

10
22
(b) 6

10
(d) 1.54

22
10
23
40.
The result of above calculation is not very accurate. Which of the following assumptions do you think can’t be the main source of error in calculating the Avogadro’s number by spreading oil in the water of the pond.
(a) The oil molecules are tiny cubes
C
OMPREHENSION

2
(b) The oil layer is one molecule thick
(c) The molecular mass is 240 g mol

1
(d) The formula used for calculating mole is wrong.
Isomers that are mirror images of each other are called enantiomers. Enantiomers have identical physical properties except for the direction of rotation of the plane of polarized light.
The two 2
 methyl

1
 butanols, for example have identical melting points, boiling points, densities, refractive indexes.
Specific rotation
Boiling point
Relative density
(+)

2
 methyl

1
 butanol
+5.90°
128.9°C
0.8193
(

)

2
 methyl

1
 butanol
 5.90°
128.9°C
0.8193
Refractive index 1.4107 1.4107
Enantiomers have identical chemical properties expect toward optically active reagents. The two lactic acids are not only acids, but acids of exactly the same strength that is dissolved in water at the same concentration, both ionize to exactly the same degree.
A mixture of equal parts of enantiomers is called a racemic modification. A racemic modification is optically inactive. They cannot be separated by ordinary methods; not by fractional distillation, because their boiling points are identical, not by fractional crystallization, because their solubilities in a given solvent are identical. The separation of a racemic modification into enantiomers is called as resolution.

Space for rough work

PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
Stereoisomers that are not mirror images of each other are called diastereomers.
Diastereomers have similar chemical properties, since they are members of the same family.
Diastereomers have different physical properties, different melting points, boiling points, solubilities in a given solvent, densities, refractive indexes and so on. Diasteromers differ in specific rotation.
As a result of their differences in boiling point and in solubility, they can in principal at least, be separated from each other by fractional distillation or fractional crystallization.
Given a mixture of all four stereoisomeric 2,3
 dichloropentanes, we could separate it by distillation, for example, into two fractions but no further. Further separation would require resolution of the racemic modification by use of optically active reagents.
A meso compound is one whose molecules are super imposable on their mirror images even though they contain chiral centers. n
 butane is allowed to undergo free radical chlorination and the reaction mixture is separated by careful fractional distillation.
Questions:
41.
How many fractions of formula C
4
H
9
Cl would you except to collect.
(a) two
(c) one
(b) three
(d) four
42.
If we were to put the sec
 butyl chloride actually prepared by the chlorination of n
 butane into a polarimeter, would it rotate the plane of polarized light?
(a) no, because it is achiral
(b) yes, because it is chiral
(c) no, because prepared as described it would consist of the racemic modification
(d) no, because it is meso isomer
43.
Monochlorination of (R)
 sec
 butylchloride at 300°C is carried out and the products are separated by careful fractional distillation. How many fractions will be collected
(a) 4
(c) 3
(b) 5
(d) 2
44.
Monochlorination of (R)
 sec
 butylchloride at 300°C is carried out and the products are separated by careful fractional distillation. How many fractions are optically active?
(a) 2
(c) 4
(b) 3
(d) 1
45.
Benzil
Ph

C

C

Ph on reduction with NaBH
4
gives (A). The number of optically
O O active stereoisomers possible for (A) are
(a) one
(c) three
(b) two
(d) four
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
C
OMPREHENSION

3
A brown powdery substance (A) when heated with concentrated HCl gives compound (B) along with the liberation of a greenish yellow gas (C), which liberates a compound (D) when passed through KI solution. Compound (D) dissolves in excess of KI, forming a yellow solution. (A) when fused with KOH in presence of atmospheric oxygen gives a green mass, which on extraction with water and on treatment with a gas (E), changes to purple. Moreover the gas (E) when passed through dry KOH at low temperature gives a deep red coloured compound. On the basis of above information answer the following questions.
Questions:
46.
Compound (A) is
(a) Fe
2
O
(c) CuS
3
(b) MnO
(d) PbS
2
47.
Compound (B) is
(a) FeCl
3
(c) CuCl
2
48.
Compound (C) is
(a) CrO
2
Cl
2
(c) Cl
2
49.
Compound (D) is
(a) NO
2
(c) O
3
(b) MnCl
(d) PbCl
(b) NO
(d) H
(b) I
2
2
(d) SO
S
2
2
2
50.
Compound (E) is
(a) O
3
(c) O
2
(b) Cl
2
(d) NO
2
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
2.
1.
for



CODE: TSF(4)20060219(

)

CH
1. b
6. a
11. b
16. a
21. d
26. c
31. a
2.
7.
12.
17.
22.
27.
32. a a a c,d c a c
36. a
41. a
46. b
37. c
42. c
47. b
3. a
8. c
13. c
18. d
23. a
28. c
33. c
38. b
43. b
48. c
4. b
9. d
14. b
19. a
24. c
29. b
34. b
39. c
44. b
49. b
5. d
10. a
15. c
20. a
25. d
30. c
35. d
40. d
45. b
50. a

In presence of HCO
3
H, antihydroxylation of alkene, takes place and triple bond remains unaffected.

(b)
Me Me
CH = CH
CH
2
COOH
 

H
2
O
O
C
CH
2
CH
C
H tautomerisation
OH
Me
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1

(a)
3.
PRACTICE TEST
H
N

H H H
109°28’ N
H H H
107°
H
3
C more than 107° & less than 109° as
N methyl group is bulkier than hydrogen and one lone pair of
CH
3
CH 3 electrons is present.
N
F F F
Less than 107°
Therefore, NH

4
> N(CH
3
)
3
> NH
3
> NF
3

(a)
4. The electroplating reaction would be
Cr
3+
+ 3e
 
Cr
Let the current is passed for t hours.

30

100
75
 t

3600
96500

1
3

52 = 39 t = 2.68 hours

(b)
5. Due to presence of active methylene group in (b) and (c), halogenation will take place at methylene carbon (–CH
2
–) rather than at methyl carbon.
In the presence of alkali (NaOH), CH
3
CH(Br)CH
3
will be converted to CH
3
CH(OH)CH
3
which will respond to iodoform test.

(d)
6.
CH
3
CH
2
– CH – CH
3



2
 
CH
3
CH
2
– CH – CH
2
F F
Since F
–
CH
3
CH
2
CH = CH
2
is a poor leaving group, attack of base (C
2
H
5
O
–
) will form stable carbanion
(primary), followed by elimination of F
–
leading to the formation of less substituted alkene.
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

7.
\

PIE TEST

1

(a)
PRACTICE TEST
Correct IUPAC name of NaMn(CO)
5
is Sodium pentacarbonylmangnate(

I)

(a)
8. hc


1
2 mv
2 hc
 o

1
2 mv
 hc


1


2
1
 o



v =


2 hc m


 o

 o
 



1 / 2

(c)
9. Inversion of configuration will occur in that reaction which has a chiral center and the nucleophilic reagent attacks from the rear side with respect to the leaving group.

(d)
10.
When the pressure is high, volume would be small. So, excluded volume ‘b’ cannot be ignored. a
But as the pressure is high, the quantity
V
2 der Waal’s equation reduces to
may be neglected in comparison with P. So the van
P(V
 b) = RT
PV

Pb = RT
V =
RT
 b
P i.e. V real
= V ideal
+ b

V real
> V ideal

(a)
11. The conjugate bases of first three diones will have the same number of resonance forms, so resonance is not expected to make a significant difference. The CF
3
group is electronwithdrawing. This stabilizes the conjugate base of (A), making this dione more acidic. Methyl is a mild electron-donating group, resulting in a small destabilization of the conjugate base of (B).
Methoxy is a moderate electron-donating group, resulting in a greater destabilization of the conjugate base of (C).
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1
(b)
PRACTICE TEST
Any 1,3-dicarbonyl compound with a conjugate base which is more stable than the most stable conjugate base from (A) will be more acidic. This can be achieved by adding more electron withdrawing groups or more resonance forms for the conjugate base as in compound (D).
(C) < (B) < (A) < (D)

12. 2H + + 2e
 
H
2
E
H

/ H
2

E o
H

/ H
2

0 .
059 log
2 p
[ H
H
2

]
2
= E o
H

/ H
2

0 .
059 log
2
1
[ H

]
2
E
'
H

/ H
2

E o
H

/ H
2

0 .
059 log
2
100
[ H

]
2

E
H

/ H
2

E
'
H

/ H
2
 
0 .
059 log
2
1
[ H

]
2

[ H

]
2
100
= 0.059

(a)
13 . Equivalents of MnO = Equivalents of Mn
4
1

5 = 1.67

(5
 x ) x + x = 2

(c)
14. Mole fraction of A in the distillate or in the vapour above the liquid mixture
=
2
5

2
5

100
100

3
5

150

4
13

Mole fraction of B in distillate =
9
13
Therefore, total vapour pressure above distillate
4
=
13

(b)

100

9
13

150 = 134.61 mm of Hg
15.
1,2
 dialdehydes will undergone intramolecular cannizaro reaction in the presence of conc. NaOH to give
 hydroxy carboxylate which on acidification furnishes
 hydroxy acids.
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
H

C

C

H
O O
H
OH

H

C

C

OH
  t ransfer
O O

H
H

C

C

OH
O

O
Proton exchange
H
HO O
H
H

C

C

OH
 
H

C

C

O

HO O

(c)
16.
Borazole, B
3
N
3
H
6
is an inorganic compound which possesses a cyclic ring structure similar to benzene with alternate single and double bonds. That is why it is called inorganic benzene.
H
N
B
H

(a)
H
B
N
H
N
H
B
H
17. The larger the number of ions and the larger the charge on each, the larger the conductivity. The compounds from lowest to highest conducting are (II)<(I) < (IV) < (III).

(c), (d)
18.
Ph—O—CH
2
CH
2
NO
2
 H


+
Ph—O—CH
2
CH
2
H

(d)
I

NO
2
 
PhOH + ICH
2
CH
2
NO
2
19. % of [In

] =
[ In

]
[ In

]

[ HIn ]

(a)

100

10
10

1

100

91%
20. NaCl + AgNO
3

AgCl

(X)

(a)
(Y)
 
4
 
[Ag(NH
3
)
2
] + Cl
–     
AgCl
(Z)
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
21. Al 3+ ion hydrolyses to give acidic solution which reacts with magnesium to evolve hydrogen gas.
Al
2
(SO
4
)
3
2Al 3+ + 3 SO
2

4
Al 3+ + 6H
2
O [Al(H
2
O)
6
] 3+
[Al(H
2
O)
6
] 3+ + H
2
O [Al(H
2
O)
5
(OH)] 2+ + H
3
O +
Mg + 2H
3
O +

Mg 2+ + 2H
2
O + H
2


(d)
22. For XeF
4
,
N
2

8

4
2

6

Structure is octahedral with 2 lone pairs, i.e. square planar.
For

TeF ,
5
N
2

6

5

1

6
2
But it has one lone pair so it is not iso-structural with XeF
4
.
For I

3
,
N
2

7

2

1

5
2
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1
For BrF

,
4
N
2

7

4

1

6
2
PRACTICE TEST
It is iso-structural with XeF
4
as it also has two lone pairs.
For XeO
3
,
23.
N
2

8
2

4

(c)
Mg
CH
2
CH
2
CH
3
Br
+ HC(OC
2
H
5
)
3
CH
3
CH
2
CH
2
CH(OC
2
H
5
)
2
+ Mg
O

C
2
H
5
Br
H
3
O
+
CH
3
CH
2
CH
2
CHO n
 butyraldehyde

(a)
24. CaC
2
+ H
2
O

Ca(OH)
2
+ C
2
H
2

(A) (B) (C)
C
2
H
2
+ 2AgNO
3
+2NH
4
OH

C
2
Ag
2
+ 2NH
4
NO
3
+ 2H
2
O

(c)
(D)
25.
P(yellow) + 5/2O
2

P(red) + 5/2O
2

Therefore, P(yellow)

(d)
P
2
O
5
(s);

H
1
=

9.91 kJ/mole
P
2
O
5
(s);

H
2
=

8.78 kJ/mole

P(red);

H = 8.78

9.91 =

1.13 kJ/mole
26. We know,
PV = nRT or P = cRT
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST

1 dP
A

K p
( P
A
) n a dt
On putting P
A
= [A]RT
 
1 a
 
1 a
Given, d [ A ]

RT dt d [ A ]
 dt

K p
[ A ] n
( RT ) n
K p
( RT ) n

1
[ A ] n …(i)

1 a d [ A ]

K c
[ A ] n dt
On comparing equation (i) and (ii) we get
K c
= K p
(RT) or K p
= K c
(RT) n

1
1
 n

(c)
…(ii)
27. The statement suggests that the equilibrium final pressure of NH
3
is 3 atm
LiCl.3NH
3
(s)

K p
=
 
2
NH
3


(a)
( 3 )
2
LiCl.NH
3
(s) + 2NH
3

9 atm
28. If the solubility of AgCl in 5

10

6
M BaCl in this solution would be 10

K sp
for AgCl = [Ag
+

5
] [Cl
and that of Cl

2
solution is 10

5
M. The concentration of Ag
+
ion would be 10

] = 2

10

10
M
2

5
+ 2

5

10

6
= 2

10

5
M
AgCl(s) Ag + (aq) + Cl

(aq)
Solubility of AgCl in water = K sp
= 2

10

5 M
Since 1 litre of saturated solution of AgCl contains 2

10

5
mole AgCl
So, 2 litre saturated AgCl solution will contain
2

10

5 
2 = 2

10

5 mole AgCl

(c)
= 2

10

5 
144

10
3
= 2.88 mg AgCl
29.
The composition of ores are
Feldspar

KAlSi
3
Cryolite

Na
3
O
8
, Fluorspar

CaF
2
AlF
6
, Mica = Al
2
O
3
.2SiO
2
.2H
2
O

(b)
30. Fractional bond order means the
 bond between the molecules has been delocalized e.g. benzene. In benzene, bond order of C is 1.5, i.e. the
 bond between the carbon atoms has been delocalized.

(c)
31. (a)
32. (c)
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
33. (c)
34. (b)
35. (d)
SECTION

II
36.
Let us suppose the edge length of the cube is l .

Volume of oil = Cross sectional area of oil film
 l
(Since the oil film is monomolecular thick)
5 = 2

10
7  l l = 2.5

10

7
cm

Volume of single oil molecule = ( l )
3
= (2.5

10

7
)
3
= 1.56

10

20
cm
3

(a)
Volume
37.
Total number of oil molecules =
Volume of of single oil oil film molecule
5
=
1 .
56

10

20

(c)
Mass
38.
Moles of oil =
Molar of mass
=
5

0 .
95
240

(b) oil of
= 1.98
 oil
10

2
39.
Avogadro’s Number =
Number of
Number of molecules moles
= 3.21

10
20
3 .
21

=
1 .
98

10
10
20

2
= 1.62

10
22

(c)
40.
(d)
41. (a)
CH
3
CH
2
CH
2
CH
3
 Cl
2
UV
 light
CH
3
CH(Cl)CH
2
CH
3
+ CH
2
(Cl)CH
2
CH
2
CH
3
42. (c)
Synthesis of chiral compounds from achiral reactants always yields the racemic modification.
43. (b)
There will be five diastereomeric fractions. Unless a bond to an original chiral center is broken, the configuration is retained at that center and if a new chiral center is generated, both possible configurations about the new center result.
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com

\

PIE TEST

1 PRACTICE TEST
44. (b)
There will be five diastereomeric fractions, two optically inactive and three optically active
CH
2
Cl CH
3
Cl
CH
3
H Cl
CH
3
H
CH
3
Cl H Cl Cl
Cl H H Cl
Cl H
CH
2
CH
3
Active
CH
2
CH
3
Achiral inactive
CH
3 meso inactive
CH
3
Active
CH
2
CH
Active
2
Cl
45. (b)
46. (b)
47. (b)
48. (c)
49. (b)
50. (a)
Hints of questions No. 46 to 50
MnO
2
+ HCl

MnCl
2
+ Cl
2

+ H
2
O
(A) (B) (C)
Cl
2
+ 2KI

2KCl + I
2
(D)
I
2
+ KI

KI
3
excess yellow sol.
2MnO
2
+ 4KOH + O
2
2K
2
MnO
4
+ 2H
2
O
green
2K
2
MnO
4
+ O
3
+ H
2
O

2KMnO
4
+ 2KOH + O
2
(E)
2KOH + 5O
3

purple
2KO
3
+ 5O
2
+ H
2
O
Deep red
PIE EDUCATION , Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:011

51828585. Fax: 26962662
Also visit us at: www.pieeducation.com