Download AIEEE 2006 Chemistry Practice Test Paper

PRACTICE TEST
for
IIT JEE 2006
CHEMISTRY
TEST –II

Time: 2 Hours
Marks: 150
INSTRUCTIONS
(i)
(ii)
(iii)
(iv)
(v)
(vi)
This paper contains 50 objective questions, which consist of multiple choice type problems.
Most questions in the chemistry test are organized into groups, each containing a
descriptive passage. After studying the passage, select the one best answer to each
question in the group. Some questions are not based on a descriptive passage and are
also independent of each other. If you are not certain of an answer, eliminate the
alternatives that you know to the incorrect and then select an answer from the remaining
alternatives. Indicate your select answer by marking the corresponding answer on your
answer sheet.
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.
Use of calculator is not allowed. A periodic table is provided for your use. You may consult
it whenever you wish.
Extra paper for rough work will not be provided.
Useful Data:
Avogadro’s constant, NAV = 6.023  1023
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, RH = 109678 cm1
Mass of electron = 9.1  1031 kg
Planck’s constant = 6.626  1034 Js
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2
PRACTICE TEST
PassageI
The enthalpy of solution (Hsoln) of a salt depends on two other quantities: the energy
released when free gaseous ions of the salt combine to give the solid salt (lattice energy:
Hlatt) and the energy released when free gaseous ions of the salt dissolve in water via
solute-solvent interactions to yield the solvated ions (enthalpy of hydration: Hsolv) where:

Hsoln = Hsolv  Hlatt
From the formal definition of the quantities, it can be seen that both Hlatt and Hsolv are
exothermic.
Although these values seem to be in competition, the factors that affect Hlatt and Hsolv
do so in the same way. Firstly, the smaller the ion, the closer the association of the ion
with either other ions in the crystal lattice, or, with water molecules and thus the more
negative Hlatt and Hsolv become. Also, the greater the charge on the ion, the greater the
increase in electrostatic forces of attraction between itself and other ions or water
molecules, and the more negative Hlatt and Hsolv become.
Although Hlatt and Hsolv undergo similar changes, the change in Hsolv up or down a
group is much more profound than that of Hlatt. A good example of this is seen in the
solubility changes of the Group II carbonates.
However, there is one exception to these general rules. If the cation of the salt is
approximately the same size as the anion, the arrangement of ions in the crystal lattice is
more uniform and hence the lattice is more stable and Hlatt is more negative.
Group II carbonates
Solubility (mol L1 in H2O)
MgCO3
1.3  103
CaCO3
0.13  103
SrCO3
0.07  103
BaCO3
0.09  103
1.
2.
The solubility product for MgCO3 is
(a) 1.3  104
(c) 1.7  106
(b) 2.6  104
(d) 6.7  108
Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater
solubility in terms of mol L1?
(a) They both have the same solubility
(b) Ca(OH)2
(c) CaSO4
(d) It depends on the temperature at the time
 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2
PRACTICE TEST
3.
The CO 32  anion is approximately the same size as
(a) Mg2+
(b) Ca2+
(c) Sr2+
(d) Ba2+
4.
The Hsolv for a doubly charged anion X2 was found to be more negative than that for the
carbonate anion. Which of the following is the most likely explanation?
(a) X2 is the same size as the carbonate anion
(b) X2 is larger than the carbonate anion
(c) X2 is smaller than the carbonate anion
(d) It depends on the Hlatt for the salt containing the anion.
5.
A solution of SrCO3 in water boils at a higher temperature than pure water. Why is this?
(a) SrCO3 increases the density of water.
(b) SrCO3 decreases the vapour pressure of the water
(c) SrCO3 has a low solubility of water
(d) SrCO3 decreases the surface tension of the water.
Passage II
The essential stages in the manufacture of H2SO4 and H2SO3 involve the burning of sulfur
or roasting of sulfide ores in air to produce SO2. This is then mixed with air, purified and
passed over a vanadium catalyst (either VO3- or V2O5) at 450 °C. Thus the following
reaction occurs.
Reaction I
2SO2(g) + O2(g)
2SO3 (g) H = -197 kJ mol-1
If the SO2 is very carefully dissolved in water, sulfurous acid (H2SO3) is obtained. The
first proton of this acid ionizes as if from a strong acid while the second ionizes as if from
a weak acid.
Reaction II
 H3O+ + HSO3
H2SO3 + H2O 
Reaction III
HSO3- + H2O
H3O+ + SO32- Ka = 5.0 x 10-6
The concentration of H2SO3 in cleaning fluid was determined by titration with 0.10 M
NaOH (strong base) as shown in Fig.1. Two equivalence points were determined using
30 ml and 60 ml of NaOH respectively:
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2
PRACTICE TEST
Figure 1
6.
Which of the following acid-base indicators is most suitable for the determination of the
first end point of the titration shown in Figure 1?
(a) Cresol red (color change between pH = 0.2 and pH = 1.8)
(b) p-Xylenol blue (color change between pH = 1.2 and pH = 2.8)
(c) Bromophenol blue (color change between pH = 3.0 and pH = 4.6)
(d) Bromocresol green (color change between pH = 3.8 and pH = 5.4)
7.
If no catalyst was used in Reaction I, which of the following would experience a change in
its partial pressure when the same system reaches equilibrium?
(a) There will be no change in the partial pressure of any of the reactants
(b) SO3 (g)
(c) SO2 (g)
(d) O2 (g)
8.
If the temperature was decreased in Reaction I, which of the following would experience
an increase in its partial pressure when the same system reaches equilibrium?
(a) There will be no change in the partial pressure of any of the reactants
(b) SO3 (g)
(c) SO2 (g)
(d) O2 (g) and SO2 (g)
9.
Reaction I is usually carried out at atmospheric pressure. During the reaction, before
equilibrium was reached, the mole fractions of SO2 (g) and SO3 (g) were 1/2 and 1/6
respectively. What was the partial pressure of O2 (g)?
(a) 0.66 atm
(b) 0.16 atm
(c) 0.50 atm
(d) 0.33 atm
10.
What is the pH of 0.01 M H2SO3?
(a) 1.0
(c) 3.0
PIE EDUCATION, Corporate Office:
(b) 2.0
(d) 4.0
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
\ PIE TEST2
PRACTICE TEST
Passage III
In the strictest sense, crystal lattice
refers to an orderly arrangement of
particles. Thus, any substance which
solidifies forms its own crystal lattice,
with orderly arrangements of atoms or
molecules. Each atom, molecule or ion
is said to occupy a lattice site as shown
below in Figure 1.
Usually, solidification or freezing occurs
when the distance between individual
particles is closer than in the liquid state.
This leads to the mutual electrostatic
attractive forces between the particles
overcoming the mutual electrostatic forces
of repulsion between the particles and the
change from the liquid state to the solid
state. In some substances, however, the
distance over which the intermolecular
forces act in the crystal lattice is greater than
in the liquid state. This is due to certain
intermolecular forces existing in the liquid
state which become "fixed" in the lattice.
An example of these types of bonds is the
hydrogen bond. Thus, solid water (ice)
floats on "liquid" water. The phenomenon
gives rise to a phase diagram similar to that
shown in Figure 2.
Figure 1
Figure 2
11.
From Figure 2, what do you expect to happen to the melting point of solid water (ice) if an
increased external pressure is applied to the system?
(a) The melting point would increase.
(b) The melting point would decrease.
(c) The melting point would remain at the same value.
(d) The direction of change in the value of the melting point depends on the magnitude of
the applied pressure.
12.
Which of the following molecules would yield a similar phase diagram to that of water?
(a) CO2
(b) CH4
(c) NH3
(d) H2
 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2
PRACTICE TEST
13.
50 grams of glucose (C6H12O6) and 50 grams of sucrose (C12H22O11) were each added to
beakers of water (beaker 1 and beaker 2, respectively). Which of the following would be
true?
(a) Boiling point elevation for beaker 1 would be greater than the boiling point elevation
for beaker 2.
(b) Boiling point elevation for beaker 1 would be less than the boiling point elevation for
beaker 2.
(c) The same degree of boiling point elevation will occur in both beakers.
(d) No boiling point elevation would be observed in either of the beakers.
14.
What would the freezing point in Kelvin of a solution which is 0.50 molal in sucrose and
0.50 molal in acetic acid be?
(Kf of water = 2.0 °C mol-1 and freezing point of water = 0 °C)
(a) -1.0 K
(b) -2.0 K
(c) 272 K
(d) 271 K
15.
The density of solid water (i.e. ice) at 25 °C is 0.98 g cm-3 while that of liquid mercury at
the same temperature is 13.60 g cm-3. What percentage of the height of the ice would be
above the surface of a container filled with mercury?
(a) 93%
(b) 81%
(c) 19%
(d) 7%
16.
A sample of N2, known to contain traces of water, occupied a volume of 200 dm3 at 25°C
and 1 atm. When passed over solid Na2SO4 (drying agent), the increase in mass of the salt
was 35 g. What was the partial pressure of the N2 in the sample? (Assume ideality and
molar volume at 25°C = 24 dm3)
(a) 0.1 atm
(b) 0.2 atm
(c) 0.4 atm
(d) 0.8 atm
These questions are not based on a descriptive passage and are independent of each
other.
17.
Which of the following is NOT characteristic of hydrogen bonding?
(a) The hydrogen atom involved must be covalently bonded to a very electronegative
atom.
(b) The hydrogen bonds are typically weaker than ionic or covalent bonds
(c) The other atom involved in the hydrogen bond (not the hydrogen atom) must be
covalently bonded to a hydrogen atom.
(d) The other atom involved in the hydrogen bond (not the hydrogen atom) must possess
at least one lone pair of electrons.
18.
A current of 2 A is passed through a wire for 1.5 minutes. How many electrons passed
through the wire? (Charge on electron = 1.6  1019 C)
(a) 1.1  1021
(b) 2.5  1019
(c) 1.1  1017
(d) 2.5  1015
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2
19.
PRACTICE TEST
The freezing point of a solution of sucrose was noted and then a dissacharidase, which
degrades sucrose, was added to the solution. The enzyme was subsequently removed. How
would the freezing point of this final solution compare to that of the original solution?
(a) It would be lower
(b) It would be higher
(c) It would be the same
(d) It cannot be determined from the information given

Passage IV
Phosphorus is a Group V element and exhibits the phenomenon known as allotropy: different forms of the same element, each form being a bit more stable than another under
particular conditions. The following is a list of some of the physical properties of two of
the allotropes of phosphorus (there is also a black phosphorus not listed in table).
Physical Property
Melting point (oC)
Boiling point (oC)
Enthalpy of formation of oxide
(Hox) kJ mol-1
Soluble in:
benzene
CS2
water
Structure
White phosphorus
44
280
Red phosphorus
Sublimes at 416
-
-3020

-2900

yes
yes
no
discrete P4 molecules
no
no
no
polymer
Red phosphorus can be obtained by carefully heating white phosphorus to 400 degrees
Celsius or via exposure of white phosphorus to ionizing radiation (usually visible light) in
the absence of air.
Phosphorus vapor contains only P4 molecules, regardless of the allotrope which was
vaporized. This is due to the breakdown of the polymer-like structure of the other
allotropes.
Although phosphorus is in Group V, black phosphorus does have a relatively significant
degree of electrical conductivity. This is similar to the graphite allotrope of carbon.
20.
Which of the following is a plausible structure for white phosphorus?
(a)
(b)
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2
PRACTICE TEST
(c)
(d)
21.
What is the standard enthalpy of formation of white phosphorus from red phosphorus?
(a) 0 kJ mol1
(b) 120 kJ mol1
(c) 2900 kJ mol1
(d) 3020 kJ mol1
22.
Which of the following values do you think is closest to the bond angle in white
phosphorus P4 molecules?
(a) 30°
(b) 45°
(c) 60°
(d) 109.5°
23.
A sample of white phosphorus was reacted with excess Cl2 gas to yield 68.75 grams of
phosphorus trichloride. How many discrete P4 molecules were there in the sample?
(a) 0.75 × 1023
(b) 1.50 × 1023
(c) 3.00 × 10-23
(d) 2.40 × 10-23
24.
At very high temperatures, P4 molecules decompose to give rise to P2 molecules. What
type of bonding would you expect in these molecules?
(a) Ionic with P3+ cations and P3- anions
(b) Covalent with no sigma bonds, 3 pi bonds and 2 lone pairs of electrons
(c) Covalent with 1 sigma bonds, 2 pi bonds and 2 lone pairs of electrons
(d) Covalent with 2 sigma bonds, 1 pi bonds and 2 lone pairs of electrons
25.
One mole of an ideal gas with a molecular mass of 124.0 grams would have a lower
density than one mole of P4 because:
(a) less particles are present.
(b) it occupies a smaller volume.
(c) it weighs less.
(d) there are less strong intermolecular forces.
26.
Which of the following species is best suited to oxidize NO
(a) Ce4+
(b) Ce3+
2+
(c) Cd
(d) Cd
Given that:
Electrochemical reaction
HNO2 + H+ + e
NO + H2O
Ce4+ + eCe3+
H2O2 + 2e2OHMnO4- + eMnO42Cd2+ + 2eCd
PIE EDUCATION, Corporate Office:
Eo value (V)
+1.00 V
+1.695
+0.880
+0.564
-0.403
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com
\ PIE TEST2
27.
PRACTICE TEST
Why is the first ionization energy of manganese (Mn) less than that of chromium (Cr)?
(a) Because chromium has a half-filled d orbital.
(b) Because manganese has a half-filled d orbital.
(c) Because the relative atomic mass of chromium is greater than that of manganese.
(d) Because the atomic radius of chromium is less than that of manganese.
Passage V
Metals can occur as ions - usually with positive oxidation states - in compounds with other
species, or more rarely, in their elemental states. The more reactive a metal is, the more
difficult it is to obtain as a free element. A number of metals have only one oxidation state,
but quite a few have multiple oxidation states. Table given below says the standard
reduction potentials for various metals in various oxidation states.
Reaction
Standard reduction potential (Eo)
Zn2+ + 2e
Zn
-0.76 V
3+
2+
V +e
V
-0.26 V
2+
SO4 + 4H + 2e
2H2O + SO2
+0.17 V
2+
+
3+
VO + 2H + e
H2O + V
+0.34 V
3+
2+
Fe + e
Fe
+0.77 V
+
+
2+
VO2 + 2H + e
H2O + VO
+1.00 V
Cr2+ + 2e
Cr
-0.91 V
3+
Cr + 3e
Cr
-0.74 V
Cr3+ + e
Cr2+
-0.41 V
2+
3+
Cr2O7 + 14H + 6e
2Cr + 7H2O
+1.33 V
Transition metals often form ions which are colored in solution. They also interact with a
variety of species known as ligands forming a complex structure with the cation centrally
located and the ligands forming bonds similar to coordinate covalent bonds with the
cation. The transition metal ion can be associated with more than one different ligand
species at a time.
28.
If the ionization energies of the first row of transition metals is examined, it is observed
that the third ionization energy of manganese is greater than expected from the trend
across the period. Why is this?
(a) Because Mn2+ has a large relative molecular mass.
(b) Because Mn2+ has a relatively small charge.
(c) Because Mn2+ is less stable then Mn3+.
(d) Because Mn2+ has a half-filled d-orbital system.
29.
If a current was passed through a solution of sulfate ions and one mole of sulfur dioxide
was obtained, how many moles of chromium metal would be obtained if the same current
was passed through a solution of chromium (III) ions in solution?
(a) 0.33 moles
(b) 0.67 moles
(c) 1.00 moles
(d) 2.00 moles
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2
PRACTICE TEST
30.
Ligands, in order to associate with transition metal ions, must possess which of the
following characteristics?
(a) A negative charge
(b) A positive charge
(c) A lone pair of electrons
(d) A vacant d-orbital
31.
How many of the species in the table given in the passage could NOT reduce Fe3+
to Fe2+ ?
(a) 2
(b) 3
(c) 4
(d) 5
32.
What would be the standard cell potential for the galvanic cell formed with the VO2+/V3+
and VO2+/VO2+ half-cells?
(a) 0 V
(b) +0.66 V
(c) +1.34 V
(d) -1.34 V
PassageVI
The following questions relate to the reaction sequence outlined below
1
2
C2H5
O
O
O
C2H5
O


/ EtOH
EtO


H3C
O
O
H3C
CH3Cl
C2H5
O
O

H 3O / 



CO2 +
CH3
H3C
33.
4
Compound 1 is a/an
(a) keto carboxylic acid
(c) keto ester
O
O
CH3
H3C
3
(b) keto carboxylic acid
(d) keto ester
34.
The first step in the sequence could be enhanced by
(a) replacing the hydrogens at the alpha position by two alkyl groups.
(b) replacing one of the hydrogens at the alpha position by a chlorine atom.
(c) replacing one of the carbonyl groups with iodine atoms
(d) using ethanoate/ethanoic acid instead of ethoxide/ethanol.
35.
Although sodium methoxide is readily available, it was not used in the first step because
(a) It is not as strong a base as ethoxide.
(b) It would cause an undesired transesterification.
(c) It would be more difficult to hydrolyze a methyl ester than an ethyl ester.
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2
PRACTICE TEST
(d) a hindered base is needed to protect the carbonyl group.
36.
The second step is a type of reaction known as
(a) SN1
(b) SN2
(c) electrophilic addition
(d) free radical substitution
37.
The third step involves
(I) protonation of O in the C = O of the ester group
(II) SN2 substitution of OC2H5 by OH
(III) heat included decarboxylation
(a) (I) and (II)
(b) (II) and (III)
(c) (II) and (III)
(d) (I), (II) and (III)
38.
What type of reaction does not occur in the sequence?
(a) Hydrolysis
(b) Acidbase
(c) Reduction
(d) Decarboxylation
PassageVII
Alkenes are characterized by C = C bonds. As such, they are subject to electrophilic
addition reactions. Most electrophilic additions obey Markovnikov’s rule; however, there
are some exceptions. Examples of alkenes are shown below
1
2
H3C
CH3
H3C
CH3
H
O
H
H3C
CH3
3
H3C
CH3
H
Cl
39.
4
H3C
CH3
H
(H3C)2N
When HBr adds to compound 2 above, the main product is
CH3 CH3
CH3 CH3
(a)
Br
H
CH3
H
CH3
CH3 CH3
(c)
Br
Br
(b) H
(d)
H
H3C
CH3
H3C
Br
Br
CH3
H
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2
PRACTICE TEST
40.
What is the order of reactivity of the compounds have to electrophilic addition of HBr?
(a) 1 > 2 > 3 > 4
(b) 3 > 1 > 4 > 2
(c) 4 > 1 > 2 > 3
(d) 2 > 4 > 1 > 3
41.
Which of the compounds is most likely to form an addition product with the general
structure shown below?
H3C
CH3
group
Br
H
H
(a) Compound 1
(b) Compound 2
(c) Compound 3
(d) Compound 4
42.
What is the major product of the following reaction?
CH3
CH
H3C
ether
CH3
CH3
CH3
CH3
CH3
HC
(b) H3C
Br
CH3
CH3
Br
CH2Br
(d)
CH2
(c) H3C
CH2
CH3
CH
(a) H3C
HBr


H3C
CH3
CH
H3C
CH3
43.
What is the major product of the following reactions?
H3C
H
Cl 2

H
CH3
H3C
H
Cl
CH3
(a) Cl
H
H3C
(c) HO
H
H 2O
H3C
(b) Cl
H
H
OH
CH3
H3C
(d) Cl
H
H
OH
CH3
H
OH
CH3
 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2
PRACTICE TEST
PassageVIII
Grignard reagents (organomagenium halides) were discovered by The French chemist
Victor Grignard in 1900. He was awarded the Nobel Prize for his discovery in 1912.
Grignard reagents behave as if they were carbanions. They have extensive use in organic
synthesis. They are usually prepared by reacting an organic halide with magnesium in an
anhydrous ether solvent as follows
 RMgX
RX + Mg 
where R is a suitable organic group and X is I, Br or Cl.
44.
45.
46.
Grignard reagents are
(I) Strong acids
(III) Strong electrophiles
(a) (I) is correct
(c) (I) and (III) are correct
(II) Strong bases
(IV) Strong nucleophiles
(b) (II) is correct
(d) (II) and (IV) are correct
In the preparation of a Grignard reagent, the reason why the ether solvent must be
anhydrous is that
(a) Hydrogen ions from any water present would immediately react with any Grignard
reagent produced.
(b) Grignard reagent are insoluble in water.
(c) Water is necessary for Grignard reagents to ionize.
(d) Water is the universal solvent.
Which of the following are suitable organic reactants for synthesizing Grignard reagents?
(I) C3H7I
(II) CH3OCH2Br
(III) HSCH2CH2I
(IV) CH3COCH2Br
(a) (I) and (II)
(b) (II) and (III)
(c) (III) and (IV)
(d) (I) and (IV)
47.
One of the products of the reaction between propylmagnesium bromide and ethyne is
(a) 1pentyne
(b) propane
(c) 1pentene
(d) propene
48.
A tertiary alcohol can be ultimately formed most directly by reacting a Grignard reagent
with
(a) Methanal
(b) an aldehyde with at lest two carbon atoms
(c) a ketone
(d) a carboxylic acid
49.
One of the products of the reaction between an ester and twice as many moles of a
Grignard reagent followed by the addition of dilute acid is
(a) a primary alcohol
(b) a secondary alcohol
(c) a tertiary alcohol
(d) another ester
50.
One of the products of the reaction below is
O
i ) Et 2 O

+ C3H7MgBr (
( ii) H 3O
(a) 1pentanol
(c) ethyl propyl ether
(b) 2pentanol
(d) pentanoic acid
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2
PRACTICE TEST
PRACTICE TEST
for
IIT JEE 2006
CHEMISTRY
TEST –II
1.
c
2.
b
3.
c
4.
c
5.
b
6.
c
7.
a
8.
b
9.
d
10.
b
11.
b
12.
c
13.
a
14.
d
15.
a
16.
d
17.
c
18.
a
19.
a
20.
d
21.
b
22.
c
23.
a
24.
c
25.
d
26.
a
27.
a
28.
c
29.
b
30.
c
31.
a
32.
b
33.
d
34.
b
35.
a
36.
b
37.
d
38.
c
39.
a
40.
c
41.
c
42.
b
43.
d
44.
d
45.
a
46.
a
47.
b
48.
c
49.
c
50.
a
1.
(c)
2.
(b)
3.
(c)
SrCO3 is actually less soluble than the salt below it i.e. BaCO 3 despite the fact that Ba is situated lower down
group II in the periodic table.
4.
(c)
5.
(b)
Because of lowering of vapour and hence elevation in boiling point.
6.
(c)
The first end point in Fig. 1 uses a volume of NaOH of 30 mL and results in a pH of 3.6. The indicator used
must change color in a range which includes the expected end point pH= 3.6
7.
(a)
Catalysts only affect the rate at which equilibrium is achieved, not the equilibrium position itself.
8.
(b)
Reaction I is exothermic. The reaction could be written as:
2SO2(g) + O2(g)
2SO3(g) + Heat
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2
PRACTICE TEST
Adding heat (increasing the temperature) adds to the right hand side of the equilibrium
forcing a shift to the left in order to compensate. The reverse occurs by decreasing the
temperature thus creating a shift to the right which would produce more SO3(g) and more
heat is released.
9.
(d)
1 1 1
 
6 2 3
1
Partial pressure of O2 = 1   0.33 atm
3
Mole fraction of O2 = 1  

10.
(b)
Since we are told to consider the first ionization of H2SO3 as if from a strong acid (Reaction II), we assume
that the first proton completely dissociates. However, the second proton ionizes as if from a very weak acid
(Reaction III). After all, a Ka≈10-6 means that the product of the reactants is about 1 000 000 (one million!)
times greater than the product of the products. The preceding fact combined with the imprecision of the
available multiple choice answers means that our answer can be estimated by assuming that H2SO3 acts as a
strong monoprotic acid like HCl.
Therefore, one proton completely dissociates while the second proton's concentration is relatively negligible;
thus [H+] = 0.01 M.
pH = log 0.01 = 2
11.
(b)
The negative slope of the solid-liquid equilibrium curve in the phase diagram depicts that increasing the
pressure causes a decrease in the melting point.
12.
(c)
Ammonia, like water, is a polar molecule. Ammonia possesses a partially negative nitrogen atom and a
partially positive hydrogen atom. Thus ammonia exhibits hydrogen bonding, which is the basis for the
negative slope of the solid-liquid equilibrium curve in the phase diagram for water.
(a)
Since glucose has a smaller relative molecular mass than sucrose, there will be a greater number of moles of
glucose present when equal masses of the two substances are used. Therefore, the molality of glucose is
greater and hence the boiling point elevation is greater.
13.
14.
(d)
From the equation Tf = Kfm, Tf = (2.0 oC mol-1)(0.5 molal + 0.5 molal) = 2.0 oC. Therefore, the freezing
point of this solution is 0 oC - 2 oC = -2 oC = (273 - 2) K = 271 K.
15.
(a)
The percentage of height of ice below the surface of mercury is = density of ice/density of mercury = 0.98 g
cm-3/13.60 g cm-3 = 1/14 approximately = 7%. Therefore, the percentage of height of ice above the surface
of mercury is = 100% - 7% = 93%.
16.
(d)
We are told that the salt Na2SO4 is a “drying agent” thus we can assume it absorbs only water.
Number of moles of water absorbed  (Mass of water)/relative molecular mass H2O
= 35/18 g mol1
= 36/18 g mol1 approximately = 2 moles
Given: 1 mole of a gas at room temperature and pressure = 24 dm3 mol1
Therefore, volume occupied by 2 moles H2O = 24 dm3 mol1  2 mol = 48 dm3
Therefore, volume occupied by N2 = Total volume of gas  Volume of H2O
17.
(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2
PRACTICE TEST
18.
(a)
Q = 2  1.5  60 = 180 C
Therefore, the total charge Q is 180 C. The charge on an electron is 1.6  1019 C per electron. Thus the
number of electrons n = Q/e
n = 180 C/(1.6  1019 C per electron) = (180/16)  1019
= (1800/16)  1019
n = (450/4)  1019 = 1.125  1021 electrons
19.
(a)
20.
(d)
Phosphorus is in Group V. It can therefore either have a valency of 3 or 5 Answer choice (d)., which has
three bonds to each phosphorous, is the only answer which fulfils this requirement.
21.
(b)
Equation: Red phosphorus  White Phosphorus Hreaction = x kJ mol1
Hformation: 2900 kJ mol-1(red), -3020 kJ mol-1(white)
Hreaction = Hproducts  Hreactants = -3020 kJ mol-1  (2900 kJ mol-1) = 120 kJ mol-1
22.
(c)
Each face of the molecule is a triangle, each "side" consisting of the same type of bond(s). Therefore, the
triangle would be expected to resemble an equilateral triangle, where each angle is 60.
23.
(a)
 4PCl3
P4 + 6Cl2 
68.75
1
1 1
mole PCl3 = mole PCl3 =  moles P4
2
4 2
137.5
1
 Number of P4 molecules =  6.023  10 23 = 0.75  1023
8
24.
(c)
Phosphorus is in Group V and therefore requires three more electrons to approach the more stable noble gas
electronic configuration. If these are shared with another phosphorus atom, three covalent bonds exist
between the two atoms. Multiple bonds usually contain one sigma bond, the other bonds being pi bonds.
Each atom will also possess a lone pair of electrons. {Notice this is the same as for N2 as nitrogen is in the
same group as phosphorus in the periodic table; when you see something you are unfamiliar with, always
look
for
similarities
with
something
you
know
well}
Answer choice A. is incorrect because if phosphorus loses three electrons, it will have a total of 12 electrons,
which does not correspond to a stable noble gas-like configuration.
25.
(d)
The molecular mass of the real gas P4 is 124 g/mol (i.e. 4  31.0 g/mol). The question is
really asking: what is the difference between a real gas and an ideal gas if the molecular
weight is the same?
An increase in density depends on an increase in the mutual forces of attraction between
particles, causing the volume occupied by the same mass of those particles to decrease
[density = (mass)/(volume)]. Ideal gases possess no intermolecular forces so its density
should be less than that of any real gas.
26.
(a)
The more positive the Eo value, the greater the tendency for that half-reaction to proceed to the right. Since
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2
PRACTICE TEST
the Eo value for the Ce4+/Ce3+ equilibrium system is more positive than that for the HNO 2/NO equilibrium
system, the latter will be shifted to the left, that is, NO will be oxidized
27.
(a)
A half-filled electron shell contains subshells each of which possesses one unpaired electron. All unpaired
electrons have parallel spins and this half-filled system confers extra stability to the atom. Thus more energy
than would be expected (from a consideration of the ionization energies of other elements in the same
period) is required to remove a valent electron from the neutral atom.
28.
(c)
A half-filled orbital has extra stability associated with it since all the electrons are unpaired and have parallel
spins. Thus it is more difficult to remove an electron from this system and hence it requires more energy.
29.
(b)
From the table we can see that each mole of sulfate ions requires two moles of electrons for reduction to one
mole of sulfur dioxide, but each mole of chromium (III) cations requires three moles of electrons for
reduction to one mole of chromium metal. Hence, the ratio of the number of moles of electrons available to
produce the chromium metal will be 2:3 producing 0.67 moles of the metal.
30.
(c)
Answer choice (c). is the only absolute requirement for ligand association with a transition metal. Though
the ligand must possess a lone pair of electrons to associate with the positive charged metal cations, no
formal charge is necessary.
31.
(a)
The key to the problem is the following: a reaction will tend to be spontaneous if the final (overall) E  for
the reaction is positive. For the reduction of Fe3+ to occur, the E of the other half-reaction used must be
more negative or less positive than that of the Fe3+/Fe2+ equilibrium system so that the equilibrium will shift
to the right, that is, reduction of Fe3+ with an overall positive E for the reaction. Only two half reactions do
not fulfil the preceding requirement: VO2+/VO2+ and Cr2O72-/Cr3+.
32.
(b)
The direction of the reactions are chosen such that a positive Eo occurs for the overall
reaction
VO2+ + 2H+ + eH2O + VO2+ Eo = +1.00 V
VO2+ + 2H+ + eH2O + V3+ (Eo = +0.34 V)
--------------------------------------------------------------------------------VO2+ + V3+ ↔ 2VO2+
33.
(d)
34.
(b)
35.
(a)
36.
(b)
37.
(d)
38.
(c)
39.
(a)
40.
(c)
Eo = +0.66 V
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2
41.
(c)
42.
(b)
43.
(d)
44.
(d)
45.
(a)
46.
(a)
47.
(b)
48.
(c)
49.
(c)
50.
(a)
PRACTICE TEST
PIE EDUCATION, Corporate Office:
44A/2, Sarvapriya Vihar, New Delhi - 16. Ph:01151828585. Fax: 26962662
Also visit us at: www.pieeducation.com