AIEEE 2007 Chemistry

8
AIEEE 2007 Chemistry 1. The equivalent conductances of two strong electrolytes at infinite dilution in H 2 O (where ions move freely through a solution) at 25 °C are given below Λ CH C ONa 2 3 Sc equiv O 0 1 91 0 = - . m Λ HCl 2 S cm equiv 0 1 426 2 = - . What additional information/quantity one needs to cal- culate Λ 0 of an aqueous solution of acetic acid? (a) Λ 0 of chloroacetic acid (ClCH 2 COOH) (b) Λ 0 of NaCl (c) Λ 0 of CH 3 COOK (d) The limiting equivalent conductance of H H + + Λ ( ) 0 Solution (b) From Kohlrausch’s law, Λ Λ Λ Λ CH COOH CH COONa HCl NaCl 3 3 0 0 0 0 = + - 2. Which one of the following is the strongest base in aque- ous solution? (a) Methylamine (b) Trimethylamine (c) Aniline (d) Diemethylamine Solution (d) In aqueous solution, the basicity order of 1° > 2° > 3° amines and with methyl group, the basicity order is 2° > 1° > 3°. In case of aniline, lone pair of nitrogen is involved in resonance, so it is weaker base than ali- phatic amines. 3. The compound formed as a result of oxidation of ethyl benzene by KMnO 4 is (a) benzyl alcohol. (b) benzophenone. (c) acetophenone. (d) benzoic acid. Solution (d) The reaction is CH 2 CH 3 Ethyl benzene Benzoic acid COOH H + KMNO 4 4. The IUPAC name of the following compound is (a) 3-ethyl-4,4-dimethylheptane. (b) 1,1-diethyl-2,2-dimethylpentane. (c) 4,4-dimethyl-5,5-diethylpentane. (d) 5,5-diethyl-4,4-dimethylpentane. Solution (a) The IUPAC name is 3-Ethyl-4,4-dimethylheptane 1 2 3 4 5 6 7 H 3 C CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CH 3 CH 3 C CH 5. Which of the following species exhibits diamagnetic behavior? (a) NO (b) O 2 2- (c) O 2 + (d) O 2 Solution (b) The electronic configurations are as follows: O 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 - * * * * = = = s s s s s p p p p s s s s p p p p p z x y x y General Instructions (a) There are objective type questions with four options having single correct answer. (b) For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted. (c) No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet. (d) The candidates are advised not to attempt such item in the answer sheet if they are not sure of the correct response. (e) More than one answer indicated against a question will be deemed as incorrect response and will be negatively marked. AIEEE 2007 Chemistry.indd 1 12/19/2011 12:44:20 PM

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AIEEE 2007

Chemistry

1. The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25 °C are given below

ΛCH C ONa2

3Sc equivO

0 191 0= −. m

ΛHCl2S cm equiv0 1426 2= −.

What additional information/quantity one needs to cal-culate Λ0 of an aqueous solution of acetic acid?

(a) Λ0 of chloroacetic acid (ClCH2COOH)

(b) Λ0 of NaCl (c) Λ0 of CH3COOK (d) The limiting equivalent conductance of H

H++ Λ( )0

Solution(b) From Kohlrausch’s law,

Λ Λ Λ ΛCH COOH CH COONa HCl NaCl3 3 0 0 0 0= + −

2. Which one of the following is the strongest base in aque-ous solution?

(a) Methylamine (b) Trimethylamine (c) Aniline (d) Diemethylamine

Solution (d) In aqueous solution, the basicity order of 1° > 2° > 3°

amines and with methyl group, the basicity order is 2° > 1° > 3°. In case of aniline, lone pair of nitrogen is involved in resonance, so it is weaker base than ali-phatic amines.

3. The compound formed as a result of oxidation of ethyl benzene by KMnO4 is

(a) benzyl alcohol. (b) benzophenone. (c) acetophenone. (d) benzoic acid.

Solution(d) The reaction is

CH2CH3

Ethyl benzene Benzoic acid

COOH

H+

KMNO4

4. The IUPAC name of the following compound is

(a) 3-ethyl-4,4-dimethylheptane. (b) 1,1-diethyl-2,2-dimethylpentane. (c) 4,4-dimethyl-5,5-diethylpentane. (d) 5,5-diethyl-4,4-dimethylpentane.

Solution(a) The IUPAC name is

3-Ethyl-4,4-dimethylheptane

1234567

H3C CH2 CH2 CH2 CH3

CH3 CH2CH3

CH3

C CH

5. Which of the following species exhibits diamagnetic behavior?

(a) NO (b) O22− (c) O2

+ (d) O2

Solution(b) The electronic configurations are as follows:

O22 2 2 2 2 2

2 2 2 2

1 1 2 2 2

2 2 2 2

− ∗ ∗

∗ ∗

=

= =

s s s s s

p p p p

s s s s p

p p p pz

x y x y

General Instructions

(a) There are objective type questions with four options having single correct answer. (b) For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted.(c) No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet.(d) The candidates are advised not to attempt such item in the answer sheet if they are not sure of the correct response.(e) More than one answer indicated against a question will be deemed as incorrect response and will be negatively marked.

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No unpaired electron, so it is diamagnetic.O2

+ ∗ ∗

=

= =

s s s s s

p p p p

1 1 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 0

s s s s p

p p p pz

x y x y*

One unpaired electron, so it is paramagnetic.O2

2 2 2 2 2

2 2 1 1

1 1 2 2 2

2 2 2 2

=

= =

s s s s s

p p p p

s s s s p

p p p pz

x y x y

∗ ∗

∗ ∗

Two unpaired electrons, so it is paramagnetic.NO =

= =

s s s s s

p p p p

1 1 2 2 2

2 2 2 2

2 2 2 2 2

2 2 1 0

s s s s p

p p p pz

x y x y

∗ ∗

∗ ∗

One unpaired electron, so it is paramagnetic. 6. The stability of dihalides of Si, Ge, Sn and Pb increases

steadily in the sequence (a) PbX SnX GeX SiX2 2 2 2<< << <<

(b) GeX SiX SnX PbX2 2 2 2<< << << (c) SiX GeX PbX SnX2 2 2 2<< << <<

(d) SiX GeX SnX PbX2 2 2 2<< << <<

Solution(d) Due to inert pair effect, the stability of +2 oxida-

tion state increases as we move down the group. Therefore, SiX GeX SnX PbX2 2 2 2<< << << .

7. Identify the incorrect statement among the following: (a) Br2 reacts with hot and strong NaOH solution to give

NaBr, NaBrO4 and H2O. (b) Ozone reacts with SO2 to give SO3. (c) Silicon reacts with aq. NaOH in the presence of air to

give Na2SiO3 and H2O. (d) Cl2 reacts with excess of NH3 to give N2 and HCl.

Solution(a) Br2 reacts with hot and strong NaOH to give NaBr,

NaBrO3 and H2O.

8. The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cat-ionic species, K+, Ca2+, Mg2+, Be2+?

(a) Ca < Mg <Be <K2+ 2+ 2+ +

(b) Mg <Be <K < Ca2+ 2+ + 2+

(c) Be < K < Ca < Mg2+ + 2+ 2+

(d) K < Ca < Mg <Be+ 2+ 2+ 2+

Solution(d) Higher is the charge/size ratio, higher is the polarizing

power. Therefore, K < Ca < Mg <Be+ 2+ 2+ 2+ .

9. The density (in g mL−1) of a 3.60 M sulphuric acid solu-tion that is 29% H2SO4 (molar mass = 98 g mol−1) by mass will be

(a) 1.45 (b) 1.64 (c) 1.88 (d) 1.22

Solution(d) Let the density of solution be r. Molarity of solution

given = 3.6 M, that is 1 L of solution contains 3.6 mol of H2SO4 or 1 L of solution contains 3.6 × 98 g of H2SO4. Since the solution is 29% by mass, 100 g solution contains 29 g H2SO4

100/r mL solution contains 29 g H2SO4

1000 mL solution contains 3.6 × 98 g H2SO4

Therefore,

3 6 98 1. × = × ×29100

000r

Solving, we get r = 1.22g mL−1.10. The first and second dissociation constants of an acid H2A

are 1.0 × 10−5 and 5.0 × 10−10, respectively. The overall dis-sociation constant of the acid will be

(a) 0.2 × 105 (b) 5.0 × 105 (c) 5.0 × 1015 (d) 5.0 × 10−15

Solution(d) From the expression K = K1 × K2, we have

K = × × = ×− − −10 5 10 5 105 10 15

11. A mixture of ethyl alcohol and propyl alcohol has a vapor pressure of 290 mm Hg at 300 K. The vapor pressure of propyl alcohol is 200 mm Hg. If the mole fraction of ethyl alcohol is 0.6, its vapor pressure (in mm Hg) at the same temperature will be

(a) 360 (b) 350 (c) 300 (d) 700

Solution(b) Let p be the vapor pressure of pure ethyl alcohol, then

according to Raoult’s law

29 2 4 60 00 0 0= × + ×. .p

p = − =29 835 mm Hg

0 00 6

0.

12. In the conversion of limestone to lime, CaCO (s) CaO(s) CO (g)3 2→ + , the values of ∆Ho and ∆So are + 179.1 k J mol–1 and 160.2 J K−1, respectively, at 298 K and 1 bar. Assuming that ∆Ho and ∆So do not change with temperature, the temperature above which conversion of limestone to lime will be spontaneous is

(a) 1118 K (b) 1008 K (c) 1200 K (d) 845 K

Solution(a) We will find out the equilibrium temperature at which

∆G = 0. We know that

∆ ∆ ∆∆ ∆

G H T SH T S

= −= −0

Therefore, ∆ ∆H T S= . Hence,

T = × =179 1 116 2

K.

.000

01118

13. The energies of activation for forward and reverse reac-tions for A +B AB2 2� 2 are 180 k J mol–1 and 200 k J mol–1, respectively. The presence of a catalyst lowers the activa-tion energy of both (forward and reverse) reactions by 100 k J mol–1. The enthalpy change of the reaction ( )A B AB2 2 2+ → in the presence of catalyst will be (in k J mol–1)

(a) 20 (b) 300 (c) 120 (d) 280

Solution(a) From Fig. 1,

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100

200

180

In absence of catalyst

In presence of catalyst

80

Reaction coordinate

Pote

ntia

l ene

rgy

Figure 1

∆r (forward) backward kJmolH E E= = =− − − −( ) 80 100 20 1

14. The cell Zn Zn M Cu M Cu; ( V)cello| || |2 21 1 1 10+ + =( ) ( ) .E , was

allowed to be completely discharged at 298 K. The rela-tive concentration of Zn2+ to Cu2+, [Zn2+]/[Cu2+] is

(a) 9.65 × 104 (b) antilog (24.08) (c) 37.3 (d) 1037.3

Solution(d) Using Nernst equation,

E Encell cell

o ZnCu

= −+

+

0 0591 2

2

.log

[ ][ ]

For complete discharge, Ecell = 0, so

Encell

o2+

2+

ZnCu

− =0 05910

.log

[ ][ ]

The reactions at anode and cathode areZn Zn→ ++ −2 2e

Cu Cu2 2+ −+ →e

So, n = 2,

Ecello

2+

2+

Zn

Cu=

0 05912

.log

Solving by substituting given value of Ecello V= 1 10. ,

we get

1 10 20 0591..

log[ ][ ]

× = ZnCu

2+

2+

37 3 1037 3. log]

[ ][ ][ ]

.= ⇒ =[ZnCu

ZnCu

2+

2+

2+

2+

15. The pKa of a weak acid (HA) is 4.5. The pOH of an aque-ous buffered solution of HA in which 50% of the acid is ionized is

(a) 7.0 (b) 4.5 (c) 2.5 (d) 9.5

Solution(d) For buffer solution (Henderson equation)

pH p logSalt]Acid]

4 5 logSalt]Acid]a= + = +K

[[

.[[

As the acid is 50% ionized, which means that [salt] = [acid]

pH 4 5 log 1 4.5= + =.

Therefore,

pOH 14 pH 14 4 5 9 5= = =− − . .

16. Consider the reaction, 2A B Products+ → . When con-centration of B alone was doubled, the half life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

(a) s–1 (b) L mol–1 s–1 (c) It has no units. (d) mol–1 s–1

Solution(b) For the reaction 2A B+ → product, when concen-

tration of B is doubled, the half life did not change, hence the reaction is of first order with respect to B because half life is independent of concentration of first-order reaction (t1/2 = 0.693/k). When the concen-tration of A is doubled, the reaction rate is doubled, hence the reaction is of first order with respect to A. So, the overall order is 1 + 1 = 2 and the units of rate constant for a second order reaction is L mol–1 s–1.

17. Identify the incorrect statement among the following (a) 4f and 5f orbitals are equally shielded. (b) d-Block elements show irregular and erratic chemical

properties among themselves. (c) La and Lu have partially filled d orbitals and no other

partially filled orbitals. (d) The chemistry of various lanthanoids is very similar.

Solution(a) 4f and 5f belongs to different energy levels, hence

shielding effect is not the same for both of them. Shielding of 4f is more than 5f. Also, 5f is less deeply buried than 4f.

18. Which one of the following has a square planar geometry? (a) [ ]PtCl4

2− (b) [ ]CoCl42−

(c) [ ]FeCl42− (d) [ ]NiCl4

2−

(Atomic numbers of Co = 27, Ni = 28, Fe = 26, Pt = 78)

Solution(a) Most of the 4d and 5d series elements show square

planar geometry.

19. Which of the following molecules is expected to rotate the plane of plane-polarized light?

(a) (b)

CHO

CH2OH

HO H

SH

(c) (d)

Ph Ph

NH2H2N

H H

COOH

H

H2N H

Solution(a) The compound has a chiral carbon atom.

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(a) CHO

CH2OH Chiral

HO H

Chiral carbon(A Carbon attachedto four di�erentsubstituents)

*

(c)

Ph Ph

NH2H2N

H H

AchiralPlane of symmetry

(b)

SH Achiral

Plane of symmetry(which divides themolecule into twoequal halves)

(d) COOH

H2N H

H Achiral

Two identicalgroups attached

20. The secondary structure of protein refers to (a) fixed configuration of the polypeptide backbone. (b) a -helical backbone. (c) hydrophobic interactions. (d) sequence of a -amino acids.

Solution(b) Secondary structure of proteins involves a -helical

backbone and b -sheet structures. These are formed as a result of hydrogen bonding between different peptide groups.

21. Which of the following reactions will yield 2,2-dibromo-propane?

(a) CH3 CH2CH + HBr

(b) CH3 C CH + 2HBr

(c) CH3CH + HBrCHBr

(d) + 2HBrCH CH

Solution(b) The reaction is as follows:

CH3 C CH HBr

HBr

CH2

Br

CH3 C+

Br

Br

CH3CH3C

Markonikov’srule

(Markonikov’srule)

2,2-Dibromo-propane

22. In the chemical reaction, CH CH NH CHCI KOH3 2 2 3 3+ +→ + +A B 3H O2 , the compounds (A) and (B) are, respectively,

(a) C2H5NC and 3KCl (b) C2H5CN and 3KCl

(c) CH3CH2CONH2 and 3KCl (d) C2H5NC and K2CO3

Solution(a) This is an example of carbylamine reaction. So, the

products will be C2H5NC and 3KCl.

23. The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly:

(a) m-chlorotoluene. (b) benzoyl chloride. (c) benzyl chloride. (d) o- and p-chlorotoluene.

Solution(d) The reaction involved is electrophilic addition reaction:

CH3 CH3

FeCl3

Cl2

Cl

CH3

Cl

+

Toluene(o, p-directing

in nature)

o-Chloro-toluene(Minor)

p-Chloro-toluene(Major)

24. Presence of a nitro group in a benzene ring (a) deactivates the ring towards electrophilic substitution. (b) activates the ring towards electrophilic substitution. (c) renders the ring basic. (d) deactivates the ring towards nucleophilic substitution.

Solution(a) Nitro group reduces the electron density in the ben-

zene ring due to strong −I effect.

NO O

NO O

+

ONO

+

NO

+

−−−

NO OO

Overall electron density on benzene ring decreases mak-ing electrophilic substitution difficult. Hence, it deacti-vates the ring towards electrophilic substitution.

25. In which of the following ionization processes, the bond order has increased and the magnetic behavior has changed?

(a) N N2 2→ +

(b) C C2 2→ +

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(c) NO NO→ +

(d) O O2 2→ +

Solution(c) The electronic configurations are as follows:

C22 2 2 2 2 21 1 2 2 2 2= =s s s s p ps s s s p px y

∗ ∗

C22 2 2 2 2 11 1 2 2 2 2+ = =s s s s p ps s s s p px y

∗ ∗

The electron has been removed from bonding molecular orbital, so the bond order decreases.In NO NO→ + ,

NO →

= =

s s s s s

p p p p

1 1 2 2 2

2 2 2 2

2 2 2 2 2

2 2 1 0

s s s s p

p p p pz

x y x y

* *

*

NO+ → =s s s s s p p1 1 2 2 2 2 22 2 2 2 2 2 2s s s s p p pz x y* *

The electron has been removed from antibonding orbital, so bond order increases and the magnetic nature changes from paramagnetic to diamagnetic.

26. The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because

(a) the 5f orbitals extend farther from the nucleus than 4f orbitals.

(b) the 5f orbitals are more buried than the 4f orbitals. (c) there is similarity between 4f and 5f orbitals in their

angular part of the wave function. (d) the actinoids are more reactive than the lanthanoids.

Solution(a) This is because of very small energy gap between 5f,

6d and 7s subshells. Hence, all their electrons can take part in bond formation.

27. Equal masses of methane and oxygen are mixed in an empty container at 25 °C. The fraction of the total pres-sure exerted by oxygen is

(a) 12 (b)

23

(c) 13

273298

×

(d) 13

Solution (d) Let the mass of methane and oxygen be x g. Then

Mole fraction of oxygenNumber of moles of oxygen

Totalnumber of mol=

ees

The number of moles of oxygen = x/32 and the total number of moles of methane = x/16. So,

Mole fraction of oxygen 32

32 16

13

=+

=

x

x x

Let the total pressure be p. Then, Partial pressure of (O2) = Mole fraction of oxygen × p

= × =13 3

pp

28. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol–1) in the same sol-vent. If the densities of both the solutions are assumed to be equal to 1.0 g m–3, molar mass of the substance will be

(a) 210.0 g mol–1 (b) 90.0 g mol–1

(c) 115.0 g mol–1 (d) 105.0 g mol–1

Solution (a) Isotonic solutions have the same osmotic pressure.

Π Π1 1 2 2= = =C RT C RT

so, C1 = C2. Let M be the molar mass of the substance. Now as the densities are same

5 25..

.MM= ⇒ = × =15

605 25

601 5

210

29. Assuming that water vapor is an ideal gas, the internal energy change (∆U) when 1 mol of water is vaporized at 1 bar pressure and 100 °C (given that molar enthalpy of vaporization of water at 1 bar and 373 K is 41 kJ mol−1 K−1) will be:

(a) 41.00 kJ mol−1 (b) 4.100 kJ mol−1

(c) 3.7904 kJ mol−1 (d) 37.904 kJ mol−1

Solution(d) The reaction is

H O(l) H O(g)2Vaporization

2 →

where ∆ng 1 1= − =0 . Now,

∆ ∆ ∆H U n RT= + g

∆ ∆ ∆U H n RT= = × × = −− − −g

341 8 3 1 373 37 9 kJ mol. .0 1

30. In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (molar mass = 283 g mol−1) the equilib-rium which sets in is

AgIO s Ag aq IO aq3 3( ) ( ) ( )� + + −

If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 × 10–8, what is the mass of AgIO3 con-tained in 100 mL of its saturated solution?

(a) 1 1 g4.0 0× −

(b) 28 3 1 g2. × 0−

(c) 2 83 1 g3. × 0−

(d) 1 1 g7.0 0× −

Solution(c) For the reaction,

AgIO s Ag IO3( )� + −3

Let S be the solubility product of AgIO3, then

K sp+Ag IO= −[ ][ ]3

1 1 1 mol L8 2 4.0 0 0 1× = ⇒ =− − −S S

Mass of AgIO3 contained in 100 mL of its saturated solution is

1 283 11

283 1 2 83 1 g per 1 ml4

5 30 00000

0 0 00−

− −× × = × = ×.

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31. A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is 10 times the permissible value, after how many days will it be safe to enter the room?

(a) 100 days

(b) 1000 days

(c) 300 days

(d) 10 days

Solution(a) We know that

Activitydd

∝N

tN

N NN

N

n n

=

⇒ =

0

0

12

12

110

12

10 2=

⇒ =n

n

Taking log on both the sides,log 1 log 20 = n

n = = =10 301

2 0 3010.

. ( log . )3 32 as

Therefore, t n t= × = × =1 2 3 32 3 99 6 days./ . .032. Which one of the following conformations of cyclohex-

ane is chiral? (a) Boat (b) Twist boat (c) Rigid (d) Chair

Solution(b) Twisted boat form of cyclohexane.

1 4

2

36

5

It is chiral because it does not have plane of symmetry. 33. Which of the following is the correct order of decreasing

SN2 reactivity? (a) R CHX R CX RCH X2 3 2> >

(b) RCH X R CX R CHX2 3 2> > (c) RCH X R CHX R CX2 2 3> >

(d) R CX R CHX RCH X3 3 2> >(where X = a halogen).

Solution(c) SN2 reactivity depends on the steric hindrance

because it involves the attack of nucleophile from the back side. So, more is the steric hindrance, lesser is the SN2 reactivity.

34. In the following sequence of reactions,

CH CH OH A B C D3 2P I Mg HCHO H O2+ 2 → → → →

the compound “D” is

(a) propanal. (b) butanal. (c) n-butyl alcohol. (d) n-propyl alcohol.

Solution(d) The reaction is

CH3 CH2 OHA

B

CH3CH2 CH2O MgICH3CH2 CH2OH

Propanol+ Mg(OH)I

CH3 CH2I

CH3CH2MgI

Mgdry ether

(Grignards reagent)

H2O/H+

+ H C O

H

P + I2

C

D

35. Which of the following sets of quantum numbers repre-sents the highest energy of an atom?

(a) n = 3, l = 0, m = 0, s = + 1/2

(b) n = 3, l = 1, m = 1, s = + 1/2 (c) n = 3, l = 2, m = 1, s = + 1/2

(d) n = 4, l = 0, m = 0, s = + 1/2

Solution(c) According to (n + l) rule, more is the value of (n + l)

more is the energy. Hence, (c) is the correct option as it has maximum value for (n + l).

36. Which of the following hydrogen bonds is the strongest? (a) O−H--F (b) O−H--N (c) F−H--F (d) O−H--O

Solution(c) The hydrogen bond in HF is the strongest because

fluorine is the most electronegative element.

37. In the reaction,

2Al s HCl(aq) Al aq Cl aq H( ) ( ) ( ) ( )+ → + ++6 2 6 332

− g

(a) 11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed.

(b) 6.0 L HCl(aq) is consumed for every 3 L H2(g) produced. (c) 33.6 L H2(g) produced regardless of temperature and

pressure for every mole Al that reacts (d) 67.2 L H2(g) at STP is produced for every mole Al that

reacts

Solution(a) The reaction is

2Al s HCl( ) Al Cl H( ) ( ) ( ) ( )+ → + ++ −6 2 6 332aq aq aq g

For each mole of HCl reacted, 0.5 mol of H2 gas is formed at STP.1 mol of an ideal gas occupies 22.4 L at STP.Volume of H2 gas formed at STP per mole of HCI reacted is 22.4 × 0.5 L.

38. Regular use of which of the following fertilizers increases the acidity of soil?

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(a) Ammonium sulphate (b) Potassium nitrate (c) Urea (d) Superphosphate of lime

Solution(a) Ammonium sulphate is a salt of strong acid and

weak base. On hydrolysis, it produces H+ ions which

increase the acidity of soil.

( )NH SO 2NH SO4 2 4 → ++ −4 4

2

NH H O NH OH H2 44+ +++ �

39. Identify the correct statement regarding a spontaneous process:

(a) Lowering of energy in the reaction process is the only criterion for spontaneity.

(b) For a spontaneous process in an isolated system, the change in entropy is positive.

(c) Endothermic processes are never spontaneous. (d) Exothermic processes are always spontaneous.

Solution(b) From the relation ∆G = ∆H − T∆S, ∆G should be nega-

tive for a spontaneous process and for that ∆S has to be positive.

40. Which of the following nuclear reactions will generate an isotope?

(a) b -particle emission (b) Neutron particle emission (c) positron emission (d) a -particle emission

Solution(b) Isotopes have same atomic number but different

mass number

AZ

AZ nX X 1 1→ +−

0

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