1. Illustration 1: One mole of a mixture of CO and CO2 requires exactly 20 gms of NaOH to convert
all the CO2 into Na2CO3. How many more gms of NaOH would it require for conversion into Na2CO3
if the mixture (one mole) is completely oxidised
to CO2.
(A) 60 gm (B) 80 gm
(C) 40 gm (D) 20 gm
Solution: Moles of NaOH =
2
1
40
20

Moles of CO2 =
4
1
2
1
2
1


2 = 2]
Moles of CO =
4
3
4
1
1 

Moles of CO2 produced =
4
3 from CO
Moles of NaOH extra =
2
3
2
4
3


Mass of NaOH extra
4
2
3

= 60
Illustration 2: One litre of 0.1 M CuSO4 solution is electrolysed till the whole of copper is
deposited at cathode. During the electrolysis a gas is released at anode. The volume of the gas is
(A) 112ml (B) 254 ml
(C) 1120 ml (D) 2240 ml
Solution: When copper is deposited at cathode, oxygen gas is released at anode
Equivalents of copper = 0.1 × 2 = 0.2
Equivalents of oxygen = 0.2
Volume of oxygen at NTP = 0.2 × 5600 = 1120 ml
Illustration 3: In a reaction
FeS2 + MnO4 + H+
⎯ → Fe3+
+ SO2 + Mn2+
+ H2O
The equivalent mass of FeS2 would be equal to
(A) Molar mass (B)
10
mass
molar
(C)
11
mass
molar (D)
13
mass
molar
Solution: Fe2+
⎯ → Fe3+
+ e–
S2
–2
⎯ → 2S+4
+ 10e–
–––––––––––––––––
FeS2 ⎯ → 2S+4
+ Fe3+
+ 11e–

2. ∴ Equivalent mass of FeS2 =
11
mass
molar
Illustration 4: Equal volumes of 0.2M HCl and 0.4M KOH are mixed. The concentration of the
principal ions in the resulting solution are
(A) [K+
] = 0.4M, [Cl–
] = 0.2M, [H+
] = 0.2M
(B) [K+
] = 0.2M, [Cl–
] = 0.1M, [OH–
] = 0.1M
(C) [K+
] = 0.1M, [Cl–
] = 0.1M, [OH–
] = 0.1M
(D) [K+
] = 0.2M, [Cl–
] = 0.1M, [OH–
] = 0.2M
Solution: Resulting solution in alkaline since M(KOH) > M(HCl), hence the molarity of
KOH after reaction
=
V
V
V
2
.
0
V
4
.
0



 = 0.1M
KOH + HCl KCl + H2O
t = 0 0.4 0.2
t = t 0.2 0 0.2
0.1 0.1 due to dilution
∴ [K+
] = 0.1 + 0.1 = 0.2M
[Cl–
] = 0.1M
[OH–
] = 0.1M
Illustration 5: To prepare a solution that is 0.5M KCl starting with 100ml of 0.4M HCl
(A) Add 0.75gm KCl (B) Add 20 ml of water
(C) Add 0.10 mol of KCl (D) Evaporate 10 ml water
Solution: a) 100ml of 0.4M KCl =
1000
100
4
.
0  mol = 0.04 mol (initial)
= 0.04 × 74.5 (initially) = 2.98gm
= 3.73g (after) = 0.05 mol in 100 ml
= 0.5M
∴ (A) is true
b) Addition of 20 ml water will decrease molarity that will be less than 0.4 M
∴ (B) is false
c) 0.04 (initial as in (a)) + 0.10 (added)
= 0.14 mol KCl in 100 ml
= 1.4 M
∴ (C) is false
d) 0.04 mol KCl in 90 ml solution (after 10 ml water evaporated) = 0.044 M
∴ (D) is also false
Illustration 6: For the reaction
mol
1
4
3PO
H +
mol
1
2
)
OH
(
Ca ⎯ → CaHPO4 + 2H2O
Which are true statements
(A) Equivalent weight of H3PO4 is 49
(B) Resulting mixture is neutralised by 1 mol of KOH
(C) CaHPO4 is an acid salt
(D) 1 mol of H3PO4 is completely neutralised by 1.5 mol of Ca(OH)2.
Solution: a) By given reaction

3. H3PO4 ≡ 2OH–
∴ Equivalent weight =
2
PO
H 4
3
= 49
∴ (A) is true
b) Only one acidic H in CaHPO4 hence neutralised by 1 mol of KOH
∴ (B) is true
c) One acidic H in CaHPO4 makes it acidic salt.
∴ (C) is true
d) H3PO4 ≡ 3H+
≡ 3OH–
≡ 1.5 Ca(OH)2
∴ (D) is true
Illustration 7: The brown ring complex compound is formulated as [Fe(H2O)5NO]SO4. The
oxidation number of iron is
(A) 1 (B) 2
(C) 3 (D) 0
Solution: [Fe(H2O)5NO]SO4 [Fe(H2O)5NO]2+
+ SO4
2–
In this complex NO transfers one of its electron to Fe (which is initially +2)
This gives +1 charge on NO and +1 charge on Fe. Fe has thus three unpaired electrons as
confirmed by its magnetic moment which is 15 B.M.
∴ (A)
Illustration 8: An organic compound contains 4% sulphur minimum molecular weight is
(A) 200 (B) 400
(C) 800 (D) 16000
Solution: 4 gm sulphurs is in 100g compound hence 32gm sulphur is in






 32
4
100
= 800 g compound.
∴ (C)
Illustration 9: If 20 ml of 0.2 M K3[Fe(CN)6] is reduced by some equivalents of N2H4, then
calculate numbers of moles of N2H4 required for the above reaction the answer is
(A) 10–5
(B) 10–3
(C) 10–6
(D) 10–2
Solution: N2H4 + K3[Fe(CN)6 + KOH ⎯ → K4[Fe(CN)6] + N2 + H2O
∴ Number of equivalents of
K3[Fe(CN)6] =








1
1000
2
.
0
20
Number of moles of
N2H4=








4
1
1000
2
.
0
20 = 10–3
∴ (B)
Illustration 10: 0.7gm of Na2CO3.XH2O were dissolved in water and the volume was made to 100
ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralisation. The value of x is
(A) 7 (B) 3
(C) 2 (D) 5

4. Solution: Meq. of Na2CO3.xH2O in 20 ml = 19.8 ×
10
1
∴ Meq. Of Na2CO3.xH2O in 100 ml = 19.8 ×
10
1 × 5
∴
1000
E
W

= 19.8 ×
10
1 × 5
or
1000
2
M
7
.
0

=
2
8
.
19
M = 141.41
∴ 23 ×2 + 12 + 3 × 16 + 18x = 141.41
x = 2
∴ (C)
Illustration 11: The Vander Wall’s constant b is equal to
(A) The molecular volume of 1 mol of the gas
(B) Two times the molecular volume of 1 mole of the gas
(C) Three times the molecular volume of 1 mole of the gas
(D) Four times the molecular volume of 1 mole of the gas
Solution: ‘b’ is the volume occupied by 1mole of a gas molecules. Excluded volume for
2 gas molecules
=
3
4 π (2r)3
=
3
4
8  πr3
r r
∴ Excluded volume / molecule = 3
r
3
4
4


∴ b = 3
r
3
4
4

 × NA
∴ b = 4 × volume of 1 mole of molecules
Illustration 12: What will be percentage of ‘free volume’ available in 1 mole of H2O(l) at 760 mm
and 373 K. Density of H2O(g) at 373 K is 0.958 gm/ml.
(A) 0.0613% (B) 99.9386%
(C) 0.0543% (D) 99.9457%
Solution: Volume occupied by 1 mole of H2O(g) at 373 K =
P
nRT
=
1
373
0821
.
0
1 
 = 30.62L
Volume of 1 mole of H2O(l)
=
958
.
0
18
d
M

= 18.789 ml

5. % of volume occupied by liquid water
=
100
64
.
30
10
79
.
18 3

 
= 0.0613%
% of free volume
= 100 – 0.0613 = 99.9386%
Illustration 13: One mole of each monoatomic, diatomic and triatomic gases are mixed, Cp/CV
for the mixture is
(A) 1.40 (B) 1.428
(C) 1.67 (D) None of these
Solution: Cv =
3
R
7
3
R
2
6
R
2
5
R
2
3









Cp =
3
R
10
3
R
2
8
R
2
7
R
2
5









7
10
C
C
v
p

= 1.428
Illustration 14: The vapour pressure of water at 20°C is 17.5 torr. What will be the no. of moles
of water present in one litre of air at 20°C and 40% relative humidity.
(A) 4.2 × 10–4
mole (B) 4.2 × 10–6
mole
(C) 3.82 × 10–4
mole (D) 3.82 × 10–6
mole
Solution: Relative humidity (RH) =
O
H
of
pressure
Vapour
air
in
O
H
of
pressure
Partial
2
2
∴ Partial pressure of H2O = RH × Vapour pressure of H2O
=
5
.
17
100
40

= 7 torr
=
atm
760
7 = 0.0092 atm
Now PV = nRT
∴
293
082
.
0
1
0092
.
0
RT
PV
n




= 0.000382 mole
= 3.82 × 10–4
mole
Illustration 15: The compressibility factor of a gas is less than unity at STP. Therefore
(A) Vm > 22.4 litre (B) Vm < 22.4 litre
(C) Vm = 22.4 litre (D) Vm = 44.8 litre
Solution: Z =
nRT
PV
Ar Z < 1 (given)
∴
1
nRT
PV

or PV < nRT

6. P = 1 atm
V = Vm
R = 0.082
T = 273
∴ Vm < 0.0821 × 273
= Vm < 22.4 litre
Illustration 16: The composition of the equilibrium mixture (Cl2 2Cl), attained at 1200°C, is
determined by measuring the rate of effusion through a narrow aperture. At 1.80 mm Hg pressure,
the mixture effuses 1.16 times as fast as Krypton under the same conditions. What will be the
fraction of chlorine molecules dissociated into atoms (Atomic mass of Kr = 84)
(A) 13% (B) 13.7%
(C) 26% (D) 26.4%
Solution:
mixture
Kr
Kr
mixure
M
M
r
r

or
M
84
1
16
.
1

or M = 62.425
for Cl2 2Cl
initially 1 0
at equilibrium (1 – α) 2α
i (Van’t Hoff factor) = 1 – α + 2α = 1 + α
.
exp
normal
M
M
i 
or 1 + α =
425
.
62
71
on solving α = 0.137
or α = 13.7%
Illustration 17: A spherical balloon of 21 cm diameter and 4.851 L volume is to be filled with
hydrogen at NTP from a cylinder containing the gas at 20 atm and 27°C the cylinder can hold 2.82
litre of water at N.T.P. what will be the number of balloons that can be filled up?
(A) 5 (B) 10
(C) 15 (D) 20
Solution: Volume of one balloon = 4.821 litre
Volume of n balloons = 4.821 n litre
Total volume of hydrogen in the cylinder at NTP
V =
1
300
273
82
.
2
20


 = 51.324 litre
Actual volume of H2 to be transferred to balloons
= 51.324 – 2.82 = 48.504 litre
(∴ 2.82 L is retained in the cylinder)
∴ NO. of balloon, n =
851
.
4
504
.
48 ≈ 10

7. Illustration 18: The numerical value of
PV
RT for a gas at critical condition is … times of
RT
PV at
normal condition
(A) 4 (B) 8/3
(C) 3/8 (D) 1/4
Solution: At critical temperature
∴ PCVC=
8
3 RTC
∴
PV
RT for a gas at critical temperature is
3
8 times of that gas at NTP
Illustration 19: What will be the molecular diameter of Helium if Vander Waal’s constant, b = 24
ml mole–1
?
(A) 2.71Å (B) 2.71mm
(C) 5.42Å (D) 542Å
Solution: b = 4Vm
(Vm is the volume occupied by one mole of gas)
or 24 = 4 × N0 ×
3
4 πr3
or r = 3
/
1
23
10
023
.
6
)
7
/
22
(
16
24
3










= 1.355 × 10–8
cm
∴ d = 2r = 2 × 1.355 × 10–8
cm
= 2.71Å
Illustration 20: The density of phosphorus vapour at 310°C and 775 torr is 2.64g dm–3
. What is
the molecular formula of phosphorus?
(A) P (B) P2
(C) P4 (D) P8
Solution:
M
dRT
P 
or M =
P
dRT
Put d = 2.64gm dm–3
R= 0.0821 dm3
atm K–1
mol–1
P =
760
775 atm
and T = 310 + 273 = 583K
We get M = 123.9g mol–1
No. of P atoms in a molecule =
31
9
.
123
P
of
mass
Atomic
9
.
123

= 3.99≈ 4
Thus the molecular formula of phosphorus = P4
Illustration 21: The nucleus of an atom is located at x = y = z = 0. If the probability of finding an
s-orbital electron in a tiny volume assumed x = a, y = z = 0 is 1 × 10–5
, what is the probability of
finding of the electron in the same sized volume around x = z = 0, y = a?

8. (A) 1 × 10–5
(B) 1 × 10–5
a
(C) 4 × 10–5
a2
(D) 1 × 10–5
× a–1
Solution: Since s-orbital is spherically symmetrical and has equi-distance from the
nucleus. Hence the probability is identical
∴ (A)
Illustration 22: What is the probability at the second site if the electron were in a pz orbital for
the data given in question number (1)
(A) 1 × 10–5
(B) 2 × 10–5
(C) 4 × 10–5
(D) 0
Solution: PZ has a node at z = 0, hence probability of finding of e–
in this volume is zero
∴ (D)
Illustration 23: Following ions will be coloured if Aufbaun rule is not followed
(A) Cu+2
(B) Fe+2
(C) Se+3
(D) (A), (B) true
Solution: Ion with unpaired electrons in d or f-orbitals will be coloured
∴ (D)
Illustration 24: If wavelength is equal to distance travelled by the electron in one second, then
(A)
p
h


(B)
m
h


(C)
p
h


(D)
m
h


Solution: Since, the distance travelled in one second by velocity = vcm = λ
mv
h
∴ λ =

m
h ⇒ λ =
m
h
∴ (D)
Illustration 25: Number of photons of light of wavelength 4000Å required to provide 1.00J of
energy is
(A) 2.01 × 1018
(B) 12.01 × 1031
(C) 1.31 × 1017
(D) None is correct
Solution:
∴ E =

hc
n
∴ n =
8
34
10
10
3
10
63
.
6
10
4000
1
hc
E








 = 2.01 × 1018
∴ (A)
Illustration 26: When a certain metal was irradiated with a light of frequency 3.2 × 1016
Hz the
photoelectrons emitted had twice the kinetic energy as did photo electrons emitted when the same
metal was irradiated with light frequency 2.0 × 1016
Hz. Hence threshold frequency is

9. (A) 1.6 × 1016
Hz (B) 0.8 × 1016
Hz
(C) 8 × 1015
Hz (D) 8 × 1016
Hz
Solution: 0 + KE
∴ KE = h(ν – ν0)
Again 2(KE)1 = (KE)2
2h(ν1 – ν0) = (ν2 – ν0) × h ⇒ 2(ν – ν0) = (ν2 – ν0)
∴ ⇒ 2ν1 – ν2= ν0
Since ν1 = 2 × 1016
Hz
ν2 = 3.2 × 1016
Hz
∴ ν0 = 8 × 1015
Hz
∴ (C)
Illustration 27: The radial distribution curve of 2s sub-level consists of ‘a’ nodes, a is
(A) 1 (B) 3
(C) 2 (D) 0
Solution: Since the radial nodes = n – – 1
Hence a = 2 – 0 – 1 = 1
∴ (A)
Illustration 28: Uncertainty in position and momentum are equal. Uncertainty in velocity is
(A)

h (B)

2
h
(C)

h
m
2
1 (D) None
Solution: Given Δp = Δx
Since Δp × Δx ≈

4
h
(Δp)2
≈

4
h
Δp ≈


h
2
1
∴ Δv =


h
m
2
1
∴ (C)
Illustration 29: If the radius of first Bohr orbit of H-atom is x, then de-Broglie wavelength of
electron in 3rd
orbit is nearly:
(A) 2πx (B) 6πx
(C) 9x (D)
3
x
Solution: 3 =

2
h
3

10. mv





 
1
x
9 =

2
h
3 ⇒ mv =
x
6
h

∴ λ =
mv
h = 6πx
∴ (B)
Illustration 30: If the shortest wavelength of H-atom in Lyman series is x, then the longest
wavelength in Balmer Series of He+
is
(A)
5
x
9 (B)
5
x
36
(C)
4
x (D)
9
x
5
Solution: For the shortest wave length of Lymann series in H-atom








 2
2
H
1
1
1
R
x
1 ⇒
x
1 = RH
Let the longest wavelength of Balmer series is λ









 9
1
4
1
4
R
1
H
max
∴ λmax =
5
x
9
∴ (A)
Illustration 31: For the reaction 2NO + Br2  2NOBr, the following mechanism has been given
fast
NO + Br2 NOBr2
NOBr2 + NO 
 
slow 2NOBr
Hence rate law is;
(A) k[NO]2
[Br2] (B) k[NO][Br2]
(C) k[NOBr2] [NO] (D) k[NO][Br2]2
Solution: For the reaction
NOBr2 + NO 
 
slow 2NOBr
 rate = k1[NOBr2] [NO] …(1)
But from the reaction
NO + Br2 NOBr2
Kequilibrium =
]
Br
][
NO
[
]
NOBr
[
2
2  [NOBr2] = Kequilibrium [NO] [Br2] …(2)
Putting equation (2) in equation (1)
rate = k1 kequilibrium [NO]2
[Br2]
= k[NO]2
[Br2], where k = k1 kequilibrium
(A)
Illustration 32: Following is the graph between (a – x)–1
and time for second order reaction,
Q = tan–1
(0.5), OA = 2L mol–1
.

11. Q
(a – x)
–1
A
O
t(min)
Hence rate at the start of the reaction is:
(A) 1.25 L mol–1
min–1
(B) 0.5 mol lit–1
min–1
(C) 0.125 L mol–1
min–1
(D) 1.25 L mol–1
min–1
Solution: For second order reaction,
Kt =
a
1
x
a
1


 (a – x)–1
= Kt +
a
1
 k = tan ( tan = 0.5 from question)
 K = 0.5 mol lit–1
min–1
and
a
1 = 2 L mol–1
 (C)
Illustration 33: Half-life period in Q.N. = (2), above is
(A) 1.386 min (B) 4 min
(C) 16 min (D) 2 min
Solution: t1/2 (for 2nd
order reaction) =
Ka
1
 (B)
Illustration 34: The rate constant for the reaction
2N2O5  4NO2(g) + O2(g) is 3  10–5
sec–1
. If the rate is 2.4  10–5
mol lit–1
sec–1
, the concentration of
N2O5 in mol L–1
is;
(A) 1.4 (B) 1.2
(C) 0.04 (D) 0.8
Solution: The decomposition of N2O5 is a first order reaction
 rate = k[N2O5]
2.4  10–5
= 3  10–5
[N2O5]
 [N2O5] =
5
5
10
3
10
4
.
2



 = 0.8 mol L–1
 (D)
Illustration 35: Following is the graph between logT50 and log a (where a = initial concentration)
for a given reaction at 27°C. Hence order is;

12. 45°
logT50
loga
(A) 0 (B) 1
(C) 2 (D) 3
Solution: T50 a(1–n)
where n = order of reaction
 T50 = k a(1 – n)
or, logT50= logk + (n – 1)log a
from question,  = tan–1
(m)
 n – 1 = 1
 n = 2
 (C)
Illustration 36: For the reaction
P
Q
R
k1
k2
a
a – y
dt
]
y
[
d is equal to
(A) k1(a – y) – k2(a – y) (B) k2(a – y) – k1 (a– y)
(C) k1(a – y) + k2(a – y) (D) –k1(a – y) – k2(a – y)
Solution:
dt
]
y
[
d

= k1(a – y) + k2(a – y)

dt
]
y
[
d = – k1(a – y) – k2(a – y)
 (D)
Illustration 37: There are two radionuclei A & B A is an -emitter and B is a -emitter, their
disintegration constant are in the ratio of 1:2. What should be the number of atoms of two at time t
= 0. So that probability of getting of  and -particles are same at time t = 0
(A) 2 : 1 (B) 1:2
(C) 1:4 (D) 4:1
Solution: Since the rate of disintegrations are same
ANA = BNB 
A
B
B
A
N
N


 









2
1
B
A
=
1
2
 (A)
Illustration 38: At radioactive equilibrium, the ratio between the atoms of two radioactive
elements A and B was found to be 3.1  109
: 1 respectively. If T50 of the element A is 2  1010
years,
then T50 of the element B is
(A) 6.2  109
years (B) 6.45 years
(C) 3  108
years (D) None

13. Solution: At radioactive equilibrium
)
B
(
50
B
)
A
(
50
A
t
N
t
N

t50(B) =
)
A
(
50
A
B
t
N
N = 10
9
10
2
10
9
.
3
1



=
9
.
3
20 years
 (B)
Illustration 39: The no. of -particles emitted during the charge
axC
 dyb
is
(A)
4
b
a  (B)





 

2
b
a
d
+ c
(C)





 

2
b
c
d
– a (D)





 

2
b
a
d
– c
Solution: axc
 dyb
+ m  2He4
+ n  –10

4
b
c
m


Hence a = d + 2m – n
n = d + 2m – a =





 

2
b
c
d
– a
 (C)
Illustration 40: One mole of A present in a closed vessel undergoes decay as
zAm
 z–4Bm – 8
+ 2  2He4
The volume of He collected at NTP after 20 days (t1/2(A) = 10 days) is
(A) 11.2 L (B) 22.4 L
(C) 33.6 L (D) 67.2L
Solution: 
N
N0
= 2n
where n = no. of half lives
=
N
1 = 22  n =
10
20 = 2
 N =
4
1 mole
 decayed moles =
4
3
4
1
1 

moles
But moles of He formed =
2
3 moles
=
2
3  22.4 L at NTP
= 33.6 L at NTP
 (C)
Illustration 41: Which one of the following does not of dissolve in conc. H2SO4?
(A) CH3 – C = C – CH3 (B) CH3 – CH2 – C ≡ CH
(C) CH ≡ CH (D) CH2 = CH2

14. Solution: If CH ≡ CH were to dissolve in H2SO4 a bisulphite salt of vinyl carbocation H2C
= C+
H would be formed. The more s-character in the positively charged ‘C’ less stable is the
carbocation and less likely to be formed.
∴ (C)
Illustration 42: Which one of the following compounds will give in the presence of peroxide a
product different from that obtained in the absence of peroxide?
(A) 1-butane (B) 1-butene, HBr
(C) 2-butene, HCl (D) 2-butene, HBr
Solution: Peroixde effect is observed when unsymmetrical alkene is treated with HBr
only (and not with HCl and HI).
∴ (B)
Illustration 43: Which of the following alkene on acid catalysed hydration form 2-methyl propan-
2-ol.
(A) (CH3)2CH = CH2 (B) CH3 – CH = CH2
(C) CH3 – CH = CH – CH3 (D) CH3 – CH2 – CH = CH2
Solution: Addition of H2O occurs according to Markownikoff’s rule.
CH3 – C = CH2 + H2O 



H
CH3 – C + CH3
CH3 CH3
CH3
∴ (A)
Illustration 44: Which of the following compounds yields only one product on
monobromination?
(A) Neopentane (B) Toluene
(C) Phenol (D) Aniline
Solution: CH3 –
|
CH
CH
|
C
3
3

CH3 has twelve equivalent 1°H. Hence H forms only one product
on monobromination.
∴ (A)
Illustration 45: Aqueous solution of the following compounds are electrolysed. Acetylene gas
is obtained from.
(A) Sodium fumarate (B) Sopdium maleate
(C) Sodium succinate (D) Both (A) and (B)
Solution:
+ 2Na
+
CH – COO
–
||
CH – COO
–
CH – COO
–
||
CH – COO
–
+ 2CO2 + 2C at cathode
CH – COO
–
||
CH – COO
–
CH
||
CH
∴ (D)
Illustration 46: Dehydration of butan-2-ol with conc. H2SO4 gives preferred product.

15. (A) but-1-ene (B) but-2-ene
(C) propene (D) ethane
Solution: CH3 –
H
|
H
C
–
|
OH
CH
– CH3



 
 4
2SO
H
CH3 – CH = CH – CH3 (80%)
+ CH3 – CH2 – CH = CH2 (20%)
This is in accordance with saytzeff rule.
∴ (B)
Illustration 47: CH3 – C ≡ C – CH3 

 
 2
NaNH ‘X’. What is X
(A) CH3CH2CH = CH2 (B) CH3CH2C ≡ CH
(C) CH3 – CH = CH – CH3 (D) CH2 = C = CH – CH3
Solution: Isomerisation occurs, when 2-butyne is treated with NaNH2, it converts into
terminal alkyne (1-butyne).
∴ (B)
Illustration 48: Identify the compound ‘Y’ in the following sequence of reactioin
HC ≡ CH Y
X
COOH
CH
/
Zn
Zn
/
O
H
)
ii
O
)
i 3
2
3




 


 

(A) O
H
OH
(B) O
OH
(C) C
H3
OAc
OAc
(D) CH3COOH
Solution: C
H CH 


 
 duction
Re
O
O



 
Ozonolysis
O
H
OH
∴ (A)
Illustration 49: Dehydration of 1-butanol gives 2-butene as a major product, by which of the
following intermediate the compound 2-butene obtained
(A)
C
H3
CH2
+
(B)
C
C
H3
CH2
CH3
+
(C) CH
C
H3
CH3
+ (D)
+
C
H3 CH2
CH3
Solution:
Shift
hydride
2
,
1



 

C
H3 OH C
H3
CH2
+
C
H3
CH2
+





H
C
H3 CH
CH3
+
2° 1°
∴ (C)
Illustration 50: The principal organic compound formed in the reaction
CH2 = CH(CH2)COOH + HBr 

 
Peroxide …………. is
(A)
CH3 –
Br
|
CH– (CH3)8COOH
(B) CH2 = CH(CH2)8COBr

16. (C)
Br
|
CH2
–CH2(CH2)8COOH
(D)
CH2 = CH(CH3)7 –
Br
|
CH – COOH
Solution: Follows the peroxide effect
∴ CH2 = CH(CH2)8COOH
Peroxide
HBr

 

Br
|
CH2
– CH2(CH2)8COOH
∴ (C)
Illustration 51: Kc for reaction [Ag(CN)2]-
Ag+
+2CN-
, the equilibrium constant at 25°C is
4.0×10-19
then the silver ion concentration in a solution which was originally 0.1 molar in KCN and
0.03 molar in AgNO3 is
(A) 7.5×1018
(B) 7.5×10-18
(C) 7.5×1019
(D) 7.5×10-19
Solution 2KCN + AgNO3 ⎯ ⎯ → Ag(CN)2
-
+ KNO3+ K+
0.1 0.03 0 0 0
(0.1-0.06) 0 0.03 0.03 0.03
[Ag(CN)2]-
= 0.03
Now [Ag(CN)2]-
Ag+
+ 2CN-
0.03 0 0.04 (left from KCN)
(0.03-a) ‘a’ (0.04+a)
a is very – very small
∴ 0.03 -a ≈ 0.03
and 0.04 + a ≈ 0.04
∴ Kc= 4×10-19
=
03
.
0
a
)
04
.
0
( 2

∴ a = 7.5×10-18
Ans
Illustration 52 Given A ⎯ → B + C, ΔH= -10 Kcals, the energy of activation of backward reaction
is 15 Kcals mol-1
if the energy of activation of forward reaction in the presence of a catalyst is 3 K
Cal mol-1
, the catalyst will increase the rate of reaction at 300 K by the number of times equal to
(A) e3.33
(B) e4.21
(C) e-2.7
(D) e2.303
Solution ΔH = Ea (for forward reaction) –Ea (of backward reaction)
-10 = Ea for FR-15
∴ Ea for FR = 5 kcals mol-1
the catalyst decreases the Ea by 2 k Cals mol-1
33
.
3
300
10
2
RT
/
E
e
e
e
K
Kc 3


 

 
Illustration 53: A vessel contains H2(g) at 2 atm pressure, when H2S(g) at a pressure of 4 atm is
introduced into the vessel. Where reaction
8H2S(g) 8H2+S8(s)
Occurs at a temperature of 1000 K. It is found that
m
equilibriu
at
2
2
)
S
H
(
n
)
H
(
n





 =
0
t
at
2
2
)
H
(
n
)
S
H
(
n






 , then
(A) maxm
wt of solid formed is 32 gm
(B) maxm
wt of solid formed is 0.32 gm

17. (C) Kp= KcRT
(D) Kc=256
Solution: Under identical conditions of volume and temperature P∝n
so initially
2
1
S
nH
nH
2
2

8H2S(g) 8H2(g) + S8(s)
Initially 2 1
at equilibrium (2-x) (1+x)
But at equilibrium
2
S
nH
nH
2
2

,
1
x
,
2
x
8
x
1




∴ Kp =  
 
 
 
256
1
2
P
P
8
8
8
S
H
8
2
2
H


Since Δn = 8-8 =0
∴ Kp = Kc = 256
Illustration 54: At a certain temperature the following equilibrium is established
CO(g) + NO2(g) CO2(g) + NO(g)
One mole of each of the four gases is mixed in one litre container and the reaction
is allowed to reach equilibrium state. When excess of baryta water is added to the equilibrium
mixture, the wt of white precipitate obtained is 236.4 gm. The equilibrium constant Kc of the
following reaction is
(A) 1.2 (B) 2.25
(C) 2.1 (D) 3.6
Solution CO(g) + NO2(g) CO2(g) + NO(g)
t = 0 1 1 1 1
t=t (1-x) (1-x) (1+x) (1+x) at equilibrium
CO2+Ba(OH)2 ⎯ → BaCO3+H2O
Moles of BaCO3 =
2
.
1
197
4
.
236

∴ moles of CO2 at equilibrium =1.2
(1+x) = 1.2
x = 0.2
∴ Kc=
25
.
2
8
.
0
2
.
1
x
1
x
1
2
2
















Illustration 55 Rate of disappearance of the reactant A at two different temperatures is given
by A B
K
300
at
]
B
[
sec
10
4
]
A
[
sec
10
2
dt
]
A
[
d 1
3
1
2 









K
400
at
]
B
[
sec
10
16
]
A
[
sec
10
4
dt
]
A
[
d 1
4
1
2 









heat of reaction in the given temperature range, when equilibrium is set up is
(A)
Cal
50
log
100
400
300
2
303
.
2 



18. (B)
Cal
250
log
100
400
300
2
303
.
2 


(C)
Cal
5
log
100
400
300
2
303
.
2 


(D) None
Solution: At 300°K, Kc =
5
10
4
10
2
K
K
3
2
b
f



 

At 400°K, Kc =
25
10
16
10
4
4
2





ΔH =
300
400
1
2
2
1
K
K
log
)
T
T
(
T
RT
303
.
2

∴ (C) is correct
Illustration 56: Liquid NH3 ionizes to a slight extent at -60°C. Its ionic product
   30
4
NH 10
NH
NH
K 2
3





The number of NH2
-
ions present per ml of pure liquid NH3 are
(A) 300 ions (B) 400 ions
(C) 600 ions (D) 500 ions
Solution: 2NH3 NH4
+
+ NH2
-
K = [NH4
+
] [NH2
-
] = x2
= 10-30
∴ x = 10-15
M = [NH2
-
]
[NH2
-
] = 10-15
×
6
10
1 ×6.023×1023
= 600 ions/ml
Illustration 57: When sulphur in the from of S8 is heated at 900 K, the initial pressure of one atm
falls by 29% at equilibrium. This is because of conversion of some S8 to S2. Find the value of
equilibrium constant for this reaction
(A) 1.16 atm3
(B) 0.71 atm3
(C) 2.55 atm3
(D) 5.1 atm3
Solution: S8(g) 4S2(g)
at t =0 1 0
at equilibrium 1-0.29 4×0.29
= 0.71 atm =1.16 atm
Kp =  
 
 
 
3
4
8
2
atm
55
.
2
71
.
0
16
.
1
PS
PS


Illustration 58: At equilibrium, constants for the reaction at 1000° K are
CoO(s) + H2(g) Co(s) + H2O(g) K1 = 65
CoO(s) + CO(g) Co(s) + CO2(g) K2= 500
The equilibrium constant for the reaction at 1000° K
CO(g) + H2O(g) CO2(g) + H2(g)
(A) 0.13 (B) 7.69
(C) 0.0179 (D) 69.3
Solution: K1 = [H2O]/[H2] = 65

19. K2 = [CO2] / [CO(g) ] = 500
Now, for the required reaction
  
  
O
H
CO
H
CO
K
2
2
2
To obtain this, divide K2 by K1, we get
69
.
7
65
500
K
K
1
2


Illustration 59: 92 gms of ethyl alcohol were treated with 120 gms acetic acid and equilibrium
was established, when 117.34 gms of ester and 24 gm of water were formed. The equilibrium
constant is
(A) 4 (B) 8
(C) 10 (D) 15
Solution: C2H5OH + CH3COOH CH3COOC2H5 + H2O
t =0 92gm 120 gm 0 0
92/46=2 126/60=2 117.34 gm 24 gm
at equi. (2-1.33) (2-1.33)
33
.
1
88
34
.
117
 33
.
1
18
24

Kc=    
   
COOH
CH
OH
H
C
O
H
H
COOC
CH
3
5
2
2
5
2
3


=
4
67
.
0
67
.
0
33
.
1
33
.
1



Illustration 60: For the reaction
2NOCl(g) 2NO(g) + Cl2(g)
the values of ΔH° and ΔS° at 298 K are 77.2 kJ mol-1
and 122 JK-1
mol-1
respectively. The standard
equilibrium constant at the same temperature is
(A) 0.695×10-8
(B) 6.95×10-8
(C) 69.5×10-8
(D) 695×10-8
Solution: Using the relation ΔG° = ΔH°
- TΔS°
ΔG° = 77200-298×122
= 40844 J mol-1
let the equilibrium constant be K°
c
we know that
ΔG° = -2.303 RT logK°
c
298
314
.
8
303
.
2
G
K
log C






298
314
.
8
303
.
2
40844



log K°
c = -7158
K°
c = 6.95×10-8
Illustration 61: An acidic indicator HIn (Kin = 10–6
) ionises as HIn H+
+ In–
. The acid colour
predominates over the basic colour when HIn is at least 10 times more concentrated than In–
ion.
On the other hand basic colour predominates over the acid colour when the In–
ion is at least 5
times more concentrated than HIn. Hence pH range of the indicator is
(A) 5.0 – 6.7 (B) 7.0 – 8.7

20. (C) 5.3 – 7.0 (D) 7.0 – 8.1
Solution: pH = pKin +
]
HIn
[
]
In
[
log
 , where pKin = 6
For acid colour to predominates
10
1
]
HIn
[
]
In
[


∴ pH ≤ 6 – 1 i.e 5.0
for base colour to predominate
]
HIn
[
]
In
[  ≥ 5
∴ pH ≥ 6 + log5 ⇒ pH ≥ 6.7
∴ (A)
Illustration 62: The correct statement amongst the following is
(A) A strong electrolyte remains completely dissociated at all dilutions
(B) Upon dilution the degree of dissociation of a weak electrolyte and number of
ions per unit volume of its solution both increase.
(C) A strong electrolyte is completely ionised at all dilutions but not completely
dissociated.
(D) pH of solution of a weak acid decreases with dilution.
Solution: “Complete dissociation” implies that interionic attraction has completely ceased
to exist. This condition in the case of a solution of strong electrolyte is achieved only at infinite
dilution when concentration of solution tends to zero.
∴ (C)
Illustration 63: 4M solution of a weak monobasic acid (x% ionized and pH = 3.0) is diluted to 1
M by adding water (distilled). Percentage ionisation and pH of solution after dilution will be
respectively.
(A) 2x and 2.7 (B) 0.25x and 3.3
(C) 0.5x and 2.7 (D) 2x and 3.3
Solution: Solution of weak acid being concentrated, we can use the approximate
α =
C
Ka i.e. α ∝
C
1
Thus, decreasing the conc. to one fourth, α will be doubled. Doubling of α means doubling of
percentage ionisation. Hence % ionisation will be 2x.
[H+
] = C
Ka  i.e. [H+
] ∝ C
If the conc. is decreased to one fourth of its original value H+
ion conc. will be halved. Thus, after
dilution [H+
] = 0.5 × 10–3
M
∴ pH = 3.3
∴ (D)
Illustration 64: pH of a buffer solution changes from 6.20 to 6.17 when 0.003 mole of acid is
added to 500 mL of the buffer. The buffer capacity of the system is, therefore
(A) 0.1 (B) 0.3
(C) 0.2 (D) 0.4
Solution: Buffer capacity =
pH
in
change
buffer
of
litre
per
added
acid
of
moles
of
.
No

21. =
03
.
0
006
.
0 = 0.2
∴ (C)
Illustration 65: Ksp of CaSO4 is 2.4 × 10–5
at 25°C. In a solution containing Ca2+
ions the
precipitation of CaSO4 begins to occur when SO4
2–
ion concentration in the solution is made just to
exceed the value of 4.8 × 10–3
M. Hence concentration of Ca2+
ion in the solution is
(A) 200 ppm (B) 40 ppm
(C) 400 ppm (D) 100 ppm
Solution: [Ca2+
] [SO4
2–
] = 2.4 × 10–5
[Ca2+
] =
3
5
10
8
.
4
10
4
.
2



 = 5 × 10–3
M
1 L solution contains 5 × 10–3
mole of Ca2+
ions
1000 L solution will contain 5 mole i.e. 200 g Ca2+
ions
Taking density of aqueous solution to be unity
1000 kg i.e. 106
g solution contains 200g Ca2+
ions
Conc. of Ca2+
ion = 200 ppm
∴ (A)
Stoichiometry
1. How many grams of phosphoric acid would be needed to neutralise 100gm of magnesium
hydroxide?
(Molecular weight of H3PO4 = 98 and Mg(OH)2 = 58.3 gm)
(A) 66.7 (B) 252gm
(C) 112gm (D) 168 gm
2. 10 L of hard water required 0.56 gm of lime (CaO) for removing hardness. Hence
temporary hardness in ppm of CaCO3is
(A) 100 (B) 200
(C) 10 (D) 20
3. 20 ml of xM HCl neutralises completely 10 ml of 0.1 M NaHCO3 solution and further 5 ml
of 0.2 M Na2CO3 solution in the presence of methyl orange at end point, the value of x is
(A) 0.167M (B) 0.133M
(C) 0.15M (D) 0.2M
4. When 10 ml of ethyl alcohol (d = 0.7893 gm/ml) is mixed with 20 ml of water (d = 0.9971
g/ml) at 25°C, the final solution has a density 0.9571 gm/ml. The percentage change in total
volume on mixing is
(A) 3.1% (B) 2.4%
(C) 1% (D) None of these

22. 5. 0.635 gm of a cupric salt was dissolved in water and excess of KI was added in the solution
and the liberated iodine required 25ml of N/5 Na2S2O3 solution. The weight percentage of copper
in the salt is
(A) 25% (B) 40%
(C) 50% (D) 63.5%
6. The chloride of a metal (M) contains 65.5% of chlorine 100 ml of the chloride of the metal
at STP weight 0.72gm. The molecular formula of the metal chloride is
(A) MCl3 (B) MCl
(C) MCl2 (D) MCl4
7. 22.4 litres of H2S and 22.4 litre of SO2 both at STP are mixed together. The amount of
sulphur precipitated as a result of chemical calculation is
(A) 16gm (B) 23gm
(C) 48gm (D) 96gm
8. In the mixture of NaHCO3 and Na2CO3 volume of a given HCl required is x ml with
phenolphthalein indicator and further y ml required with methyl orange indicator. Hence volume
of HCl for complete reaction of NaHCO3 is
(A) 2x (B) y
(C)
2
x (D) (y – x)
9. A sample of oleum is labelled 109%. The % of free SO3 in the sample is
(A) 40% (B) 80%
(C) 60% (D) 9%
10. 7.65 × 10–3
moles of NaNO3 require 3.06 × 10–2
moles of a reducing agent to get reduced
to NH3. What is the n-factor of the reducing agent?
(A) 2 (B) 3
(C) 4 (D) 5
11. How many grams of sodium bicarbonate are require to neutralise 10ml of 0.902 M
vinegar?
(A) 8.4 gm (B) 1.5gm
(C) 0.758 gm (D) 1.07gm
12. 1 gm mole of oxalic acid is treated with conc. H2SO4. The resultant gaseous mixture is
passed through a solution of KOH. The mass of KOH consumed will be
(A) 28gm (B) 56gm
(C) 84gm (D) 112 gm
13. NH3 + OCl–
⎯ → N2H4 + Cl–
On balancing the above equation in basic solution using integral coefficients, which of the
following whole number will be the coefficient of N2H4?
(A) 1 (B) 2
(C) 3 (D) 4
14. A certain compound has the molecular formula X4O6. If 10 gm of X4O6 has 5.72g X, atomic
mass of X is
(A) 32 amu (B) 37 amu
(C) 42 amu (D) 98 amu

23. 15. In the following reaction
28NO3
–
+ 3As2S3 + 4H2O ⎯ → 6AsO4
3–
+ 28NO + 9SO4
2–
+ 8H+
equivalent weight of As2S3 (with molecular weight M) is
(A)
2
M (B)
4
M
(C)
24
M (D)
28
M
Gaseous State
16. At constant pressure what would be the percentage decrease in the density of an ideal
gas for a 10% increase in the temperature.
(A) 10% (B) 9.1%
(C) 11% (D) 12.09%
17. A small hole is pricked in a container containing a mixture of CH4 and O2 and placed in
vacuum. What should be the molar ratio of methane to oxygen in the container when the two
gases are effusing out in the ratio of 2:1
(A) 1
:
2 (B) 2
:
1
(C) 2:1 (D) 1:2
18. The Van der Wall’s equation is









V
a
n
P
2 (V – nb) = nRT for n moles where ‘a’ and ‘b’ are
Van der Wall’s constant which of the following statements are true about ‘a’ and ‘b’ when the
temperature of the gas is too low
(A) Both remains same (B) ‘a’ remains same, b varies
(C) ‘a’ varies, ‘b’ remains same (D) both varies
19. The tube in the figure is shielded at both end and heated upto double the original
temperature both side of Hg column gases are packed with increasing temperature the Hg column
Hg
10 cm
5 cm
A B
(A) Shift towards ‘B’ (B) Shifts towards A
(C) Remain same (D) Start to vibrate
20. The molar specific heat at constant volume of mixture of gases A and B is 4.33 cal. A is
monoatomic and B is diatomic then the ratio of moles of A and B is
(A) 1:1 (B) 2:1
(C) 1:2 (D) 3:1
21. Rate of diffusion of two gases are rA and rB, molecular weights are MA and MB then partial
pressure PA is (if number of moles are nA and nB).
(A)
total
B
A
B
B
A
B
A
P
n
n
n
M
M
r
r




(B)
total
A
B
B
A
B
A
P
n
n
M
M
r
r



(C)
total
B
A
a
B
A
B
A
P
n
n
n
M
M
r
r




(D) None of these

24. 22. There are two gases A and B having degree of freedom fA and fB, the difference in energy
supplied for increment in temperature of one mole gas by 50°C is (in cal)
(A) 50 (fA – fB) (B) 100 (fA– fB)
(C) 200 (fA + fB) (D) 50










B
A
B
a
f
f
f
f
23. When an ideal gas undergoes unrestricted expansion no cooling takes place because the
molecules
(A) Exert no attractive forces on each other (B) Do work equal to loss of KE
(C) Colide without loss of energy (D) Are about the inversion temperature
24. If two gases of molecular weights MA and MB at temperatures TA, TB and TAMB = TBMA, then
which property has the same magnitude for both the gases.
(A) Density (B) Pressure
(C) KE per mole (D) Vrms
25. At low pressure Vander Wall’s equation for 3 moles of a real gas will have its simplified
form
(A)
3
V
a
3
RT
PV


(B)
3
Rb
RT
PV


(C)
1
Pb
3
RT
PV


(D)
3
V
9
RT
PV


26. for two gases A and B with molecular weights MA and MB, its observed that a certain
temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus the
mean velocity of A can be made equal to the mean velocity of B if
(A) A is at temperature T, and B at T′⋅ T > T′
(B) A is lowered to a temperature T2 =

3
8 T
(C) Both A and B are raised to a higher temperature
(D) Both A and B are placed at lower temperature
27. The quantity
T
K
PV
B
represents the
(A) Number of molecules in the gas (B) Mass of the gas
(C) Number of moles of the gas (D) Translational energy of the gas
28. The rms velocity of hydrogen is 7 times the rms velocity of nitrogen. If the temperature
of the gas
(A) T(H2) = T(N2) (B) T(H2) > T(N2)
(C) T(H2) < T(N2) (D) T(H2) = 7 T(N2)
29. 1 lt of N2 and
8
7 l of O2 at the same temperature and pressure were mixed together what
is the relation between the masses of two gases in the mixture.
(A)
2
2 O
N M
3
M  (B)
2
2 O
N M
8
M 

25. (C)
2
2 O
N M
M  (D)
2
2 O
N M
16
M 
30. Consider a mixture of SO2 and O2 kept at room temperature compared to the oxygen
molecule, the SO2 molecule will hit the wall with
(A) Smaller average speed (B) Greater average speed
(C) Greater kinetic energy (D) Greater mass
Atomic Structure
31. Wave function vs distance from nucleus graph of an orbital is given below:

r
–
+
The number of nodal sphere of this orbital is
(A) 1 (B) 2
(C) 3 (D) 4
32. For an electron in a hydrogen atom the wave function, ψ is proportional to
0
a
r
e
 , where a0
is the Bohr’s radius. What is the ratio of probability of finding the electron at the nucleus to the
probability of finding it at a0.
(A) e (B) e2
(C)
2
e
1 (D) Zero
33. What transition in He+
ion shall have the same wave number as the first line in Balmer
series of H-atom
(A) 3 → 2 (B) 6 → 4
(C) 5 → 2 (D) 7 → 5
34. The potential energy of the electron in an orbit of H-atom would be
(A) – mv2
(B)
2
2
r
e

(C) 2
mv
2
1

(D)
r
2
e2

35. An electron is moving with a kinetic energy of 4.55 × 10–25
Joules. What will be the de-
Broglie wave length for this electron?
(A) 5.28 × 10–7
m (B) 7.28 × 10–7
m
(C) 2 × 10–10
m (D) 3 × 10–5
m
36. If each orbital can hold a maximum of 3-electrons. The number of elements in 4th
period
of periodic table is:

26. (A) 48 (B) 57
(C) 27 (D) 36
37. The no. of orbital in a sub-shell is equal to
(A) n2
(B)
(C) (D) m
38. Which of the following curves may represent the speed of the electron in a hydrogen atom
as a function of principal quantum no. n.
V
n
b
a
d
c
(A) a (B) b
(C) c (D) d
39. The ratio of the energy of the electron in ground state of hydrogen to that of the electron
in the first excited state of Be+3
is
(A) 1 : 4 (B) 1 : 8
(C) 1 : 16 (D) 16 : 1
40. The electronic transition from n = 2 to n = 1 will produce the shortest wavelength in
(A) H-atom (B) D-atom
(C) He+
ion (D) Be+3
ion
41. The wave number of first line of Balmer series of hydrogen is 152,000 cm–1
.
The wavelength of first Balmer line of Li+2
ion is
(A) 15,200 cm–1
(B) 60,800 cm–1
(C) 76,000 cm–1
(D) 1,36,8000 cm–1
42. The dissociation energy of H2 is 430.53 kJ mol–1
. If H2 is dissociated by
illumination with radiation of wavelength 253.7 nm. The fraction of the radiant energy which will
converted into kinetic energy is given by
(A) 8.86% (B) 2.33%
(C) 1.3% (D) 100%
43. The orbital angular momentum of an electron in 2s-orbital is
(A)

4
h (B) zero
(C)

2
h (D)

2
h
2
44. Photoelectric emission is observed from a surface for frequencies v, and v2 of
the incident radiation (where v1  v2). If the maximum kinetic energy of the photoelectrons in two
cases are in the ratio of 1:k, then the threshold frequency v0 is given by
(A)
1
K
v
v 2
1

 (B)
1
K
v
Kv 2
1


(C)
1
K
v
Kv 1
2

 (D)
K
v
v 1
2 

27. 45. The difference between nth and (n + 1)th Bohr’s radius of H-atom is equal to
its (n – 1)th Bohr’s radius. The value of n is
(A) 1 (B) 2
(C) 3 (D) 4
Chemical Kinetics
46. A radioactive isotopes x with half-life of 1.37  109
years decays to y which is
stable. A sample of rock from the moon was found to contain both the elements x and y in the
ratio 1:7. What is the age of the rock.
(A) 1.96  108
years (B) 3.85  109
years
(C) 4.11  109
years (D) 3.06  109
years
47. A radioactive mixture containing a short lived species A and short lived species
B. Both emitting -particles at a given instant, emits at rate 10,000 -particles per minute. 10
minutes later, it emits at the rate of 7000 particles per minute. If half lives of the species are 10
min and 100 hours respectively, then the ratio of activities of A : B in the initial mixture was
(A) 3:7 (B) 4:6
(C) 6:4 (D) None
48. 92U238
(IIIB) undergoes following emissions:
92U238
C
B
A 



 


 
 





Which is/are correct statements
(A) A will be of IB group (B) A will be of III B group
(C) B will be of IA group (D) C will be of IIIA group
49. The electron in a hydrogen atom makes transition from M shell to L, the ratio
of magnitude of initial to final acceleration of electron is:
(A) 9:4 (B) 81:16
(C) 4:9 (D) 16:81
50. In order to determine the volume of blood in an animal without killing it, a 1.00
ml sample of an aqueous solution containing tritium is injected into the animal blood stream. The
sample injected has an activity of 1.8  106
cps (counts per second). After sufficient time for the
sample to be completely mixed with the animal blood due to normal blood circulation, 2.00 ml of
blood are withdrawn from animal and the activity of the blood sample withdrawn is found to be
1.2  104
cps. Calculate the volume of the animal blood.
(A) 300ml (B) 200ml
(C) 250 ml (D) 400 ml
51. A radioactive isotope is being produced at a constant rate x. Half-life of the radioactive
substance is y. After sometimes no. of radioactive nuclei becomes constant, the value of this
constant is
(A)
2
ln
xy (B) xy
(C) (ln2)xy (D)
y
x

28. 52. Number of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at
time = 2 sec. Number of nuclei at t = 4s will be
(A) 800 (B) 810
(C) 790 (D) 700
53. There are two radioactive substances A and B. Decay constant of B is twice
that of A. Initially both have equal no. of nuclei. After n-half lives of A, the rate of disintegration of
both becomes equal, the value of n is
(A) 1 (B) 2
(C) 4 (D) None
54. A radioactive nuclide is produced at a constant rate of -per second. It decay
constant is . If N0 be the no. of nuclei at time t = 0, then maximum no. of nuclei possible are
(A)

 (B) N0 +


(C) N0 (D)

 + N0
55. For the reaction
R – X + OH–
 R – OH + X–
, the rate expression is given as rate
= 4.7  10–5
[R – X] [OH–
] + 0.24  10–5
[R – X]
What % of R – X react by SN2 mechanism when [OH–
] = 0.001 M
(A) 1.9 (B) 4.7
(C) 2.8 (D) 4.9
56. The mechanism of the reaction
2NO + O2  2NO2 is
k1
NO + NO N2O2 (fast)
k–1
N2O2 + O2 

 2
k 2NO2 (slow)
the rate constant for the reaction is
(A) k2 (B) k2k1k–1
(C) k2k1 (D)








1
1
2
k
k
k
57. The inversion of cane sugar proceeds with half-life of 600 minutes at pH = 5
for any concentration of sugar. However if pH = 6, the half-life changes to 60 minute. The rate law
expression for sugar inversion can be written as
(A) r = k[sugar]2
[H+
]0
(B) r = k [sugar]1
[H+
]0
(C) r = k [sugar]1
[H+
]1
(D) r = k [sugar]0
[H+
]1
58. Rate of chemical reaction is nA  Product, is doubled when the concentration
of A is increased four times. If the half-life time of the reaction at any temperature is 16 minutes,
then the time required for 75% of the reaction to complete is
(A) 24.0 minutes (B) 27.3 minutes
(C) 48 minutes (D) 49.4 minutes

29. 59. The rate of a chemical reaction generally increases rapidly even for small temperature
increase because of rapid increase in the
(A) Collision frequency
(B) Fraction of molecules with energies in excess of the activation energy
(C) Activation energy
(D) Average kinetic energy of the molecule
60. Rate constant, k = 1.8 × 104
mol–1
Ls–1
and Ea = 2 × 102
kJ mol–1
, when T → ∞, then the
value of A is
(A) 1.8  104
kJ mol–1
(B) 1.8  104
mol–1
L sec–1
(C) 1.8  104
mol L–1
sec–1
(D) 2.4  103
kJ mol–1
sec–1
Chemical Equilibrium
61. For the reaction A + B C + D, equilibrium concentrations of [C] = [D] = 0.5M if we start
with 1 mole each of A and B. Percentage of A converted into C if we start with 2 mole of A and 1
mole of B is
(A) 25% (B) 40%
(C) 66.66% (D) 33.33%
62. CH3 – CO – CH3(g) CH3 – CH3(g) + CO(g)
Initial pressure of CH3COCH3 is 100mm when equilibrium is set up, mole fraction of CO(g) is
3
1
hence Kp is
(A) 100mm (B) 50mm
(C) 25mm (D) 150mm
63. For the reaction
CaCO3(s) CaO(s) + CO2(g)
Equilibrium constant Kp is 1.642 atm at 1000K, if 40 gm of CaCO3 was put into a 10 L flask,
percentage of calcium carbonate would remain unreacted at the equilibrium.
(A) 66% (B) 34%
(C) 50% (D) 25%
64. If equilibrium constant of
CH3COOH + H2O CH3COO–
+ H3O+
Is 1.8 × 10–5
, equilibrium constant for
CH3COOH + OH–
⎯ → CH3COO–
+ H2O is
(A) 1.8 × 10–9
(B) 1.8 × 109
(C) 5.55 × 10–10
(D) 5.55 × 1010
65. PCl5 is 40% dissociated when pressure is 2 atmosphere it will be 80% dissociated when
pressure is approximately
(A) 0.2 atm (B) 0.5 atm
(C) 0.3 atm (D) 0.6 atm
66. For the reaction: 2HI(g) H2(g) + I2(g), the degree of dissociated (α) of HI(g) is related to
equilibrium constant Kp by the expression

30. (A)
2
K
2
1 p
 (B)
2
K
2
1 p

(C)
p
p
K
2
1
K
2

(D)
p
P
K
2
1
K
2

67. For which of the following reactions, the degree of dissociation cannot be calculated from
the vapour density data.
i) 2HI(g) H2(g) + I2(g)
ii) 2NH3(g) N2(g) + 3H2(g)
iii) 2NO(g) N2(g) + O2(g)
iv) PCl5(g) PCl3(g) + Cl2(g)
(A) (i) and (iii) (B) (ii) and (iv)
(C) (i) and (ii) (D) (iii) and (iv)
68. Van’t Hoff equation giving the effect of temperature on chemical equilibrium is represented
as
(A)
2
RT
H
dT
F
ln
d 

(B)
R
HT
dT
K
ln
d 2
p 

(C)
2
p
RT
H
dT
K
ln
d 

(D)
H
RT
dT
K
ln
d 2
p


69. Pure ammonia is placed in a vessel at temperature where its degree of dissociation (α) is
appreciable at equilibrium.
(A) Kp does not change with pressure (B) α does not change with pressure
(C) (NH3) does not change with pressure (D) [H2] < [N2]
70. Steam starts reacting with iron at high temperature to give hydrogen gas and Fe3O4. The
correct expression for equilibrium constant is
(A)
2
O
H
2
H
2
2
p
p (B)
]
Fe
[
p
]
O
Fe
[
p
4
O
H
4
3
4
H
2
2
(C)
4
O
H
2
H
2
2
p
p (D)
]
Fe
[
]
O
Fe
[ 4
3
71. X nY, X decomposes to give Y (in one litre vessel) if degree of dissociation is α then
KC and its unit.
(A)
)
1
(
n


 ,moln–1
litn–1
(B)
)
1
(
)
n
(


 ,moln
litn
(C)
)
1
(
n


 ,KC is unit less (D)
)
1
(
)
n
( n


 ,KC is unit less
72. Solubility of a solute in a solvent (say H2O) is dependent on temperature as given by
S = Ae–ΔH/RT
where ΔH is heat of reaction
Solute + H2O Solution, ΔH = ±X
For a given solution variation of log S with temperature is shown graphically. Hence solute is
(A) CuSO4.5H2O (B) NaCl

31. (C) Sucrose (D) CaO
73. For the reaction A(g) B(g) + C(g)
(A) Kp = α3
P (B) Kp = α2
(Kp + P + 1)
(C) Kp = α2
(KP + P) (D) Kp = α2





 
P
P
KP
74. The values of KC for the following reactions are given as below
A B, KC =1, B C, KC = 3 and C D, KC = 5
Evaluate the value of KC for A D
(A) 15 (B) 5
(C) 3 (D) 1
75. For the reaction (1) and (2)
A B + C
D 2E
Given
1
P
K :
2
P
K : : 9 : 1
If the degree of dissociation of A and D be same then the total pressure at equilibrium (1) and (2)
are in the ratio.
(A) 3:1 (B) 36:1
(C) 1:1 (D) 0.5:1
Ionic Equilibrium
76. If the degree of ionization of water be 1.8 × 10-9
at 298K. Its ionization
constant will be
(A) 1.8 × 10-16
(B) 1 ×10-14
(C) 1 × 10-16
(D) 1.67 × 10-14
77. When a solution of benzoic acid was titrated with NaOH the pH of the solution when half
the acid neutralized was 4.2. Dissociation constant of the acid is
(A) 6.31 × 10-5
(B) 3.2 × 10–5
(C) 8.7 × 10–8
(D) 6.42 × 10–4
78. 10-2
mole of NaOH was added to 10 litre of water. The pH will change by
(A) 4 (B) 3
(C) 11 (D) 7
79. For an aqueous solution to be neutral it must have
(A) pH = 7 (B) [H+
]=[OH–
]
(C) [H+
] =
w
K (D) [H+
] < [OH–
]
80. If an aqueous solution at 25°C has twice as many OH-
as pure water its pOH will be
(A) 6.699 (B) 7.307
(C) 7 (D) 6.98
81. What would be the pH of an ammonia solution if that of an acetic acid solution of equal
strength is 3.2? Assume dissociation constant for NH3 & acetic acid are equal.
(A) 3.2 (B) 6.4

32. (C) 9.6 (D) 10.8
82. The pH of an aqueous solution of 0.1M solution of a weak monoprotic acid which is 1%
ionised is
(A) 1 (B) 2
(C) 3 (D) 11
83. pH of a 10–10
M NaOH is nearest to
(A) 10 (B) 7
(C) 4 D) 10
84. The weight CH3COONa needed to add in 1 litre
10
N CH3COOH to obtain a buffer solution
of pH value 4 is (Given Ka = 1.8 × 10–5
)
(A) 1.2 g (B) 1.47g
(C) 1.8 g (D) 4 g
85. A dilute HCl solution saturated with H2S has pH value 3 then the conc. of S– –
is (Given
K1 = 1 × 10–7
, K2 = 1.3 × 10–13
)
(A) 2 × 10–13
(B) 2.4 × 10–13
(C) 3 × 10–15
(D) 1.3 × 10–15
M
86. Which is the strongest acid
(A) H3AsO4 (B) H2AsO4
-
(C) HAsO4
--
(D) AsO4
---
87. In an aqueous solution of triprotic acid H3A which is true
(A) [H+
] = 3[A3–
] (B) [H+
] > 3[A3–
]
(C) [H+
] < 3[A3–
] (D) [H+
] = [OH–
]
88. If K1 & K2 be first and second ionisation constant of H3PO4 and K1>> K2 which is incorrect.
(A) [H+
] = [H2 PO4
–
] (B) [H+
] =  
4
3
1 PO
H
K
(C) K2 = [HPO4
– –
] (D) [H+
]= 3[A3
-
]
89. Which one of the following anion does not hydrolyse
(A) H–
(B) CN–
(C) NO2
– –
(D) S– –
90. If the ionic product of water varies with temperature as follows and the density of water be
nearly constant for this range of temperature the process
H+
+ OH H2O is
Temp. °C 0 10 25 40 50
Kw
0.114×10–
14
0.292×10–
14
1.008 ×10–
14 2.91×10–14
5.474 ×10–14
(A) Exothermic (B) Endothermic
(C) Can’t say (D) Ionization

33. Stoichiometry
1. C 2. B
3. C 4. A
5. C 6. A
7. C 8. D
9. A 10. A
11. C 12. D

34. 13. A 14. A
15. D
Gaseous State
16. B 17. A
18. D 19. C
20. C 21. A
22. A 23. A
24. D 25. A
26. B 27. A
28. C 29. C
30. A
Atomic Structure
31. A 32. D
33. B 34. A
35. B 36. B
37. C 38. C
39. A 40. D
41. D 42. A
43. B 44. B
45. B
Chemical Kinetics
46. C 47. C
48. B 49. B
50. A 51. A
52. B 53. A
54. A 55. A
56. D 57. C
58. B 59. B

35. 60. B
Chemical Equilibrium
61. D 62. B
63. B 64. B
65. A 66. D
67. A 68. C
69. A 70. C
71. D 72. D
73. C 74. A
75. B
Ionic Equilibrium
76. A 77. A
78. A 79. B
80. A 81. D
82. C 83. B
84. B 85. D
86. A 87. B
88. D 89. A
90. A