Physics by Shiv R Goel, (B.Tech, IIT–Delhi) WAVES
Chemistry by Dr. Sangeeta Khanna, Ph.D. (CCC)
MOCK TEST - 1
NAME:
ROLL NO.
Time: 3 Hr.
Date 04.08.2013
BOOKLET–A
Max. Marks:298
Paper
Paper – I PHYSICS
Paper – II
II CHEMISTRY
Subject-Chemistry
TEST CODE–1370
TEST CODE–1371
Part : 2
SECTION-A
(Single Answer is Correct)
Q.1
The vapour pressure of pure A and B are 0.02 and 0.04 bar respectively. Their liquid mixture
behaves as an ideal solution. If the mole fraction of B in the liquid phase is 0.50, then its mole
fraction in the vapour phase in equilibrium with the liquid phase is equal to:
(a.)
0.50
(b.)
0.33
(c.)
0.67
(d.)
0.25
Ans. C
Q.2 The actual freezing point of the solution is -4.46°Ct. What percentage of the KI is dissociated if it is 1.24
M with density 1.15 gm/ml, Kf = 1.86 K Kg/mole [At. Mass of I = 127, K = 39]
(a.)
83%
(b.)
92%
(c.)
73%
(d.)
50%
Sol. A
Mass of solvent = 1150 – 1.24 × 166
= 1150 – 207.5
= 942.5
1.24 × 1000
molality =
= 1.31
942.5
∆Tf = i Kf m
4.46
=i
1.86 × 1.31
i = 1.83
i=i+α
α = 0.83 i.e. 83%
Q.3
For oxidation reaction: Fe0.95 O → Fe2O3
If molecular weight of Fe0.95O be M, then the equivalent weight of Fe0.95O will be:
Sol.
(a.)
M
1
(b.)
M
2
(c.)
M
0.95
(d.)
M
0.85
D
Fe0.95o
→
Fe 2 O 3
+3
0.95 mole iron in +2 state
1 mole Fe in + 3
0.95 Fe in 3 × 0.95 = 2.85
Change in oxidation state = 0.85
x-factor = 0.85
Q.4 In a cubic closest packed structure of mixed oxides, the lattice is made up of oxide ions, 20% of
tetrahedral voids are occupied by divalent X2+ ions and 50% of the octahedral voids are occupied
by trivalent Y3+ ions. The formula of the oxide is :
(a.)
Sol.
B
x
0.4
X4
Q.5
X2YO4
X4Y5O10
(b.)
y
1
2
y5
(c.)
X5YO10
(d.)
XY2O4
O
1
O10
The standard potentials are shown in the following diagram:
I
0.54
0.45
BrO–3
BrO–
1/2
1.07
Br–
II
III
What is the value of E° for the reaction path (III) shown in the above diagram?
(b.) 1.61 volt
(c.) 1.45 volt
(d.) 0.61 volt
(a.) 0.97 volt
Sol. D
BrO 3− + 4e − → BrO − ; E10 = 0.54V
1
BrO − + 1e − → Br2 ; E 02 = 0.45V
2
1
Br2 + 1e − → Br − ; E 03 = 1.07V
2
Over all reaction BrO 3− + 6e − → Br −
Q.6
∆G° = -4 F × 0.54
∆G° = -1F × 0.45
∆G°
E0 = ?
= -1F × 1.07
∆G° = -6F E°
25 ml of 0.1 M solution of transition metal M reacts completely with 50 ml of 0.02 M KMnO4 in
presence of H2SO4. Which is correct representation of change in oxidation state of M:
(a.)
M+ → M+2
(b.)
M+2 → M+3
Sol. C
25 x 0.1 × x = 50 x 0.02 x 5
50 × 0.02 × 5
x=
25 × 0.1
x=2
Q.
For the reaction: 4KClO3 → 3KClO4 + KCl
(c.)
M+2 → M+4
(d.)
M+2 → M+5
− d[KClO 3 ]
= k 1[KClO 3 ] 4
dt
d[KClO 4 ]
= k 2 [KClO 3 ] 4
dt
d[KCl]
= k 3 [KClO 3 ] 4 , the correct relation between k1, k2 and k3 is:
dt
7
If
(a.)
k1 = k2 = k3
Sol. C Rate = −
1 d[KClO 3 ]
4
dt
(b.)
=
4k1 = 3k2 = k3
1 d[KClO 4 ]
3
dt
=
d[KCl]
dt
or
3k1 = 4k2 = 12k3
(c.)
k1
4
[KClO 3 ] 4 =
k2
3
(d.)
None of these
[KClO 3 ] 4 = k [KClO 3 ] 4
3
or 3k1=4k2=12k3
Q.8
A radioactive isotope has a half – life of 25 days. Starting with 4g of the isotope, what will be mass
remaining after 75 days?
(a.)
1.00 g
(b.)
0.50 g
(c.)
0.58 g
(d.)
1.58 g
1
× 0.4 = 0.5
8
Sol. B After 3 half life amount left is =
Q.9
In some reduction process, the concentration of a substance changes from 0.24M to 0.12 M in 10
minute and to 0.06 M in 20 minute. What is the order and the half-life of the reaction?
(a.)
order = 1, t 1 = 5 minute
(b.)
order = 1, t 1 = 10 minute
2
(c.)
2
order = 0, t 1 = 10 minute
(d.)
order = 2, t 1 = 20 minute
2
2
Ans. B
Q.1 The decomposition of ozone is believed to occur by mechanism:
O3
O2 + O (fast)
O + O3 → 2O2 (slow)
When the concentration of O2 is increased, the rate :
0
(a.)
Increases
(b.)
Decreases
(c.)
Remains the same
(d.)
Cannot be answered
Sol. B
For rate expression of this reaction see solved example 3. This is clear from the rate expression that rate
decrease with increase in the conc. of O2.
Q.1
B
1
A
If all the atoms, on the shaded plane are removed then the molecular formula of the solid will be:
(a.)
A5B7
(b.)
A7B5
Sol. C Effective number of atoms of B = 10 ×
Effective number of atoms of A
(c.)
1 5
=
4 2
AB
(d.)
A3B4
1
1 1
5
+ 4× = + 2 =
8
2 2
2
5 5
A : B = : 
2 2  Simplest formula of solid = AB
= 1: 1 
=4×
Q.1
2
Which one of the following pairs of solution can we expect to be isotonic at the same temperature?
(a.)
0.1 M urea and 0.1 M NaCl
(b.)
0.1 M urea and 0.2 M MgCl2
(c.)
0.1 M NaCl and 0.1 M Na2SO4
(d.)
0.1 M Ca(NO3)2 and 0.1 M Na2SO4
Sol. D
π[Ca(NO3)2] = iCRT
= 3 × 0.1RT = 0.3 RT
π[Na2SO4] = iCRT
= 3 × 0.1 RT = 0.3 RT
Q.1
3
In the following reaction,
2SO2(g) + O2(g) → 2SO3(g)
The rate of formation of SO3 is 100 g min–1. Hence, the rate of disappearance of O2 is:
(a.)
50 g min-1
(b.)
20 g min-1
(c.)
100 g min-1
(d.)
200 g min-1
Sol. B
In the reaction,
2SO 2 (g) + O 2 (g) → 2SO 3 (g)
1 mol
32g
2 mol
160 g
∵ 160 g of SO3 is formed by 32 g of O2
32
× 100 or 20 g of O2
160
∴ Rate of disappearance of O2 = 20 g min–1
∴ 100 g of SO3 will be formed by
Q.1 Bombardment of aluminium by α - particle leads to its artificial disintegration in two ways, (i) and (ii)
as shown. Products X, Y and Z respectively are:
4
(ii)
30
27
Al
15 P + Y
13
(i)
30
14 Si + X
30
14 Si +
Z
(a.)
proton, neutron, position
(b.)
neutron, positron, proton
(c.)
proton, positron, neutron
(d.)
positron, proton, neutron
Sol. C
30
27
13
Al + 24 He →14 Si + X
27
13
Al + 24 He →15 P + Y
(X =11 H)
30
30
15
(Y =10 H)
30
P →14 Si + Z
(Z =1+ e)
Q.1 If Ca3 (PO4)2 and H3PO3 contain same number of ‘P’ atom then the ratio of oxygen atoms in these
5
compounds respectively is:
(a.)
8/3
(b.)
2/3
(c.)
3
Sol. D
Let number of moles of Ca3(PO4)2 and H3PO3 are x and y respectively.
(d.)
4/3
x 1
=
y 2
Moles of ' O' in Ca 3 (PO 4 )2 8 x 4
=
=
Moles of ' O' in H3PO 3
3y 3
Q.1 A non stoichiometric compound Cu1.8S is formed due to incorporation of Cu2+ ions in the lattice of
6
cuprous sulphide. What percentage of Cu2+ ion in the total copper content is present in the
compound?
(a.)
88.88
2+
(b.)
11.11
(c.)
99.8
(d.)
89.8
and (L8 – x) Cu ions are present in the compound Cu1.8S. Compound is electricially
+
Sol. B Let x Cu
neutral.
∴
+2x + (1.8 – x) – 2 = x = 0.2
0.2
% Cu 2 + =
× 100 = 11.11
1.8
Q.1 At 300 K, solubility of a gas in a liquid was measured at different partial pressures.
7
Mole fraction of Gas
0.010
0.015
0.020
Partial pressure of Gas (kPa)
82
122
166
Which of the following graph is correct in accordance with the Henry’s law?
(a.)
(b.)
Partial
pressure
Partial
pressure
kH= 8000 kPa
kH= 8234.5 kPa
Mole fraction
(c.)
Mole fraction
(d.)
kH= 8234.5 kPa
Partial
pressure
Partial
pressure
kH= 8234.5 kPa
Mole fraction
Mole fraction
Ans. B
Q.1 Plots showing the variation of the rate constant ‘k’ with temperature (T) are given below. The plot
8
that follows Arrhenius equation is:
(a.)
(b.)
K
K
T
T
(c.)
(d.)
K
K
T
Sol. A
Arrhenius equation is:
k = Ae −Ee / RT
T
Rate constant k will increase exponentially with increase in temperature (as shown below)
K
T
SECTION-B
(Multiple Answer is Correct)
Q.1
Which of the following are incorrect statements
(a.)
Spontaneous adsorption of gases on solid surface is an exothermic process as entropy
decreases during adsorption
(b.)
Formation of micelles takes place when temperature is below Kraft Temperature (Tk) and
concentration is above critical micelle concentration (CMC)
(c.)
A colloid of Fe(OH)3 is prepared by adding excess of NaOH in FeCl3 solution, the particles of
this sol will move towards cathode during electrophoresis
(d.)
According to Hardy-Schulze rules, the coagulation (flocculating) value of Fe3+ ion will be
more than Ba2+ or Na+
Sol. B,C,D
(A)
(B)
(C)
(D)
Q.2
∆G = ∆H - T∆S < O as ∆S < O so ∆H has to be negative
micelles formation will take place above Tk and above CMC
this sol will be negatively charged & will move toward anode
Fe3+ ions will have greater flocculatibility power so smaller flocculating value
In which of the following pairs of solutions will the values of the vant Hoff factor be the same?
(a.)
0.05 M K4 [Fe(CN)6] and 0.10 M FeSO4
(b.)
0.10 M K4[Fe(CN)6] and 0.05 M FeSO4 (NH4)2SO4.6H2O
(c.)
0.20 M NaCl and 0.10 M BaCl2
(d.)
0.05 M FeSO4 (NH4)2SO4.6H2O and 0.02 M KCl. MgCl2.6H2O
Sol. B,D
Number of particles from K4[Fe(CN)6] = 6
Number of particles from FeSO4 (NH4)2SO4.6H2O = 5
Number of particles from KCl.MgCl2.8H2O = 5
Q.3
An electrochemical cell is set up as follows:
Pt (H2, 1 atm.) | 0.1 M HCl | 0.1 M CH3COOH| (H2, 1 atm.) Pt
E.M.F. of the cell will not be zero because
(a.)
the temperature is constant
(b.)
the pH of 0.1 M HCl and 0.1 M CH3COOH is not the same
(c.)
acids used in the cell have different Ka values
(d.)
E.M.F. of above cell depends on molarities of the acids used
Ans. B,C
Q.4 Which of the following statement is incorrect?
(a.)
doping of As in silicon makes it a semi-conductor of p-type
(b.)
doping of B in silicon make it a semi-conductor of n – type
(c.)
doping of Cd & Se in silicon forms a binary compound which act as semiconductor
(d.)
doping of Ge with As make it a semi-conductor of n – type
Ans. A,B
Q.5 The vapour of an organic compound requires three times its own volume of oxygen for complete
combustion and produces twice its own volume of carbon dioxide. Which of the following
compounds would give these results?
(a.)
CH3CHO
(b.)
C2H5OH
(c.)
CH3CO2H
(d.)
C2H4
Ans. B,D
SECTION-C – (Comprehension Type Questions)
Comprehension – 1 ( Next 3 Questions)
The minimum concentration of the electrolyte in millimoles per litre of the solution, required to cause
coagulation of a particular sol is called coagulation value of flocculation value of the electrolyte for the sol.
The ions carrying opposite charge to that of sol particles are effective in causing coagulation of the sol.
Coagulating power of an electrolyte is directly proportional to the valency of the active ions i.e., ions
causing coagulation. Coagulation or flocculation is the precipitation of a colloid through induced aggregate
of its particles. There are several methods for coagulation of colloidal solution like by repeated dialysis, by
adding electrolyte, by electrophoresis, etc.
Q.1
Which of the following will have highest coagulating power for As2S3 colloid?
(a.)
Al3+
(b.)
Na+
(c.)
SO 24 −
(d.)
PO 34 −
Sol. A As2S3 is negative colloid hence Al3+ will be most effective for its coagulation
Q. In the coagulation of positive sol, the flocculation power of Cl–, SO 2 − , PO 3 − , [Fe(CN) ] 4 − are in
4
4
6
2
the order of :
(a.)
Cl − > SO 24 − > PO 34 − > [Fe(CN) 6 ] 4 −
(b.)
[Fe(CN) 6 ] 4 − > PO 34 − > SO 24 − > Cl −
(c.)
Cl − > PO 34 − > SO 24 − > [Fe(CN) 6 ] 4 −
(d.)
Cl − > SO 24 − > [Fe(CN) 6 ] 4 − > PO 34 −
Sol. B Greater is the charge of ion, more effective is the coagulation of oppositely charged colloid
Q.3 Flocculating value of ion depends on:
(a.)
the shape of flocculating ion
(b.)
the amount of flocculating ion
(c.)
nature of the charge on the flocculating ion
(d.)
both, the nature and magnitude of the charge of the flocculating ion
Ans. D
Comprehension – 2 ( Next 3 questions)
Faraday’s law of electrolysis are used for quantitative estimation of products of electrolysis. According to
which, mass deposited at electrode depends on quantity of electricity passed. Product of electrolysis
depends on discharge potential, concentration & nature of electrode.
Answer the following questions:
Q.4
If same amount of electricity is passed through aqueous solution of AgNO3 and CuSO4 and number
of Cu and Ag atoms deposited are x and y respectively, then
x=y
(a.)
(b.)
x = 3y
(c.)
x = y/2
(d.)
x = 2y
Sol. C 2Ag + 2e → 2Ag; Cu + 2e → Cu
Thus, number of Ag atoms will be twice the number of Cu atoms.
Q.5 The passage of 25 milliampere of current through molten CaCl2 for 60 seconds will cause the
+
–
2+
–
deposition of N calcium atoms on cathode. The value of N is
(a.)
6.02 × 1019
(b.)
2 × 1018
(c.)
3 × 1018
(d.)
4.68 × 1018
Sol. D Q = 0.025 (amp) × 60(s) = 1.5 C
2 × 96500 C deposit Ca atoms = 6.02 × 1023
6.02 × 10 23 × 1.5
1.5 C deposit Ca atoms =
= 4.68 × 1018
2 × 96500
Q.6 The time required for the complete decomposition of two moles of water using a current of 2
ampere is
(a.)
1.93 × 105 sec
(b.)
2.93 × 105 sec
(c.)
0.93 × 105 sec
(d.)
4.93 × 105 sec
Sol. A
2H2O
2H2 + O2
Electrons involved = 4
∴
4 × 96500 = 2 × t
4 × 96500
t=
= 1.93 × 105 sec.
2
Comprehension – 3 ( Next 3 questions)
The rates of reaction between the reactants A and B were studied starting with different initial
concentrations. The following data were obtained:
1.
2.
3.
Q.7
[A], mol lit –1
[B], mol lit –1
Initial rate,
mol lit-1 s-1 at 300 K
2.5 × 10–4
5.0 × 10–4
1.0 × 10–3
3.0 × 10–5
6.0 × 10–5
6.0 × 10–5
5.0 × 10–4
4.0 × 10–3
1.6 × 10–2
(mol ℓ–1 s–1) at
320 K
2.0 ×10–3
-------
The order of reaction with respect to A and B are respectively
(a.)
1, 1
(b.)
1, 2
(c.)
2, 1
(d.)
2, 0
Sol. C using (ii) & (iii) from table we get
4.0 × 10 −3
= [0.5] α
−2
1.6 × 10
[0.5]2 = [0.5]α
⇒α=2
from (i) & (ii) we get
α
β
 5 × 10 −4   2.5 × 10 −14   3 × 10 -5 
=

−3 
−4  
−5 
 4 × 10   5 × 10
  6 × 10 
1.25 = (0.5)2 (0.5)β
(0.5)β = 0.5
⇒β=1
Rate = KI [A]2 [B]
Order wit h respect to A = 2
B=1
Q.8
The rate constant of the reaction at 320 K when concentrations are in mol L–1 and time in seconds
will be
(a.)
2.67 × 108
(b.)
1.33 × 105
(c.)
1.6 × 104
(d.)
10.6 × 108
2 × 10-3
(d.)
30 × 10-3
Sol. D
K2 =
=
2 × 10 −3
(2.5 × 10 − 4 ) 2 (3 × 10 − 5 )
2 × 10 −3
18.75 × 10 −13
= 0.106 × 1010
K2 = 10.6 × 108
Q.9
What is rate of reaction at 320° in 2 nd experiment?
(a.)
Sol.
15 × 10-5
(b.)
16 × 10-3
(c.)
B
Rate = 10.6 × 108 [5 × 10-4]2 [6 × 10-5]
= 15.9 × 10-3
SECTION-D
Matrix Match Type Questions
Q.1
Match column – I with column II.
Column-I
(i)
Column- II
(a.)
1
(b.)
2
(c.)
3
x
t
(ii)
Rate
conc. [A]
(iii)
1
[A]
(iv)
(d.)
0
Rate
[A]
3
Ans. i → s; ii → p; iii → q; iv → r
Q.2 Match Column - I with Column - II and select the correct answer using codes given below the
lists.
Column-I
(i)
(ii)
(iii)
(iv)
Column- II
S1
S2
0.1M sucrose
0.1M NaCl
0.1M urea
0.1M KCl
(a.)
0.1M Na2SO4
0.1M glucose
0.1M KCl
0.1M CuSO4
(c.)
(b.)
(d.)
S1 and S2 are isotonic
No osmosis takes place when S1 and
S2 are separated using semipermable
membrane
S1 is hypertonic to S2
S1 is hypotonic to S2
Ans. i → AB; ii → AB; iii → C ; iv → D
Q.3 Match Column- I with Column – II
Column-I
(i)
(ii)
(iii)
(iv)
Decay process
Column- II
206
234
90 Th → 82 Pb
206
238
U

→
Pb
82
92
(a.)
221
235
U

→
Pb
84
92
208
232
90 Th → 82 Pb
(c.)
(b.)
(d.)
No. of particles emitted
7α
6β
8α
6α
Ans. i → AB; ii → AB; iii → C ; iv → D
(a)
234 = 206 + 4x
234 – 206 = 4x
28 = 4x
x=7
90 = 82 + 14 – y
y = 82 + 14 – 90
y=6
SECTION-E
(Integer Type Questions)
Q.1
The rate expression for reaction A(g) + B(g) → C(g) is rate = k[A]1/2 [B]2. What will be the rate if
initial concentration of A and B increase by factor 4 and 2 respectively?
Ans. 8
Q.2
Density of lithium atom is 0.53 g/cm3. The edge length of Li is 3.5 Å. The number of lithium atoms in
a unit cell will be ____. (Atomic mass of lithium is 6.94)
Ans. 2
What is the valency of an element of which the equivalent weight is 12 and specific heat is 0.25
Cal/gm?
6.4
25.5
Ans. atomic mass ≈
= 25.5 ; Valency=
=2
0.25
12
Q.3
Q.4
The shaded plane represents diagonal plane of symmetry. How many such planes are possible in
the cubic system?
A
B
D
C
Ans. 6
Thus conductivity of 0.01 mol/dm3 aqueous acetic acid at 300K is 19.5 × 10–5 ohm–1 cm–1 and the
limiting molar conductivity of acetic acid at the same temperature is 390 ohm–1 cm2 mol–1. The
percentage degree of dissociation of acetic acid is
1000
1000
19.5
Sol. Λm = K ×
= 19.5 × 10–5 ×
=19.5,
α=
= 0.05
M
0.01
390
Percentage α = 5
Q.5