Chapter wise important questions in Chemistry for Karnataka 2 year PU Science students. This is taken from the PU board website and compiled together.

1.
UNIT-1 SOLIDS
QUESTIONS CARRYING ONE MARK:
1. Which type of solid is anisotropic in nature?
Ans: Crystalline solids are anisotropic in nature
2. Which type of solids is called as super cooled liquids or pseudo solids?
Ans:Amorphous solids are called super cooled liquids
3. A solid has a sharp melting point, and then to which type of solids does it
belong?
Ans: Crystalline solids
4. Which type of solids has long range orderly arrangement of constituent
particles?
Ans: Crystalline solids
5. Sodium chloride and quartz belong to which type of solid?
Ans: Crystalline solids
6. A solid shows different values for refractive index when measured in different
directions. - Identify the type of solid
Ans: Crystalline solids
7. When a solid is cut with a sharp edged tool, they cut into two pieces and the
newly generated surfaces are plain and smooth. – Identify the type of solid.
Ans: Crystalline solids
8. Which type of force of attractions is present between the molecules in polar
molecular solids?
Ans: Dipole –dipole interactions
9. Which type of force of attractions is present between the molecules in
non-polarmolecular solids?
Ans: London forces or Dispersion forces
10. Which type of force of attractions is present between the particlesin
ionic solids?
Ans: Electrostatic force of attraction or coulombic force of attraction
11.Solid SO2 and solid NH3belong to which type of molecular solids?
Ans: Polar molecular solids
12. What is crystal lattice?
Ans: The regular three dimensional array of lattice points in space is called
crystal lattice
13. What is a unit cell?
Ans: It is the smallest repeating unit which when arranged in three dimension
gives the crystal lattice.
14. How many types of primitive unit cells are present?
Ans: Three types
15. What is a primitive cubic unit cell?
Ans: The cubic unit cell in which the particles/atoms are present only at the
eightcorner of the cube is called primitive cubic unit cell.
16. Define the co-ordination number of a particle in solids.
Ans:It is the total number of nearest neighboring particles to a given particle.
17. What is the number of octahedral voids generated, if the number of close
packed spheres is N?
Ans: N
18. What is the number of tetrahedral voids generated, if the number of close
packed spheres is N?
Ans: 2N
19. What is the co-ordination number of a particle in a tetrahedral void?
Ans:Four
20. Among Schottky and Frenkel defect, which type of defect decreases
the density of the crystal?
Ans: Schottky defect

2.
21. What are point defects?
Ans: Deviations from the ideal arrangement around a particular point or an
atomin a crystalline solid
22. What are F-centers?
Ans: The anionic sites occupied by the unpaired electrons are called F- centre.
23. To which colour potassium chloride crystal turns, when excess potassium
ionis present?
Ans: Violet
24. Name the type of non-Stoichiometric defect observed when white ZnO
turns yellow on heating.
Ans: Metal excess defect
25. Name the non-Stoichiometric defect responsible for the composition of ferrous
oxide to be Fe0.95O1.
Ans:Metal deficiency defect
26. Which type of point defect is observed when NaCl containing little SrCl2 is
crystallized?
Ans:Impurity defect
27. Which defect is also called as dislocation defect?
Ans:Frenkel defect
28. What is doping?
Ans: The process of increasing the conductivity of an intrinsic semiconductors
by adding asuitable impurity is called doping
29. What type of semiconductors are obtained when silicon doped with
boron impurity?
Ans: p-type semiconductor
30. Name the unit used to measure magnetic moment.
Ans: Am2( 1Bohr magneton= 9.27x10-24 Am2)
31. What are diamagnetic substances?
Ans: These are the substances which are repelled by the magnetic field
32. What are ferromagnetic substances?
Ans: These are the substances which are strongly attracted by the magnet
33. How body diagonal and radius of a sphere(r) are related in bcc unit cell?
Ans:4r =√2a
34. Give an example for Ferromagnetic substance.
Ans: Fe Co Ni Gd CrO2
35. Give an example for Diamagnetic substance.
Ans: H2O, NaCl, and C6H6
QUESTIONS CARRYING TWO MARKS:
1. How crystalline solids differ from amorphous solids in their melting point?
Ans:Crystalline solids have sharp melting point whereas amorphous solid do not
have a sharp melting point
2. Write any two differences between crystalline solids and amorphous solids?
Crystalline solid Amorphous solid
3-D long range orderly arrangement
of particles
No orderly arrangement of constituent
particles
Sharp Melting point Do not have sharp M P
( Softening temperature)
True solids having definite shape Pseudo solids having irregular shapes
They have a well-defined cleavage
planes
Do not have cleavage planes
Anisotropic in nature Isotropic in nature
3. What is meant by anisotropy? What type of solids show this nature?
Ans: The physical properties like refractive index, coefficient of thermal expansion,
when measured in different directions gives different value for a crystalline
solid hence it is anisotropic in nature.
Crystalline solids
4. What are the nature of particles and the force of attractions between
the particles in non-polar molecular solids?
Ans: In a non-polar molecular solids, the constituent particles are non-polar
moleculeslike H2, Cl2, I2 and even atoms like Ar, Ne, Xe etc.
The nature of force of attraction is weak dispersion force or London force.

3.
5. What are the nature of particles and the force of attractions between
the particles in polar molecular solids?
Ans: In a polar molecular solids, the constituent particles are formed by polar covalent
bond like HCl, SO2.
The nature of force of attraction is dipole-dipole attractions
6. What are the nature of particles and the force of attractions between
the particles in hydrogen bonded molecular solids?
Ans: In a hydrogen bonded molecular solids, the constituent particles are polar
molecules capable of forming hydrogen bond like water.
The nature of force of attraction is hydrogen bonding
7. What are point defects? Mention the types
Ans: Point defects are the irregularities in the arrangement of constituent
particlesaround a point or a lattice site in a crystalline substance.
These are of three types.
1. Stoichiometric defects.
2. Non-stoichiometric defect
3. Impurity defect.
8. What are the differences between Schottky and Frenkel defect?
Schottky defect
a. Shown by ionic solidscontaining
similar-sized cationsand anions
(having high coordination number)
b. An equal number of cations and
anions are missing to maintain
electrical neutrality
c. Decreases the density of the
substance
d. Example,
NaCl, KCl , CsCl, and AgBr
Frenkel defect
a. Shown by ionic solids containing
largedifferences in the sizes of ions,
(having less coordination number)
b. Created when the smaller ion
(usually cation) is dislocated from its
normal site to an interstitial site
c. No change in density of the crystal.
creates a vacancy defect as well as an
interstitial defect .Also known as
dislocation defect
d. Example:
AgCl, AgBr, AgI and ZnS
9. What are the nature of particles and the force of attractions between
the particles in ionic solids?
Ans: The nature of the particles is ions (both cation and anion). The nature
of the force of attraction is electrostatic force or coulombic force
10. What are the nature of particles and type of bonding in network solids?
Ans: The nature of the particles is atoms. The bonding is covalent bond.
11. Classify the following into polar and non-polar molecular solids:
Ar, HCl, I2 and SO2
Ans: Non-polar molecular solids: Ar, I2
Polar molecular solids:HCl, SO2
12. Calculate the number of particles present per unit cell in an FCC unit cell.
Ans:Contribution of corner particle = 8 x 1/8 = 01
Contribution of a particle at the centreof face = 6 x ½ = 03
Total number particle /unit cell = 04
13. Calculate the number of particles present per unit cell in a BCC unit cell.
Ans:Contribution of corner particle = 8 x 1/8 = 01
Contribution of a particleat the centre = 1 x 1 = 01
Total number particle /unit cell = 02
14. Calculate the number of particles present per unit cell in a simple cubic
unit cell.
Ans:Contribution of corner particle = 8 x 1/8 =01
Total number particle /unit cell = 01
15. Mention the two characteristics of a unit cell.
Ans: Two characteristics of unit cells are
a. Edge length
b. Axial angles
16. What is the relation between edge length (a) and radius of the sphere (r)
infcc unit cell? What is itspacking efficiency?
Ans: The relationship between edge length and radius of the sphere are a=2 2 r
Packing efficiency is 74%
17. What is the relation between edge length (a) and radius of the sphere (r)
in bcc unit cell? What is its packing efficiency?
Ans: The relationship between edge length and radius of the sphere are a=
!!
!
Packing efficiency is 68 %
18. How many tetrahedral and octahedral voids is present, if the number
of sphere is N?
Ans: The number of tetrahedral void is 2N
The number of octahedral void is N
19. Explain Schottky defect. Give an example.
Ans: The defect which arises due to missing of equal number of cations and anions
from the crystal lattice is called Schottky defect. Ex. NaCl, KCl ,CsCl, AgBr

4.
20. Explain Frenkel defect. Give an example.
Ans: The defect in which an ion (generally cation) leaves the original site and
occupies the interstitialsite is called Frenkel defect. E. AgCl, AgBr, AgI
21. How Schottky defect and Frenkel defect affect the density of the crystal?
Ans: In Schottky defect density of the crystal decreases.
In Frenkel defect the density of the crystal remains same.
22. Mention the two types of Non-stoichiometric defects in solids?
Ans: Metal excess defect and metal deficiency defect.
23. What is F- center? What colour is imparted to the NaCl crystal, due to the
presence of excess sodium?
Ans: The anionic sites occupied by the unpaired electrons are called F- Centre
The colour of NaCl crystal is Yellow
24. Write the formula to calculate the density of the unit cell and explain the
terms.
Ans:
z = number of particles present per unit cell
d =
𝒛𝑴
𝒂 𝟑 𝑵 𝑨
M = Molecular mass , d = density NA = Avogadro’s number
a = Edge length.
25. What are n-type and p-type semiconductors?
Ans:
n-type semiconductor is obtained by doping of the crystal of a group 14 element
such as Si or Ge, with a group 15 element such as P or As(pentavalent).
Conductivity increases due to negatively charged electrons.
p-type semiconductor is obtained by doping of the crystal of a group 14 element
such as Si or Ge, with a group 13 element such as B, Al or Ga( trivalent).
Conductivity increases as a result of electron hole
26. An ionic compound is formed by two elements A and B. The cat ions A are in
ccp arrangement and those of anions B occupy all the tetrahedral voids.
What is the simplest formula of the compound?
Ans:
Since cations are in ccp arrangement, the total number cat ions A = 4
The number of tetrahedral voids is double the number of particles = 8
All the tetrahedral voids are occupied by anions B.
The number of elements of B = 8
Hence the formula of the ionic compound is A4B8 or AB2
27. A compound is formed by two elements X and Y. The element X forms ccp and
atoms of Y occupy 1/3 rd of tetrahedral voids. What is the formula of the
compound?
Ans:
Since element X are in ccp arrangement, the number of X per unit cell = 4
The number of tetrahedral void = 8
But only 1/3 rd is occupied by Y, therefore 8 x1/3 = 8/3
Hence the formula of the compound is X4Y8/3 = X12Y8 or X3Y2
28. Gold(atomic radius=0.144nm)crystallizesin a face centered unit cell.
What is the length of the side of the cell?
Ans:
For FCC the edge length and radius of sphere arerelated by the equation,
r = 0.144nm a=2 2 r
a = ? = 2 2 x 0.144 nm
= 2x1.414 x 0.144
= 0.40723nm.
29. Silver forms ccp lattice and X- ray studies of its crystals show that the edge
lengthof its unit cell is 408.6pm. Calculate the density of silver
(atomic mass = 107.9 u)
Ans:
d =
!"
!!!!
d= 4 x 107.9/(4.08)3 x10-24 x 6.022 x1023
d = 431.6/40.899
d = 10.5528g/cm3
30. X- ray diffraction studies show that copper crystallizes in an fcc unit cell with
cell edge of 3.6 x10-8cm. In a separate experiment, copperis determined to
have a density of 8.92g/cm3,calculate the atomic mass of copper.
Ans:
d =
!"
!!!!
M = d a3 NA/Z
= 8.92 x(3.6)3x10-24 x 6,022 x1023/4
= 250.61/4
M = 62.6525 u
31. The edge of fcc unit cell of platinum is 392 pm and density is 21.5 g/cm3,
calculate the Avogadro number.
Ans:
d =
!"
!!!!
NA = Z x M/ d a3
= 4 x 195.08/21.5 x (3.92)3x 10—24
= 780.32/1295.08 x10—24
NA= 6.025 x1023

5.
32. A unit cell of sodium chloride has four formula units. The edge length of the
unit cell is 0.564 nm. What is the density of sodium chloride?
Ans:
d =
!"
!!!!
d = 4 x 58.5/(5.64)3 x10-24 x 6.022 x1023
d = 234/108.038
d = 2.165 g/cm3
33. A body centered cubic element having density 10.3 g/cm3, has a edge length
of 314pm. Calculate the atomic mass of the element
(Avogadro’s number= 6.023x1023/mol)
Ans:
d =
!"
!!!!
M = d x a3 xNA/Z
= 10.3 x (3.14)3x 10-24 x 6.022 x1023/2
M = 96.01u
34. Calcium metal crystallizes in a face centered cubic lattice with edge length
of 0.556nm. Calculate the density of the metal.
(Atomic mass of calcium = 40g/mol and Avogadro number= 6.022 x1023mol-1)
Ans:
d =
!"
!!!!
d = 4 x 40/(5.56)3 x10-24 x 6.022 x1023
d = 160/103.50
d = 1.54 g/cm3
35. Copper crystallizes into afcc lattice with edge length 3.61 x10-8cm.
Calculate the density of the of the crystal
(Atomic mass of copper =63.5g/mol and Avogadro number= 6.022 x1023mol-1 )
Ans:
d =
!"
!!!!
d = 4 x 63.5/(3.61)3 x10-24 x 6.022 x1023
d = 254/28.33
d = 8.9 g/cm3
36. Silver crystallizes in a face centered cubic structure. If the edge length is
4.077 x10-8cm and density is 10.5 g/cm3, calculate the atomic mass of silver.
Ans:
d =
!"
!!!!
M = d a3 NA/Z
= 10.5 x (4.077)3x10-24 x 6,022 x1023/4
= 103.57/4
The atomic mass of silver M = 107.09 u
37. The density of Li atoms is 0.53g/cm3.The edge length of Li is 3.5 A0. Find out
the number of Li atoms in a unit cell (N0= 6.022 x1023/mol& M= 6.94)
Ans:d =
!"
!!!!
Z = d x 𝑎!
𝑁!/𝑀
= 0.53 x (3.5)3 x10–24x 6.022 x1023/6.94
= 2
The number of lithium atoms in unit cell is 2
Questions carrying THREE marks
1. Calculate the packing efficiency in simple cubic unit cell
Edge length of the cube = a = 2r
Volume of the cubic unit cell= a3 = (2r)3= 8r3
volume of one particle(sphere) =
!
!
𝝅r3
The number of particles per unit cell =1
Total volume occupied by one sphere =
!
!
𝜋𝑟!
Packing efficiency=
!"#$% !"#$%& !""#$%&' !" !"# !"!!"!
!"#$%& !" !"#$! !"#$ !"##
× 100
=
!
!
!!!×!
! !! ×100
= 52.4%

6.
2. Calculate the packing efficiency in face centered cubic unit cell
edge length of the cube
be ‘a’
In ABC
AC2 = BC2 + AB2
b2 = a2 + a2
b2 = 2a2
b = 2 a
Let the radius of the
atom = r
Length of the diagonal
of ABC, b= 4r
2 a = 4r
a = 2 2 r
Edge length of the cube =a=2 2 r
Volume of the cubic unit cell= a3 = 2 2 r
!
volume of one particle(sphere) =
!
!
𝝅r3
The number of particles per unit cell =4
Total volume occupied by four spheres = 4 ×
!
!
𝜋𝑟!
packing efficiency=
!"#$% !"#$%& !""#$%&' !" !"#$ !"!!"!#
!"#$%& !" !"#$! !"#$ !"##
× 100
=
!
!
!!!×!
! !!
! × 100
=
!"
!
!!!
!" !!! × 100 = 74 %
3. Calculate the packing efficiency in body centered cubic unit cell
In ABG,
b2 = a2 + a2 ⇒ b2 = 2 a2
In, AGD,
C2 = a2 + b2
C2 = a2 + 2a2
C2 = 3a2⇒ C = 3𝑎
Radius of the atom = r.
Length of the body
diagonal, C=4r
3𝑎 = 4r
a =
!!
!
Edge length of the cube =a=
!!
!
Volume of the cubic unit cell= a3 =
!!
!
!
volume of one particle(sphere) =
!
!
𝝅r3
The number of particles per unit cell =2
Total volume occupied by two spheres = 2 ×
!
!
𝜋𝑟!
packing efficiency=
!"#$% !"#$%& !""#$%&' !" !"# !"!!"!#
!"!"#$ !" !"#$! !"#$ !"##
× 100
=
!
!
!!!×!
!
!
!
! × 100
=
!
!
!!!
!"
! !
!!
× 100 = 68%
4.Based on band theory explain conduction in metals, insulators and
semiconductors
Conduction of electricity in metals: In metals, the valence shell is partially filled, so
this valence band overlaps with a higher energy unoccupied conduction band so that
electrons can flow easily under an applied electric field.
Conduction of electricity in insulators: In insulators, the valence shell is empty, so
the gap between the valence band and conduction band is very large. so that electrons
cannot flow under an applied electric field.
Conduction of Electricity in Semiconductors In semiconductors, the gap between
the valence band and conduction band is so small that some electrons may jump to
the conduction band. Electrical conductivity of semiconductors increases with
increase in temperature. Substances like Si, Ge show this type of behaviour, and are
called intrinsic semiconductors.
5. How are solids classified on the basis of the force of attraction?
Ans:
a. Molecular solids: Particles are held by
a. London forces (in non-polar solids) ex : Benzene, Argon, P4O10, I2, P4
b. Dipole - dipole interaction ( in polar solids) ex: Urea, Ammonia
c. Hydrogen bonding (in hydrogen bonded solids) ex: ice
b. Ionic solids
a. Particles are held by ionic bond
b. Conduct electric current in aqueous solution or molten state
c. Examples: NaCl, MgO, ZnS
d. In solid state, ions are held together by strong electrostatic forces and are
not free to move about within the solid. Hence, ionic solids do not conduct
electricity in solid state. However, in molten state or in solution form, the
ions are free to move and can conduct electricity.
c. Covalent or network solids:
a. Particles are held by covalent bonding. Examples SiO2 (quartz), diamond,
d. Metallic solids:
a. Particles are held by metallic bond.
b. These are electrical conductors, malleable, and ductile. Examples: Fe, Cu,
6. What are point defects? Explain the types.
Ans: Point defects are the irregularities in the arrangement of constituent particles
around a point or an atom in a crystalline substance. These are of three types.
1. Stoichiometric defects: Do not disturb stoichiometry of the solid.
These are also called intrinsic or thermodynamic defects
Ex : Frenkel defect, Schottky defect
2. Non-stoichiometric defects: This defect alters the stoichiometric ratio of the
constituent elements
i) Metal excess defect
a. Metal excess defect due to anionic vacancies:
b. Metal excess defect due to the presence of extra
cations at interstitial sites:
ii) Metal deficiency defect
a. By cation vacancy
3. Impurity defect.

7.
7. What are diamagnetic, paramagnetic and ferromagnetic substances?
1. Paramagnetic substance: The substance which are attracted by the magnet.
The magnetic character is temporary and is present as long as the external
magnetic field is present. Ex; O2, Cu2+, Fe3+, Cr3+ NO.
2. Diamagnetic substance: The substance which are weakly repelled by the magnetic
field TiO2, H2O,NaCl.This property is shown by those substance which contain
fully –filled orbitals (no unpaired electrons)
3. Ferro magnetic substance: The substance which are strongly attracted by the
magnet. They show permanent magnetism even in the absence of magnetic field.
Ex : Fe Co Ni Gd& CrO2
8. An element with molar mass 2.7 x 10-2 kg/mol forms a cubic unit cell with edge
length 405pm. If its density is 2.7 x 103 kg/m3, what is the nature of the cubic
unit cell
Ans:
d =
!"
!!!!
Z = d x 𝑎!
𝑁!/𝑀
= 2.7 x103 x(405)3 x 10—27 x 6.022 x1023/2.7 x 10-2
= 4
Since there are 4 atoms of the element present per unit cell. Hence, the cubic
unit cell must be face centered or cubic close packed structure (ccp)
9. Niobium crystallises in body-centered cubic structure. If density is 8.55g/cm3,
calculate atomic radius of niobium, given that its atomic mass is 93 u.
Ans: d =
!"
!!!!
a3 =
!"
! !!
= 2 x 93/8.55 x6.022 x1023
= 36.1 x106
a = (36.1)1/3 x102
=330 pm
For BCC r =
!
!
a
r =
!
!
x 330
r = 143pm
10. An element has a body-centered cubic (bcc) structure with cell edge of 288pm.
The density of the element is 7.2 g/cm3. How many atoms are present in 208g
of the element?
Ans:
d =
!"
!!!!
M = d a3 NA/Z
= 7.2 x (2.88)3x10-24 x 6,022 x1023/2
= 103.57/2
M = 51.78 u
51.78 g (1mole) contains 6.022 x1023 atoms
Therefore 208g contains 4.01 x 6.022 x1023 = 24.187 x1023 atoms.

8. nBM
V
=
=
A
nBM
W
Unit -2
THEORY OF DILUTE SOLUTIONS
1) What
is
solution?
[1]
A:
It
is
a
homogenous
mixture
of
two
or
more
compounds.
2) What
is
dilute
solution?
[1]
A:
It
is
a
solution
in
which
solute
concentration
is
very
less.
3) Give
an
example
for
solid-­‐solid
solution
[1]
A:
Copper
dissolved
in
gold.
4) Give
an
example
for
gas-­‐gas
solution
[1]
A:
Mixture
of
oxygen
and
nitrogen
gases.
5) Give
an
example
for
gas-­‐solid
solution
[1]
A:
Solution
of
hydrogen
in
palladium.
6) Give
an
example
for
liquid-­‐solid
solution
[1]
A:
Amalgam
of
mercury
with
sodium.
7) Give
an
example
for
liquid-­‐liquid
solution
[1]
A:
Ethanol
dissolved
in
water.
8) Give
an
example
for
solid-­‐gas
solution
[1]
A:
Camphor
in
nitrogen
gas.
9) Define
mole
fraction
and
give
the
equation
to
calculate
it.
[2]
A:Mole
fraction
is
the
ratio
of
number
of
moles
of
one
component
to
the
total
number
of
moles
of
all
the
components
in
the
solution.
A
B
n
AX
n n
A
=
+
B
B
nBX
n n
A
=
+
10)Define
molarity
and
give
the
equation
to
calculate
it.
[2]
A:
Number
of
moles
of
the
solute
present
per
liter
solution
is
known
as
molarity.
11)Define
molality
and
give
the
equation
to
calculate
it.
[2]
A:
Number
of
moles
of
the
solute
present
perkgsolvent
is
known
as
molality.
12)Define
the
term
solubility
of
a
substance.
[1]
A:
Solubility
of
a
substance
is
its
maximum
amount
that
can
be
dissolved
in
a
specified
amount
of
solvent
at
a
specified
temperature
13)State
Henry’s
law.
[2]
A:
Henry’s
Law:
At
constant
temperature
solubility
of
a
gas
in
a
liquid
is
directly
proportional
to
the
partial
pressure
of
gas
present
above
the
solution.
OR
At
constant
temperature
the
partial
pressure
of
the
gas
in
vapor
phase
(p)
is
proportional
to
the
mole
fraction
of
the
gas
(x)
in
the
solution.
Mathematically p ∝ x ; p = KH x.
Where KH is Henry’s law constant. KH depends on the nature of the gas.
14)Write
the
plot
which
shows
relation
between
partial
pressure
of
a
gas
v/s
its
mole
fraction.
[2]
A:
15)Mention
the
factors
affecting
solubility
of
a
gas
in
liquid.
[2]
A:
1.
Temperature
2.
Pressure
16)Explain
how
temperatures
effect
the
solubility
of
a
gas
in
liquid.
[2]
A:
Solubility
of
gases
in
liquid
decreases
with
rise
in
temperature.
According
to
Le
Chatelier’s
Principle,as
dissolution
is
an
exothermic
process,
the
solubility
should
decrease
with
increase
of
temperature.
17)Explain
how
pressure
effects
the
solubility
of
a
gas
in
liquid.
[1]
A:
The
solubility
of
gases
increases
with
increases
of
pressure.
18)Mention
the
applications
of
Henry’s
law.
[3]
A:
(a)
To
increase
the
solubility
of
CO2
insoft
drink
and
soda
water,
the
bottle
is
sealed
under
high
pressure.
Mole
fraction.
Partial
pressure
of
a
gas

9. (b)
To
avoid
bends,
as
well,
the
toxic
effects
of
high
concentration
of
nitrogen
in
the
blood,
the
tanks
used
by
scuba
divers
are
filled
with
air
dilute
with
helium.
(c)
At
high
altitudes
the
partial
pressure
of
oxygen
is
less
than
that
at
the
ground
level.
This
leads
to
low
concentrations
of
oxygen
in
the
blood
and
tissues
of
people
living
at
high
altitudes
or
climbers.
19)State
Raoult’slaw
of
liquid-­‐liquid
dilute
solutions.
[2]
A:
The
partial
vapour
pressure
of
each
component
of
the
solution
is
directly
proportional
to
its
mole
fraction
present
in
solution.
Thus,
for
component
1
P1
⍺
x1
And
p1
=
p1
0x1
20)What
are
ideal
solutions?
[1]
A:
The
solution
which
obey
Raoul’s
law
over
the
entire
range
of
concentration
are
known
as
ideal
solution
21)Mention
the
characters
of
ideal
solutions.
[3]
A:
22)What
are
non-­‐ideal
solutions?
[1]
A:
When
a
solution
does
not
obey
Raoult’s
law
over
the
entire
range
of
concentration,
then
it
is
called
non-­‐ideal
solution.
23)Mention
the
types
of
non-­‐ideal
solutions.
[1]
A:
There
are
two
types
(a)
Non-­‐ideal
solution
with
positive
deviation
from
Raoult’s
law
(b)
Non-­‐ideal
solution
with
negative
deviation
from
Raoult’s
law
24)Give
an
example
for
non-­‐ideal
solution
with
positive
deviation
from
Raoult’s
law.
[1]
A:
Mixtures
of
ethanol
and
acetone
Ideal
I.
It
obeys
Raoults
law
is
obeyed
at
all
temperature
and
concentration
P
=
PA
+
PB
II. ∆
V
mix
=
O
i.e.,
there
is
no
change
in
volume
on
mixing
III. ∆Hmix
=
O
i.e.,
there
is
no
enthalpy
change
when
ideal
solution
formed
IV. It
doesn’t
form
azeotropic
mixture
V. Force
of
attraction
between
A―A,
B―B is similar
as A―B
25)Give
an
example
for
non-­‐ideal
solution
with
negative
deviation
from
Raoult’s
law.
[1]
A:
An
example
of
this
type
is
a
mixture
of
phenol
and
aniline.
26)What
are
azeotropes?
Give
example.
[2]
A:
Azeotropes
are
binary
mixtures
having
the
same
composition
in
liquid
and
vapour
phase
and
boil
at
a
constant
temperature.
For
example:
ethanol-­‐water
mixture
27)State
Raoult’s
law
of
relative
lowering
of
vapour
pressure.
[1]
A:
Relative
lowering
of
vapour
pressure
is
equal
to
the
mole
fraction
of
the
solute.
28)Define
colligative
property.
[1]
A:
The
properties
depend
on
the
number
of
solute
particles
irrespective
of
their
nature
relative
to
the
total
number
of
particles
present
in
the
solution.
Such
properties
are
called
colligative
properties
29)Mention
four
colligative
properties
of
dilute
solutions.
[2]
A:
Relative
lowering
of
vapour
pressure
I. Elevation
in
Boiling
point
II. Depression
in
Freezing
point
III. Osmotic
pressure
30)Define
the
term
relative
lowering
of
vapour
pressure.
[2]
A:
It
is
the
ratio
of
lowering
of
vapour
pressure
to
the
vapour
pressure
of
the
pure
solvent
o
o
P P
Relativeloweringof V.P
P
−
=
31)What
is
elevation
in
boiling
point?
[1]
A:
Elevation
in
boiling
point
is
the
difference
between
the
boiling
point
of
the
solution
containing
non-­‐volatile
solute
and
the
boiling
point
of
the
pure
solvent
∆Tb
=
T
–
To
32)Give
the
relation
between
elevation
in
boiling
point
and
molecular
mass
of
solute.
[2]
A:
ΔTb = Kb
2
1 2
w 1000
w M
×
×
Where w2 is mass of solute, w1 is the mass of the solvent; M2 is molar mass of the solute

10.
33)Give
the
S.I.unit
of
ebullioscopic
constant
or
boiling
point
elevation
constant
or
molal
elevation
constant.
[1]
A:
The
unit
of
Kb
is
K
kg
mol-­‐1
34)What
is
depression
infreezing
point?
[1]
A:
It
is
the
decrease
in
the
freezing
point
of
solution
when
non-­‐volatile
solute
is
added
into
solvent.
35)Give
the
relation
between
depression
infreezing
point
and
molecular
mass
of
solute.[2]
A:
ΔTf = Kf
2
2
1
w
M
w
1000
∴ M2 = f 2
f 1
K 1000 w
T w
× ×
Δ ×
where M2 is molar mass of the solute.
Note: Values of Kf and Kb of the solvent depends on their molecular mass and ΔHfusion and ΔHvap
of the solvent respectively.
36)Give
the
S.I.unit
of
cryoscopic
constant.
[1]
A:
The
unit
of
Kf
is
K
kg
mol-­‐1
37)Draw
the
plot
showing
elevation
in
boiling
point
in
a
solution.
[2]
A:
―∆Tb
―
Temperature/K
Vapour
pressure
Tb
Tb
0
Solution
Solvent
38)Draw
the
plot
showing
depression
in
freezing
point
in
a
solution.
[2]
―∆Tf―
Temperature/K
39)Define
osmosis.
[1]
A:
The
process
of
movement
of
solvent
particles
from
lower
concentration
to
higher
concentration
through
semi-­‐permeable
membrane
to
attain
equilibrium
is
called
osmosis.
40)What
is
osmotic
pressure
and
give
its
relation
with
concentration
of
solution.
[2]
A:
The
amount
of
external
pressure
required
to
stop
the
osmosis.
=
CRT
Where:
=
osmotic
pressure,
R
=
gas
constant,
T
=
temperature,
C
=
concentration
of
solution.
41)
What
are
isotonic
solutions?
[1]
A:
Two
different
solutions
having
sameosmotic
pressure
are
called
isotonic
solutions
42)What
are
hypertonic
solutions?
[1]
A:
The
solution
having
more
osmotic
pressure
than
other
43)What
are
hypotonic
solutions?
[1]
A:
The
solution
having
less
osmotic
pressure
than
other
Tf
Tf
o
Vapour
pressure
Solution
Liquid
solvent
Frozen
solvent

11. 44)Explain
the
application
of
reverse
osmosis
in
desalination
of
water.
[2]
A:
When
pressure
more
than
osmotic
pressure
is
applied,
pure
water
is
squeezed
out
of
the
sea
water
through
the
membrane.
A
variety
of
polymer
membranes
are
available
for
this
purpose.
The
pressure
required
for
the
reverse
osmosis
is
quite
high.
A
workable
porous
membrane
is
a
film
of
cellulose
acetate
placed
over
a
suitable
support.
Cellulose
acetate
is
permeable
to
water
but
impermeable
to
impurities
and
ions
present
in
sea
water.
45)What
is
reverse
osmosis?
[1]
A:
Movement
of
solvent
particles
from
higher
concentration
to
lower
concentration
through
a
semi
permeable
membrane,
when
pressure
is
applied
greater
than
osmotic
pressure
46)What
is
abnormal
molar
mass?
[1]
A:
A
molar
mass
that
is
either
lower
or
higher
than
the
expected
or
normal
value
is
called
as
abnormal
molar
mass.
47)
Define
Vant
hoff
factor
Van’t Hoff factor ‘i’ to account for the extent of association or dissociation of a solute in a solvent
is
i =
Normal molar mass
Abnormal molar mass
or
i =
observed colligative property
calculated colligative property
or
i =
total number of moles of particles after association or dissociation
Number of moles of particles before association or dissociation
48)What
is
the
value
of
i
for
NaCl.
[1]
A:
2
49)What
is
the
value
of
i
for
K2SO4.
[1]
A:
3
50)What
is
the
value
of
i
for
sugar.
[1]
A:
1
51)What
is
the
value
of
i
for
glucose.
[1]
A:
1
52)On
what
factor
the
colligative
property
depends
on.
[1]
A:
It
depends
on
number
of
moles
of
solute
particles
but
not
on
the
nature
of
the
solute.
53)Write
the
mathematical
equation
of
Raoults
law
in
case
of
non-­‐volatile
solute.
[1]
A:
If
one
of
the
components
(solute)
is
non-­‐volatile
then
the
equation
of
Raoults
law
is.
PB=
O
P
=
PA
+
PB
P
=
PA
+
O
P
=
PA
54)Write
the
differentiate
between
non-­‐ideal
solutions
with
positive
deviation
and
negative
deviation
from
Raoult’s
law
[2]
55)Define
lowering
of
vapour
pressure?
[1]
A:
It
is
defined
as
the
difference
between
the
vapor
pressure
of
the
solvent
in
pure
state
and
the
vapour
pressure
of
the
solution
∆P
=
Po
–
P
56)State
Roult’s
law
of
relative
lowering
of
vapour
pressure
[1]
A:
It
states
that
the
relative
lowering
of
vapour
pressure
is
equal
to
the
mole
fraction
of
the
solute
Positive
deviation
(a)In
this
solution
solvent
–
solute
interaction
is
weaker
than
solvent
–
solvent,
solute-­‐solute
interactions
(b)
P
>
PA
+
PB
(c)
∆
V
>
O
(d)
∆H
=
positive
(e)
It
forms
azeotrope
with
minimum
boiling
point
Negative
deviation
(a)
In
this
solution
solvent
–
solute
interaction
is
stronger
than
solvent
–
solvent,
solute-­‐solute
interactions
(b)
P
<
PA
+
PB
(c)
∆
V
<
O
(d)
∆H
=
negative
(e)It
forms
azeotrope
with
maximum
boiling
point
P
=
PA
O.
XA

12. 57)Why
sea
water
freezes
below
00C?
[1]
A:
Sea
water
freezes
below
00C
due
to
the
presence
of
the
non-­‐volatile
solute
dissolved
in
the
water.
58)Derive
the
equation
to
calculate
molecular
mass
of
unknown
solute
using
Raoult’s
law
of
relative
lowering
of
V.P
[3]
A:
According
to
Raoult’s
law
relative
lowering
of
vapour
pressure
is
equal
to
the
mole
fraction
of
the
solute.
o
Bo
P P
X
P
−
=
o nP P B
o n nP BA
−
=
+
nB<<<nA
for
dilute
solution
So
we
can
neglect
nB
in
denominator
o nP P B
o nP A
−
=
B
B
A
A
W
o MP P
o WP
M
−
=
o MWP P B A
o W MP BA
−
=
B A
B
A
oW .M P
M oW P P
⎛ ⎞
= ⎜ ⎟
−⎝ ⎠
Numerical
problems
1. A
solution
containing
2.56
g
sulphur
in
100
g
CS2
gave
a
freezing
point
lowering
of
0.383
K.
Calculate
the
molar
mass
of
sulphur
molecules.
Given
Kf
of
CS2
=
3.83
K
kg
mol−
1
.
Ans.
ΔTf
=
0.383
K,
Kf
=
3.83
K
kg
mol−
1
ΔTf
=
Kf
×
m
;
ΔTf
=
Kf
×
2
2
1
W
M
W
1000
M2
(molar
mass
of
sulphur
molecules)
=
2.56 1000 3.83
100 0.383
× ×
×
=
256
g
mol−
1
2. 100
g
of
water
has
3g
of
urea
dissolved
in
it.
Calculate
the
freezing
point
of
the
solution.
Kf
for
water
=
1.86
K
kg
mol−
1
,
molar
mass
of
urea
=
60
g
mol−
1
,
freezing
point
of
water
=
273.15
K
(0°C)
Ans.
ΔTf
=
Kf
×
m
;
ΔTf
=
1.86
×
2
2
1
W
M
W
1000
ΔTf
=
1.86
×
3 1000
60 100
×
×
=
0.93
ΔTf
=
0
f fT T−
∴
Tf
=
273.15
−
0.93
=
272.22
or
−0.93°C
3. Human
blood
has
osmotic
pressure
of
7.2
atm
at
body
temperature
of
37°C.
Calculate
the
molar
concentration
of
solute
particles
in
blood.
Given
R
=
0.0821
L
atm
K−
1
.
Ans.
π
=
CRT
;
C
=
RT
π
T
=
273
+
37
=
310
K
C
(molar
concentration)
=
7.2
0.0821 310×
=
0.2828
M
4. Vapour
pressure
of
benzene
is
200
mm
of
Hg.
2g
of
a
non-­‐volatile
solute
in
78
g
benzene
has
vapour
pressure
of
195
mm
of
Hg.
Calculate
the
molar
mass
of
the
solute.
Molar
mass
of
benzene
=
78
g
mol−
1
.
Ans.
2
1
nP P
P n
°−
=
°
;
2
2
1
1
W
MP P
WP
M
°−
=
°
200 195
200
−
=
2
2
M
78
78
;
Molar
mass
of
solute
(M2)
=
200 2
5
×
=
80
g
mol−
1

13. 5. 500
g
of
water
containing
27
g
of
a
non-­‐volatile
solute
will
boil
at
100.156°C.
Calculate
the
molar
mass
of
the
solute.
Given
boiling
point
of
water
=
100°C,
Kb
=
0.52
K
kg
mol−
1
.
Ans.
ΔTb
=
Kb
×
m
;
ΔTb
=
Kb
×
2
2
1
W
M
W
1000
Molar
mass
of
solute
(M2)
=
0.52 27 1000
500 0.156
× ×
×
=
180
g
mol−
1
.
Unit
3
Electrochemistry
One
mark
questions
1. What
is
an
electrolyte?
An
electrolyte
is
a
compound
which
conducts
electricity
either
in
its
aqueous
solution
or
in
its
molten
state.
e.g
Acids
HCl,
CH3COOH,
HNO3
Bases
NaOH,
NH4OH
Salts
CuSO4,
NaCl
etc
2. Define
conductivity
of
an
electrolytic
solution.
Conductivity
of
a
solution
of
an
electrolyte
is
the
conductance
of
a
solution
placed
between
two
electrodes
each
of
one
square
meter
area
kept
at
a
distance
of
1
meter
apart.
3. Write
the
S.I
unit
for
conductivity.
SI
unit
for
conductivity
is
Sm-­‐1
.
4. Give
the
S.I
unit
for
molar
conductivity.
Sm2
mol-­‐1
5. State
Kohlrausch
Law.
The
limiting
molar
conductivity
of
an
electrolyte
can
be
represented
as
the
sum
of
the
individual
contributions
of
the
anion
and
cation
of
the
electrolyte.
6. Define
electrode
potential.
The
potential
difference
developed
between
the
electrode
(metal)
and
the
electrolyte
(solution
containing
its
own
ions)
when
both
the
metal
and
the
solution
are
in
equilibrium
is
called
electrode
potential.
7. Define
standard
electrode
potential.
Standard
electrode
potential
is
the
electrode
potential
when
the
concentrations
of
all
the
species
involved
is
unity
(1M)
and
if
a
gas
is
involved
its
pressure
should
be
1
bar.
8. Write
Nernst
Equation.
⎡ ⎤⎣ ⎦
n+ n+
o
10 n+( M / M) ( M / M)
0.059 1
E = E - log
n M
9. State
Faradays
second
law
of
electrolysis.
The
amounts
of
different
substances
liberated
by
the
same
quantity
of
electricity
passing
through
the
electrolytic
solution
are
proportional
to
their
chemical
equivalent
weights.

14. 10. Define
cell
potential.
Cell
potential
is
the
potential
difference
between
the
two
electrodes
of
the
galvanic
cell.
11. Define
EMF
of
the
cell.
It
is
the
difference
between
the
electrode
potential
of
the
cathode
and
anode
when
no
current
is
drawn
through
the
cell.
12. What
is
Fuel
cell?
Galvanic
cells
that
are
designed
to
convert
the
energy
of
combustion
of
fuels
like
hydrogen,
methane
etc
directly
into
electrical
energy
are
called
fuel
cells.
13. Give
a
method
to
prevent
rusting.
Rusting
may
be
prevented
by
barrier
protection
like
painting,
metal
plating
etc.
14. Write
the
relationship
between
cell
potential
and
Gibb’s
energy
o o
r cellG nFEΔ = −
15. Write
the
relationship
between
equilibrium
constant
and
Eo
cell
0.059
logo
cell c
V
E K
n
=
2
mark
questions
1. What
are
redox
reactions?
Give
an
example.
Reactions
in
which
both
oxidation
and
reduction
taken
place
simultaneously
are
called
redox
reactions.
e.g
⎯⎯→2+ 2+
Zn+Cu Zn +Cu
In
this
Zn
is
oxidised
to
Zn2+
Cu2+
is
reduced
to
Cu
2. Mention
any
two
factors
on
which
the
conductivity
of
an
electronic
conductor
depends.
The
electronic
conductance
depends
on
(i) The
nature
and
structure
of
the
metal
(ii) The
number
of
valence
electrons
per
atom.
(iii)
Temperature
(it
decreases
with
increase
in
the
temperature)
(any
two)
3. Mention
any
two
factors
on
which
the
conductivity
of
an
electrolytic
conductor
depends.
The
conductivity
of
electrolytic
solution
depends
upon
(i)
The
nature
of
the
electrolyte
(ii)
Size
of
the
ions
produced
and
their
solvation.
(iv) The
nature
of
the
solvent
and
its
viscosity.
(iv)
Concentration
of
the
electrolyte
and
(v)
Temperature
(increases
with
increase
in
temperature
(any
two)
4. Give
two
difference
between
the
conductivity
of
an
electronic
conductor
and
electrolytic
conductor.
1.
On
passing
direct
current
composition
of
electronic
conductor
does
not
change
but
that
of
electrolytic
conductor
changes.
2.
On
increasing
the
temperature
in
case
of
electronic
conductor
conductivity
decreases
in
case
of
electrolytic
conductor
conductivity
increases.
5. What
is
a
strong
electrolyte?
Give
an
example.
A
strong
electrolyte
is
an
electrolyte
that
dissociates
completely
into
ions
at
moderate
concentrations
of
its
aqueous
solution
Ex:
acids
HCl,
H2SO4,
HNO3
Base
NaOH,
KOH
Salts
NaCl,
CuSO4
(any
salt)
6. What
is
a
weak
electrolyte?
Give
an
example.
A
weak
electrolyte
is
an
electrolyte
that
dissociates
partially
into
ions
in
its
aqueous
solution.
Ex:
CH3COOH,
NH4OH
7. Define
molar
conductivity.
How
is
it
related
to
conductivity?
Molar
conductivity
of
a
solution
at
a
given
concentration
is
the
conductance
of
the
volume
V
of
a
solution
containing
one
mole
of
electrolyte
kept
between
two
electrodes
with
area
of
cross
section
A
and
distance
of
unit
length.
It
is
represented
by
λm
λm
=
kv
where
k
is
conductivity
and
v
is
volume
of
the
solution
containing
1
mole
of
the
electrolyte
or
If
λm
is
in
Sm2
mol-­‐1
and
k
in
Sm-­‐1
m
k
λ =
1000C
where
C
is
conc.
in
mol
L-­‐1
or
When
λm
is
in
S
cm2
mol-­‐1
and
k
is
in
Scm-­‐1
m
1000k
λ =
C
8. How
does
conductivity
of
a
solution
change
with
change
in
concentration
of
the
solution?
Give
reason.
Conductivity
of
a
solution
decreases
with
decrease
in
concentration
of
the
solution
due
to
decrease
in
the
number
of
ions
per
unit
volume
of
the
solution.

15. 9. Define
limiting
molar
conductivity.
Write
the
relationship
between
molar
conductivity
and
limiting
molar
conductivity.
Limiting
molar
conductivity
is
the
molar
conductivity
of
a
solution
when
concentration
approaches
zero
or
molar
conductivity
at
infinite
dilution.
1
o 2
m mλ = λ - AC
where
λm
is
molar
conductivity
and
λo
m
is
limiting
molar
conductivity,
C
is
concentration
in
mole/L
and
A
is
constant
which
depends
on
nature
of
the
electrolyte,
solvent
and
temperature.
10. Draw
a
graph
of
molar
conductivity
verses
square
root
of
the
molar
concentration
for
KCl
and
CH3COOH
mentioning
clearly
each.
11. How
is
limiting
molar
conductivity
for
a
strong
electrolyte
found
out
by
extrapolation
method?
Prepare
four
solutions
of
given
strong
electrolyte
of
different
concentrations.
Measure
the
conductivities
of
each
solutions
using
conductivity
cell
and
calculate
the
molar
conductivities
of
each
solution.
Plot
a
graph
of
molar
conductivity
verses
square
root
of
the
molar
concentration
for
these
solutions.
A
straight
line
is
obtained
which
is
to
be
extrapolated
back
so
as
to
touch
the
vertical
axes
.This
point
of
intersection
on
the
vertical
axes
gives
the
limiting
molar
conductivity.
12. State
and
illustrate
Faradays
first
law
of
electrolysis.
The
amount
of
chemical
reaction
which
occurs
at
any
electrode
during
electrolysis
by
a
current
is
proportional
to
the
quantity
of
electricity
passed
through
the
electrolyte
either
through
its
aqueous
solution
or
molten
state.
If
w
is
the
mass
of
the
substance
deposited
and
Q
is
the
current
passed
in
coulombs
w
∝
Q
But
Q
=
I
t
where
I
is
the
current
strength
in
ampere
and
t
is
time
in
seconds.
13. Conductivity
of
0.01
M
NaCl
solution
is
0.12
Sm-­‐1
.
Calculate
its
molar
conductivity.
-2 2
m
k 0.12
λ = = =1.2 ×10 Sm / mol
1000C 1000 × 0.01
14.
The
molar
conductivity
of
0.1M
nitric
acid
is
630
S
cm2
/mol.
Calculate
its
conductivity.
m
-1
1000k
λ =
C
1000k
630 =
0.1
630 × 0.1
∴ k = = 0.063 Scm
1000
15. A
solution
of
Ni(NO3)2
is
electrolysed
between
platinum
electrodes
using
a
current
of
5
amperes
for
20
minutes.
What
mass
of
nickel
is
deposited
at
the
cathode?
(Mol
mass
of
Ni
=
58.7)
Q
=
I
t
=
5×20×60
=
6000C
⎯⎯⎯⎯⎯→2+ -
Ni + 2e Ni
2 × 96500C
193000C 58.7g
For
193000C
of
electricity
mass
of
nickel
obtained
=
58.7g
For
6000C
of
electricity
6000 × 58.7
=1.812g
193000
16. How
long
it
will
take
for
the
deposition
of
0.2g
of
silver
when
silver
nitrate
solution
is
electrolysed
using
0.5
ampere
of
current
(Mol
mass
of
Ag
=
108)
⎯⎯→+ -
Ag + e Ag
96500C 108g
For
108g
of
silver
to
be
deposited
current
required
is
96500C.
For
0.2g
of
Ag
But
Q
=
I
t
0.2 × 96500
=178.7C = Q
108
Q 178.7
t = = = 357.4 se
I 0.5
17.
The
cell
in
which
the
following
reaction
occurs
3 2
( ) ( ) ( ) 2( )2 2 2aq aq aq sFe I Fe I+ − +
+ ⎯⎯→ +
Has
Eo
cell
=
0.236V
at
298K.
Calculate
the
standard
Gibb’s
energy
and
the
equilibrium
constant
for
the
cell
reaction.
n
=
2
Δ.Go
=
-­‐nFEo
=
-­‐
2×96500×0.236
=
-­‐
45548
J
0.059
logcellE K
n
=

16.
0.059
0.236 log
2
K=
2 0.236
log 8
0.059
K
×
= =
Taking
the
antilog
K
=
108
18. Write
the
reaction
taking
place
at
cathode
and
anode
when
aqueous
solution
of
copper
sulphate
is
electrolysed
using
copper
electrodes.
2
( ) ( )
2
( ) ( )
t 2
t 2
anode
cathode
oxdn
s aq
redn
aq s
A Cu Cu e
A Cu e Cu
+ −
+ −
⎯⎯⎯→ +
+ ⎯⎯⎯→
Thus
copper
from
anode
dissolves
and
an
equivalent
amount
of
pure
copper
is
deposited
on
cathode.
This
technique
is
used
in
electrolytic
refining
of
crude
copper.
19. Write
the
reaction
taking
place
at
anode
and
cathode
when
molten
NaCl
is
electrolysed.
When
molten
sodium
chloride
is
electrolysed
using
inert
electrodes
22 2t anode oxdn
redn
A Cl Cl e
At cathode Na e Na
− −
+ −
⎯⎯⎯→ +
+ ⎯⎯⎯→
Thus
chlorine
gas
is
liberated
at
anode
and
Sodium
metal
is
formed
at
cathode.
20. Write
the
reaction
taking
place
when
aqueous
solution
of
NaCl
is
electrolysed.
When
aqueous
solution
of
NaCl
is
electrolysed,
2
NaCl Na Cl
H O H OH
+ −
+ −
⎯⎯→ +
+à àÜá àà
The
reaction
taking
place
at
cathode
is
2( )
1
2
aq gH e H+ −
+ ⎯⎯→
The
reaction
taking
place
at
anode
is
⎯⎯→- -
aq 2 aq
1
Cl Cl + e
2
21. What
is
a
primary
battery/cell
?Give
an
example.
Primary
battery
is
one
in
which
reaction
occurs
only
once
and
cannot
be
recharged.
Eg
Dry
cell
or
Leclanche
cell
and
Mercury
cell
22. What
is
a
secondary
battery/cell
?
Give
an
example.
Secondary
battery
is
one
which
can
be
recharged
by
passing
current
through
it
in
opposite
direction,so
that
it
can
be
Reused.
Eg:
Lead
storage
battery
and
Nickel
cadmium
cell.
23. Eo
Cu
=
+0.34V
and
Eo
Zn
=-­‐0.76V.
Daniel
cell
is
obtained
by
coupling
these
two
electrodes.
(i)
represent
the
cell
symbolically
(ii)
calculate
the
EMF
of
the
cell
(i)
Daniel
cell
can
be
represented
as
Zn/
Zn2+
(aq)
||
Cu2+
(aq)
/Cu
(ii)
EMF
of
Daniel
cell
Eo
cell
=
Eo
R
-­‐
Eo
L
=
Eo
Cu
-­‐
Eo
Zn
=
0.34-­‐(-­‐0.76)
=
1.10V
24. Calculate
the
molar
conductivity
of
a
solution
of
MgCl2
at
infinite
dilution
given
that
the
molar
ionic
conductivities
of
2+ -
o 2 -1 o 2 -1
( Mg ) ( Cl )
λ =106.1 Scm mol and λ = 76.3 Scm mol
2+ -
2
o o o
MgCl Mg Cl
2 -1
λ = λ + 2λ
=106.1+ 2( 76.3)
= 258.7 Scm mol
25. The
resistance
of
a
conductivity
cell
containing
0.001
M
KCl
solution
at
298K
is
1500Ω.
What
is
the
cell
constant
if
the
conductivity
of
0.001M
KCl
solution
at
298K
is
0.146×10-­‐3
Scm-­‐1
?
Cell
constant
G*=
Rk
=resistance
×
conductivity
=0.146×10-­‐3
Scm-­‐1
×1500S-­‐1
=
0.219
cm-­‐1
Question
carrying
3
or
4
marks
1. Explain
the
construction
of
Daniel
cell.
Write
the
reaction
taking
place
at
anode
and
cathode
and
the
net
cell
reaction.
(3
mark)
To
prepare
Daniel
cell
get
a
zinc
electrode
by
dipping
zinc
rod
in
1M
ZnSO4
solution.
Get
a
copper
electrode
by
dipping
a
copper
plate
in
1
M
CuSO4
solution.
Couple
these
two
electrodes
using
a
salt
bridge
to
get
Daniel
cell.
Reactions
taking
place
2
2
2 2
( ) ( ) ( ) ( )
t anode 2
cathode 2
oxdn
redn
s aq aq s
A Zn Zn e
At Cu e Cu
Net cell reaction Zn Cu Zn Cu
+
−
+ −
+ +
⎯⎯⎯→ +
+ ⎯⎯⎯→
+ ⎯⎯→ +

17. 2. With
a
labeled
digram
explain
standar
hydrogen
electrode.
Represent
it
symbolically.
Write
the
reduction
reaction
at
the
anode.
What
is
its
electrode
potential?
(4
marks)
It
consists
of
a
platinum
electrode
coated
with
platinum
black.
The
electrode
is
dipped
in
1M
HCl.
Pure
hydrogen
gas
is
bubbled
through
it
under
a
pressure
of
1
bar.
S.H.E
is
represented
as
Pt(s)
|H2
(g)(1bar)
|H+
(aq)(1M)
The
reduction
reaction
taking
place
is
2
1
( ) ( )
2
H aq e H g+ −
+ ⎯⎯→
S.H.E
is
assigned
an
electrode
potential
of
0.0
V
at
all
temperatures.
3. Explain
the
use
of
standard
hydrogen
electrode
in
measuring
the
standard
electrode
potentials
of
copper
and
zinc
electrode
(4
mark)
Construct
a
standard
electrode
of
the
given
metal
by
dipping
the
pure
metal
in
1M
solution
of
its
own
ion
at
25o
C
Couple
this
standard
electrode
with
SHE
using
a
salt
bridge
to
get
galvanic
cell.
Measure
the
emf
of
the
cell
using
suitable
instrument
like
potentiometer.
Eo
=
Eo
R
–
Eo
L
One
of
the
electrodes
of
the
cell
is
SHE
and
its
electrode
potential
is
0.0V.
So
the
electrode
potential
of
the
given
electrode
will
be
the
emf
of
the
cell
in
magnitude.
If
reduction
takes
place
at
the
given
electrode
its
Eo
will
be
+ve
but
if
oxidation
takes
place
at
the
given
electrode
is
Eo
will
be
–ve.
e.g
if
SHE
is
coupled
with
standard
copper
electrode
reduction
takes
place
at
copper
electrode
cell
can
be
represented
as
Pt
(s)
|H2(g.
1bar)|H+
(aq1M)||Cu2+
(aq.1M)|Cu
2+ +
2
o o o
cell Cu / Cu H / H
E = E - E
2+ 2+
o o
Cu / Cu Cu / Cu
0.34 = E - 0 ∴E = 0.34V
If
SHE
coupled
with
standard
zinc
electrode
oxidation
takes
place
at
zinc
electrode.
Cell
can
be
represented
as
2
( .1 ) ( ) ( ) 2( .1 ) ( .1 )/aq M s s g bar aq MZn Zn Pt H H+ +
⏐⏐ ⏐ ⏐
2
2/ /
o o o
H H Zn Zn
E cell E E+ += −
2 2 //
0.76 0 0.76o
ZnZn Zn Zn
E E V+ += − ∴ = −
4. How
is
Kohlrausch
law
helpful
in
finding
out
the
limiting
molar
conductivity
of
a
weak
electrolyte?
(3
m)
Let
us
try
to
calculate
λo
m
for
a
weak
electrolyte
CH3COOH.
Select
three
strong
electrolytes
whose
λo
m
can
be
found
by
extrapolation
method
in
such
a
way
that
if
we
subtract
λo
m
for
one
electrolyte
from
the
sum
of
λo
ms
of
the
remaining
two
electrolyte
λo
m
for
CH3COOH
can
be
obtained.
The
three
electrolytes
to
be
selected
are
CH3COONa,
HCl
&
NaCl
3 3
o o o o
CH COOH CH COONa HCl NaClλ = λ + λ - λ
5. The
values
of
limiting
molar
conductivities
(λo
m)
for
NH4Cl,
NaOH
and
NaCl
are
respectively
149.74;
248.1
and
126.4
Scm2
mol-­‐1
.
Calculate
the
limiting
molar
conductivity
of
NH4OH
(3M)
4 4
o o o o
NH OH NH Cl NaOH NaClλ = λ + λ - λ
=
149.74+248.1-­‐126.4
=
271.44
Scm2
mol-­‐1
6. Calculate
the
equilibrium
constant
for
the
reaction
at
298K
2
( ) ( ) ( )2 ( ) 2s aq sCu Ag aq Cu Ag+ +
+ ⎯⎯→ +
Given
that
Eo
Ag+
/Ag
=
0.80V
and
Eo
(Cu2+
/Cu)
=
0.34V
0.059
logo
cell cE K
n
=
log
0.059
o
c
nE cell
K∴ =
2
( / ) ( / )
o o o
cell Ag Ag Cu Cu
E E E+ += −
=0.80-­‐0.34=0.46V
2 0.46
log 15.59
0.059
cK
×
= =
Taking
the
antilog
Kc
=3.92×1015
7. In
Leclanche
cell
(dry
cell)
what
are
anode
and
cathode?
What
is
the
electrolyte
used?
Write
the
reactions
at
each
electrode.
What
is
the
role
of
zinc
chloride?
It
consists
of
a
zinc
container
as
an
anode.
A
graphite
rod
surrounded
by
a
mixture
of
manganese
dioxide
and
carbon
powder
is
cathode.
The
space
between
the
electrodes
is
filled
with
electrolyte
a
moist
paste
of
ammonium
chloride
and
zinc
chloride

18. Reaction
taking
place
⎯⎯→ 2+ -
( s)At anode Zn Zn + 2e
⎯⎯→+ -
2 4 3At cathode MnO + NH + e MnO( OH) + NH
NH3
produced
in
the
reaction
forms
a
complex
with
Zn2+
to
form
[Zn(NH3)4]2+
.
8. What
are
the
anode
and
cathode
of
lead
acid
battery?
What
is
the
electrolyte?
Write
the
reactions
taking
place
at
anode
and
cathode
and
the
overall
reaction
during
discharging
of
the
battery.
(3
M)
It
consists
of
lead
anode
and
a
grid
of
lead
packed
with
lead
dioxide
(PbO2)
as
cathode.
Electrolyte
is
38%
solution
of
sulphuric
acid.
The
reactions
taking
place
when
the
battery
is
in
use
are
2
( ) 4 ( ) 4( )
2
2 4 ( ) ( ) 4 ( ) 2 ( )
2
( ) 4 2 2
s aq s
aq aq s l
Anode Pb SO PbSO e
Cathode PbO s SO H e PbSO H O
− −
− + −
+ ⎯⎯→ +
+ + + ⎯⎯→ +
The
overall
reaction
is
( ) 2 ( ) 2 4 ( ) 4 ( ) 2 ( )2 2 2s s aq s lPb PbO H SO PbSO H O+ + ⎯⎯→ +
9. In
Hydrogen
oxygen
fuel
cell
(i)
Draw
the
schematic
diagram
mentioning
the
anode
and
cathode.
What
is
the
electrolyte?
Write
the
reaction
taking
place
at
each
electrodes
and
the
net
cell
reaction.
(4M)
In
this
hydrogen
and
oxygen
gases
are
bubbled
through
porous
carbon
electrodes
into
concentrated
aqueous
sodium
hydroxide
solution.
Catalyst
like
finely
divided
platinum
or
palladium
is
incorporated
into
the
electrodes
for
increasing
the
rate
of
electrode
reaction
Reaction
taking
place
are
⎯⎯→
⎯⎯→
- -
2 ( g) 2 ( l) ( aq)
- -
2( g) ( aq) 2 ( l)
Cathode O + 2H O + 4e 4OH
Anode 2H + 4OH 4H O + 4e
Overall
reaction
is
⎯⎯→2 ( g) 2 ( g) 2 ( l)2H +O 2H O
10. What
is
corrosion?
During
rusting
of
iron
write
the
anodic
and
cathodic
reactions.
Give
the
composition
of
rust.
(3M)
When
a
metal
is
exposed
to
the
atmosphere
it
is
slowly
attacked
by
the
constituents
of
the
environment
as
a
result
of
which
the
metal
is
slowly
lost
in
the
form
of
its
compound
.
This
is
called
corrosion.
Reaction
taking
place
are
⎯⎯→
⎯⎯→
2+ -
( s)
-
2 ( g) 2 ( l)
At Anode 2Fe 2Fe + 4e
At Cathode O + 4H +( aq)+ 4e 2H O
H+
are
produced
from
H2CO3
formed
due
to
dissolution
of
carbon
dioxide
from
air
into
water
The
Fe2+
ions
are
further
oxidised
by
atmospheric
oxygen
to
ferric
ion
which
are
ultimately
converted
to
hydrated
ferric
oxide
called
rust.
Composition
of
rust
is
(Fe2O3.xH2O).
11. A
conductivity
cell
when
filled
with
0.01M
KCl
has
a
resistance
of
747.5
ohm
at
25o
C.
When
the
same
cell
was
filled
with
an
aqueous
solution
of
0.05M
CaCl2
solution
the
resistance
was
876
ohm.
Calculate
(i)
Conductivity
of
the
solution
(ii
)Molar
conductivity
of
the
solution
(given
conductivity
of
0.01M
KCl
=
0.14114
sm-­‐1
)
(3M)
Cell
constant
G*
=
Rk
=
747.5×0.14114
=0.105.5m-­‐1
-1
-1cell constant 105.5m
Conductivity k = = = 0.1204Sm
R 876 ohm
2 -1
m
k 0.1204
Molar conductivity λ = = = 0.00241sm mol
1000C 1000 × 0.05
12. The
electrical
resistance
of
a
column
of
0.05M
NaOH
solution
of
diameter
1cm
and
length
50cm
is
5.55×103
ohm.
Calculate
its
(i)
resistivity
(ii)
conductivity
(iii)
molar
conductivity
(3M)
Cell
constant
o l
G =
a
l
=
50
cm
Diameter
=
1
cm
∴
radius
=
0.5
cm
Area
of
cross
section
A
=
πr2
=
3.14×(0.5)2
=
0.785
cm3
* -150
G = = 63.694 cm
0.785

19. -2
1 1
Resistivity ρ = = = 87.135 Ω
k 1.148 ×10
m
-2
2 -1
1000k
Molarconductivityλ =
C
1000 ×1.148 ×10
=
0.05
= 229.6 S cm mol
13. Calculate
the
emf
of
the
cell
in
which
the
following
reaction
takes
place.
2
( ) ( )2 (0.002 ) (0.160 ) 2s sNi Ag M Ni M Ag+ +
+ ⎯⎯→ +
Given
that
Eo
cell
=
1.05V
2 2
( )
10 2
( )
[ ][ ]0.059
log
2 [ ][ ]
so
cell cell
s
Ni Ag
E E
Ni Ag
+
+
= −
But
[M]
for
any
element
is
taken
as
unity
⎡ ⎤⎣ ⎦
⎡ ⎤⎣ ⎦
2+
o
cell cell 10 2+
Ni0.059
E = E - log
2 Ag
( )
2
0.059 0.160
= 1.05 - log
2 0.002
=
0.914V
Unit-4
CHEMICAL KINETICS
Number of Hours of Teaching-9
Marks allotted-8
In part-D, 5mark question is split preferable in the form of 3+2
Definition :- The branch of chemistry which deals with study of reaction rate and their
mechanism is called chemical kinetics
Rate of a chemical reaction
Q. 1 What is rate of reaction? (1m)
Ans: Change in molar concentration of reactant or product in per unit time is called rate of
reaction.
Types of rate of reactions
For reaction R à P
Average rate = decrease in conc. R
Of reaction time taken
rav = - ∆[R]
∆t
Average rate = increase in conc. Of P
Reaction time taken
rav = + ∆[P]
∆t
8
Marks
Part –A
1x1=1
Part –B
1x2=2
Part-C
1x5=5

20. Q2:- For the reaction RàP, the conc. of reactant changes from 0.03M to 0.02M in 25 min.
calculate average rate of the reaction using the unit of time in seconds.
rav= - ∆[R]= - (0.02-0.03 )
∆t 25x60
=-[-0.01]
1500
= 6.66x10-6
M/s
Q3: What is the SI Unit of rate of reaction ? (1m)
Ans: Mol /L /s
Factors influencing Rate of reaction
Q4 :- Mention any two factors which influence the rate of reaction . 2M
Ans 1) Pressure or conc. of reactants
2) temperature
3) catalyst.
Dependence of rate on concentration .
Q5.) What is rate law ? (1 m)
Ans: Representation of rate of reaction in terms of concentration of reactants is called rate law.
Rate expression and rate equation
Q.6) Define rate equation or rate expression (2m)
Ans: Expression in which reaction rate is given in terms of molar conc. of reactants with each
term raised to some power which may or may not be same as the stoichiometric coefficient of
the reacting species in a balanced chemical equation.
Q.7) Define rate constant of a reaction. (1m)
Ans: Rate constant is equal to rate of reaction when the product of the molar conc. of
reactants is unity.
Order of a Reaction
Q.8) Define order of a reaction. 1M
Ans: Sum of the powers of the concentration of the reactants in the rate equation is called order
of reaction.
Q.9) Calculate the overall order of a reaction which has the rate expression. 1M
Rate= K [A]1/2
[B]3/2
Ans: Order of reaction = 1/2 + 3/2
= 2
Q.10) What is elementary reaction ? (1m)
Ans: Reaction taking place in one step is called elementary reaction.
Q.11)What are complex reactions?(1m)
Ans: Reactions taking place in more than one step are called complex reaction.
Q.12) What is SI Unit of rate constant of nth order reaction ? (1m)
Ans: (mol)1-n
. Ln-1.s-1
Q.13) What is SI unit of rate constant of zero order reaction? (1m)
Ans: Mol/L/s
Q.14) What is the order of reaction whose unit of rate constant and rate of reaction are same ?
(1m)
Ans: Zero order.
Q.15) Identify the reaction order from the rate constant K=2.3x10-5
mol-1
.L.S-1
(1m)
Ans: Comparing the unit of rate constant with general unit
Mol-1
.L.S-1
with ( Mol)1-n
.Ln-1
.S-1
1-n= -1
n=2
Molecularity of a reaction
Q.16) Define molecularity of a reaction . (1 m)
Ans: The number of reacting species taking part in an elementary reaction which must colloid
simultaneously in order to bring about a chemical reaction is called molecularity of reaction.
Q.17) In a complex reaction which step controls the overall rate of reaction and
what is it called? (2m)
Ans: Slowest step, which is called rate determining step.
Q.18) The conversion of molecules X to Y follows second order kinetics .If conc. of X
Increased to three times,how will it affect the rate of formation of Y ? (1m)
Ans: Increased rate =( Increased conc.)n
=32
=9
Rate of formation of Y increases
by 9 times
Integrated rate equations
Q.19) Derive rate constant of zero order reaction (3m)
Ans: Consider a zero order reaction R--> P
Rate =-d[R] = K[R]o
dt
= - d[R] = K
dt
= d[R] = -kdt -------(1)

21. Integrating equation (1) both sides
[R]= -kt+I ----------(2)
Where “I” is integration constant
At t=0 [R]=[R]o where
[R]o is initial concentration of reactant.
∴Eqn (2) becomes
I=[R]o
Substituting I in eqn-------- (2)
[R] = -Kt + [R]o
-Kt = [R]-[R]o
Kt = [R]o-[R]
K = [R]o-[R]
t
Q:20) Derive integrated rate equation for first order reaction? (4m)
Ans-Consider a first order reaction.
RàP
Rate = - d[R] = K[R]
dt
d[R]= - K[R]
dt
d[R] = - K. dt. -------(1)
[R]
Integrating eqn.(1)on both side
ln [R] = - Kt + I --------(2)
Where “I” is integration constant
At t=o [R]=[R]o which called initial Concentration reactant
Substituting the values in
ln[R]o= I
Equation (2) can be written as
ln[R] = -Kt + ln[R]o
Kt = ln[R]o – ln [R]
Kt = ln [R]o
[R]
Kt =2.303 log[R]o
[R]
K= 2.303 x log[R]o
t [R]
Log [R]o Slope = K
[R] 2.303.
Or
o time K= 2.303xSlope
Half life of a reaction
Q21)Define halfe life of a reaction . (1m)
Ans: The time in which the conc.of a reactant is reduced to one half of its initial conc. is called
half life of a reaction (t1/2)
Q:22) Show that half life of a zero order reaction is directly proportional to initial
concentration of reactant from integrated rate equation.
OR
Derive the relation between half life and rate constant of zero order reaction .(2m).
Ans:-Rate constant of zero order reaction is
K= [R]o – [R]
t
At half life t =t ½ & [R] = ½ [R]o
.: K= [R]o – ½ [R]o
t1/2
K=[R]o
2t½
t ½ =[R]o
2 K
OR
t ½ ∝ [R]o
Q.23) Show that half life of a first order reaction is independent of initial Conc. of reactant
from integrated rate equation (2m)
Or
Derive the relation between half life of a first order reaction and its rate constant . (2m)
Ans: Rate constant of first order reaction is
K= 2.303 x log [R]o
t [R]
At half life t=t ½ ,[R]=[R]o
2
.: K = 2.303 x log [R]o
t½ [R]o/2
K =2.303 x log 2
t ½
K= 2.303 x0.3010
t½
t½ = 0.693
K
K= 2.303 x0.3010
t½
t½ = 0.693
K

22. Q.24)A first order reaction is found to have a rate constant 5.5x10-14
/s .Calculate the half life
of the reaction (2m)
Soln. : K= 5.5x10-14
/s t½ =?
t½ = 0.693
K
= 0.693
5.5x10-14
t½ = 1.26x1013
sec
Q:25) Show that the time required for 99/. Completion of a first order reaction is twice the
time required for the completion of 90% of reaction( 4m)
I set : [R]o= 100, [R]=[100-90]=10 t=t90%
IIset : [R]o= 100 [R]= [100-99]1 t=t99%
To be proved t99%= 2t90%
K= 2.303 x log [R]o
t [R]
Sub. I set values .
K= 2.303x log 100
t 90% 10
K = 2.303x log 10
t90%
K= 2.303 X 1 - (1)
t90%
Substituting II set values
K=2.303 x log 100
t99% 1
K= 2.303 x 2 --------------(2)
t99%
Comparing equations (1) & (2)
2.303x 1 = 2.303x2
t90% t99%
t99% = 2t90%
Pseudo first order reaction
Q:26 Define pseudo first order reaction . Give an example. (2m)
Ans: Chemical reactions which are not first order but behave as fist order reaction under
suitable conditions are called pseudo first order Reactions. Ex: Inversion of cane sugar.
C12 H22O11+H2O àC6H12O6 + C6H12 O6
Temperature dependence of the rate of a reaction
Q:27)How does rate of reaction vary with temperature? (1m).
Ans: Rate of reaction increases with increase of temperature.
Q:28) What happens to the rate constant of a reaction when temperature is increased by 10o
.?
Ans: Rate constant increases nearly by two times.
Q.29) Write Arrhenius equation which relates the rate constant , activation energy and
temperature . (1m)
Ans K= A e-Ea/RT
Energy of activation
Q.30) Define energy of activation (1m).
Ans: The minimum energy required for the reactants to form activated complex is called
Activation energy.
Q.31) How is activation energy related to rate of reaction? (1m)
Ans: Rate of reaction is inversely proportional to activation energy.
ie r ∝ 1
Ea
Q.32) How is activation energy affected by presence of positive catalyst? (1m)
Ans: Activation energy of a reaction decreases in presence of catalyst.
Q.33) On increasing 100
K temperature rate of reaction becomes double,
explain from the max well Boltzmann distribution curve. (2m)
On increasing 100
K temperature, substance Increases the fraction of molecules double,hence
rate of reaction doubles.
Q:34) How does positive catalyst increases the rate of reaction? (2m)
Ans positive catalyst decreases the activation energy by changing the
Path of the reaction,which increases the rate of reaction

23. Collision theory of chemical reactions.
Q:35) What is effective collision? How is it related to rate of reaction? (2m).
Ans. Collision in which molecules colloid with sufficient kinetic energy and proper orientation
so as to form products is called effective collision. It is directly proportional to the rate of
reaction.
Q:36)How is activation energy calculated by plotting graph ln K against 1/T ? (2m)
Q:37)Write Arrhenius equation at different Temperature and rate constants.
Ans: log K2/K1 = Ea X T2-T1
2.303RT T1 T2
Q:38)The rate constants of a reaction at 500K. and 700K are 0.02s-1
and 0.07s-1
respeetively
calculate the activation energy. (3m)
Ans: log K2/K1 = Ea X T2-T1
2.303RT T1 T2
log 0.07 = Ea x 700- 500
0.02 2.303x8.314 500x 700
0.544 = Ea x 5.714 x10-4
19.15
Ea= 0.544x19.15
5.714 x10-4
Ea= 18230.8 J = Ea= 18. 2308 KJ.
UNIT -5
SURFACE CHEMISTRY
A. Short answer questions carrying 1 mark
1. What is adsorption
A surface phenomenon wherein there is accumulation of molecules on the surface
(than in the bulk) of a solid or a liquid.
2. Why solids in finely divided state are good adsorbent?
Solids in finely divided state have large surface area, as surface area increases
adsorbing power increases.
3. What is desorption?
The process of removing an adsorbed substance from a surface on which it is
adsorbed is called desorption.
4. Name the substance used to decolour the solution of raw sugar.
Animal charcoal.
5. Name of the phenomenon in which both the adsorption and desorption takes
place simultaneously.
Sorption
6. Why is adsorption always exothermic?
During adsorption there is always decrease in residual forces on the surface, hence
adsorption is always exothermic.
Or
There is decrease in surface energy which appears as heat, hence adsorption is always
exothermic.
7. Name catalyst used in the conversion of alcohols into gasoline (petrol)
Zeolite ZSM-5 (Zeolite Sieve of molecular porosity-5)
8. Name the colloidal system in which dispersed phase is solid and dispersion
medium is liquid
Sol
9. Name the dispersed phase in gel
Liquid
10. Give an example for oil in water emulsion
Milk, Vanishing cream

24. 11. What type of colloidal emulsion is present in butter
Water in oil (W/O)
12. What is the dispersion medium in gel?
Solid
13. Between Na2SO4 and Na3PO4 which has greater power to coagulate a positively
charged colloid?
Na3PO4
14. Alum is added to muddy drinking water. Why?
Alum is added to muddy drinking water to coagulate
15. What is the dispersed phase in milk?
Oil or liquid
16. A liquid is dispersed in a gas. Name the type of colloid obtained.
Liquid aerosal
17. Name the instrument designed by Zigmondy.
Ultramicroscope
18. Movement of the dispersion medium in an electric field by preventing the
movement of colloidal particles by suitable method.Name the phenomenon
Electroosmosis
19. The process by which colloidal particles aggregate, become bigger and settle
down. Name the phenomenon
Coagulation
20.What happens when an electrolyte is added to lyophobic sol?
Coagulation or precipitation
21. Name the phenomenon, when an electrolyte having a common ion is added to
freshly prepared precipitate?
Peptization
B. Answer questions carrying 2 marks
ADSORPTION
1. What are adsorbate and adsorbent? Give an example.
Molecules (substances) that accumulates on the surface is called adsorbate.
The material on the surface of which adsorption takes place is called adsorbent.
Example: Ni adsorbs H2. Ni is the adsorbate, H2 is the adsorbent
2. Give two examples for adsorption.
i) When animal charcoal is added to methylene blue, charcoal adsorbs the dye.
ii) Air becomes dry in the presence of silica gel because silica gel adsorbs water
molecules on the surface
iii) A small pillow of silica gel in a box adsorbs moisture in the box keeps the air
dry. (Any two)
3. Give differences between adsorption and absorption.
Adsorption Absorption
1. A substance gets
concentrated on the surface
of a solid or liquid.
2. It increases with increase in
surface area.
Example: adsorption of water
by silica gel.
A substance gets uniformly distributed
through the bulk of solid or liquid.
It remains unaffected by increase in surface
area.
Example: Absorption of water by anhydrous
CaCl2.
4. Of SO2 (critical temperature 630K) and CH4 (critical temperature 190K) which
gas will be adsorbed readily on the surface of 1 gram of activated charcoal.
Justify the answer.
SO2 gas
Easily liquefiable gases with higher critical temperature are readily adsorbed as the
theVander Waal’s forces are stronger near critical temperature.
5. What is the effect of temperature on physical and chemical adsorption?
Physical adsorption decreases with increase in temperature. Chemical adsorption
increases with increase intemperature.
6. Mention any two applications of adsorption.
i) In the production of high vaccum
ii) In gas mask, to adsorb poisonous gases
iii) In the separation of noble gases using activated charcoal
iv) Removal of colouring matter from solutions
v) In adsorption chromatography to analyse a given

25. CATALYSIS
1. What is catalysis? Give an example.
A substance that accelerates the rate of a reaction without itself remaining unchanged
chemically and quantitatively is a catalyst. The phenomenon is catalysis.
E.g.: 2KClO3
2MnO
⎯⎯⎯→2KCl + 3O2
MnO2 is a catalyst.
2. What are promoters and poisons with respect to a catalytic process?
Promoters are substance that increases the activity of a catalyst. E.g.: In Haber’s
process molybdenum acts as a promoter for iron used as a catalyst. A catalytic poison
is one that decreases the efficiency or activity of a catalyst. E.g.: In Haber’s process
CO if present in the mixture of H2 and N2, poisons the iron catalyst.
3. What is homogeneous catalysis? Give an example.
When reactants and catalyst are in the same phase the process is homogeneous
catalysis. E.g.:
a) 2SO2(g) + O2(g)
( )gNO
⎯⎯⎯→ 2SO3(g)
Here the reactants (SO2 and O2) and catalyst (NO) are all gases.
b) Acid hydrolysis of cane sugar is also an example for homogeneous catalysis. Here
the reactants sugar solution, water and the catalyst dil. HCl are in the same phase
(aqueous solution)
C12H22O11(aq) + H2O(l)
H+
⎯⎯→C6H12O6 + C6H12O6 (both are in aq solution)
Sucrose glucose fructose
4. What is heterogenous catalysis? Give an example.
A catalytic process in which reactants and catalyst are in different phases are known as
heterogenous catalysis.
E.g.: 1. N2(g) + 3H2(g)
(s)Fe
⎯⎯⎯→2NH3
Here the reactants are gases, catalyst iron is a solid
2. Vegetable oil (l) + H2(g)
( )sNi
⎯⎯⎯→Vanaspathi ghee
Here reactants and catalyst are in different phases.
5. Write a note on
a) activity b) selectivity of solid catalysts.
a) Activity: The activity (efficiency) of a solid catalyst depends on how strongly the
reactants are chemisorbed on it. It is found that elements (metals) in group 7-9 of
the periodic table show greater catalytic activity for hydrogenation reactions.
E.g.: 2H2(g) + O2(g)
Pt
⎯⎯→2H2O (l)
b) Selectivity: For a given set of reactants, different catalyst may yield different
products. This is selectivity of a catalyst.
E.g.: CO(g) + 3H2(g)
Ni
⎯⎯→CH4(g) + H2O(g)
CO(g) + H2(g)
Cu
⎯⎯→H−CHO
Ni is selective to convert water gas to CH4 whereas Cu converts water gas into
formaldehyde. In otherwords Ni catalyses the conversion of water gas to CH4 but
cannot catalyse to convert water gas to formaldehyde.
Catalyst is highly selective in nature i.e a given substance can act as a catalyst only in
a particular reaction and not for all the reactions.
6. What is shape selective catalysis? Give an example.
A catalytic reaction that depends on pore structure of the catalyst and size of the
reactant and product molecules is called shape selective catalysis. E.g.: zeolites.
7. Write a note on zeolites as shape selective catalysts.
Zeolites are aluminosilicates with 3D nework of Al-O-Si frame with honey comb like
structure. This structure makes them to act as shape selective catalyst depending on
pore size in them and on the size of reactant and products. Many zeolites are
synthesized for selective catalytic activity.
E.g.: 1) Zeolite ZSM-5 (Zeolite Sieve of molecular porosity- 5) converts alcohols
into gasoline (petrol) by dehydrating alcohols.
2) Many zeolites are used in petroleum industry in cracking of hydrocarbons
and in isomerisation.
8. What are enzyme catalysis or biochemical catalysis? Give an example for
enzyme catalysis.
Enzymes are proteins, which catalyse large number of reactions that maintain life
processes in both plants and animals. Hence they are biochemical catalysts and the
phenomenon is called as biochemical catalysis.
Inversion of cane sugar in the presence of enzyme invertase into glucose and fructose
9. Give two examples for enzyme-catalysed reaction.
a) Conversion of starch into maltose
2(C6H10O5)n (aq) + nH2O (l) diastase enzyme
⎯⎯⎯⎯⎯→nC12H22O11 (aq)
Starch maltose
b) Urea into ammonia and carbon dioxide
NH2CONH2(aq) + H2O (l)
urease
⎯⎯⎯→2NH3(g) + CO2(g)
c) In human beings enzyme pepsin converts proteins into peptides and pancreatic
trypsin enzyme converts proteins into amino acids.