CLASS XI FINAL TERM EXAMINATION
2012
1ST SHIFT
Chemistry
Paper II
Time allowed:
2 hrs 20 minutes
Max. Marks 55
(Total 8 marks)
Q.1 a) Calculate
the Percentage yield of the reaction when the 6 g hydrogen react with 96g oxygen
to form 90 g of water . (2 mark)
Total mass of reactant = 6g +96g
=
102 g
Actual yield =90g
Percentage yield = 90g ÷102 x 100
= 88.2
(b) Determine
the molecular formula the compound which contain 83.3% of Carbon and
17.7% of Hydrogen. The molecular formula mass is 58 g. (2 mark)
Element
|
Carbon
|
Hydrogen
|
%
|
83.3
|
17.7
|
Mole ratio
|
83.3 ÷12=6.9
|
17.7÷1-17.7
|
Simplest mole ratio
|
1
|
2.46
|
Empirical ratio
|
2
|
5
|
n =molecular formula mass ÷
empirical formula mass
58÷29
=2
Molecular formula =(Empirical formoula) x n
= (C2H5)2
= C4H10
(c) Desicrbe the
Mosely’s experiment with reference to production of X-rays. What are the
types of X-rays (2
mark)
Reference: From Punjab Text book
(d) Differentiate
between Orbit and Orbital of an atom. (1
mark)
e) State
and explain the Hund’s Rule
(1 mark)
According to this rule “When the orbitals of same energy levels are
available, then electrons are distributed in orbitals in such a way as to give
the maximum numbers of unpaired electrons . Only when the orbital are
separately occupied then the pairing of electrons commences”.
This rule explain the filling of electrons in degenerate (having same
energy) orbitals like p,d,f. In simple, it state that the electrons remain
unpaired as far as possible. i.e. If there are available orbitals of equal
energy to the electrons, the electrons would lie in separate orbitals and have
same spin rather than to lie in the same orbital and have paired spin.
 |
 |
||||
 |
|||||
e.g. (N(Z=7) = 1s2, 2s2,2Px
, 2Py , 2Pz
is true |
and
not 1s2, 2s2,2Px,
2Py, 2Pz
Q.2 (Total
8 marks)
a) Define
Dipole Moment. How it explain the Percentage
ionic character in covalent bond in the diatomic molecules.
(2 mark)
DEFINITION:
“The quantitative
measurement of concentration of polarity of Charge of in a molecule is known as dipole moment or it is
the tendency of a molecule to orient in an electric field”
The dipole moment
measures the concentration of positive and negative charges in different part
of the molecule and is equal to the product of ionic charge and distance
between the center of positive and negative charges.
Here q = charge on molecule
d = distance between the center of
positive and negative charges.
Dipole
moment is represented by (→) along with (-) and the direction of arrow is
towards high electronegative atom in the molecule.
EXPLAINTION:
Dipole
moment is expressed in Debye after the name of introducer but in S.I system the
unit of dipole moment is C.m(coulomb meter).
The relation between debye and C.m is given below
1
Debye = 10-18 esu X cm = 3.335 x 10-30 sm
Dipole moment is
a measurement of degree of polarity of the molecule greater the value of dipole
moment, more polar will be the molecule. Does by knowing the value of dipole
moment
(b) Define the
Bond order . Determine the bond order of the He. (2 mark)
Bond
order = ___2-2_______________
= ___________________
(c) Define the hybridization. Draw the shape of BF3 on the basis
of Hybridization theory.
Hybridization: (2
mark)
The hypothetical process of
mixing of different atomic orbital to produce the same number of equivalent
orbital Having same shape and energy is
known as hybridization, the orbital so formed are called hybrid orbital”
d) Explain absolute Zero on the basis of
Charles’ Law (2 Marks)
It is Zero Kelvin temperature at which the
volume , pressure and temperatures is
zero . According
Charle’s “volume
is directly proportional to temperature at constant pressure” this statement only
applicable to absolute temperature Or temperature in Kelvin scale which start from Absolute Zero because at 0oC
temperature of gas the volume of is greater than zero.
Viscosty:
“The internal resistance to the flow of
a liquid is called its viscosity”
It
is common observation that some liquids flow more readily than the other. For
example water moves over a glass plate more quickly than glycerine. Similarly,
honey and mobil oil flow more slowly than water. Hence, liquids which flow
easily are called “MOBILE
Surface Tension:
“The inter-molecular force that drawn
the molecules on the surface of a liquid together causing the surface to act
like a thin elastic skin, this phenomenon is called SURFACE TENSION”.
OR
“The force per unit length or energy per
unit area of the surface of a liquid is called SURFACE TENSION”.
Surface tension of the liquid is
represented by g ORs, and its units are dynes
/ cm OR erg / cm2
b) Differentiate between Crystalline solids and amorphous solids (2 Marks)
Crystalline Solid
|
Amorphous Solid
|
Geometry
|
|
Particles
of crystalline solids are arranged in an orderly three dimensional network
called crystal, hence they have definite shape.
|
Particles
of amorphous solids are not arranged in a definite pattern, hence they do not
have a definite shape.
|
Melting Point
|
|
Crystalline
solids have sharp melting point, this is because attractive forces between
particles long range and uniform.
|
Amorphous
solids melt over a wide range of temperature i.e. they do not have sharp
melting point, because the inter-molecular forces vary from place to place
|
Cleavage
|
|
The
breakage of a big crystal into smaller crystals of identical shape is called
cleavage.
|
Amorphous
solids do not break down at fixed cleavage planes.
|
c) Apply the Le-Chateleir principle on to increase the production of
ammonia (4 mark)
N2
+ 3H2 ↔ 2NH3
∆H= Negative
i)
Effect of Temperature :
Forward direction is an exothermic reaction i.e. heat is evolved
during the reaction. According to Le-Chatelier’s principle, the law temperature
shifts the equilibrium to the right ion i.e.
ii ) Effect
of Pressure:
Volume of products is less than reactants. The increase in pressure Equilibrium
position moves towards right i.e. more NH3 is formed
iii) Effect of concentration:
According to Le-Chatelier’s Principle, the
addition of more N2 or H2 or both will move the reaction
to the right, thus more NH3 gas will be produced till the
equilibrium state is reached again.
Similarly addition of NH3 at the
equilibrium state will move the reaction to the left. Thus, a part of NH3
gas will decompose into N2 & H2 gases in order to
reach the equilibrium state again.
iv)
Effect of Catalyst :
There is no effect of
catalyst on reversible reaction at Equilibrium state
Q.4. (Total
8 marks)
a) Calculate the PH
of 0.25 M NH4Cl (4
Marks)
NH4+
+ H2O ↔ NH3 + H3O+
The Dissociation
constant Kb is 1.76 x10-5 and Kw is 1x10-14
Solution:
We
Know that
Kw
=Ka x Kb
Ka
= Kw÷ Kb
Ka =1x10-14 ÷1.76 x10-5
Ka = 5.68 x10-10
According
the equation
NH4+ + H2O ↔ NH3 + H3O+
t=0 0.25M Nil Nil
t=eq. 0.25M-x x x
Ka = [ NH3][ H3O+]
/ [ NH4+]
5.68 x10-10= X.X /0.25-X
5.68 x10-10= X2
/0.25-x
X2 = 5.68 x10-10(0.25-X)
|
X2
=1.4x x10-10-5.68 x10-10X
0 =
X2 +5.68 x10-10X - 1.4x x10-10
Find
the vlaue of X by solution set
X=1.2x10-5
The
concentration of H3O+ is 1.2x10-5
pH=-log
[ H3O+]
pH=
-log [1.2x10-5]
pH=
4.9
|
b) Discuss the relation of activation energy
rate of chemical reaction.
(2Marks)
The rate of reaction depends
upon the energy of activation of the reaction. The reaction which have low
value of energy of activation are called fast reaction. Reactions whose energy
of activation is high are called slow reactions.
In simple the rate reaction is
inversely proportional to the activation energy
c)
Define the following terms (2
Marks)
i)
Order of Reaction
The sum of exponent
of the concentration of reactant in rate law expression
ii)
Rate determining Step in reaction
The slowest step of
multiple step chemical reaction is the rate determining step
Q.5 (Total 8 marks)
a) Find the Freezing point of a solution of
92.1 g of iodine, in 800g of Chloroform.
Assuming that the iodine nonvolatile at
that the solution is an ideal.
Kf of chloroform is 4.68 oC/mol/kg an
freezing point is -63.5 oC (02 mark)
Data:
Mass of iodine
=92.1g
Mass of
chloroform=800g
Kf
=4.68
Freezing point = - 63.5 0C
The atomic mass
is 127 amu
|
∆Tf =Kf x molality
∆Tf =4.68
x 92.1g÷254g/mol ÷0.8 Kg
∆Tf =2.1 0C
Tf(sol) =
-63.50C - (2.10C)
|
(b) Distinguish between colloids and
suspension of solution. (2
Marks)
Collidal
|
Suspension
|
Colloidal Solution is a
heterogeneous mixture in which particle size of substance is intermediate of
true solution and suspension i.e. between 1-1000 nm. Smoke from a fire is
example of colloidal system in which tiny particles of solid float in air.
Just like true solutions, Colloidal particles are small enough and cannot be
seen through naked eye. They easily pass through filter paper. But colloidal
particles are big enough to be blocked by parchment paper or animal membrane.
|
Suspension is a heterogeneous
mixture in which particle size of one or more components is greater than
1000nm.When mud is dissolved in water and stirred vigorously,
particles of mud are distributed evenly in water. After some time, the
particles of this solution settle under water due to influence of gravity.
This solution is an example of Suspension (see picture below). Contrary to
True Solution, particles of suspension are big enough to be seen with naked
eye.
|
(c) If 1800 J work is done on a system which
gives off 1500 J heat.What change will
occur in internal energy of the system?
(2 mark)
Data:
Work done on the system = +1800 J
Heat gives off = - 1500 J
|
∆E =q + work
∆E = -1500J + 1800 J
∆E = +300J
|
(d) State the Faraday’s 1st and 2nd
1st
Law :
It
states that the amount of any substance that is liberated at an electrode during electrolysis is directly
proportional to the quantity of electricity passed through the electrolyte. W
= Z ´ I ´ t
|
2nd Law:
It states
that when the same amount of electricity is passed through different
electrolytes, the amount of different substances deposited or liberated are
directly proportional to the equivalent weight of the substances.
|
well dont sir. thumbs up to u:)
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