Solution of CRQ's of First Year AKUEB Chemistry (First Shift )

CLASS XI FINAL TERM EXAMINATION 2012

1^ST 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, 2s²,2P_x , 2P_y , 2P_z        is true

       and  not                   1s², 2s²,2P_x, 2P_y, 2P_z

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 0^oC temperature of gas the volume of is greater than zero.

 Q.3                                                                                                                             (Total 8 marks)            a)State explain the following ;                                                                             (2 Marks)

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” & such liquids which do not flow easily are known as “Viscous”. The resistance of a liquid to flow is expressed in terms of viscosity,

            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. Crystals cleavage along particular direction.	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 P^H of 0.25 M NH₄Cl                                                                       (4 Marks)

                        NH₄⁺ + H₂O ↔ NH₃ + H₃O⁺

The Dissociation constant K_b is 1.76 x10^-5 and K_w 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.

K_f  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 1^st and 2^nd 

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.

Solid State

solid state

Behaviors of Solid:

           

            The solids have the following properties and characteristics due to which they are easily distinguished from liquid and gases.

i. Compressibility:

            The compressibility of solids is nearly zero, because the particles in solid are closely packed and so tightly bound together that no                inter-spaces are left. Hence the density of solids is much higher than gases and liquids.

ii. Deformity:

            Solids are deformed or shattered by higher pressure. This is because, when some particles are dislocated the forces of attraction is so strong that the rearranged atoms are held equally well to their new neighbours.

iii. Diffusibility:

            Diffusion is very slow in solids when compared with liquids and gases. There is no free movement of the particles but there is only vibration about the mean position e.g. Zn & Cu sheets when placed in close contact for a long time, they diffuse into each other in very slight ratio.

iv. Melting:

            On heating the solids change to liquid state and they melts.

In terms of Kinetic Theory when solids are heated, vibrational energy of their particles increases; until at melting point, some particles are vibrating with sufficient energy to overcome the forces holding them, hence they become mobile i.e. Solid fuse.

v. Sublimation:

            There are some solids substances as camphor iodine, naphthalene etc, which change directly to vapour on heating without passing through liquid phase. This phenomenon is called Sublimation.

In terms of Kinetic Theory, the inter-molecular force in which solids is less ordinary solids, hence high energy molecules at solid surface overcome the attractive forces and directly pass into vapours.

vi. Latent Heat of Fusion:

            Latent Heat of fusion is the heat energy which is required to change  1 gram of a solid into liquid at its melting point e.g. 1 gm of ice at 0 ^oC requires 334 J of heat energy to convert ice completely into water. Hence 334 J is called Latent Heat of fusion of ice.

Classification of Solids:

            There are two main types of solids which are as follows:

§     Crystalline Solid

§     Amorphous Solid

1. Crystalline Solid:

            The solids which have very orderly arrangement of their particles are called “CRYSTALLINE SOLID”. In this type of solid, the particles are arranged in layers and plane. Due to this they have definite geometric shapes e.g. Diamond, Graphite, NaCl etc are crystalline solids.

2. Amorphous Solid:

            These are the solids which do not have definite geometrical shape. The particles of such solids have a random i.e. non repetitive three dimensional arrangements. If a substance in liquid state is cooled rapidly, the particles are unable to arrange themselves in an orderly fashion, hence an amorphous solids results. Examples are glass, plastic, rubber etc.

Difference b/w Crystalline & Amorphous Solids

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. Crystals cleavage along particular direction.	Amorphous solids do not break down at fixed cleavage planes.
Anisotropy & Isotropy
Physical properties of crystals such as electrical conductivity, refractive index, , are different in different direction. This property is called Anisotropy. For Example Graphite can conduct electricity parallel to its plane of layers but not perpendicular to plane.	Amorphous solids are isotropic, i.e. their physical properties are same in directions.
Symmetry
When crystalline solids, are rotated about an axis their appearance does not change i.e. they possess symmetry.	Amorphous solids are not symmetrical.

Types of Crystals:

            Crystals are classified on the basis of kind of bond, by which atoms, ions or molecules are held together in solid. They are classified into the following four types.

1) Ionic Crystals:

            Such crystals which have ionic bonds are called Ionic Crystals. In ionic crystals, there are electrostatic forces of attraction between positive and negative ions.

e.g. NaCl, KCl, BaCl₂­ etc.

Properties:

1. The ionic crystals are brittle.

2. The ionic crystals are bad conductors of electricity.

3. They have high melting point.

4. They have very high values of lattice energy.

5. They have high melting point.

2) Covalent Crystals:

            In covalent crystals the atoms or molecules are held together by covalent bonds. Non – metals usually form this type of crystals e.g. Diamond, Sic etc.

Properties:

1. They are very hard.

2. The covalent crystals have high melting points.

3. They have low coefficients of expansion.

4. They have high refractive indices.

5. They are non – conductors of electricity.

3) Molecular Crystals:

            In molecular crystals, the molecules are held together due to Vander Wall’s forces. Vander Wall’s forces result from dipole-dipole interaction.

e.g. H₂O, CO₂, NH₃ etc.

Properties:

1. Molecular crystals are soft, wax like solids.

2. They have low melting point.

3. They are non-conductor of electricity.

4. They are non-conductor of heat.

5. They are usually brittle.

4) Metallic crystals:

            In metal, valence electrons are loosely attached in a crystal and these electrons can jump to the other atoms. When one or more electrons detach themselves from an atom a positive charge on the atom is produced. Thus, free electrons serve as an atmosphere of evenly distributed negative charge and positive ions are immersed in it. Examples may be taken as Na, Cu or Fe etc.

Properties:

1. The metals are good conductor of heat and electricity.

2. They have high tensile strength.

3. They are malleable and ductile.

4. They have luster.

5. They have high lattice energies.

Isomorphism:

            When two different substances have same crystal structure, they are said to be isomorphous and the phenomenon is called Isomorphism.e.g.

K₂Cr₂O₇ & K₂SO₄ are orthorhombic.

CaCO₃ & NaNO₂ are Trigonal.

ZnSO₄ & NiSO₄ are orthorhombic.

Properties:

            Isomorphic substances have following properties:

1. They have different physical and chemical properties.

2. They have empirical formula.

3. When their solutions are mixed, they from mixed crystals.

4. They show the property of over growth.

Polymorphism:

            The substance which can exist in more than one crystalline form, under different condition, is called Polymorphous and the phenomenon is called Polymorphism. e.g. CaCO₃ exist in nature in two crystalline form

1. Calcite , which is Trigonal      2. Aragonite, which is orthorhombic

UNIT CELL:

            Unit cell is a basic structural of a crystal having a definite geometrical shape and containing a definite no. of atoms or ions. The different arrangement of these unit cells in three dimensions, give different external shape of the crystal.

CRYSTAL SYSTEM:

            These are the group of crystals whose external shapes are built up by only one kind of unit cell. A crystal unit cell is three dimensional therefore it has three axis and three angles b/w three axis. The length, breath and height a, b & c, while angle between these lengths  µ ,b and g.   The crystal are classified into a following even crystal system formed from seven types of unit cell.

1. CUBIC SYSTEM:

            In this system, all length are equal and all the angle are of 90˚. i.e.

a=b=c

& a=b=g=90˚

e.g.                  Nacl, NaBr, Diamond etc.

2-TETRAGONAL SYSTEM:

            In this system, all the length are different but the angles are equal and of 90˚.

i.e.       a=b≠ c

&         µ=b=g=90˚

e.g.      SnO₂, BaSO₄,4H₂O

3-orthorhombic SYSTEM:

            In this system, all the length are different but the angle are equal to 90˚.

i.e.       a ≠ b ≠ c

& µ=b=g=90˚

e.g.      FeSO₄.7H₂O, ZnSO₄.7H₂O etc.

4-TRIGONAL SYSTEM: (RHOMBOHEDRAL)

            In this system, all the length are equal and all the angles are equal but more than 90^o & less than 120^o i.e.

a = b = c

&

µ=b=g¹90˚ (Angles > 90^o < 120^o)

e.g. KNO₃, AgNO₃ etc.

5. Hexagonal System:

            In this systems tow lengths are equal but third length is different. Also, two angles are of 90^o but third angle is 120^o i.e.

a = b ≠ c

& µ=b =90˚

g = 120^o

            e.g.      SiO₂, Graphite etc

6. Mono clinical System:

            In this system, all the lengths are different. Two angles a^’ and g^’ are of 90^o and third angle b^’ is different.

i.e. a ≠ b ≠ c

& a = g = 90^o

b ≠ 90^o

e.g.

CuSO₄.5H₂O, Na₂CO₃.10H₂O

7. Triclinic System:

In this system, all the lengths and all the angles are different.

i.e. a ≠ b ≠ c

a ≠ b ≠ g ≠ 90^o

e.g.

CuSO₄.5H₂O, K₂Cr₂O₇ etc.