Solution of AKUEB Paper II of XI year (CRQs And ERQS) 2014

CLASS XI FINAL TERM EXAMINATION 2014

1^st  SHIFT

Chemistry Paper II

                 

Time allowed: 2 hrs 20 minutes                                                                     Max. Marks 55

Solution

                                                                                                                              (Total 9 marks)

Q.1(a)  You are provided with 6.0 gm of C and excess amount  of O₂

Calculate the amount of CO₂ prepared by reacting them. ?                                                                          (2 mark)   

Step I

Mole of carbon =0.5 mol

Step II

Mole of CO₂ =0.5 mol

Step III

Amount of C O₂= mol x molar mass of CO₂=22 g

(b)        Diethyl zinc is a chemical used in the library to protect   the books from the worms. Its composition is 53% zinc, 38.9% carbon and 8.1% hydrogen; find the empirical formula of the compound.            (   Atomic mass Zn= 65.5 amu     )                       (2 mark)                                                                     

Element	Zn	C	H
%	53	38.9	8.1
Mole ratio	0.809	3.24	8.1
Simple mole ratio	1	4	10

(c) Write down  two differences  between  the following  spectrums.         (2 mark)

Line (or ) Discontinuous	Continuous
“In the line spectrum, the bands of colours are separated by dark spaces”. This type of spectrum may be obtained when light emitted from a gas source pass through a prism.	It is spectrum in which different colours are different into each other. There are no dark spaces separating these colours”.
If the light from the discharge tube is allowed to pass through the prism, some discrete sharp lines on an otherwise complete dark background are obtained such spectrum is called “LINE SPECTRUM”. .	The visible portion of spectrum obtained from ordinary sunlight and lights from incandescent sources are the examples of continuous spectrum.

         

                                                                                  

   (d)    Differentiate between Atomic orbital and Molecular orbital.(Only one )                  (1 mark)

Atomic Orbital

“Orbitals  are  the  regions  around  the nucleus in which the probability of finding the electron is maximum.”

            Molecular Orbital

The linear combination of atomic orbital gives two type of molecular orbital called as bonding and anti bonding molecular orbital.

   

e)      Define the following terms:                                                                                  (2 mark)

         i) Hydrolysis                                      

The reaction of cation and anion in which water is dissociates & combined with the ions so as to change its pH is known as hydrolysis".

            Hydrolysis is a chemical reaction between the ions of electrolyte and water molecules during which the pH of solution becomes change. This reaction is associated with weak electrolyte i.e. salts of weak acids and base.

            ii)Hydration

The process in which water molecules surround and interact with the ions of an electrolyte is called “Hydration” and the ions is called hydrated ions.

When an aqueous solution of a salt is evaporated the salt is crystallized with a definite no. of molecules

Is due to the attraction of positive ions for the –ve terminal of water molecules and negative ion for the +ve terminal of water molecules

                                               

Q.2                                                                                                                                          (Total  6 marks)

  a) Define Hybridization. How it explain the Shape of SiH₄                                                 (2 mark)

    

Definition:      

The hypothetical process of mixing of different atomic orbital to produce the same number of equivalent orbitals having same shape  and same energy is known as hybridization, the orbital so formed are called hybrid orbital” 

                        The concept of hybridization was introduced by pauling. The type of hybridization depends upon the number of mixing orbital i.e.SP³,SP²,SP,dSP²,d²SP³ etc                                                                                    

Explanation:   

The electronic configuration of carbon is given below.

   Si(z=14)=1s²,2s²,2p_x²,2p_y 2Pz², 3s¹,3p_x¹,3py¹ ,3Pz¹

                        The 3s and three 3p orbital of carbon mixed together to get a set of four equivalent sp3 hybrid orbital which are located at the corner of regular tetrahedron and each sp3 orbital of carbon over laps with s orbital of hydrogen from four sigma bonds.

Other molecules which show sp3 hybridization are CCl_4, SiCl₄,SnCl₄ etc.                                                                                                     

_______________________________________________________________________________

(b) Define the Bond order . Determine the bond order of the He.                                     (2 mark)

Bond order  is represent the  no of  number covalent bond

           Energy Diagram:

                                                                                                           

                                                                                                            Bond order = 2-0 /2

                                                                                                                               = 1

(c)    In Ammonia, water and methane the central atom has same sp³ hybridization but they form different geometrical shape. Explain your answer in detail .                                                               (2 Marks)             

                                              

·         Methane is a tetravalent molecule. In order to create four separate but equivalent orbitals, a single 2s and three 2p orbitals of the carbon hybridize into four sp3 orbitals. These four hybrid orbitals extend toward the corners of a tetrahedron centered on the carbon atom.

·         In the ammonia molecule, 2s and 2p orbital’s create four sp3 hybrid orbital’s. Three form covalent bond with hydrogen. The fourth sp3 orbital has a non-bonding lone pair of  electron, which can create ammonium ion  when they coordinate with hydrogen ion in acidic solution.

·         In a water molecule, two sp3 hybrid orbital’s  are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding with hydrogen. The H-O-H bond angle is less than the tetrahedral angle because of the repulsion  of non-bonding electrons on the other orbital’s.

Q.3                                                                                                                              (Total 7 marks)          a) Define the following  terms
                                                                                                                                (2 Marks) 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    OR   s, and its units are dynes / cm   OR      erg / cm².

          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, which may be defined as,

“The internal resistance to the flow of a liquid is called its viscosity”

Viscosity is represented by ‘h’ and its unit is “POLSE”. Normally smaller units “CENTIPOISE” or “MILLIPOISE” are used.

1 POISE          =          1 gm/cm

& 1 POISE      =          100 CENTIPOISE

                        =          1000 MILLIPOISE

b)      Differentiate between the following (3 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.

                                                                                                  

  

Q.3 c)

Draw a graph between volume and temperature on the basis of Charles’ Law        

           and explain  absolute zero                                                                                             (2 Marks)

 

            The absolute zero and Charles  refer to an experiment that gas tend to expand when heated. Charles law can be used to determined with volume and temperature readings when plotted T on the horizontal axis. You have to make sure you have the volume at zero and then extrapolate the data until it goes through the zero volume point.

Absolute Zero:                                                                                                                                                                                                                                                                                                              

The lowest temperature that is theoretically possible, at which the motion of particles which constitutes heat would be minimal. It is zero on the Kelvin scale, equivalent to −273.15°C  

It is Zero Kelvin temperature at which the volume , pressure and temperatures is  zero  .   According Charles’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.4                                                                                                                                    (Total 8 marks)

a)   In industry Sulphuric is manufacture by contact process. The production of Sulphric acid  depend the amount of SO_3.   Apply the Le-Chateleir  principle to increase the production of Sulphur trioxide in contact process .       

                              2SO₂ + O₂       ↔ 2SO₃                          ∆H= Negative               (4 mark)          

            i) Effect of Temperature :

If we increase the temperature of above reaction at equilibrium state the equilibrium state will shift backward direction.  

If we decrease the temperature of above reaction at equilibrium state the equilibrium state will shift forward  direction.  

            ii ) Effect of Pressure:

If we increase the pressure of above reaction at equilibrium state the equilibrium

state will shift forward direction.  

If we decrease the pressure of above reaction at equilibrium state the equilibrium state will shift backward direction.

            iii) Effect of concentration:

If concentration of reactant or concentrations of or reactants are increase at equilibrium state then equilibrium will shift to words forward direction.

If concentration of product or concentrations of products are increase at equilibrium state then equilibrium will shift to words Backward direction   .

If concentration of product or concentrations of  products are removed at equilibrium state then equilibrium will shift to words Backward direction   .

            iv) Effect of Catalyst:

The catalyst does not effect on equilibrium state 

b) Calculate the pH 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

Method I

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      x is very small value  therefore  negligible  in 0.25 -x           Ka              =   [ NH3][ H3O+] / [ NH4+]       5.68 x10-10= X.X /0.25       5.68 x10-10= X2 /0.25                    X2 = 5.68 x10-10(0.25)

X=1.191 x10-5 The concentration of H3O+ is 1.19x10-5 pH=-log [ H3O+] pH= -log [1.191x10-5] pH= 4.92

Method  II

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

                                                                                     

Q.5                                                                                                                              (Total 4 marks)

   a) Discuss the relation of activation energy and rate of chemical reaction.                         (2Marks)

Energy Of Activation:

            “The excess energy which is required for the reaction in addition to the average energy of the molecules is called Energy of Activation”. It is denoted by Ea and is measured in K.J/mole.

            According to the “collision theory” chemical reaction can only be possible when the effective collision among the molecules takes place and it happens only when the molecules of reactants it happens only when the molecules of reactants acquire threshold energy.

            “The minimum energy which required for the reaction is called Threshold Energy”. It is denoted by E_l.

b)Define the following terms                                                                                (2 Marks)

i)                    Order of Reaction

The sum of order of reactants in a chemical reaction which effect the rate of reaction

___________________________________________________________________________

ii)                  Rate determining Step in reaction

The slowest step of multiple step reaction is known as rate deterring step

_______________________________________________________________________________

Q.6                                                                                                                           (Total 6 marks)

a) Find the boiling 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_b  of chloroform is 3.63 ^oC/mol/kg an freezing point is  61.26 ^oC                                                                                                                                            (2 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) Identify the set up in the above diagram.

Write down its application as well.       (2 mark)                                                               

The apparatus is know is standard hydrogen electrode it is reference electrode which is used to determine the electrode potential of every element 

________________________________________________

(c)  Calculate the heat of formation of N₂O₄ from the following data:       (2 mark)          

               i)     2NO ₂           →      N₂O₄   ∆H=?

              ii) ½  N₂ + O₂        →      NO₂    ∆H= +333.95 KJ /mol

              iii)  N₂  +2 O₂        →      N₂O₄  ∆H= +9.3 KJ/mol                            

Solution : Equation (ii) Multiple by  2 2x {½  N2 + O2        →      NO2  }         ∆H=  2x {+333.95 KJ /mol}   iv)   N2 + 2O2        →      2NO2            ∆H=  667 KJ /mol Equation  (iv)   Subtract form Equation  (iii)            N2  +2 O2        →      N2O4                      ∆H= +9.3 KJ/mol               N2 + 2O2        →      2NO2                       ∆H=  667 KJ /mol  -             -                         -                                      -    2NO 2             →      N2O4                        ∆H=  - 658  KJ/mol

                                                                             

 Q.7                                EITHER                                                     (Total 7 marks )

a)   i)Discuss different spectrum Series   observed in spectrum study of hydrogen atom.                                                     

      Explain the different energies of these series on the basis of Bohr’s theory 

 ii) Discuss two defects of Bohr’s atomic theory      

                                                            OR

b)  i) Al₂S₃  can be prepared by the reaction        2Al    +    3S                                 Al₂S₃

When 20.0 g of  Aluminum (Al)  and 30.0 g of Sulphur(S) react in the reaction . Determine the limiting reactant and amount of Al₂S₃ produce in the reaction. How much non limiting reactant was present unused or in excess.  

     ii)  Differentiate between theoretical and actual yield      ( Two Difference )                 

(a)

 SPECTRUM OF HYDROGEN ATOM:

Although Hydrogen atoms contains only one electron, its spectrum gives a large number of series. Balmer in 1885, studied the spectrum of Hydrogen . he found that , when energy is supplied to the sample of hydrogen gas, individual atoms absorb different amount of energy. The electrons in higher energy levels are unstable & drop back to the lower energy levels &during this process energy is emitted in the from of line spectrum containing various lines of particular frequency & wave length.

                 Balmer observed that a series of lines appeared in  the visible region when the electrons drop from 3^rd , 4^th ,----nth energy levels to the second orbit . These spectral line are known as “BALMER SERIES”. He proposed an empirical formula to find wave no.(u) of each line.

where n₂ = 3,4,5,…………

Lyman later on discovered another series in uv-region . wave no. of each line was sound by similar formula.

where n₂ = 2,3,4,…………

Paschen discovered another such series in infrared region. Wave no. of each line was given by:-

Þ             

where n₂ = 4,5,6,……………

Bracket found another series infar-infrared region. Wave no. of such lines was given by.    

Þ             

where n₂ = 5,6,7,……………

Pfund also found another series far-infrared region. Wave no. of each line was given by:-

Þ             

where n₂ = 6,7,8,……………

iii)                Spectral line , which correspond to the transition of an element other than Hydrogen  from one energy level to another , have from one  energy level to another have for the most part , a fine structure , each line actually  consist of several separate close lying lines as doublet , triplet and so indicating that some of the electron of the given energy level have different energies .That is to say , that the electrons belonging to same energy level may differ in their energy 

iv)                Does not explain  sub-shell and sub energy level of electron in an atom  

Q.8                                                                                                                    (Total 8 marks)

                                                                

 b)  i) Define  of Raoult’s  Law (All three definition )and derive an expression between lowering of vapour pressure with molality of solution

       ii) Write  down the two  differences  between  colloids and suspension of solution.                                           

Solution

b) According to Raoult’s law

                            

1^st definition

“Vapour pressure solvent in solution is directly proportional to mole fraction of solvent”

P^o    α   mole fraction of solvent

2^nd definition

“Lowering vapour pressure of solvent in solution is directly proportional to mole fraction of solute”

ΔP α   mole fraction of solute

3^rd definition

Relative change in vapour pressure of solution is equal the mole fraction of solute

Where

            P is vapour pressure of pure solvent

             

          ΔP is  change of vapour pressure of solution

Derivation

Where   mole of solute = 

              Total mole of solution  = no. of mole of solute + no. of mole  solvent

     

             mole of solvent  =

                             

Where

mass of solute = w

mass of solvent = W

Molar mass of solute =m

Molar mass of solvent =M

Fro very dilute solution w/m is very small value therefore it is neglected

ii)

Properties	Colloidal Solutions	Suspension
Size	1– 1000nm	>1000nm
Nature	Heterogeneous	Heterogeneous
Filterability(Diffusion through parchment paper)	Colloidal particles pass through filter paper but not through parchment paper.	Suspension particles do not pass through filter paper and parchment paper.