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Chemistry 2004 Set 1 Sol

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PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS

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									2/4/2011                                          Subjective Test Paper - Chemistry - Meri…




  Chemistry 2004 Set 1                                                                        Close


 Subjective Test




      (i)    All questions are compulsory.
     (ii)    This question paper consists of four sections A, B, C and D.
              Section A contains 8 questions of one mark each.
              Section B is of 10 questions of two marks each.
              Section C is of 9 questions of three marks each and
              Section D is of 3 questions of five marks each.
     (iii)   There is no overall choice. However, an internal choice has been provided.
     (iv)    Wherever necessary, the diagrams drawn should be neat and properly labelled.



   Section A


    Question 1 ( 1.0 marks)
   Mention a large scale use of the phenomenon called ‘reverse osmosis’.

   Solution:
   Reverse osmosis is used on a large scale in the desalination of sea water.



    Question 2 ( 1.0 marks)
   Give an example of a pseudo first-order reaction.

   Solution:
   Inversion of cane sugar is a pseudo first-order reaction.




    Question 3 ( 1.0 marks)
   Mention two properties of acetonitrile because of which it acts as a good solvent.

   Solution:
   Two properties of acetonitrile because of which it acts as a good solvent are −

   (i) Due to its high polar nature, it can dissolve variety of solvents.

   (ii) It does not react with mild acidic and basic substances.



    Question 4 ( 1.0 marks)
   What are the types of lattice imperfections found in crystals?

   Solution:
   Two types of lattice imperfections are found in crystals.

   (i) Point defects (irregularities in arrangement around a point or an atom)

   (ii) Line defects (irregularities in arrangement in entire rows of lattice points)



   Question 5 ( 1.0 marks)
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    Question 5 ( 1.0 marks)
   What is the principal ore of Iron?

   Solution:
   The principal ore of iron is haematite (Fe 2 O 3 ).



    Question 6 ( 1.0 marks)
   Write the chemical reaction to transform butanal to butanoic acid.

   Solution:




    Question 7 ( 1.0 marks)
   Name the type of structure possessed by a unit cell of CsCl.

   Solution:
   A unit cell of CsCl possesses body-centred cubic structure.



    Question 8 ( 1.0 marks)
   Write the IUPAC name of the following compound:




   Solution:




   Section B


    Question 9 ( 2.0 marks)
   Explain Brownian movement.

   Solution:
   When colloidal solutions are observed under a powerful ultra microscope, the colloidal particles
   appear to be in continuous motion in a zig-zag path. This motion of colloidal particles is know n as
   Brownian movement.

   The unbalanced bombardment of particles by the molecules of the dispersion medium causes
   Brownian movement. This movement is independent of the nature of the colloid, but varies inversely
   with respect to the size of the particles and the viscosity of the solution. It is responsible for the
   stability of sols.




    Question 10 ( 2.0 marks)
   Distinguish between reaction rate and reaction rate constant (specific reaction rate) of a reaction.

…meritnation.com/…/sNlZgWFT847bbfY…                                                                         2/15
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…

   Solution:

                       Rate of a reaction                             Rate constant of a reaction

    The rate of a reaction is the rate of                    The rate constant of a reaction is the rate of
    disappearance of the reactant, or the rate of            the reaction when the concentration of each
    appearance of the product.                               reactant is unity.


    The rate of a reaction can be determined by              Rate constant can be calculated using the
    determining the rate of decrease in the                  rate law.
    concentration of the reactant.




    Question 11 ( 2.0 marks)
   Write the cell reactions which occur in lead storage battery (i) when the battery is in use and (ii)
   when the battery is on charging.

   Solution:
       i. Cell reactions which occur in lead storage battery when it is in use −

   At anode:


   At cathode:

   Overall reaction:




      ii. Cell reactions which occur in lead storage battery when it is being charged −

   At anode:


   At cathode:

   Overall reaction:




    Question 12 ( 2.0 marks)
   Give chemical reaction in support of each of the following statements:

   (i) The +1 oxidation state gets stabilised progressively from Ga to Ti in Group 13.

   (ii) All the bonds in PCl5 molecule are not equivalent.


   Solution:
   (i) The outer electronic configuration of group-13 elements is ns 2 np1 . On moving down the group
   (i.e., from Ga to Ti), due to inert pair effect, there is a decrease in the tendency of s electrons of the
   valence shell to participate in bond formation. This is due to the poor shielding of the ns 2 electrons
   of the valence shell by the intervening d or f electrons. So, only the p electrons can participate in
   bonding. Hence, the oxidation state is restricted to +1.

   (ii) PCl5 exists as a salt in solid state. It contains tetrahedral cations          and octahedral anions

           . Thus, all the bonds in PCl5 molecule are not equivalent.



    Question 13 ( 2.0 marks)
   Expand DDT. Write its structure.

   Solution:
   DDT is dichlorodiphenyltrichloroethane.




…meritnation.com/…/sNlZgWFT847bbfY…                                                                             3/15
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    Question 14 ( 2.0 marks)
   Write the names of reagents and equations in the conversion of

   (i) phenol to salicyl aldehyde

   (ii) anisole to p-methoxyacetophenone

   Solution:
   (i)




   (ii)




    Question 15 ( 2.0 marks)
   Write the modes of free-radical polymerisation of an alkene.

   OR

   Differentiate between addition and condensation polymers based on the mode of polymerisation.
   Give one example of each type.

   Solution:
   The polymerisation of an alkene is carried out in the presence of a free-radical initiator such as
   benzoyl peroxide, acetyl peroxide, tert-butyl peroxide, etc. The steps involved in the free-radical
   polymerisation of alkene are as follows:

   Chain-initiation steps




   Chain-propagation step




…meritnation.com/…/sNlZgWFT847bbfY…                                                                      4/15
2/4/2011                                            Subjective Test Paper - Chemistry - Meri…
   Chain-termination step

   One of the modes of termination is given below.




   Or


               Addition polymers                                   Condensation polymers

    Addition polymers are the polymers            Condensation polymers are the polymers formed by the
    formed by the process of repeated             repeated condensation reactions between tw o bi-
    addition of monomers possessing               functional or tri-functional monomers by the loss of
    double or triple bonds.                       molecules such as H2 O, HCl, etc.

    Example: Buna-S, Buna-N, polythene            Example: Nylon 6, 6, terylene



    Question 16 ( 2.0 marks)
   Differentiate between Lanthanides and Actinides series.

   Solution:

                      Lanthanides                                           Actinides

    1.     Besides the +3 oxidation state,          Higher oxidation states are more common in actinides.
           which is the most common, the +2         Besides the +3 state, which is the most common, they
           and +4 states are also displayed.        show the +4, +5, +6 and +7 states as well.

    2.     They are not radioactive, except for     They are all radioactive.
           promethium.

    3.     The ions formed by these metals          The ions formed by these metals are mostly coloured.
           are mostly colourless.

    4.     They do not form complexes easily.       They form complexes easily.




    Question 17 ( 2.0 marks)
   Calculate the density of silver which crystallises in the face-centred cubic structure. The distance
   between the nearest silver atoms in this structure is 287 pm.

   Solution:
   Given, molar mass of Ag, M = 107.87 g mol−1

   NA = 6.02 × 10 23 mol−1

   In case of fcc lattice, number of atoms per unit cell, z = 4

   Distance between two nearest Ag atoms = 287 pm




   Thus, edge length = 406 pm (approx)

   = 406 × 10−12 m

   Therefore, density of silver is given by



…meritnation.com/…/sNlZgWFT847bbfY…                                                                         5/15
2/4/2011                                        Subjective Test Paper - Chemistry - Meri…




    Question 18 ( 2.0 marks)
   Two elements A and B form compounds having formula AB2 and AB4 . W hen dissolved in 20 g of
   benzene (C 6 H6 ), 1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4 lowers it by 1.3
   K. The molar depression constant for benzene is 5.1 Kkg mol−1 . Calculate atomic masses of A and B.

   Solution:
   We know that




   Then,


   = 110.87 g mol−1




   = 196.15 g mol−1

   Now, we have the molar masses of AB2 and AB4 as 110.87 g mol−1 and 196.15 g

   mol−1 respectively.

   Let the atomic masses of A and B be x and y respectively.

   Now, we can write:




   On subtracting equation (i) from equation (ii), we have

   2y = 85.28

   ⇒ y = 42.64

   On putting the value of y in equation (i), we have

   x + 2 × 42.64 = 110.87

   ⇒ x = 25.59

   Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.



   Section C


    Question 19 ( 3.0 marks)
   Explain the Froth Floatation Method.

   Solution:
   Froth floatation method is used for the concentration of sulphide ores. In this process, a suspension
   of powdered ore is made with water. To this suspension, collectors (such as pine oils, xanthates,
   etc.) are added to enhance the non-wettability of the mineral particles. Froth stabilisers (such as
   cresols, aniline, etc.) are also added so as to stabilise the froth.




…meritnation.com/…/sNlZgWFT847bbfY…                                                                        6/15
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   The gangue particles become wet by water and the mineral particles become wet by oils. There is a
   rotating paddle which agitates the mixture and draws air into it. As a result of this, froth is formed.

   The forth carries the mineral particles, and is skimmed off. It is then dried to recover the ore
   particles.



    Question 20 ( 3.0 marks)
   Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1 . Calculate its molar conductivity and if
      for acetic acid is 390.5 S cm2 mol−1 , what is its dissociation constant?


   Solution:
   Given,

   κ = 7.896 × 10 −5 S m−1

   c = 0.00241 mol L−1


   Then, molar conductivity,




   = 32.76 S cm2 mol−1

   Again,      = 390.5 S cm2 mol−1



   Now,


   = 0.084


     Dissociation constant,




   = 1.86 × 10 −5 mol L −1



    Question 21 ( 3.0 marks)
   A reaction is first order in A and second order in B.

   (i) Write differential rate equation.

   (ii) How is the rate affected if the concentration of B is tripled?

   (iii) How is the rate affected when the concentrations of both A and B are doubled?

   What is the significance of rate constant in the rate expression?

   Solution:

   (i) The differential rate equation is



   (ii) Rate of the reaction is given by

   If the concentration of B is tripled, then the rate is given by




…meritnation.com/…/sNlZgWFT847bbfY…                                                                          7/15
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…




   Therefore, the rate of the reaction is increased by 9 times.

   (iii) If the concentrations of both A and B are doubled, then the rate is given by




   Therefore, the rate of the reaction is increased by 8 times.

   Significance of the rate constant: Rate constant of a reaction is the rate of the reaction when the
   concentration of each reactant is unity.



    Question 22 ( 3.0 marks)
   How are the colloids classified on the basis of the nature of interaction between dispersed phase
   and dispersion medium? Describe an important characteristic of each class. Which of these sols need
   stabilising agents for preservation?

   Or

   What are detergents? Give their scheme of classification. Why are the detergents preferred over
   soaps?

   Solution:
   On the basis of the nature of interaction between the dispersed phase and the dispersion medium,
   colloids can be classified as lyophilic (solvent-attracting) colloids and lyophobic (solvent-repelling)
   colloids.

   Lyophilic colloids can be directly formed by mixing substances like gum, gelatin, starch, rubber, etc.,
   with the dispersion medium. An important characteristic of lyophilic colloids is that even if the
   dispersion medium is separated from the dispersed phase, the colloidal form can be re-obtained
   simply by adding more of the dispersion medium. So, these sols are named as reversible sols.

   Colloidal sols cannot be formed by simply mixing substances like metals, their sulphides, etc., with
   the dispersion medium. Their colloidal sols can be prepared only by special methods. Such sols are
   called lyophobic sols. An important characteristic of lyophobic colloids is that once precipitated, these
   colloids cannot be re-obtained simply by adding more of the dispersion medium. So, these sols are
   named as irreversible sols.

   Lyophilic colloids are quite stable and cannot be precipitated easily. Lyophobic colloids are not stable
   and are readily precipitated. Hence, lyophobic colloids need stabilising agents for preservation.

   Or

   Detergents are surfactants used for cleaning purposes. They do not contain any soap, but possess
   all the properties of soap.

   Detergents are mainly classified into three categories depending on the charge carried by the polar
   part.

   (i) Anionic detergents: Sodium salts of sulphonated long-chain alcohols or hydrocarbons are known as
   anionic detergents. These detergents have the large part as the anion.

   For example,




   (ii) Cationic detergents: Quaternary ammonium salts of amines with acetates, chlorides or bromides
   as anions are known as cationic detergents. These detergents have the large part as the cation.

   For example,




        i. Non-ionic detergents: Non-ionic detergents do not contain any ions in their constitution.
…meritnation.com/…/sNlZgWFT847bbfY…                                                                            8/15
2/4/2011                                            Subjective Test Paper - Chemistry - Meri…
   For example,




   Soaps work in soft water. However, they are not effective in hard water. In contrast, detergents
   work in both soft water and hard water. Therefore, detergents are preferred over soaps.



    Question 23 ( 3.0 marks)
   Draw a figure to show splitting of degenerate d orbitals in an octahedral crystal field. How does the
   magnitude of ∆0 decide the actual configuration of d orbitals in a complex entity?


   Solution:




   In                   systems, the electrons occupy a t2g orbital.

   In      system, there are two possibilities.

   If crystal-field splitting (∆0 ) < pairing energy (P), then the fourth electron enters one of the t g
   orbitals. The resulting configuration will be       .

   On the other hand, if ∆0 > P, then the fourth electron enters a t2g orbital. The resulting configuration
   will be    .


   In                   systems, there are two possibilities yet again.

   In                   systems, electrons occupy an eg orbital.

   Therefore, we can say that the higher the value of ∆0, the more is the pairing of electrons.



    Question 24 ( 3.0 marks)
   Explain the following terms with suitable examples:

   (i) Frenkel defect

   (ii) Interstitials

   Solution:
   (i) Frenkel defect:Ionic solids containing large differences in the sizes of ions show this type of defect.
   When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel
   defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also
   known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.




   (ii) Interstitials:Interstitial defect is show n by non-ionic solids. This type of defect is created when
   some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density
   of a substance increases because of this defect.

…meritnation.com/…/sNlZgWFT847bbfY…                                                                              9/15
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    Question 25 ( 3.0 marks)
   Draw the structures of monomers for the following polymers. Also draw the structures of the
   polymers and uses of:

   (i) Teflon

   (ii) PMMA

   (iii) Buna-S

   Solution:
   (i) The monomer of teflon is




   The structure of teflon is




   Teflon is used for making oil seals, gaskets and utensils having non-stick surface coating.

   (ii) The monomer of PMMA is




   The structure of PMMA is




   PMMA is used as a substitute for glass, and for making decorative materials.

   (iii) The monomers of Buna-S are




   The structure of Buna-S is




   Buna-S is used for making automobile tyres and footwear.



    Question 26 ( 3.0 marks)

…meritnation.com/…/sNlZgWFT847bbfY…                                                              10/15
2/4/2011                                     Subjective Test Paper - Chemistry - Meri…
   (a) Identify A and B in the following:

   (i)




   (ii)




   Solution:
   (i)




   (ii)




    Question 27 ( 3.0 marks)
   Write the IUPAC names of the following:

   (i)




   (ii)

   (iii)




   Solution:
   (i)




   (ii)




   Pentan − 2 − one

   (iii)




   Section D
…meritnation.com/…/sNlZgWFT847bbfY…                                                      11/15
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    Question 28 ( 5.0 marks)
   (a) Assign appropriate reason for each of the following observations:

   (i) Anhydrous AlCl3 is used as a catalyst.

   (ii) Phosphinic acid behaves as a monoprotic acid.

   (iii) SF6 is not easily hydrolysed whereas SF4 is readily hydrolysed.

   (iv) No form of elemental silicon is comparable to graphite.

   (b) Draw the structure of XeOF4 or BrF3.

   Or

   Account for the following:

   (i) Ammonia is a stronger base than phosphine

   (ii) The tendency to exhibit +2 oxidation state increases with increasing atomic number in group 14.

   (iii) HF is a weaker acid than HI.

   Solution:
   (a)

   (i) The central atom Al in AlCl3 is electron deficient, and hence, AlCl3 acts as a Lewis acid catalyst. For
   example, it acts as a catalyst in Friedel Crafts reaction and helps in generating electrophiles for
   attack.

   (ii)




   Phosphinic acid has one P(OH) group and two hydrogen atoms directly attached to phosphorus, i.e.,
   it has only one ionisable hydrogen. This is why it behaves as a monoprotic acid.

   (iii)




   In SF6 , sulphur atom is stearically protected due to the presence of 6 F atoms. On the other hand,
   the S atom in SF4 is less stearically hindered. So, SF6 is not easily hydrolysed, whereas SF4 is readily
   hydrolysed.

   (iv) Carbon enters into pπ − pπ bonds with itself and with other elements. However, due to its large
   size, silicon does not show this characteristic. This is the reason why no form of elemental silicon is
   comparable to graphite. Elemental sulphur exists only in the diamond structure.

   (b)




   Or

   (i) Nitrogen is more electronegative than phosphorus. So, there is a greater charge separation in an
…meritnation.com/…/sNlZgWFT847bbfY…                                                                              12/15
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…
   (i) Nitrogen is more electronegative than phosphorus. So, there is a greater charge separation in an
   ammonia (NH3 ) molecule than in a phosphine (PH3 ) molecule. As a result, the ammonia molecule has
   a larger concentrated region of negative charge than the phosphine molecule.




   Hence, the ammonia molecule has a greater tendency to attract a positive proton. This is why
   ammonia is a stronger base than phosphine.

   (ii) The general outer electronic configuration of group-14 elements is ns 2 np2 . On moving down the
   group, the ‘ns’-pair of electrons is held more tightly. This is due to the inert pair effect. As a result,
   only 2p-orbital electrons are available for taking part in bonding. This is why as we move down the
   group, the tendency to exhibit the +2 oxidation state increases.

   (iii) F is more electronegative than I. Hence, HF undergoes stronger H bonding than HI. As a result, it
   is difficult for HF to donate proton. Hence, HF is a weaker acid than HI.



    Question 29 ( 5.0 marks)
   (a) How would you account for the following:

   (i) The transition elements exhibit high enthalpies of atomisation.

   (ii) The 4d and 5d series of the transition metals have more frequent metal−metal bonding in their
   compounds than do the 3d metals.

   (iii) There is a greater range of oxidation states among the actinoids than among the lanthanoids.

   (b) Write the complete chemical equation for each of the following:

   (i) An alkaline solution of KMnO 4 reacts with an iodide.

   (ii) An excess of SnCl2 solution is added to a solution of mercury (II)

   Or

   An aqueous solution freezes at 272.4 K, while pure water freezes at 273.0 K. Determine

   (i) the molality of the solution

   (ii) the boiling point of the solution

   (iii) the lowering of vapour pressure of water at 298 K

   [Given Kf = 1.86 K kg mol−1 , Kb= 0.512 K kg mol−1 and vapour pressure of water at 298 K = 23.756
   mm Hg]

   Solution:
   (a)

   (i) Transition elements have high effective nuclear charge and a large number of valence electrons.
   Therefore, they form very strong metallic bonds. As a result, they exhibit high enthalpies of
   atomisation.

   (ii) The transition metals of 4d and 5d series have greater enthalpies of atomisation than their
   corresponding metals of 3d series. Hence, the transition metals of 4d and 5d series have more
   frequent metal−metal bonding in their compounds than do the 3d metals.

   (iii) There is a greater range of oxidation states among the actinoids than among the lanthanoids.
   This is because of the fact that the 5f, 6d and 7s levels in actinoids are of comparable energies, and
   hence, they can take part in bond formation.

   (b)

   (i)

   (ii) SnCl2 reduces HgCl2 to Hg 2 Cl2 first, and then, to Hg.




   Or

   (i) It is known that


…meritnation.com/…/sNlZgWFT847bbfY…                                                                             13/15
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   Therefore, molality,


   (ii) It is known that




   Therefore, ∆Tb = 0.512 K kg mol−1 × 0.323 mol kg −1

   = 0.165 K

   Hence, boiling point of the solution, Tb = (373 + 0.165) K

   = 373.165 K

   (iii) It is known that




   Here, vapour pressure of pure water at 298 K = 23.756 mm Hg

   m = 0.323 kg mol −1

   Let the amount of water be 1000 g.




    Question 30 ( 5.0 marks)
   Name the products obtained on complete hydrolysis of DNA. Enumerate the structural differences
   between DNA and RNA. In what way is a nucleotide different from a nucleoside? Illustrate with
   examples.

   Or

   Define and classify vitamins. Name the diseases caused due to lack of any three of them.

   Solution:
   The following products are obtained on complete hydrolysis of DNA.

   (i) Pentose sugar − deoxyribose

   (ii) Phosphoric acid

   (iii) Purine and pyrimidine bases (nitrogen-containing heterocyclic compound), e.g., thymine, cytosine
   etc.

…meritnation.com/…/sNlZgWFT847bbfY…                                                                         14/15
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   The structural differences between DNA and RNA are −

   (i) The sugar in DNA is deoxyribose, while that in RNA is ribose.

   (ii) DNA contains the base thymine, while RNA contains uracil.

   (iii) DNA has a double-stranded helical structure, while RNA has a single-stranded helical structure.

   Difference between nucleoside and nucleotide:

   Nucleoside is the unit formed by the attachment of a purine or pyrimidine base to the 1′-position of a
   pentose sugar.

   Nucleotide is the unit formed by the attachment of a nucleoside to phosphoric acid at the 5′-position
   of sugar moiety.




   Or

   Vitamins are essential dietary substances required in small quantities to perform specific biological
   functions for the normal maintenance of the optimum growth and health of the organism. Their
   absence can cause specific deficiency diseases.

   Vitamins are classified into two categories depending on their solubility in water or fat.

   (i) Water-soluble vitamins: These vitamins are soluble in water, e.g., B-group vitamins and vitamin C.

   (ii) Fat-soluble vitamins: These vitamins are soluble in fats and oils, but insoluble in water, e.g.,
   vitamins A, D and K

           Vitamin                              Deficiency disease

           Vitamin A   Xerophthalmia (hardening of the cornea of the eye), night blindness

           Vitamin D   Rickets, osteomalacia

           Vitamin K   Delaying of blood clotting




…meritnation.com/…/sNlZgWFT847bbfY…                                                                         15/15

								
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