Chemistry 2009 Set 1 Sol by nehalwan

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




  Chemistry 2009 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)
   Which point defect in crystals does not affect the density of the relevant solid?

   Solution:
   Frenkel defect in crystals does not affect the density of the relevant solid.



    Question 2 ( 1.0 marks)
   Define the term ‘Tyndall effect’.

   Solution:
   Tyndall effect is observed when a fine beam of light enters a room through a small hole. It happens
   because of the scattering of light by particles of colloidal and suspension solutions.



    Question 3 ( 1.0 marks)
   Why is the froth flotation method selected for the concentration of sulphide ores?

   Solution:
   Froth flotation process is selected for the concentration of sulphide ores as in this process, sulphide
   ore particles are preferentially wetted by oil whereas gangue particles are wetted by water.



    Question 4 ( 1.0 marks)
   Why is Bi (V) a stronger oxidant than Sb (V)?

   Solution:
   Bismuth and antimony both belong to the nitrogen family and exhibit the +5 oxidation state.
   However, on moving down the group, i.e., from antimony to bismuth, the stability of the +5 oxidation
   state decreases. This is due to the inert pair effect. Thus, Bi (V) is a stronger oxidant than Sb (V).



    Question 5 ( 1.0 marks)
   Give the IUPAC name of the following compound:




   Solution:
   2 − Bromo-3-methyl-but-2-ene-1-ol.


…meritnation.com/…/Cuy6WobDThiILNv…                                                                          1/14
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…

    Question 6 ( 1.0 marks)
   Write the structure of 3-oxopentanal.

   Solution:




    Question 7 ( 1.0 marks)
   Why is an alkylamine more basic than ammonia?

   Solution:
   An alkylamine is more basic than ammonia because of inductive effect (+I effect). Alkyl group or ‘R’
   has an electron-releasing effect, which increases electron density over nitrogen atom. This increases
   its basicity.



    Question 8 ( 1.0 marks)
   Give an example of elastomer.

   Solution:
   Natural rubber is an example of elastomer.



   Section B


    Question 9 ( 2.0 marks)
   A reaction is of second order with respect to a reactant. How will the rate of reaction be affected if
   the concentration of this reactant is

   (i) Doubled,

   (ii) Reduced to half?

   Solution:
   (i) When the concentration of the reactant is doubled, the rate of reaction will become four times.

   (ii) When the concentration of the reactant is reduced to half, the rate of reaction will become one-
   fourth.



    Question 10 ( 2.0 marks)
   Explain the role of

   (i) Cryolite in the electrolytic reduction of alumina.

   (ii) Carbon monoxide in the purification of nickel.

   Solution:
   (i) Cryolite is used in the electrolytic reduction of alumina so as to reduce its melting point and make
   it a good conductor of electricity.

   (ii) Carbon monoxide is used in the purification of nickel because it reacts with nickel to give a volatile
   complex called nickel tetracarbonyl, which on heating, decomposes to gives pure nickel metal.



    Question 11 ( 2.0 marks)
   Draw the structures of the following molecules:

   (i) XeF4

   (ii) BrF3


   Solution:




…meritnation.com/…/Cuy6WobDThiILNv…                                                                              2/14
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…




    Question 12 ( 2.0 marks)
   Complete the following chemical reaction equations:

   (i)


   (ii)


   Solution:
   (i)


   (ii)



    Question 13 ( 2.0 marks)
   Differentiate between molality and molarity of a solution. What is the effect of change in temperature
   of a solution on its molality and molarity?

   Solution:
   Molarity of a solution is defined as the number of gram moles of solute present in l L of solution,
   while molality of a solution is defined as the number of gram moles of solute present in 1 kg of
   solvent.

   Molality of a solution decreases with increase in temperature, while molality of a solution is not
   affected by temperature.



    Question 14 ( 2.0 marks)
   Which ones in the following pairs of substances undergoes SN2 substitution reaction faster and
   why?

   (i)




   (ii)


                  or

   Solution:



   (i)      undergoes SN2 substitution reaction faster than           . This is because the alkyl group
   present in benzyl chloride increases its basicity due to +I effect. Stronger the base, lesser is its




   leaving ability. So,     reacts faster.

   (ii) Iodide is a weaker base than chloride. Weaker the base, greater is its leaving ability. So,

                  undergoes SN2 substitution reaction faster.



    Question 15 ( 2.0 marks)
   Complete the following reaction equations:

   (i)



…meritnation.com/…/Cuy6WobDThiILNv…                                                                         3/14
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…

   (ii)




   Solution:
   (i)




   (ii)




    Question 16 ( 2.0 marks)
   Explain what is meant by

   (i) A peptide linkage

   (ii) A glycosidic linkage

   Solution:
   (i) A peptide linkage (−CO−NH−) holds together amino acid units in proteins. It is an amide bond
   formed between −COOH of one amino acid and −NH2 group of another amino acid by the loss of
   water molecule.

   (ii) The linkage formed by the reaction of the −OH group of anomeric carbon of a monosaccharide
   with the −OH group of other monosaccharide is called glycosidic linkage.



    Question 17 ( 2.0 marks)
   Name two water soluble vitamins, their sources and the diseases caused due to their deficiency in
   diet.

   Solution:
   Thiamine (vitamin B1 ) & riboflavin (vitamin B2 ) are soluble in water.

   Thiamine is found in unpolished rice, whole cereals, yeast, egg yolk, milk, green vegetables, etc. The
   deficiency of thiamine causes beriberi and loss of appetite.

   Riboflavin is found in egg yolk, liver, milk and green leafy vegetables. The deficiency of riboflavin
   causes cracked lips, sore tongue and skin disorders.



    Question 18 ( 2.0 marks)
   Draw the structures of the monomers of the follow ing polymers:

   (i) Teflon

   (ii) Polythene

   OR

   What is the repeating unit in the condensation polymer obtained by combining HO 2 CCH2 CH2 CO 2 H
   (succinic acid) and H2 NCH2 CH2 NH2 (ethylene diamine)?


   Solution:
   (a) The monomer present in Teflon is Tetrafloroethylene. The structure of the monomer of teflon is




   (b) The monomer present in polythene is ethene. The structure of the monomer of polythene is




                                                       OR

   The repeating unit of the condensation polymer obtained by combining succinic acid and ethylene
…meritnation.com/…/Cuy6WobDThiILNv…                                                                         4/14
2/4/2011
    The repeating unit of the                    Subjective Test Paper - Chemistry - and ethylene
                                condensation polymer obtained by combining succinic acid Meri…
    diamine is




    Section C


     Question 19 ( 3.0 marks)
    Iron has a body-centred cubic unit cell w ith a cell edge of 286.65 pm. The density of iron is 7.87 g
    cm−3 . Use this information to calculate Avogadro’s number.

    (At. Mass of Fe = 56 g mol−1 )

    Solution:
    In a body-centred cubic unit cell, number of atoms present = 2

    At mass of iron = 56 g mol−1

    Density of iron = 7.87 g cm−3

    Mass of iron = 7.87 × Volume

    Volume in BCC = (a)3

    = (286.65)3 pm

    = 2.34 × 10 −23 cm

    Mass = 7.87 × 2.34 × 10−23 g




    ∴ Avogadro’s number = 6.022 × 1023



     Question 20 ( 3.0 marks)
    100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution
    has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?

    (R = 0.0821 L atm mol−1 K−1 and 760 mm Hg = 1 atm)

    Solution:
    We know that


    Osmotic pressure (π)




    π = 13.3 mm Hg


    V = 10 mL = 10/1000 L = 0.01 L

    R = 0.0821 L atom mol−1 K−1

    T = 25ºC = 273 + 25 K = 298 K

    W = 100 mg = 100/1000 g = 0.1 g

    On putting the above values in the formula, we get




…meritnation.com/…/Cuy6WobDThiILNv…                                                                         5/14
2/4/2011                                              Subjective Test Paper - Chemistry - Meri…


   Thus, the molar mass of the protein is                        .



    Question 21 ( 3.0 marks)
   A first-order reaction has a rate constant of 0.0051 min −1 . If we begin with 0.10 M concentration of
   the reactant, what concentration of reactant will remain in the solution after 3 hours?

   Solution:
   The integrated rate equation of the first-order reaction is




   Here,

   A0 = Initial concentration = 0.10 M

   k = 0.0051 min-1

   t = 3 hours = 3 × 60 min




   Log 0.1 − log A = 0.4

   Log A = −1 − 0.4

   A= antilog (−1 − 0.4)

   A = 0.0398



    Question 22 ( 3.0 marks)
   How are the following colloids different from each other with respect to dispersion medium and
   dispersed phase? Give one example of each type.

   (i) An aerosol (ii) A hydrosol (iii) An emulsion

   Solution:
   (i) An aerosol is a suspension of a solid or a liquid in a gas.

   Example: smog, smoke

   (ii) Hydrosol is a colloidal suspension of essential oils. It is obtained from steam distillation of
   aromatic plants,

   Example: rose water used as facial toner

   (iii) An emulsion contains one liquid dispersed in another liquid.

   Example: butter and margarine; milk and cream



    Question 23 ( 3.0 marks)
   Account for the following:

   (i) NH3 is a stronger base than PH3 .

   (ii) Sulphur has a greater tendency for catenation than oxygen.

   (iii) Bond dissociation energy of F2 is less than that of Cl2 .

                                                          OR

   Explain the following situations:

   (i) In the structure of HNO 3 molecule, the N − O bond (121 pm) is shorter than the N − OH bond (140
   pm).

   (ii) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.

   (iii) XeF2 has a straight linear structure and not a bent angular structure.


   Solution:
…meritnation.com/…/Cuy6WobDThiILNv…                                                                         6/14
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…
   (i) On moving from nitrogen to phosphorus, i.e., down the group, the atomic size increases. As the
   size of the central atom increases, the lone pair of electrons occupy a larger volume. Consequently,
   the electron density, and hence, the basic strength decrease. Thus, NH3 is a stronger base than
   PH3 .

   (ii) The tendency for catenation depends upon the bond energy. The bond energy of a sulphur
   molecule is more than that of an oxygen molecule. Thus, the sulphur − sulphur bond strength is
   higher, and as a result, the tendency of catenation is also higher. Sulphur shows catenation up to
   eight atoms.

   (iii) Bond dissociation energy of fluorine is less than that of chlorine. It is due to the low value of
   electron affinity of small-sized fluorine. Also, the value of enthalpy of hydration of fluorine is much
   higher than that of chlorine.

                                                       OR

   (i) The structure of nitric acid is




   The N − O bond has a double-bond character. On the other hand, N − OH bond is a single bond.
   Since a double bond is shorter than a single bond, the N − O bond (121 pm) is shorter than the N −
   OH bond (140 pm).

   (ii) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed. SF6 is an inert, non- toxic gas. This
   is due to the presence of stearically protected six fluorine atoms. As a result, reactions like hydrolysis
   are not allowed to take place.

   (iii)




   According to VSEPR theory, XeF2 has trigonal bipyramidal geometry with sp3d hybridisation. It has
   two bond pairs and three lone pairs of electrons. The bond pairs occupy axial positions and the lone
   pairs occupy the equatorial positions. This is the most favourable arrangement as far as stability is
   concerned. The tw o fluorine atoms if placed at the equatorial positions will face minimum repulsion,
   and this would consequently increase the stability. Hence, the structure is linear.



    Question 24 ( 3.0 marks)
   For the complex [Fe(en)2 Cl2 ] Cl, (en = ethylene diamine), identify

   (i) The oxidation number of iron,

   (ii) The hybrid orbitals and the shape of the complex,

   (iii) The magnetic behaviour of the complex,

   (iv) The number of geometrical isomers,

   (v) Whether there is an optical isomer also, and

   (vi) Name of the complex. (At. No. of Fe = 26)

   Solution:
   (i) Let the oxidation number of Fe in                     be x.


   The oxidation state of Fe can be calculated as follows:

   x + 2 (0) + 2 (-1) = + 1

   Or, x − 2 = 1

   Or, x = 3
…meritnation.com/…/Cuy6WobDThiILNv…                                                                             7/14
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…
   Or, x = 3

   (ii) The electronic configuration of Fe 3+ is 1s 2 2s 2 2p6 3s 2 3p6 3d5




   So, hybridisation of Fe 3+ is sp3 d2 , i.e., one s, three p and two d orbitals hybridise. It has octahedral
   geometry

   (iii) The complex is para magnetic due to the presence of 5 unpaired electrons.

   (iv) This complex exists as cis-trans isomers. Thus, it has 2 geometrical isomers.




   (v) The cis isomer exhibits optical activity.

   (vi) The name of the complex is Dichlorobis-(ethylenediamine) iron (III) chloride.



    Question 25 ( 3.0 marks)
   Explain the mechanism of the following reactions:

   (i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by
   hydrolysis.

   (ii) Acid catalysed dehydration of an alcohol forming an alkene.

   (iii) Acid catalysed hydration of an alkene forming an alcohol.

   Solution:
   (i) Grignard’s reagent is an alkyl magnesium halide. The alkyl group has a partial negative charge,
   whereas the magnesium group has a partial positive charge. The alkyl group attacks the carbon of
   the carbonyl group to form an addition compound.




   Grignard’s reagent acts as a nucleophilic agent and attacks electrophilic carbon atoms to yield a
   carbon − carbon bond. The addition to the nucleophile is an irreversible process due to the high pka
   value of the alkyl group.




   (ii) When heated with concentrated sulphuric acid, phosphoric acid or boric acid, alcohols undergo
   dehydration to form alkenes. The mechanism of this reaction involves the protonation of alcohol,
   followed by loss of a water molecule and a proton.


   (a)




   (b)



…meritnation.com/…/Cuy6WobDThiILNv…                                                                              8/14
2/4/2011                                        Subjective Test Paper - Chemistry - Meri…

   (c)



   During the dehydration of alcohol, the intermediate carbocation may undergo re-arrangement,
   resulting in the formation of a stable carbocation.

   (iii) Some reactive alkenes like 2 − methyl propene undergo direct hydration in the presence of
   mineral acids which act as catalysts. The addition of water to the double bond takes place in
   accordance with Markonikoff’s rule.




    Question 26 ( 3.0 marks)
   Giving an example for each, describe the following reactions:

   (i) Hofmann’s bromamide reaction

   (ii) Gatterman reaction

   (iii) A coupling reaction

   Solution:
   (i) Hofmann’s bromamide reaction: It involves the reaction of bromine with an acid amide in the
   presence of an alkali. It results in the formation of a primary amine with one carbon less than the
   parent compound. Here, the alkyl group migrates from carbonyl, with the elimination of CO 2 . For
   example:




   (ii) Gatterman reaction: This is a modification of Sandmeyer reaction in which benzenediazonium
   chloride is treated with copper powder and halogen acid to form aryl halides.




   (iii) Coupling reaction: It is the reaction of diazonium salts with phenols and aromatic amines to form
   azo compounds of the general formula Ar − N = N − Ar. The coupling of phenol takes place in a mildly
   alkaline medium.




    Question 27 ( 3.0 marks)
   Explain the following types of substances with one suitable example, for each case:

   (i) Cationic detergents

   (ii) Food preservatives

   (iii) Analgesics

   Solution:
   (i) Cationic detergents are quaternary ammonium salts such as chlorides and acetates.

   Example − Cetyl trimethyl ammonium chloride




…meritnation.com/…/Cuy6WobDThiILNv…                                                                          9/14
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   These detergents are very good cleansing agents and are used as germicidals.

   (ii) Food preservatives are chemical substances used for inhibiting the growth of micro-organisms in
   food materials so as to prevent their spoilage.

   Example − benzoic acid and sulphur dioxide

   Benzoic acid is used for preserving fruits, fruit juices, jams, etc.,as it is soluble in water, while sulphur
   dioxide is used for the preservation of colourless food materials.

   (iii) Analgesics are the chemical substances used for relieving pain. They are also used for alleviating
   fever.

   Example − Aspirin, Analgin, Novalgin




   Section D


    Question 28 ( 5.0 marks)
   (a) Define molar conductivity of a substance and describe how for weak and strong electrolytes,
   molar conductivity changes with concentration of solute. How is such change explained?

   (b) A voltaic cell is set up at 25°C with the following half cells:

   Ag + (0.001 M) | Ag and Cu 2+ (0.10 M) | Cu

   What would be the voltage of this cell?

   (                  )

                                                           OR

   (a) State the relationship amongst the cell constant of a cell, the resistance of the solution in the cell
   and the conductivity of the solution. How is molar conductivity of a solute related to the conductivity
   of its solution?

   (b) A voltaic cell is set up at 25°C with the following half-cells:

   Al | Al3+ (0.001 M) and Ni | Ni2+ (0.50 M)

   Calculate the cell voltage.




   Solution:
   (a) Molar conductivity is defined as the conductance of a solution containing 1 g molecule or 1 mol of
   electrolyte such that the entire solution is placed between two electrodes one cm apart. It is
   denoted as ^m. In the case of strong electrolytes, when the concentration is zero, molar
   conductivity attains a definite value known as limiting molar conductivity. It is denoted by ^m, which
   can be calculated.

   For weak electrolytes, when concentration reaches zero, the graph of ^m vs.                       becomes
   parallel to the y-axis. Thus, the limiting molar conductivity cannot be calculated.


   (b)



                                                           OR

         1. The conductivity (   ) of the solution in a cell is the reciprocal of its resistivity.




   The quantity       is cell constant.
…meritnation.com/…/Cuy6WobDThiILNv…                                                                                10/14
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…

   l → Distance between 2 electrodes

   a → Area of cross section

   R → Resistance


   Also, conductivity,


   Thus, ^m =

   (b) For the given cell,

   Eº ce ll = Eº right − Eº le ft




   Also, E ce ll = Eºcell +



   The net reaction is

   ∴ n =6




    Question 29 ( 5.0 marks)
   (a) Complete the follow ing chemical reaction equations:

   (i)


   (ii)

   (b) Explain the following observations about the transition/inner transition elements:

   (i) There is in general an increase in density of element from titanium (Z = 22) to copper (Z = 29).

   (ii) There occurs much more frequent metal−metal bonding in compounds of heavy transition
   elements (3 rd series).

   (iii) The members of the actinoid series exhibit a larger number of oxidation states than the
   corresponding members of the lanthanoid series.

                                                      OR

   (a) Complete the follow ing chemical equations for reactions:

   (i)


   (ii)

   (b) Give an explanation for each of the following observations:

   (i) The gradual decrease in size (actinoid contractions) from element to element is greater among the
   actinoids than among the lanthanoids (lanthanoid contraction).

   (ii) The greatest number of oxidation states are exhibited by the members in the middle of a
   transition series.

   (iii) With the same d-orbital configuration (d4 ) Cr2+ ion is a reducing agent but Mn 3+ ion is an
   oxidising agent.
…meritnation.com/…/Cuy6WobDThiILNv…                                                                        11/14
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…

   Solution:
   (a)

   (i)


   (ii)

   (b)

   (i) On moving from titanium (Z = 22) to copper (Z = 29), electrons get added to the 3d orbital. So, the
   mass per unit volume increases. As a result, density also increases.

   (ii) In the third transition series, due to the introduction of the d orbital, the shielding effect
   decreases. Therefore, the effective nuclear charge increases. Consequently, the atomic volume
   increases. Due to an increase in mass by volume ratio, density increases.

   (iii) The lanthanoids and actinoids exhibit a principal oxidation state of +3. Some lanthanoids also
   exhibit +4 and +2 oxidation states. However, actinoids exhibit oxidation states of +2, +4 ,+5, +6 and
   +7. They exhibit greater range of oxidation states because the 5f, 6d and 7s sub-shells present in
   actinoids are of comparable energy. Thus, they can take part in bonding, giving rise to different
   oxidation states.

                                                      OR

   (a)

   (i)


   (ii)

   (b)

   (i) The gradual decrease in size (actinoid contraction) from element to element is greater among the
   antinoids than among the lanthanoids. This is because of the poor shielding effect of electrons in the
   5f orbital of actinoids

   (ii) The elements at the beginning of a series exhibit fewer oxidation states because they have less
   number of electrons which they can lose or contribute for bond formation. The elements at the end
   of a series exhibit fewer oxidation states because they have too many d electrons, and hence, fewer
   vacant d orbitals that can be involved for bonding.

   (iii) For Cr, the +3 oxidation state is more stable than the +2 state. Thus, Cr2+ changes to Cr3+ and
   it behaves as a strong reducing agent. However, for Mn, the +2 state is more stable than the +3
   state. So, Mn3+changes into Mn2+ and it behaves as a strong oxidising agent.



    Question 30 ( 5.0 marks)
   (a) Illustrate the following name reactions by giving example:

   (i) Cannizzaro’s reaction

   (ii) Clemmensen reduction

   (b) An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The
   molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition
   compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation it
   gives ethanoic and propanoic acids. Derive the possible structure of compound A.

                                                      OR

   (a) How are the following obtained?

   (i) Benzoic acid from ethyl benzene

   (ii) Benzaldehyde from toluene

   (b) Complete each synthesis by giving the missing material, reagent or products:

   (i)

   (ii)




   (iii)


…meritnation.com/…/Cuy6WobDThiILNv…                                                                          12/14
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…




   Solution:
   (a)

   (i) In Cannizzaro’s reaction, aldehydes which do not have α-H atom undergo self oxidation and
   reduction on treatment with a concentrated alkali, an alcohol and an acid.




   Thus, 1 molecule is oxidised to alcohol and the other is reduced to carboxylic acid salt.

   (ii) In Clemmenson’s reduction, the carbonyl group of aldehydes and ketones is reduced to CH2
   group on treatment with zinc amalgam and conc. HCl.




   (b) Since the compound does not reduce Tollen’s reagent, it has a ketonic group.

   Also, as it forms an addition compound with NaHSO 3 and gives positive iodoform test, the presence
   of methyl ketone is confirmed.

   On oxidation, it gives ethanoic acid and propanoic acid.

   So, the compound can be




   This is because in unsymmetrical ketone, the point of cleavage is such that the keto group stays with
   the smaller alkyl group.




   The compound is pentan − 2 − one, with molecular mass 86.

   (a)

   (i)




   (ii)




   (b)


   (i)


   (ii)




   (iii)

…meritnation.com/…/Cuy6WobDThiILNv…                                                                        13/14
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…meritnation.com/…/Cuy6WobDThiILNv…                                               14/14

								
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