Chemistry : Part II 2010 CBSE [ Delhi ] Set II

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  • Q1

    What is an emulsion?

    Marks:1
    Answer:

    Emulsion is a colloid in which the dispersed phase and dispersion medium are liquids. If a mixture of two immiscible or partially miscible liquids is shaken, a coarse dispersion of one liquid in the other is obtained, which is called emulsion.

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  • Q2

    Define ‘order of a reaction’.

    Marks:1
    Answer:

    The sum of the exponents (powers) of the concentration of reactants in the rate law expression is termed as order of that chemical reaction.

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  • Q3

    A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl. Which one of these is more easily hydrolysed?

    Marks:1
    Answer:

    CH3CH2CH2CH2Cl undergoes hydrolysis more easily than CH3CHClCH2CH3.CH3CH2CH2CH2Cl being a primary alkyl halide has less steric hindrance than CH3CHClCH2CH3 which is secondary alkyl halide.

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  • Q4

    Draw the structural formula of 1-phenylpropan-1-one molecule.

    Marks:1
    Answer:

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  • Q5

    Give the IUPAC name of


    H2N – CH2 – CH2 – CH = CH2

    Marks:1
    Answer:


    But-3-en-1-amine

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  • Q6

    Which point defect in crystals of a solid does not change the density of the solid?

    Marks:1
    Answer:

    Frankel defect does not change the density of a solid. This defect arises when ion is missing from its lattice site and it occupies the interstitial site.

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  • Q7

    What is the oxidation number of phosphorus in H3PO2 molecule?

    Marks:1
    Answer:

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  • Q8

     Give an example of coordination isomerism.

    Marks:1
    Answer:

    [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]

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  • Q9

    Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type.

    Marks:2
    Answer:

    These deviations are caused due to difference in intermolecular forces of attraction between solute molecules in pure solute, solvent molecules in pure solvent and solute and solvent in a solution.

    Non ideal solutions showing positive deviation

    Non ideal solutions showing negative deviation

    The intermolecular attractive forces between solute-solvent molecules are weaker than those between solute-solute and solvent-solvent molecules, i.e., A-B < A-A and B-B interactions.

    The intermolecular attractive forces between solute-solvent molecules are stronger than those between solute-solute and solvent-solvent molecules, i.e., A-B > A-A and B-B interactions.

    Example:

    Mixture of ethanol and acetone. In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them. Due to weakening of interactions, the solution shows positive deviation from Raoult’s law.

     

    Example:

    Mixture of chloroform and acetone. This is because chloroform molecule is able to form hydrogen bond with acetone molecule. This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting in negative deviation from Raoult’s law.

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  • Q10

    A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times (ii) the concentrations of A as well as B are doubled?

    Marks:2
    Answer:

    Since it is given that a reaction is first order in reactant A and second order in reactant B; therefore,

     r = k [A] [B]2 …………………………. (i)

    Where r is the rate of reaction and k is the rate constant of the reaction.

    Case 1

    When concentration of B alone is increased three times, let the new rate be r1

    r1= k [A] [3B]2  = 9k [A] [B]2  …………………………. (ii)

    Dividing eq.(ii) by eq.(i), we get

    r1= 9r

    Therefore, the rate of the reaction would become 9 times the initial rate when the concentration of B alone is increased three times.

    Case 2

    When the concentration of both the reactants is doubled, then the rate of the reaction would be r2

    r2= k [2A] [2B]2  = 8k [A] [B]2  …………………………. (iii)

     

    Dividing eq.(iii) by eq.(i), we get

    r2= 8r

     

    Therefore, the rate of the reaction would become 8 times the initial rate when the concentration of both A and B is doubled.

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