Chemistry:2010:CBSE:[Delhi]:Set-I
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Q1
What is an emulsion?
Marks:1View AnswerAnswer:
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
Write a feature which will distinguish a metallic solid from an ionic solid.
Marks:1View AnswerAnswer:
Metallic solids are conductors of electricity in solid state as well as in molten state, but the ionic solids are insulators in solid state and conductors in molten and liquid state.
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Q3
Define ‘order of a reaction’.
Marks:1View AnswerAnswer:
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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Q4
Why does NO2 dimerise?
Marks:1View AnswerAnswer:
NO2 contains odd number of valence electrons. It behaves as an odd electron molecule and therefore undergoes dimerisation to form stable N2O4 molecule with even number of electrons.
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Q5
Give an example of linkage isomerism?
Marks:1View AnswerAnswer:
[Co(NH3)5ONO]Cl2 and [Co(NH3)5NO2]Cl2 are linkage isomers. In [Co(NH3)5ONO]Cl2 , the bonding is through oxygen (–ONO) whereas in
(–NO2). -
Q6
A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl. Which one of these is more easily hydrolysed?
Marks:1View AnswerAnswer:
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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Q7
Draw the structural formula of 1-phenylpropan-1-one molecule.
Marks:1View AnswerAnswer:
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Q8
Give the IUPAC name of
H2N – CH2 – CH2 – CH = CH2
Marks:1View AnswerAnswer:
But-3-en-1-amine -
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:2View AnswerAnswer:
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:2View AnswerAnswer:
Since it is given that a reaction is first order in reactant A and second order in reactant B; therefore,
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.
