CHEMISTRY (THEORY) – 2004
SET- I (DELHI BOARD)
Class- XII Delhi Board Papers
Time allowed: 3 hours
General instructions:
Maximum Marks: 70
(i)
All questions are compulsory.
(ii)
Marks for each question are indicated against it.
(iii)
Question numbers 1 to 8 are very short-answer questions carrying 1 mark
each. Answer these in one word or about one sentence each.
(iv)
Question numbers 9 to 18 are short-answer questions, carrying 2 marks
each. Answer these in about 30 words each.
(v)
Question numbers 19 to 27 are short answer questions, carrying 3 marks
each. Answer these in about 40 words each.
(vi)
Question numbers 28 to 30 are long answer questions of carrying 5
marks each. Answer these in about 70 words each.
(vii) Use Log Tables, if necessary. Use of calculators is not permitted.
______________________________________________________________________
Q. 1. Mention a large scale use of the phenomenon called ‘ reverse osmosis ’.
1
Ans. Reverse osmosis is used in the desalination of sea water.
Q. 2. Give an example of a pseudo first order reaction.
1
Ans. CΗ3CΟΟC2 Η5 + Η 2Ο → CΗ3CΟΟΗ + C2 Η5ΟΗ
1
0
Rate = k [ CΗ3CΟΟC2 Η5 ] [ Η 2 Ο]
Q. 3. Mention two properties of acetonitrile because of which it acts as a good
solvent.
1
Ans. Acetonitrile is a good solvent because.
(i)
it is non reactive in mild acidic and basic conditions.
(ii)
Having high polarity it is capable of dissolving a variety of reactants.
*Q. 4. What are the types of lattice imperfections found in crystals?
Ans. Three types of lattice imperfections are possible.
(a)
Schottky defect
(b)
Frenkel defect
(c)
Interstitial defects
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1
*Q. 5. What is the principal ore of Iron?
1
Ans.
Haematite: Fe2Ο3 .
Q. 6.
Write the chemical reaction to transform butanal to butanoic acid.
1
Ammonical ΑgΝΟ
3
Ans. CΗ3CΗ 2CΗ 2CΗΟ 
→ CΗ3CΗ 2 CΗ 2CΟΟΗ
(Tollen's reagent)
Butanal
Q. 7.
Butanoic acid
Name the type of structure possessed by a unit cell of CsCI.
1
Ans. Body centred cubic structure.
Q. 8. Write the IUPAC name of the following compounds:
1
CΗ3 - CΟ - CΗ - CΗ 2 - CΗ 2CΙ
CΗ 3
Ans. IUPAC Name: 5 chloro 3 methyl 12-pentanone.
*Q. 9. Explain Brownian Movement.
2
Ans. Brownian Movement: When the colloidal particles are observed under the
ultramicroscope, the particles are seen to be in constant motion in zig-zag path.
This zig-zag motion of dispersed phase particles is called Brownian Movement.
Importance: Avogadro’s number can be calculated with the help of Brownian
movement.
*Q.10. Distinguish between reaction rate and reaction rate constant (Specific
reaction rate) of a reaction.
Ans.
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Reaction rate
Reaction rate constant
1. The rate of reaction is the rate
of disappearance of reactant
or rate of the appearance of
product.
1. The rate of constant or specific
reaction rate in equal to the rate
of reaction when the
concentration of the reactant is
unity.
2. The rate of reaction can be
calculated by calculating the
decrease in concentration of
the reactant in unit time.
For reaction Α → Β rate of
reaction
2. Rate constant can be calculated
with the help of rate law.
rate = K[A]
where K = rate constant
[A] = concentration of reactant
The value of K depends upon the
order of reaction and
temperature.
=-
d [ Α]
d [ Β]
=+
dt
dt
Q.11. 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
Ans. (i)
2
The cell reactions during the use (discharge) of lead storage battery are
At anode
: Ρb ( s ) + SΟ 2- 4 → ΡbSΟ 4 ( s ) + 2e At cathode : ΡbΟ 2 + 4Η + + SΟ 2- 4 + 2e - → ΡbSΟ 4 ( s ) + 2Η 2Ο ( l )
Net reaction : Ρb + ΡbΟ 2 + 4Η + + 2SΟ 2- 4 → 2ΡbSΟ 4 ( s ) + 2Η 2Ο ( l )
(ii)
During charging of the battery, the above reactions are reversed.
At anode
: ΡbSΟ 4 ( s ) + 2e - → Ρb ( s ) + SΟ 2- 4 ( aq )
At cathode : ΡbSΟ 4 ( s ) + 2Η 2Ο → ΡbΟ2 ( s ) + 4Η + + SΟ 2- 4
Net reaction : 2ΡbSΟ 4 ( s ) + 2Η 2Ο → Ρb ( s ) + ΡbΟ 2 ( s ) + 4Η + + 2SΟ 2- 4
Q.12. Give chemical reaction in support of each of the following statements:
(i)
The +1 oxidation state gets stabilized progressively from Ga to TI in
Group 13.
(ii)
All the bonds in Ρ C Ι 5 molecule are not equivalent.
Ans. (i)
2
Of GaX and TIX, the gallium halide undergoes disproportion while thallium
halide not do so.
3GaΧ ( s ) → 2Ga ( s ) + Ga 3+ ( aq ) + 3Χ - ( aq )
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(ii)
When heated with water, ΡCΙ3
ΡCΙ5 + Η 2 Ο → ΡΟCΙ3 + 2ΗCΙ
Or,
∆
ΡCΙ5 
→ ΡCΙ3 + CΙ 2
*Q.13. Expand DDT. Write structure.
2
Ans. DDT is Dichloro-Diphenyl-trichloroethane.
2, 2- Di (p-chlorophenyl) –1, 1, 1-trichloroethane.
Q.14. Write the names of the reagents and equations in the conversion of
(i)
phenol to salicyl aldehyde
(ii)
anisole to p-methoxyacetophenone
Ans. (i)
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2
(ii)
Q.15. Write the modes of free radical polymerization of an alkene.
2
Ans. Free radical polymerization of an alkene accurs through radicals generated by an
initiator. Initiators are molecules, which decompose easily to provide free
radicals.
Chain initiating steps:
•
Initiator → Ι n
(organic peroxide)
•
•
Ι n+ CΗ 2 = CΗ 2 → Ιn - CΗ 2 - C Η 2
Chain propagation Step:
•
•
Ιn - CΗ 2 - C Η 2 + CΗ 2 = CΗ 2 → Ιn - CΗ 2 - CΗ 2CΗ 2 - C Η 2
Chain terminating Step:
•
Ιn - CΗ 2 - CΗ 2CΗ 2 - C Η 2 → Ιn(CΗ 2 - CΗ 2 )n
Polymer
Differentiate between addition and condensation polymers based on mode
of polymerization. Give one example of each type.
2
Ans. Addition polymers: The polymers formed by the direct addition of a large
number of monomers are called addition polymers. For example polythene,
polystyrene.
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Condensation polymers: The polymers formed by the condensation of two or
more than two monomers by the loss of molecules such as Η 2 Ο, ΗCΙ, are called
condensation polymers. For example, nylon-66, bakelite.
*Q.16.
Differentiate between Lanthanides and Actinides series. 2
Ans.
Lanthanides
1. Last electron goes to 4f
orbitals. The general
electronic.
Configuration is
4 f −14 5d 0−1 6 s 2
2. They show only +2, +3 and
+4 oxidation states.
Actinides
1. Last electron goes to 5f orbitals.
The general electronic configuration
is 5 f 0 −14 6 d 0 −1 7 s 2 .
2. They show a variable oxidation
states from +2 to +7.
3. Most of their ions are coloured.
3. Most of their ions are
colourless.
4. They have much tendency to form
complexes.
4. They do not form complexes
easily.
5. Their compounds are more basic.
5. Their compounds are less
basic.
7. All are radioactive.
6. Their paramagnetic character can
not be explained easily.
6. Their paramagnetic character
can be explained easily.
7. Except promethium, they are
non-radioactive.
Q.17. Calculate the density of silver which crystallizes in the face-centered cubic
structure. The distance between the nearest silver atoms in this structure
is 287pm.
2
-1
23
-1
(More mass of Αg = 107.87g mol , Ν Α = 6.02×10 mol )
Ans. Distance between two nearest neighbour Ag atoms.
d=
Face diagonal
2a
a
=
=
2
2
2
∴Edge length, a = 2d = 1.414× 287 = 406 pm
In f.c.c. unit cell of silver, number of atoms = 4.
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4× Atomic mass of Αg
Avogadro's number
4 × 107.87
=
= 7.17 × 10−22 g
6.02 × 1023
∴Mass of unit cell =
3
Volumeof unit cell = a 3 = ( 406×10-10 ) cm3
= 6.692 × 10−23 cm3
Mass of unit cell
7.17 ×10-22 g
Density =
=
= 10.71g cm-3
-23
Volumeof unit cell 6.692×10
Q.18. Two elements A and B form compounds having molecular formula ΑΒ 2
and ΑΒ4 . When dissolved in 20g of benzene, 1g of ΑΒ 2 lowers the
freezing point by 2.3K, whereas 1 g of ΑΒ 4 lowers it by 1.3K? The molar
depression constant for benzene is 5.1 K kg mol-1 . Calculate the atomic
masses of A and B.
Ans. Let a and b be the atomic masses of A and B.
Molecular mass of ΑΒ2 = a + 2b
Molecular mass of ΑΒ4 = a + 4b
1000Κ f × WΒ
∆Τ f × WΑ
1000 × 5.1 × 1
For ΑΒ2 , a + 2b =
= 110.87
2.3 × 20
1000 × 5.1 × 1
For ΑΒ4 ,a + 4b =
= 196.15
1.3 × 20
As Μ Β =
Solving (1) and (2), we get,
∴
Or
2b = 196.15 − 110.87 = 85.28
b = 42.64
a = 110.87 − 2 × 42.64 = 25.89
Atomic mass of A = 25.89
Atomic mass of B = 42.64
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…… (1)
…… (2)
2
*Q.19. Explain the Froth Floatation Method.
Ans. Froth Floatation Method-
The method has been in use for removing gangue from sulphide ores. In this
process a suspension of the powdered ore is made with water. To it, collectors
and forth stabilizers are added. Collectors (e.g., pine oils, fatty acids, xanthates,
etc.) enhance non-wattability of the mineral particles and froth stabilizers (e.g.,
Cresols, aniline) stabilize the froth.
The mineral particles become wet by oils while the gangue particles by water. A
rotating paddle agitates the mixture and draws air in it. As a result, froth is
formed which carries the mineral particles. The froth is light and is skimmed off.
It is then dried for recovery of the ore particles.
Q.20. Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 Scm-1 . Calculate its
molar conductivity. If ∧° for acetic acid is 390.5Scm 2 mol-1 , what is its
dissociation constant?
3
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Ans.
∧ acetic acid =
Κ ×1000 7.896×10-5 ×1000
=
C
0.00241
= 32.76Scm 2 mol-1
Degree of dissociation of acetic acid,
∝=
Λ 32.76
=
= 0.084
Λ 0 390.5
Dissociation constant,
2
C ∝2 0.000241× ( 0.084 )
Κa =
=
= 1.836 ×10−5
1− ∝
1.0.084
Q.21. 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?
1
3
2
Ans. (i) rate = k [ Α ] [ Β]
1
2
(ii) r0 = k [ Α] [ Β]
1
2
1
2
r1 = k [ Α ] [3Β] ⇒ r1 = Κ [ Α] .9 [ Β]
∴ r1 = 9× r0 .
Hence rate becomes 9 times the initial rate.
1
2
(iii) r0 = k [ Α] [ Β]
1
∴
2
1
2
r2 = k [ 2Α ] [ 2Β] ⇒ r1 = Κ 2 [ Α ] .4 [ Β]
r2 = 8× r0 .
Hence rate becomes 8 times the initial rate.
Rate constant of a reaction is equal to the rate of the reaction when
concentration of each of the reactant is unity.
Q.22. How are the colloids classified on the basis of the nature of interaction
between depressed phase and dispersion medium? Describe an important
characteristic of each class. Which of these sols need stabilizing agents of
preservation?
3
Ans. Depending upon the nature of interaction between the dispersed phase and the
dispersion medium, colloids are divided into two categories:
(i)
Lyophilic sols: The colloids in which the particles of the dispersed phase
have a strong affinity for the dispersion medium are called lyophilic sols.
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(ii)
These colloidal solutions, even if precipitated, change back to the colloidal
form simply by adding dispersion medium. So lyophilic sols are reversible
in nature e.g., glue, starch, rubber, etc.
Lyophobic sols: The colloids in which particles of the dispersed phase
have no or very little affinity for dispersion medium are called lyophobic
sols. These are irreversible in nature i.e., once precipitated, they have
little tendency to get back into the colloidal form on simply adding
dispersion medium e.g., Αs 2S 2 sol.
Lyophobic sols need stabilizing agents for their preservation.
Or
What are detergents? Give their scheme of classification. Why are the
detergents preferred over soaps?
Ans. A detergent is a surface-active agent used for cleaning dirty surfaces. It contains
a nonpolar hydrocarbon chain (hydrophobic part) and polar group (hydrophilic
part) with in the molecules and thus shows adsorption and micellization.
On the basis of charge on polar part, detergents are classified as follows:
(i)
Anionic detergents in which large part of the molecule acts as anionic
e.g., alkyl benzene-sulphonate.
(ii)
Cationic detergents, which are mostly acetates or chlorides of
quaternary amines e.g., Cetyltrimethyl ammonium chloride.
(iii)
Nonionic detergents like esters of high molecular mass formed by
reaction between polyethylene glycol and stearic acid.
Detergents are preferred over soaps as they work even in hard water and acidic
phase where soaps become insoluble. They have powerful cleansing action.
Q.23. 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?
Ans. Greater the value of ∆ 0, the greater are the electron pairing. Figure shows the
splitting of degenerate d orbitals in an octahedral crystal field.
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The actual configuration adopted is decided by the relative values of ∆ 0 and P.
Here P represents energy required for electron pairing in a single crystal.
(i)
If ∆ 0 < Ρ, we have the so called weak field, high spin situation. The
fourth electron enters one of the e g orbitals giving the configuration
t 32 g e1g . If now a fifth electron is added to a weak field coordination entity,
the configuration becomes t 32g e 2 g .
(ii)
If ∆ 0 > Ρ, we have the strong field, low spin situation and pairing will
occur in the t 2g level with the e g level remaining unoccupied in entities of
d1 of d 6 ions.
*Q.24. Explain the following terms with suitable examples:
(i)
Frenkel defect
(ii)
Interstitials
Ans. (i) Frenkel defect: (a) This defect is caused when a cation leaves its normal
lattice side and occupies an interstitial site.
(b) This defect does not affect the density of the solid.
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(c) This defect is shown by ionic solids having low coordination number
and in which anions are larger in size than cation.
Examples : Zn S, AgCI
(ii) Interstitials: When the atoms of some substance occupy the vacant sites (
tetrahedral or octahedral voids) in the crystal of a particular substance, then the \
defect is called interstitials.
*Q.25. Draw the structures of monomers for the following polymers. Also draw
the structures of the polymers and use of:
(i)
Teflon
(ii)
PMMA
(iii) Buna-S.
3
Ans. (i) Teflon : (Polytetrafluoro ethylene):
Monomer name : Tetrafluoroethylene
Structure:
Structure of Polymer: ( CF2 - CF2 ) n
Uses: It is used as lubricant, insulator and making cooking wares.
(ii) PMMA (Polymethyl metha acrylate or Plexi glass)
Monomer Name: Methyl metha acrylate
Structure : CΗ 2 - C - CΟΟCΗ 2
CΗ 3
CΗ 3
Structure of Polymer: ( CΗ 3 - C ) n
CΟΟCΗ 3
Uses: It is used as substitute of glass and making decorative materials.
(iii) Buna-S (Styrene butadiene rubber)
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Monomer Name: CΗ 2 = CΗ - CΗ = CΗ 2
1, 3, - Butadiene
and
CΗ = CΗ 2 (styrene)
C6 Η 5
Polymer name: ( CΗ 2 - CΗ = CΗ - CΗ 2 - CΗ - CΗ 2 ) n
C6Η 5
Uses: It is used in making automobile tyres and footwear.
Q.26. (a) Identify A and B in the following:
(i)
(ii)
(b) Explain why the aromatic amines are less basic then ammonia and
aliphatic amines.
Ans. (a)
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3
ΝΗ
Νi / Η
3
2
(i) R 2CΟ 
→ R - C = ΝΗ 
→ R - CΗ - ΝΗ 2
-Η Ο
2
R
(A)
R
(B)
b) Aromatic ring has electron withdrawing nature due to resonance. This
decreases the availability of electron pair on nitrogen, thereby making it less
basic then ammonia and aliphatic amines.
*Q.27. Write the IUPAC names of the following
(i)
(ii) CΗ3 CΟ CΗ 2 CΗ 2 CΗ3
(iii) CΗ3 - CΗ - Βr
CΗ 3
Ans. (i) 2-hydroxy-benzaldehyde
(ii) Pentan-2-one
(iii) 2-Bromopropane
Q.28. (a) Assign appropriate reason for each of the following observations:
(i)
Anhydrous ΑΙCΙ 3 is used as a catalyst.
(ii)
Phosphinic acid behaves as a monoprotic acid.
(iii)
SF6 is not easily hydrolysed whereas SF4 is readily nydrolysed.
(iv)
No form of elemental silicon is comparable.
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3
(b)
Draw the structure of ΧeΟF4 or ΒrF3 .
5
Ans. (a) (i) The central AI atom is ΑΙCΙ3 is electron deficient. Because of its lewis
acid character, ΑΙCΙ3 acts as a catalyst in the Friedel Craft’s alkylation and
acylation of benzene and helps in generating electrophile for attack on the ring.
(ii) Phosphinic acid, Η 3ΡΟ 2 , has one Ρ(ΟΗ ) group and two hydrophilic directly
attached to phosphorous. It has only one ionisable hydrogen and hence
benzenes as a monoprotic acid.
(iii) SF6 is inert due to the presence of sterically protected sulphur atom which
does not allow thermodynamically favourable reactions like hydrolysis to take
place. On the other hand, the less sterically hindered SF4 undergoes hydrolysis.
(iv) The size of silicon atom is larger than that of carbon atoms. So pπ - pπ
bonding is not so effective in silicon atoms. However, this is present in carbon
atoms in graphite of smaller size. So elemental silicon cannot give graphite like
structure. But it resemble diamond and is very hard poor conduct of electricity.
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Or
Account for the following:
(i)
Ammonia is a stronger base than phosphine
(ii)
The tendency of exhibit +2 oxidation state increase with increasing
atomic number in group 14.
(iii) HF is weaker acid than HI.
3
Ans. Account for the following:
(i)
Ammonia is stronger base than phosphine: Both possess a lone pair
of electrons on their central atom. They donate this lone pair to some
other species and thus behave as Lewis bases. Since atomic size of
phosphorous is more than that of nitrogen therefore, negative charge on
phosphorous gets diffused, hence ammonia is stronger base than
phosphine.
(ii)
As we move down the group ‘ns’ pair of electrons becomes more and
more reluctant in bond formation. Electronic configuration of the elements
of group 14 is ns 2 np 2 . Thus only 2 p-orbital electrons are left for bond
formation. Therefore, as we more down the group 14 tendency to exhibit
‘+2’ oxidation state increases.
(iii)
‘F’ is more electronegative element. Therefore, hydrogen bonding among
HF molecules is very strong. Thus liberation of proton becomes very
difficult. In case of HI the state of affair is just opposite. Hence HF is a
weaker acid than HI.
Q.29. (a) How would you account for the following:
(i)
The transition elements exhibit high enthalpies of atomization.
(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
that in the lanthanoids.
(b)
Write the complete chemical equation for each of the following:
(i)
Al alkaline solution of ΚΜnΟ 4 reacts with an iodide.
(ii)
An excess of SnCΙ 2 solution is added to a solution of mercury (II)
chloride.
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5
Ans. (a) (i) Because of having large number of unpaired electrons in their atoms,
they have stronger inter atomic interaction and hence stronger bonding
between atoms. Consequently, transition elements exhibit high enthalpies
of atomization.
(ii) The transition metals of 4d and 5d series have greater enthalpies of
atomization then the corresponding elements of 3d series. This indicates
that the metals of 4d and 5d series have more frequent metal-metal bonding
in their compounds than do the 3d metals.
(iii) Lenthanoids mainly exhibit +3 oxidation state. In addition, they show
lower oxidation state of +2 and higher oxidation state of +4, Actinoids also
exhibit +3 oxidation state. In addition, they show a wide range of higher
oxidation states from +4 to +7. This is due to comparable energies of 5f, 6d
and 7s orbitals in actinoids.
(b) (i) 2ΚΜnΟ 4 + ΚΙ + Η 2 Ο → ΚΙΟ3 + ΜnΟ 2 + 2ΚΟΗ
(ii) SnCΙ 2 first reduces mercury (II) chloride into mercurous chloride (white)
and then into mercury (black).
2ΗgCΙ 2 + SnCΙ → Ηg 2 CΙ2 + SnCΙ4
White
Ηg 2 CΙ 2 + SnCΙ 2 → 2Ηg + SnCΙ 4
Black
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)
boiling point of the solution
(iii) lowering of vapour pressure of water at 298 K.
[Given Κ f = 1.86 Κ kg mol-1 , Κ b = 0.512 Κ kg mol-1 and vapour pressure of
water at 298 K = 23.756 mm Hg].
Ans. (i) ∆Τ f = k f .m
⇒
1
kf
1.86Κ mol-1
=
=
m ∆Τ f ( 273Κ - 272.4Κ )
Or
1 1.86 Κ kg mol-1
=
= 3.1kg mol-1
m
0.6 Κ
Or
Or
1
= mol kg -1
3.1
m = 0.3225mol kg -1
m=
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(ii) ∆Τ f = k b × m
= 0.512 Κ kg mol-1 × 0.3225mol kg
= 0.1612 Κ
Τb = 373 Κ + 0.1612 Κ = 373.1612 Κ .
Q.30. 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.
5
Ans. The products of complete hydrolysis of DNA are
(i)
a pentose sugar
(ii)
two types of heterocylic nitrogenous bases and
(iii)
phosphoric acid.
The structural differences between DNA and RNA are
(i)
DNA has deoxyribose while RNA has a ribose sugar.
(ii)
DNA contains thymine and RNA has uracil.
(iii)
DNA is double stranded while RNA is single stranded.
Nucleosides: The N-glycosides of purine or pyrimidine bases with pentose
sugars are know as nucleosides.
Nitrogen Base + Sugar = Nucleoside
Nucleotides: A nucleotide is a phosphate ester of nucleoside and consists of a
purine or pyrimidine base, the 5-carbon sugar and one or more phosphate
groups.
Nitrogen Base + Sugar + Phosphate = Nucleotide.
Or
Define and classify vitamins. Name the diseases caused due to lack of any
three of them.
Ans. Vitamins are defines as essential dietary substances required by an
organism in minute quantities and their absence causes specific
deficiency diseases.
Vitamins cannot be synthesized by the body and have to be supplied in
the food.
Vitamins are classified into two types:
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(i
i) Fat soluble vitamins: These are oily substances not readily soluble in
water. For example, vitamins A, D, E and K.
(ii) Water soluble vitamins: This group included remaining vitamins
e.g., vitamin B group (B-Complex) vitamin C etc. These are stored in the
cells in much smaller amounts.
Name of vitamins
Vitamin A
Vitamin Β1
Vitamin C
Deficiency caused
Xerophthamia
Beriberi
Scurvy
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CHEMISTRY (THEORY) – 2004
SET- II (DELHI BOARD)
Class- XII Delhi Board Papers
Time allowed: 3 hours
General instructions :
Maximum Marks: 70
(i)
All questions are compulsory.
(ii)
Marks for each question are indicated against it.
(iii)
Question numbers 1 to 8 are very short – answer questions, carrying 1 mark
each. These are to be answered in one or two sentences.
(iv)
Question numbers 9 to 18 are short – answer questions, carrying 2 marks.
Answers to them should not exceed 30 words each.
(v)
Question numbers 19 to 27 are also short – answer questions, each carrying
3 marks. Answer to them should not exceed 50 words each.
(vi)
Question numbers 28 to 30 are long – answer questions, each carrying 5
marks. Answers to them should not normally exceed 150 words each.
(vii)
Use Log Tables, if necessary
Note: Except for the following questions, all the remaining questions have been
asked in Set-I.
Q.19. (a) Give the IUPAC name of [ΡtCΙ ( ΝΗ 2CΗ 3 )( ΝΗ 3 ) 2 ]CΙ.
(b) Compare the magnetic behaviour of the complex entities  Fe ( CΝ )6 
3-
and [ FeF6 ] . ( Fe = 26 ) .
Ans. (a) Diamminechloro (methyl amine) platinum (II) chloride.
4-
(b)  Fe ( CΝ )6  ion
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4-
Fe3 + ion hybridized:
4-
 Fe ( CΝ )6  ion formation :
Since the complex ion does not contain any unpaired electron, so it is
diamagnetic.
3-
[ FeF6 ]
ion
Fe3+ion :
As the complex ion contains five unpaired electrons, it is highly paramagnetic in
nature.
Q.20. Describe the following with suitable examples:
(i)
Preservatives
(ii)
Transquilizers
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3
Ans. (i) Preservatives: The chemical substances which are used to protect food
against bacteria, yeasts and moulds are called preservatives. The most common
preservative used is sodium benzoate, C6 Η 5CΟΟΝa .
(ii) Transquilizers: These medicines act on central nervous system and help in
reducing anxiety and relieve tension on the nerves. For example, valium.
Q.28. (a) How would you account for the following:
(i)
Cobalt (II) is stable in aqueous solution but in the presence of strong
ligands it is easily oxidized. (Co = 27).
(ii)
The transition metals form interstitial compounds.
(iii)
Silver halides find use in photography.
(b) What is lanthanoid contraction? Mention its main consequence.
5
Ans. (a) (i) Co (III) is stabilized because of higher crystal field splitting energy [or
strong ligand causes pairing of electrons to give more stable Co (III) ions].
(ii) Transition metals form a large number of interstitial compounds because
small atoms of certain non metallic elements (H, B, C, N etc.) get trapped in
voids or vacant spaces of lattice of the transition metal.
As a result of filling up of the interstitial spaces, such interstitial compounds are
hard and rigid.
(iii) Under the action of light, decomposition of silver bromide takes place in
traces where it is illuminated. The number of silver atoms formed is proportional
to the quantity of light falling on the surface.
(b) The steady decrease in the ionic radium from La 3+ to Lu 3+ is termed as
lenthanoid contraction.
Consequences of lenthanoid contraction are
(i)
Similar physical chemical properties among lenthanoids.
(ii)
Zr and Hf exist together.
Q.29. How does DNA replicate? Describe the mechanism of replication. How is
the process responsible for preservation of heredity?
5
Ans. Replication is the process by which a single DNA molecule produces two
Identical copies of itself, during cell division. During this process, the two strands
of the double helix first separate and each separated strand then serves as a
template for the synthesis of new strand. Due to specificity of base-pair, the
sequence in the two new strands is complementary to the original. In other
words, replication leads to the production of two identical copies of DNA from a
single parent DA and a copy of each then passes on to the new cells resulting
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from cell division. In this way, the hereditary effects are transmitted from one cell
to another.
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