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CBSE
XII
Board Paper – Set 1 (Delhi)
Chemistry Theory
[Time Allowed: 3 Hours]
[Maximum Marks: 70]
General
(i)
(ii)
(iii)
Instruction:
All the questions are compulsory.
Marks for each question are indicated against it.
Question numbers 1 to 8 are very short answer questions and carry
one mark each.
(iv) Question numbers 9 to 18 are short questions and carry two marks
each.
(v)
Question numbers 19 to 27 are also short answer question and
carry three marks each.
(vi) Question numbers 28 to 30 are long answer questions and carry
five marks each.
(vii) Use Log Tables, if necessary. Use of calculation is not allowed.
1.
2.
3.
4.
5.
6.
7.
8.
9.
‘Crystalline solids are anisotropic in nature’. What does this statement
mean?
Express the relation between conductivity and molar conductivity of a
solution held in a cell.
Define ‘electrophoresis’.
Draw the structure of XeF2 molecule.
Write the IUPAC name of the following compound:
(CH3)3CCH2Br
Draw the structure of 3-methylbutanal.
Arrange the following compounds in an increasing order of their
solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2
What are biodegradable polymers?
The chemistry of corrosion of iron is essentially an electrochemical
phenomenon. Explain the reactions occurring during the corrosion of
iron in the atmosphere.
10. Determine the values of equilibrium constant (Kc) and ∆G o for the
following reactions:
Ni (s) + 2Ag+ (aq) 
→ Ni2+ (aq) + 2Ag (s) , Eo = 1.05V
(1F = 96500 C mol−1)
11.
12.
Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction.
State reasons for each of the following:
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The N – O bond is NO-2 is shorter than the N – O bond in NO3(ii)
SF6 is kinetically an inert substance.
OR
State reason for the each of the following:
(i)
All the P-Cl5 molecule are not equivalent.
(ii)
Sulphur has greater tendency for catenation than oxygen.
13.
Assign reasons for the following:
(i)
Copper (I) ion is not known in aqueous solution.
(ii)
Actinoids exhibits greater range of oxidation states than
Lanthanoids.
14.
Explain the following giving one example for each:
(i)
Reimer-Tiemann reaction.
(ii)
Friedel Craft’s acetylation of anisole.
15.
How would you obtain:
(i)
Picric acid (2, 4, 6-trinitrophenol) from phenol,
(ii)
2-Methylpropene from 2-methylpaopanol?
16.
What is essentially the difference between α -form of glucose and β form of glucose? Expain.
17.
Describe what do you understand by primary structure and secondary
structure of proteins.
18.
Mention two important uses of each of the following:
(i)
Bakelite
(ii)
Nylon 6
(i)
19.
20.
Silver crystallizes in face-centred cubic unit cell. Each side of this
unit cell has a length of 400 pm. Calculate the radius of the silver
atom. (Assume the atoms just touch each other on the diagonal
across the face of the unit cell. That is each face atom is touching
the four corner atoms.)
Nitrogen pentaoxide decomposes according to equation:
2N 2O5 (g ) 
→ 4NO2 (g ) + O 2 (g )
This first order reaction was allowed to proceed at 40 oC and the
data below were collected:
[N2O5] (M)
Time (min)
0.400
0.00
0.289
20.0
0.209
40.0
0.151
60.0
0.109
80.0
(a) Calculate the rate constant. Include units with your answer.
(b) What will be the concentration of N2O5 after 100 minutes?
(c) Calculate the initial rate of reaction.
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21.
22.
23.
Explain how the phenomenon of adsorption finds applications in
each of the following processes:
(i)
Production of high vacuum
(ii)
Heterogeneous cataklysis
(iii). Froth floation process
OR
(i)
Micelles
(ii)
Peptization
(iii)
Desorption
Describe the principle behind each of the following processes:
(i)
Vapour phase refining of a metal.
(ii)
Electrolytic refining of a metal.
(iii)
Recovery of silver after ore was leached with NaCN.
Complete the following chemical equations:
+
2−
→
(i) MnO−
4 + C2 O 4 + H 
heated
(ii) KMnO 4 →
→
(iii) Cr2O72− + H 2S + H+ 
24.
Write the name, stereochemistry and magnetic behaviour of the
following:
(At. Nos. Mn = 25, Co = 27, Ni = 28)
(i)
K4[Mn(CN)6]
(ii)
[Co(NH3)5Cl] Cl2
(iii)
K2[Ni(CN)4]
25.
Answer the following:
(i)
Haloalkanes easily dissolve in organic solvents, why?
(ii)
What is known as a racemic mixture? Give an example.
(iii)
Of the two bromoderivatives, C6H5CH(CH3)Br and
C6H5CH(C6H5)Br, which one is more reactive in SN1
substitution reaction and why?
26.
(a)
Explain why an alkylamine is more basic than ammonia.
(b)
How would you convert:
(i)
Aniline to nitrobenzene
(ii)
Aniline to iodobenzene?
27
Describe the following giving one example for each:
(i)Detergents
(ii)Food preservatives
(iii)Antacids
28.
(a)
Difference between molarity and molality for a solution. How
does a change in temperature influence their values?
(b)
Calculate the freezing point of an aqueous solution containing
10.50 g of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g)
(Kf for water = 1.86 K kg mol-1)
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29.
30.
OR
(a) Define the terms osmosis and pressure. Is the osmotic pressure of
a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g
of NaCl to 250.0 g of water.(Kb for water = 0.512 K kg mol-1),
Molar mass of NaCl = 58.44 g.
(a)
Give the chemical test to distinguish between
(i)
Propanal and propanone,
(ii)
Benzaldehyde and acetophenone.
(b)
How would you obtain
(i)
But-2-enal from ethanal,
(ii)
Butanoic acid from butanol,
(iii)
Benzoic acid from ethylbenzene?
OR
(a)
Describe the following giving linked chemical equations:
(i)
Cannizzaro reaction
(ii)
Decarboxylation
(b)
Complete the following chemical equations:
(a)
(b)
Explain the following:
(i)
NF3 is an exothermic compound whereas NCl3 is not.
(ii)
F2 is most reactive of all the four common halogens.
Compete the following chemical equations:
(i)
→
C + H 2SO 4 (conc) 
→
(ii) P4 + NaOH + H 2O 
(iii) C 2 + F2 
→
(access)
OR
(a)
Account for the following:
(i)
The acidic strength decreases in the order HCl>H2S >PH3
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(b)
(ii)
Tendency to form pentahalides decreases down the group
in group 15 of the periodic table.
Complete the following chemical equations:
(i)
→
P4 + SO2Cl2 
→
(ii) XeF2 + H 2O 
→
(iii) I 2 + HNO3 
(conc)
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CBSE
XII
Solution to Board Paper – Set 1 (Delhi)
Chemistry Theory
Solution 1: This statement means that some of the physical properties of
crystalline solids such as electrical resistance or refractive index show
different values when measured along different directions in the same
crystals.
Solution 2. The Molar conductivity of a solution at a given concentration is
related to conductivity of that solution, by the following relation.
A
Where, ∧m is molar conductivity and k is the conductivity of the solution.
∧m = k
Solution 3. Electrophoresis is the phenomenon of movement of colloidal
particles under the applied electric potential.
Solution 4. XeF2 is a linear molecule and adopts the following structure:
Solution 5.
The IUPAC name of the given structure is 2, 2-Dimethylbromopropane.
Solution 6.
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Solution 7. The solubility order of the given amines is as follows
C6H5NH2 < (C2H5)2NH < C2H5NH2
Reason:
The more extensive the H-bonding, the higher is the solubility. C2H5NH2
contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus
C2H5NH2 undergoes more extensive H-bonding than (C2H5)2NH. Hence, the
solubility in water of C2H5NH2 is more than that of (C2H5)2NH.
Further, the solubility of amines decreases with increase in the molecular
mass. This is because the molecular mass of amines increase with an
increase in the size of the hydrophobic part. The molecular mass of C6H5NH2
is greater than that of C2H5NH2 and (C2H5)2NH. Thus, the solubility of
C6H5NH2 is less than that of C2H5NH2 and (C2H5)2NH.
Solution 8. A polymer that can be decomposed by bacteria is called a
biodegradable polymer.
For example: poly- β -hydroxybutyrate-CO- β -hydroxyvalerate (PHBV) is
biodegradable aliphatic polyester.
 O − CH − CH 2 − C − O − CH − CH 2 − C −




|
||
|
||




CH
O
CH
CH
O
3
2
3

n
PHBV
Solution 9. The process of corrosion is a redox reaction that involves
simultaneous oxidation and reduction reactions. It can therefore be referred
to as an electrochemical reaction.
In the process of corrosion, due to the presence of air and moisture,
oxidation takes place at a particular spot of an object made of iron. That spot
behaves as the anode. The reaction at the anode is can be written as follows.
Anodic reaction: Fe (s) 
→ Fe2+ (aq) + 2e−
Electrons released at the anodic spot move through the metallic object and
go to another spot of object. There, in the presence of H+ ions, the electrons
reduce molecular oxygen. This spot behaves as the cathode. There H+ ions
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come either from H2CO3 , which are formed due to the dissolution of carbon
dioxide from air into water or from the dissolution of other acidic oxides from
the atmosphere in water.
The reaction corresponding at the cathode is written as follows.
Cathodic reaction: O 2 (g) + 4H+ (aq) + 4e− 
→ 2H 2O(l)
Thus, the overall reaction is:
2Fe(s) + O2 (g) + 4H + (aq) 
→ 2Fe2+ (s) + 2H 2O (l)
Also, ferrous ions are further oxidized by atmosphere oxygen to ferric ions.
These ferric ions combine with moisture, present in the surroundings, to form
hydrated ferric oxide (Fe2O3, xH2O) i.e., rust.
Solution 10.
Ni(s) + 2Ag+(aq) 
→ Ni2+(aq) + 2Ag(s), Eo = 1.05V
The galvanic cell of the given cell reaction is depicted as :
Ni(s) | Ni2+ || Ag+ | Ag(s)
Now, the s tan dard cell potential is
Eocell = 1.05 V
n=2
F = 96500 C mol−1
∆rGo = − nFEocell
∆rGo = − 2 × 96500 C mol−1 ×1.05 V
= − 202650 Jmol−1
= − 202.65 kJmol−1
Solution 11. The rate expression can be defined as an expression in which
the rate of reaction is given as the product of the molar concentration of the
reactants, with each term raised to some power, which may or may not be
the stoichiometric coefficients of the reacting species in a balanced chemical
equation.
The rate constant can be defined as the rate of reaction when the
concentration of each of the reactant is taken as unity.
2NO(g) + O2 (g) 
→ 2NO 2 (g)
Example:
The rate expression for the above reaction can be written as follows:
Rate = k[NO]2 [O 2 ] (Experimentally determined)
Now, if the concentration of NO and O2 is taken to be unity, then the rate
constant is found to be equal to the rate of the reaction.
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Solution 12.
(i) The Shorter N – O bond in NO−
2 is due to the existence of resonance
in NO−
2 . The resonating structure can be drawn as follows.
Due to resonance in NO−
2 , the two bonds are equivalent. This leads to a
decrease in bond length. Thus, the N – O bond length in NO−
2 resembles a
double bond.
Now, the resonating structure for NO−
3 can be drawn as:
−
As seen from the above resonating structure of NO3 , the three oxygen
atoms are sharing two single bonds and one double bond. So the real N-O
bond length resembles a single bond closely.
This explains the existence of shorter bond length of the N-O bond in NO−
2
than in NO−
3 .
(ii)
The kinetic inertness of SF6 can be explained on the basis of its
structure.
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As seen from the above structure, the six fluorine (F) atoms protect the
sulphur atom from attack by the regents to such an extent that even
thermodynamically most favourable reactions like hydrolysis do not occur.
OR
(i)
(ii)
In gaseous and liquid state, PCl5 has a trigonal bipyramidal
structure. In this structure, the two axial P – Cl bonds are longer
and less stable than the three equatorial P – Cl bonds. This is
because of the greater bond pair – bond pair repulsion in then axial
bonds. Hence, all the bonds in PCl5 are not equivalent.
Because of stronger S-S bonds as compared to O-O bonds, sulphur
has a greater tendency for catenation than oxygen.
Solution 13.
(i)
In aqueous solution, Cu+ ion undergoes oxidation to Cu2+ ion. The
relative stability of different oxidation states can be seen from their
electrode potentials.
Cu + (aq) + e− 
→ Cu (s),
Cu 2+ (aq) + 2e− 
→ Cu(s),
(ii)
E o(red) = 0.52V
E o(red ) = 0.34V
Due to more reduction electrode potential value of Cu+, it
undergoes oxidation reaction quite feasibly. Hence, Copper (I) ion
is not known in aqueous solution.
The actinoids show a larger number of oxidation states because of
very small energy gap between the 5f, 6d and 7s sub-shells.
Hence all their electrons can take part in bond formation.
Solution 14.
Riemer-Tiemann reaction: Riemer-Tiemann reaction
involves the treatment of phenol with chloroform in the presence of aqueous
sodium hydroxide at 340 K followed by hydrolysis of the resulting product to
give 2-hydroxybenzaldehyde (salicylaldehyde). The chemical reaction can be
represented as follows.
(i)
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(ii)Friedel-Crafts acetylation of anisole: Friedel-Crafts acetylation of
anisole involves the treatment of anisole with either acetyl chloride or acetic
anhydride to give 2-methoxyacetophenone (as a mirror product) and 4methoxyacetophenone (as a major product), the chemical reaction can be
represented as follows.
Solution 15.
(i)
Phenol on reaction with concentrated HNO3 results in the formation
of picric acid.
(ii)
2-Methyl propene can be obtained from 2-methyl propanol by the
reaction of the later with alc. KOH
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Solution 16. The α -form of glucose and β -form of glucose can be
distinguished by the position of the hydroxyl group on the first carbon atom.
In open chain α -glucose, the hydroxyl group on the first carbon atom is
towards the right whereas, in the closed ring α - glucose, the hydroxyl group
on the first carbon atom is below the plane of the ring.
On the other hand, in open chain β -glucose, α -glucose, the hydroxyl group
on the first carbon atom is towards the left whereas, in the closed ring α glucose, the hydroxyl group on the first time atom is above the plane of the
ring.
The structures of open and closed α -form and β - form of glucose can be
drawn as follows.
Solution 17.
I.
Primary structure of proteins: Each polypeptide chain in
a protein has amino acids linked with each other in a specific
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II.
sequence. This sequence of amino acids is said to be the
primary structure of proteins.
Secondary structure of proteins: The secondary structure
of proteins refers to the shape in which a long polypeptide
can exist. The two different secondary structure possible are
α -Helix structure and β – pleated sheet structure.
• α -Helix structure: In α -Helix structure, a polypeptide
chain forms all possible hydrogen bonds by twisting into a
helix with – NH group of each amino acid residue and
hydrogen bonded to >C=O of an adjacent turn of helix.
• β -Helix structure: In a β -pleated structure, all peptide
chains are stretched out of nearly maximum extensions
and then laid side by side which are held together by
intermolecular hydrogen bonds.
Solution 18. (i)
Uses of Bakelite:
(a)
It is used for making combs.
(b)
It is used for manufacturing electrical switches.
(ii) Uses of Nylon 6:
(a)
It is used for making tyre cords.
(b)
It is used for making fabrics and mountaineering
ropes.
Solution 19.
Given, silver crystallizes in fcc unit cell
So, r =
a
2 2
Where e is the radius of the silver atom and a is the edge length
Now, edge length = 400 pm = 400 × 10−10 cm
Thus,
r=
400×10−10 cm
2×1.414
= 141.44×1010 cm
= 141.4 pm
Thus, the radius of the silver atom was found to be 141.4 pm
Solution 20.
(a)The plot of [N2O5] v/s t is as follows
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[N2O5] (M)
Time (min)
0.400
0.289
0.209
0.151
0.109
0.00
20.0
40.0
60.0
80.0
log[N2O5]
-.03979
-0.5391
-0.6798
-0.8210
-0.9625
From the plot, log [N2O5] v/s t, we obtain
Slope =
=
−0.70 −(−0.60)
40 − 20
−0.70 + 0.60 −0.10
=
20
20
Also, slope of the line of the plot =
−k
2.303
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0.10
−k
=
2.303
20
0.10
or k =
× 2.303 = 0.01151
20
∴
or, k = 1.15 × 10−3 min−1
(b)
After 100 min
k=
[N O ]
2.303
log 2 5 o
t
[N 2O5 ]t
After 100 min
k=
2.303
0.400
log
t
0.098
= 0.1406 min−1
(c)
The initial rate of reaction
r = k[N 2O5 ]
= 1.15 × 10−3 × 0.400
= 4.6 × 10−4 s−1
Solution 21.
(i)
Production of high vacuum: Traces of air can be adsorbed by
charcoal from a vessel, evacuated by a vacuum pump to give a
very high vacuum.
(ii)
Heterogeneous catalysis: The gaseous reactants are adsorbed
on the surface of the solid catalysts. As a result, the concentration
of the reactants increase on the surface and hence the rate of the
reaction increases.
(iii) Froth floatation process: This process is used to remove gangue
from sulphide ores. The basic principle involved in this process is
adsorption.
In this process, a mixture of water pine oil is taken in tank. The
impure powdered sulphide ore is dropped in through hopper and
the compressed air is blown in through the agitator is rotator is
rotated several times. As a result, froth is formed and the sulphide
ores get adsorbed in the froth. The impurities settled down and are
let out through an outlet at the bottom. The froth formed is
collected in froth collector tank. After sometime, the ore particles in
the froth collecting tank start settling gradually, which are then
used for further metallurgical operations.
OR
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(i)
(ii)
(iii)
A micelle is an aggregate of surfactant molecules dispersed in a a
liquid. A micelle in aqueous solution forms as aggregate such that
the hydrophilic “head” regions are in the centre of micelle.
Peptization is the process of conversion of a precipitate into a
colloidal sol by shaking it with the dispersion medium in the
presence of an electrolyte. The electrolyte used in this reaction is
known as a peptizing agent.
Desorption is the process of removing an adsorbed substance from
the surface through which it was adsorbed.
Solution 22.
(i)
Vapour phase refining : Vapour phase refining is the process of
refining metal by converting it into its volatile compound and then,
decomposing it to obtain a pure metal. The basic principle involved
in this process are:
(a) The metal should form a volatile compound with an available
reagent, and
(b) The volatile compound should be easily decomposable so that
the metal can be easily.
Nickel, zirconium, and titanium are refining using this method.
(ii)
Electrolytic refining of a metal is the process of refining impure
metals by using electricity. In this process, impure metal is made
the anode and a strip of pure metal is made the cathode. A solution
of a solution salt of the same metal is taken as the electrolyte.
When an electric current is passed, metal ions from the electrolyte
are deposited at the cathode as pure metal and the impure metal
from the anode dissolves into the electrolyte in the form of ions.
The impurities present in the impure metal gets collected below the
anode. This is known as anode mud.
Anode :
Cathode :
M 
→ M n + + ne−
M n + + ne− 
→M
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(iii)
In the process of leaching, the finely divided silver is treated with
dilute solution of sodium cyanide while a current of air is
continuously passed. As a result, silver pass into the solution
forming solution dicyanoargenate(I) while the impurities remain
unaffected which are filtered off.
Ag 2S + 4NaCN 
→
2Na[Ag (CN )2 ]
+
Na 2S
Sodium dicyano arg enate (I)
Solution 23.
(i)
+
5C2 O 24− + 2MnO−
→ 2Mn 2+ + 8H 2O + 10CO 2
4 + 16H 
(ii)
2KMnO4 
→ K 2 MnO4 + MnO2 + O2
(iii)
Cr2O72− + 8H+ + 3H 2S 
→ 2Cr3+ + 7H 2O + 3S
Heated
Solution 24.
(i)
K4[Mn(CN)6]
Name: Potassium hexacyanomanganate(II)
Stereochemistry – Does not show geometric or optical isomerism
Magnetic behaviour – Paramagnetic
(ii)
[Co(NH3)5Cl] Cl2
Name: Pentaamminedchloridocobalt (III) chloride
Stereochemistry – Does not geometric isomerism but is optically active
Magnetic behaviour – Paramagnetic
(iii)
K2[Ni(CN)4]
Name: Potassium tetracynoinickelate (II)
Stereochemistry – Does not show geometric or optically isomerism
Magnetic behaviour – Diamagnetic
Solution 25.
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(i)
(ii)
(iii)
Haloalkanes can easily dissolve in organic solvents of low polarity
because the new forces of attraction set up between haloalkanes
and the solvent molecules are of same strength as the forces of
attraction being broken.
A mixture of equal amounts of two enantiomers is known as
racemic mixture.
For example: When a 3o halide undergoes substitution with KOH,
the reaction proceeds through SN1 mechanism forming the racemic
mixture in which one of the products has the same configuration as
a reactant, while the product has an inverted configuration.
The SN1 substitution reaction involves the formation of carbocation,
which is not affected by the presence of bulky groups.
Thus, C6H5CH(C6H5)Br will be more reactive towards SN1
substitution reaction forming racemic mixture.
Solution 26.
(a)
The basicity of amines depends on the +I effects of the alkyl group.
The presence of –CH3 group in alkylamine increases the electron
density on the nitrogen atom and thus increases the basicity.
Hence, alkylamine is more basic than ammonia
CH3NH2 > NH3
(b)
(i)
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(ii)
Solution 27.
(i)
Detergents:
A detergents is a surfactant or a mixture of
surfactants having cleaning properties in dilute solution. Commonly,
detergent refers to alkylbenzenesulphonates. For example: Sodium
dodecylbenzene sulphonate
(ii) Food preservatives:
Food preservatives are chemicals that
prevent food from spoilage due to microbial growth. Table salt, sugar,
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vegetable oil, sodium benzoate (C6H3COONa), and salts of propanoic acid are
some examples of food preservatives.
(iii) Antacids: Any drug that is used to counteract the effects of excess
acid in the stomach and raise the pH to an appropriate level is called an
antacid.
Example: Omeprazole
Solution 28.
(a) Molartiy is defined as the number of moles of solute
dissolved per litre of solution.
Mathematically M =
Number of moles of solute
Volume of solution in litres (dm3 )
Molality of a solution is defined as the number of moles of solute
dissolved in 1000 grams of solvent.
Mathematically, m =
Number of moles of the solute
Mass of solvent in kg
While molarity decreases with an increase in temperature, molality
is independent of temperature. This happens because molality
involves mass, which deos not change with a change in
temperature, while molarity involves volume, which is temperature
dependent.
(b) Given w2 = 10.50 g
w1 = 200g
Molar mass of MgBr2 (M2) = 184 g
Using the formula,
∆Tf =
1000 × k f × w2
w1 × M2
=
1000 × 1.86 × 10.50
200 × 184
=
19.530
= 0.53
200 × 184
Now, Tf = To - ∆Tf
= 273 – 0.53 = 272.47 K
OR
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(a) Osmosis : The process of flow of solvent molecules from pure
solvent to solution or from solution of lower concentration of
solution of higher concentration through a semi – permeable
membrane is called osmosis.
Osmotic pressure : The pressure required to just stop the flow of
solvent due to osmosis is called osmotic pressure (π) of the
solution. Yes, the osmotic pressure of a solution is colligative
property. The osmotic pressure is expressed as.
π =
n
RT
V
Where, π = osmotic pressure
n = number of moles of solute
V = volume of solution
T = temperature
From the equation, it is clear that osmotic pressure depends upon
the number of moles of solute ‘n’ irrespective of the nature of the
solute. Hence, osmotic pressure is a colligative property.
(b) Given, Kb = 0.512 k kg mol-1
w2 = 15.00 g
w1 = 250.0 g
M2 = 58.44 g
Using the formula,
∆Tb =
1000 × Kb × w2
w1 × M2
=
1000 × 0.512 × 15.00
250.0 × 58.44
=
7.680
= 0.52
14.600
Now, Tb = To + ∆Tb
= 373 + 0.53 = 373.53 K
Ans29.
(a)
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(i) Propanal (CH3CH2CHO) can be distinguished from propanone
(CH3COCH3) by iodoform test.
Being a methyl ketone, propanone on treatment whith I2/NaOH
undergoes iodoform reaction to give a yellow ppt. of iodoform
→
CH3COCH3 + 3NaOI 
Propanone
CHI3
+ CH3COONa + 2NaOH
Iodoform
Propanal on the other hand does not give this test.
NaOI
CH3CH2CHO →
No yellow ppt. of
Propanal
Iodoform
(ii) Benzaldehyde (C6H5CHO) and acetophenone (C6H5COCH3) can
be distinguished by iodoform test.
Acetophenone, being a methyl ketone on treatment with I2/NaOH
undergoes iodoform reaction to give a yellow ppt. of iodoform. On
the other hand, benzaldehyde does not give this test.
C6H5COCH3 + 3NaOI
Acetophenone
C6H5COONa + CHI3
+ 2NaOH
Iodoform
NaOI
C6H5CHO →
No yellow ppt of iodoform
Benzaldehyde
(b)
(i)
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(i)K Cr O / H SO
(ii) CH3CH2CH2CH2OH 
→ CH3CH2CH2COOH
(ii) Dil.H SO
2
2
7
2
2
Butanol
4
4
Butanoic acid
(iii)
OR
(i) Cannizaro reaction
In this reaction, the aldehydes which do not have an α hydrogen atom, undergo self oxidation and reduction
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(disproportionation) reaction on treatment with a concentrated
alkali.
Example:
(ii) Decarboxylation
The decarboxylation reaction can be carried out either by using
soda lime or by electrolysis
•
Using soda lime
Sodium salts of carboxylic acids when heated with soda lime
(NaOH + CaO) in the ratio 3:1 undergoes decarboxylation
reaction to yield alkanes.
NaOH − CaO
R – COONa →
R – H + Na2CO3
Heat
(Alkane)
•
Electrolytic decarboxylation
Electrolysis of aqueous solutions of sodium of potassium salts
of carboxylic acids give alkanes having twice the number of
carbon atoms present in the alkyl group of acid. This is
known as Kolbe’s decarboxylation.
+
→ 2RCOO + 2Na
2RCOONa 
+
→ 2OH + 2H
H2O 
At Anode:→ H2
2H+ + 2e- 
(b)
(i)
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(ii)
(iii)
H3 O+
heat
C6H5CONH2 
→ C6H5COOH
Benzoic acid
Ans30.
(a)
(i) As we move down the group 17, the size of the atom increases
from fluorine to chlorine. The larger difference in the size of N and Cl
results in the weakness of strength of N – Cl bond. On the other hand,
the difference in size of N and F is small; consequently the N – F bond
is quite strong. As a result, NF3 is an exothermic compound.
(ii) Due to the small size of F atom, the three lone pair of electrons on
each F atom F – F molecule repels the bond pair. As a result, F – F is
most reactive of all the four common halogens.
(b)
→ 2SO2 + CO2 + 2H2O
(i) C + 2H2SO4 
Sulphur dioxide
∆,CO2
(ii) P4 + 3NaOH + 3H2O → PH3 + 3NaH2PR2
Phosphine
(iii) Cl2
+
→ 2ClF3
3F2 
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(excess)
Chlorine trifluoride
OR
(a)
(i) In a period, the electro negativity decreases in the order Cl > S >
P. As a result, the loss of H+ ions decreases.
Thus, the acidic strength of the hydrides decreases in the following
order:
HCl > H2S > PH3
(ii) The tendency to form pentahalides decreases down the group 15
due to inert pair effect i.e., in Bi the s-electrons remain inert and do not take
part in bonding.
(b)
(i) P4 + 10SO2Cl2 → 4PCl5 + 10SO2
(ii) 2XeF2 + 2H2O → 2Xe + 4HF + O2
→ 2HIO3 + 10NO2 + 4H2O
(iii) I2 + 10HNO3 
(conc).
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