XI
CHEMISTRY
SAMPLE PAPER 4
Time: Three Hours
Max. Marks: 70
General Instructions
1. All questions are compulsory.
2. Question nos. 1 to 8 are very short answer questions and carry 1 mark
each.
3. Question nos. 9 to 18 are short answer questions and carry 2 marks
each.
4. Question nos. 19 to 27 are also short answer questions and carry 3
marks each
5. Question nos. 28 to 30 are long answer questions and carry 5 marks
each
6. Use log tables if necessary, use of calculators is not allowed.
Q 1: Write the formula of the compound Nickel (II) sulphate?
Q 2: Temporary hardness in water is due to presence of which salts?
Q 3: What is the mass of one atom of oxygen?
Q 4: What are silicones?
Q 5: Give the values for principal quantum number and magnetic
quantum number for 19th electron of K (Potassium).
Q 6: What shapes are associated with sp3d and sp3d2 hybrid orbitals ?
Q 7: Why is an organic compound fused with sodium for testing halogens,
nitrogen, sulphur and phosphorous??
Q 8:
2A(g) + B(g)  4C(g) + Heat
What is the effect of adding He at a constant volume on above
equilibrium?
Q 9: Which of the following has largest size? Mg, Mg2+, Al3+, Al
Q10: Give IUPAC name and symbol of element with atomic number 110
and 115.
Q11: Which of the two is more concentrated and why? 1M or 1m aqueous
solution of a solute
+
Q12: N2  N2
+
O2  O2
Why is there an increase in bond order in going from O2 to O2+ while a
decrease in going from N2 to N2+ ?
Q13: Calculate the total pressure of a mixture of 8g of O2 (g) and 4g of H2
(g) in a vessel of 1dm3 at 270C. (Atomic mass of O=16gmol-1and H=1
gmol-1 .R=0.083 bar dm3/K mol).
Q14: Give equations to prove amphoteric nature of water.
Q15: Balance the following equation in an alkaline medium by half
reaction method.
2Cr(OH)3 + IO3
 I
+ CrO4
Q16: The wavelength of first spectral line in Balmer series is 6561 Å .
Calculate the wavelength of second spectral line in Balmer series.
Q17: On a ship sailing in Pacific Ocean where temperature is 23.4°C, a
balloon is filled with 2 L air. What will be the volume of the balloon
when the ship reaches Indian Ocean, where temperature is 26.1°C?
Q18: All C-O bond lengths in CO32- are equal. Account for this
observation.
Q19: Give reasons:
(i) Anhydrous AlCl3 is covalent but hydrated AlCl3 is electrovalent.Explain
(ii)Boric acid behaves a Lewis acid? Explain
(iii) Boron is unable to form BF63– ion. Explain
Q20: At 60°C, dinitrogen tetroxide is fifty percent dissociated. Calculate
the standard free energy change at this temperature and at one
atmosphere.
Q21: What type of isomerism is exhibited by following pair of compounds?
(i) Ethanol and Methoxy methane
(ii) o-cresol and m-cresol
(iii) Pentan-3-one and pentane-2-one
Q22: When a metal X is treated with sodium hydroxide, a white
precipitate A is obtained, which is soluble in an excess of NaOH to give
soluble complex B. Compound A is soluble in diluted HCl to form
compound C. The compound A when heated strongly gives D, which is
used to extract metal. Identify X, A, B, C and D. Write suitable equations
to support their identities.
Q23:
Calculate the energy associated with the first orbit of He +. What is the
radius of this orbit?
Q 24: What are the necessary conditions for any system to be aromatic?
Q 25: Calculate the enthalpy of combustion of glucose from the following
data
θ
C (graphite) +O2  CO2 (g) ;
rH = -395kJ
H2 (g) +
1
O (g)
2 2

H2 O(l)
θ
rH = -269.4kJ
;
6C (graphite) + 6H2 (g) + 3 O2 (g)
 C6H12O6 (s) ;
θ
r H = -1169.9kJ
Q 26:
(a) Fish do not grow as well in warm water as in cold water. Why?
(b) Why does rain water normally have a pH about 5.6?
(c) Name two major greenhouse gases.
Q27: 0.2325g of an organic compound was analysed for nitrogen by
Duma’s method. 31.7mL of moist nitrogen was collected at 250C and
755.8mm Hg pressure. Calculate the percentage of N in the sample. (Aq.
Tension of water at 25oC is 23.8mm).
OR
(a) Why cannot sulphuric acid be used to acidify sodium extract for
testing S using lead acetate solution?
(b)Which of the carbocations is most stable and why?
+
+
+
(CH3 ) 3C , CH3CH2CH2 , CH3CHCH2CH3
(c) Why does a liquid vaporize below its boiling point in steam
distillation process?
Q 28
(a) Convert:
(i) Propene to propane-1,2-diol
(ii) Isopropylbromide to n-propylbromide
(b) An alkene on ozonolysis gives butan-2-one and 2-methylpropanal.Give
the structure and IUPAC name of Alkene. What products will be obtained
when it is treated with hot, concentrated KMnO4?
OR
Q 28: Complete the equations:
NaNH2
Red hot iron tube
 A 
B
(i) CH2 =CHBr 
873K
Anhdrous
(ii) C6H6 +CH3COCl 
 A+B
AlCl
3
NaOH(aq)
Sodalime
B
(iii) CH3COOH  A 

No peroxide
(iv) CH2  CH  CH2  CH2  CH2  CH3  HBr 
Peroxide
(v) CH2  CH  CH2  CH2  CH2  CH3  HBr 

Q29: Calculate the pH of a 0.10M ammonia solution. Calculate the pH
after 50.0 mL of this solution is treated with 25.0 mL of 0.10M HCl. The
dissociation constant of ammonia, Kb = 1.77 x 10-5.
OR
Q29: Calculate the pH of the resultant mixtures:
10mL of 0.2M Ca(OH)2 + 25 mL of 0.1 M HCl
Q30: Give reasons for the following
(a)
Unlike Na2CO3, K2CO3 cannot be prepared by Solvay process.
Why?
(b)
Why are alkali metals not found in nature?
(c)
Sodium is less reactive than potassium why?
(d)
Alkali metals are good reducing agents. Why?
(e)
Alkali metals are paramagnetic but their salts are diamagnetic.
Why?
OR
Q30: Complete the following reactions:
(a) Why does the solubility of alkaline earth metal carbonates and
sulphates in water decrease down the group?
(b)Arrange the following alkali metal ions in decreasing order of their
mobility: Li+, Na+ , K+, Rb+, Cs+ .Explain
(c) NaOH is a stronger base than LiOH. Explain
(d) Why are alkali metals kept in paraffin or kerosene ?
(e) Why does lithium show properties uncommon to the rest of the alkali
metals?
SOLUTIONS TO SAMPLE PAPER 4
Ans 1:
NiSO4 (1 mark)
Ans 2: Temporary hardness in water is due to salts of magnesium and
calcium in form of hydrogen carbonates. (1 mark)
(1 mark)
Ans 3:
23
NA or 6.023 × 10 atoms of oxygen have a mass of 16g
where NA is Avogadro's number
So, mass of one atom of oxygen
=
16
23
6.023 × 10
= 2.65  10
23
g
(1 mark)
Ans 4: Silicones are a group of organosilicon polymers containing
repeated (R2SiO) units. (1 mark)
Ans 5:
Electronic configuration of K = 1s22s22p63s23p64s1
This means that the 19th electron lies in the s- subshell, Thus it is 4s, so
values for principal quantum number ,n=4 and magnetic quantum
number, ml =0 (1 mark)
Ans 6:
1

(a) sp3d hybrid orbitals -Trigonal bipyramidal  mark 
2

1

(b) sp3d2 hybrid orbitals-Octahedral
 2 mark 


Ans 7: Organic compound is fused with sodium metal to convert N, S, P
and halogens present in organic compound to their corresponding sodium
salts. (1 mark)
Ans 8: There will be no effect of adding He at constant volume. This is due
to the fact that when an inert gas is added to equilibrium at constant
volume, the total pressure will increase. But the concentration of
reactants and products will not change.
(1 mark)
Ans 9: Atomic radii decrease across a period.Cations is smaller than their
parent atoms. Among isoelectronic species, the one with the larger
positive nuclear charge will have a smaller radius. Hence the largest
species is Mg; the smallest one is Al3+. (2 marks)
Ans 10:
(a) Name and symbol of element with atomic number 110
1

Name: Ununnillium  mark 
2

1

Symbol: Uun
 2 mark 


(b) Name and symbol of element with atomic number 115
1

Name: Ununpentium  mark 
2

1

Symbol: Uup
 2 mark 


Ans 11:
1 molar solution contains 1mole of solute in 1 L of solution while
1

1 molal solution contains 1 mole of solute in 1000g of solvent.  mark 
2

Considering density of water as almost 1g/mL, then 1mole of solute is
present in 1000mL of water in 1molal solution while 1mole of it is present
in less than 1000mL of water in 1 molar solution
(1000mL solution in molar solution = Volume of solute + Volume of
solvent). (1 mark)
1

Thus 1M solution is more concentrated than 1m solution.  mark 
2

Ans 12:
Electronic configuration of O2 is
2
2
2
2
2
2
2
1
1
(1s) ( * 1s) (2s) ( * 2s) (2pz ) (2px  2py )( * 2p x   * 2p y )
1

 2 mark 


In going from O2 to O2+ , an electron is lost from an antibonding
molecular Orbital . Bond order is calculated as one half of the difference in
number of electrons in bonding and antibonding molecular orbital. Thus
1

bond order increases  mark 
2

Electronic configuration of N2 is
2
2
2
2
2
2
2
(1s) ( * 1s) (2s) ( * 2s) (2px  2py )(2pz )
In going from N2 to N2+ , the electron is lost from a bonding molecular
Orbital. Bond order is calculated as one half of the difference in
number of electrons in bonding and antibonding molecular orbital. Thus
bond order decreases.
1

 2 mark 


Ans 13:
No. of moles of O2 =
No. of moles of H2 =
8
= 0.25
32
4
=2
2
1

 2 mark 


pV=nRT
 pO 
2
pH =
2
0.25  0.083  300
 6.23 bar
1
2  0.083  300
1
 49.8 bar
1

 2 mark 


1

 2 mark 


Total pressure = pO + pH
2
2
= 6.23 + 49.8
= 56.03 bar
Ans 14:
1

 2 mark 


Amphoteric nature means H2O can act as an acid as well as a base
H2 O

+ NH3
+
-
NH4 + OH
(1mark)
(Acid)
H2 O
+
HCl

+
-
H3O +
Cl
(1mark)
(Base)
Ans 15:
a) First, we will write down the oxidation number of each atom
Cr(OH)3
3
+
2
IO3
5
-

I
1
+
2
CrO4
6
b) Write separately oxidation & reduction half reactions
Oxidation half reaction:
Cr(OH)3
3

CrO4
6
2-
Reduction half reaction:
IO3 + 6e

I
5
1
1

 2 mark 


-
 3e
c) Balance O atoms by adding H2O molecules to the side deficient in 'O'
atoms and then balancing H atoms
Cr(OH)3 + H2 O 
-
IO3
-
+ 6e
CrO4
-

I
2-
-
 3e
+ 3H2O
1

 2 mark 


d) Balance H atoms. Since the medium is alkaline, therefore H2O molecules
are added to the side deficient in H atoms and equal no. of OH
ions to the other side :
-
5OH 
Cr(OH)3 +
2-
-
 3e
-
 5OH 
(  Cr(OH)3 + H2 O
-
CrO4
-
-
IO3 + 6e  3H2 O 
I
(  IO3 + 6e  6H2 O 
CrO4
+4H2O
2-
-
 3e
+5H2O)
-
+ 6OH
I + 3H2O + 6OH )
1

 2 mark 


e) Equalise the electrons lost and gained by multiplying the oxidation half
reaction with 2.
2Cr(OH)3 +
-
10OH 
2CrO4
2-
-
 6e
+8H2O
Adding the oxidation half reaction and reduction half reaction we get
2Cr(OH)3 +
-
-
-
10OH 
2CrO4
-
2-
-
 6e
-
+8H2O
IO3 + 6e  3 H2 O 
I + 6OH
_________________________________________________
22Cr(OH)3  IO3  4OH  2CrO4  I +5 H2O
1

 2 mark 


Ans 16 : According to Rydberg equation,
R = 109,677 which is the Rydberg's constant
1
v=
Wavelength
For first line in Balmer series, n1 = 2, n2 = 3
Given wavelength of 1st spectral line = 6561 Å
1
 1
1 
 5 
1

Therefore,
= R  2  2  =R 
(i)  mark 

6561
 36 
2

3 
2
For second line in Balmer series, n1=
1
1
 1
Therefore,
= R 2  2
Wavelength
4
2
2, n2 =4

 3 
 = R  16 



1

(ii)  mark 
2

Dividing eq. (i) by (ii), we get:
Wavelength 5  16

6561
36  3
1

 2 mark 


1

 2 mark 


 Wavelength=4860 Å
Ans 17:
V1 = 2 L
T1 = (23.4 + 273) K
= 296.4 K
T2 = 26.1 + 273
= 299.1 K
From Charles law
V1
T1

V2
1

 2 mark 


T2
V T
 V2  1 2
T1
 V2 
2 L  299.1 K
296.4 K
1

 2 mark 


 2 L  1.009
 2.018 L
1mark 
Ans 18: All C-O bond lengths in CO32- are equal because of resonance. All
bonds have partial double bond character
(1 mark)
(1 mark)
Ans 19 :
(i)Anhydrous AlCl3 is covalent but hydrated AlCl3 is electrovalent because
when it is dissolved in water the high heat of hydration is sufficient to
break the covalent bond of AlCl3 into Al3+ and Cl- ions .(1 mark)
(ii) Boric acid behaves as Lewis acid by accepting a pair of electron from
1

OH- ion (in water) .  mark 
2

-
B(OH)3 + 2H-O-H  [B(OH)4 ] + H3O

1

 2 mark 


(iii)Due to non-availability of d orbitals, boron is unable to expand its
octet. Therefore, the maximum covalence of boron cannot exceed 4 and
it cannot form BF63– ion.
(1 mark)
Ans 20.


N2O4 (g) 
 2NO2 (g)
If N2O4 is 50% dissociated, the mole fraction of both the
subs tan ces is given by
1  0.5 0.5
xN O 

;
1  0.5 1.5
2 4
xNO
2

1

 2 mark 


1

 2 mark 


2  0.5
1

1  0.5 1.5
0.5
0.5
pN O 
 1 atm 
1.5
1.5
2 4
pNO
2

1

 2 mark 


1

 2 mark 


1
1
 1 atm 
1.5
1.5
The equilibrium cons tan t KP is given by
pNO2 

2
KP
pN2O4

1.5
2
(1.5) (0.5)
1

 2 mark 


 1.33 atm

Since r G  RTln KP

1
1
1
1
r G  (8.314 JK mol
 (8.314 JK mol
1
 789.98 kJ mol
)  (333 K)  (ln(1.33))
)  (333 K)  (2.303)  (0.1239)
1

 2 mark 


Ans 21:
(i) Functional isomerism or functional group isomerism
(ii) Position isomerism
(iii) Metamerism
(1 × 3)
Ans 22:
Since metal X reacts with NaOH to first give a white ppt. A which dissolves
in excess of NaOH to give a soluble complex B , therefore metal X is
aluminium ;ppt A is Al(OH)3 and complex B is sodium
teterahydroxoaluminate (III)
+ 3NaOH 
Al
X

Al(OH)3
3Na
Aluminium hydroxide
Na [Al(OH)4 ]
Al(OH)3 +NaOH 
A
Sodium tetrahydroxoaluminate (III)
B
1mark 
Since A is amphoteric in nature, it reacts with dilute HCl to form
compound C which is AlCl3
Al(OH)3  3HCl 
 AlCl3  3H2O
A
C
1mark 
Since A on heating gives D which is used to extract metal, therefore, D
must be alumina (Al2O3)

2Al(OH)3 
 Al2O3  3H2O
A
D
1mark 
Ans 23:
2.18  10 J Z

18
En
2
1
atom
2
n
1

 2 mark 



For He ,n  1, Z  2
2.18  10 J2 

18
E1
2
2
1
 8.72  10
18
J
1mark 
The radius of the orbit is given by equation
rn 
0.0529 nm n2
Z
1

 2 mark 


S in ce n  1, and Z  2
rn 
0.0529 nm 12
2
 0.02645 nm
.
1mark 
Ans 24: Necessary conditions for a compound to be aromatic is
(i)
The molecule should contain a cyclic cloud of delocalised  electrons above and below plane of molecule 1mark 
(ii)
For delocalisation of electrons the ring must be planar to
allow cyclic overlap of p-orbitals 1mark 
(iii)
The compound should contain (4n+2)  -electrons where
n=0,1,2 .This is known as Huckel’ rule 1mark 
Ans 25.
Given
C (graphite) +O2
H2 (g) +
1
O (g)
2 2
 CO2 (g)

H2 O(l)
θ
; rH = -395kJ
(Eq.1)
θ
; rH = -269.4kJ
C6H12 O6 (s)  6C (graphite) + 6H2 (g) + 3 O2 (g)
;
(Eq.2)
θ
r H = 1169.9kJ
(Eq.3)
Combustion of graphite is given by the equation shown below
C6H12 O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O (g)
(Eq.4)
Thus enthalpy for combustion reaction can be obtained
1

 2 mark 


by multiplying Eq.1 and Eq.2 by 6 and adding equation Eq.3
6C (graphite) + 6O2
6H2 (g) +
3O2 (g)
 6CO2 (g)
 6 H2 O(l)
;
;
θ
r H = -2370 kJ
θ
rH = -1614.4 kJ
1

 2 mark 


(Eq.6)
1

 2 mark 


(Eq.7)
1

 2 mark 


1

(Eq.8)  mark 
2

______________________________________________________________________
C6H12 O6 (s) + 3O2 (g)  6CO2 (g) + 6H2O (g)
C6H12 O6 (s)  6C (graphite) + 6H2 (g) + 3 O2 (g)
Reaction enthalpy
= (-2370 kJ-1614 kJ) + (1169.9 kJ+0)
= -2814.1kJ
;
θ
r H = 1169.9kJ
1

 2 mark 


Ans 26:
(a) As the temperature increases solubility of gas in water decreases. Due
to high temperature of water, amount of dissolved oxygen is less, which
creates a problem for fish. So, fish do not grow well in warm water
(1 mark)
(b) Rain water is acidic due to dissolution of CO2 in it, leading to formation
of H2CO3 which lowers the pH. Hence the pH is about 5.6
CO2 + H2O 
+
H2CO3
(1 mark)
2-
H2CO3  2H
+ CO3
(c) Carbon dioxide and methane (1 mark)
Ans 27:
Calculation of volume of nitrogen at S.T.P
Experimental conditions
Pressure of dry gas P1 = 755.8 - 23.8 = 732 mL
V1 = 31.7 mL
T1 = 25+273 = 298 K
1

 2 mark 


S.T.P condition
P2 = 760 mm
V2 = ?
T2 = 273 K
Applying gas equation,
P1 V1 P2 V2
=
T1
T2
732×31.7 760×V2

=
298
273
1

 2 mark 


1

 2 mark 


 V2 = 27.97 mL
% of Nitrogen =
=
28× Volume of N2 at S.T.P×100
22400×Mass of compound
28×27.97×100
22400×0.2325
1

 2 mark 


1

 2 mark 


1

 2 mark 


= 15.04
OR
(a) This is because, in the presence of sulphuric acid, lead acetate will
react with it forming white precipitate of PbSO4, thus interfering with the
test. (1 mark)
(b) (CH3) 3C+ is most stable because it has the maximum number of alkyl
groups i.e. three. Greater the number of alkyl groups on the carbon
carrying positive charge, greater would be the dispersal of charge and
hence more will be the stability of carbocation. So, tertiary carbocation is
the most stable due to maximum dispersal of charge. (1 mark)
(c) In steam distillation, water and organic substance vaporize together
and total pressure becomes equal to atmospheric pressure. Thus organic
substance vaporizes and distils at a temperature lower than its boiling
point.
(1 mark)
Ans 28:
(a)
(i)
(ii)
CH3
|
CH3 -CHBr
(1 mark)
alc. KOH
HBr

 CH3CH=CH2 


Peroxide effect
CH3CH2 CH2 Br
(b) From the given products of Ozonolysis,
CH3 -CH2 - C- CH3
||
O
and
CH3 H
|
|
CH3  CH-C=O
The alkene would be
CH3
|
CH3 -CH2 - C=CH-CH-CH3
|
CH3
1 mark 
IUPAC Name: 2, 4-dimethylhex-3-ene
KMnO
4
CH3 -CH2 - C=CH-CH(CH3 )2 
 CH3 -CH2 -C=O
|
|
CH3
CH3
Butane-2-one
1

 2 mark 


OR
(1 mark)
+
HOOC-CH(CH3 )2
2-Methylpropanoic acid
1

 2 mark 


NaNH
2
CH2 =CHBr 
 HC  CH
(i)
A
1

 2 mark 


C6H6 +CH3COCl
Anhdrous


AlCl
3
(ii)
(iii)
Red hot iron tube

 C6H6
873K
NaOH (aq)
CH3COOH 

(iv)
C6H5COCH3
A
1


 2 mark 


CH3COONa
A
1

 2 mark 


B
1

 2 mark 


+
HCl
B
1


 2 mark 


Sodalime



B
CH4
1

 2 mark 


No peroxide
CH2  CH  CH2  CH2  CH2  CH3  HBr  CH3  CH  CH2  CH2  CH2  CH3
|
Br
Hex  1  ene
2  Bromohexane
(v)
peroxide
CH2  CH  CH2  CH2  CH2  CH3  HBr  CH2  CH2  CH2  CH2  CH2  CH3
|
Br
Hex  1  ene
1  Bromohexane
Ans 29:
NH3
+
+
Kb =
+

H2O
NH4
-
[NH4 ][OH ]
= 1.77 x 10-5
[NH3 ]
Before neutralization,
+
[NH3 ] = 0.10 - x  0.10
Kb =
-
OH
1

 mark 
2

-
[NH4 ] = [OH ] = x
Thus
+
1

 mark 
2


2
x
-5
= 1.77 x 10
0.10
-5
 x = 1.33 x 10
+
-
= [OH ]
1

 mark 
2

-
Now Kw  [H ][OH ]
+
 [H ] = Kw / [OH ]
=
-14
10
-5
1.33 x 10
-12
= 7.51 x 10
-12
pH = -log(7.5 x 10
) = 11.12
1

 mark 
2

On addition of 25 mL of 0.1M HCl solutions (i.e., 2.5 mmol of HCl)
to 50 mL of 0.1M ammonia solution (i.e., 5 mmol of NH 3), 2.5 mmol
of ammonia molecules are neutralized. The resulting 75 mL solution
contains the remaining unneutralized 2.5 mmol of NH3 molecules
and 2.5 mmol of NH4+.
NH3
+
2.5
At equilibrium

HCl
2.5
0
0
+
NH4
+
Cl
0
0
2.5
2.5
1

 2 mark 


+
The resulting 75 mL of solution contains 2.5 mmol of NH4 ions
(i.e., 0.033 M) and 2.5 mmol (i.e., 0.033 M) of uneutralized NH3
molecules. This NH3 exists in the following equilibrium.
+


NH4OH
NH4 +


0.033M - y
y
+
Where, y = [OH ] = [NH4 ]
-
OH
y
The final 75 mL solution after neutralization already contains 2.5 m mol
+
+
NH4 ions (i.e., 0.033M), thus total concentration of NH4 ions is given as:
+
[NH4 ] = 0.033 + y
As y is small, [NH4OH] = 0.033 M and [NH4 +] = 0.033M.
+
Kb
=
-
[NH4 ][OH ]
1

 2 mark 


[NH4OH]
y(0.033)
=
(0.033)
-5
= 1.77 x 10
1

 2 mark 


1

 2 mark 


M
-5
Thus, y = 1.77 x 10 = [OH-]
-14
10
+
[H ] =
-5
1.77 x 10
= 0.56 x 10-9
1

 2 mark 


Hence, pH = 9.24
1

 2 mark 


OR
Ans29:
10 mL of 0.2 M Ca(OH)2 = 10 ×
0.2
1000
-3
=2 ×10
25 mL of 0.1 M HCl = 25 ×
1

 2 mark 


moles
0.1
1000
=2.5 ×10
-3
1

 2 mark 


moles
Ca(OH)2 + 2 HCl  CaCl2 + 2H2O
2 moles of HCl reacts with 1 mole of Ca(OH)
 2.5 ×10
-3
2
-3
moles of HCl will react with 1.25 × 10
-3
Therefore moles of Ca(OH)2 left unreacted= 2 ×10
-3
=0.75 ×10
Total volume = 10+ 25 = 35 mL
1

 2 mark 


moles of Ca(OH)2
-3
- 1.25 × 10
moles
1

 2 mark 


-3
0.75  10  1000
35
Molarity of Ca(OH)2 =
1

 2 mark 


= 0.0214 M
-
-2
[OH ] in Ca(OH)2 = 2 ×0.0214 =4.28 ×10
1

 2 mark 


M
-2
pOH = -log(4.28 × 10 )
= -(-2+0.6314)
= 1.368
1 mark 
pH = 14-1.368 = 12.632
1 mark 
Ans 30:
(a) Unlike NaHCO3, the intermediate KHCO3 formed during reaction, is
highly soluble in water and thus cannot be taken out from solution
to obtain K2CO3.Hence, K2CO3 cannot be prepared by Solvay
process. (1 mark)
(b)Alkali metals are highly reactive because of low ionization enthalpy
value and therefore are not found in nature. They are present in
combined state only in form of halides, oxides etc. (1 mark)
(c) Ionization Energy of potassium is less than sodium because of large
size or less effective nuclear charge. Thus, potassium is more
reactive than sodium. (1 mark)
(d)Alkali metals are strong reducing agents due to their greater ease
to lose electrons. (1 mark)
(e) Alkali metals have one unpaired electrons (ns1) and are
paramagnetic. However, during the salt formation, the unpaired
electron is lost by alkali metals to other atom forming anion. Their
salts have all paired electrons and show diamagnetic nature.
OR
Ans 30:
(a) The size of anions being much larger compared to cations, the lattice
enthalpy will remain almost constant within a particular group. Since the
hydration enthalpies decrease down the group, solubility will decrease as
found for alkaline earth metal carbonates and sulphates. (1 marks)
(b) Cs+ > Rb+> K+> Na+> Li+
Smaller the size of the ion greater is the degree of hydration. Lithium
being small in size is hydrated to a large extent and cesium being large in
size is hydrated to small extent. (1 mark)
(c)The M-OH bond in hydroxides of alkali metal is very weak and can
easily ionize to form M+ ions and OH- ions. This accounts for the basic
character. Since ionization energy decreases down the group bond
between metal and oxygen becomes weak. Therefore basic strength of
hydroxides increases accordingly. Thus NaOH is a stronger base than
LiOH. (1 marks)
(d) Alkali metals are highly sensitive towards air and water and are kept
therefore in kerosene or paraffin oil.
(1 mark)
(e) This is because of exceptionally small size of Lithium and high charge
to radius ratio of Li+.
(1 mark)