1. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
50
SCQ
1. For the electrolytic production of NaClO4
from NaClO3
as per the equation,
NaClO3
+ H2
O  NaClO4
+ H2
, how many faradays of electricity will be required to produce
0.5 mole of NaClO4
, assuming 60% efficiency?
(A) 0.835 F (B*) 1.67 F (C) 3.34 F (D) 1.6 F
2. Electrolytic reduction of 6.15 g of nitrobenzene using a current effeciency of 40% will require which of the
following quantity of electricity. [C = 12, H = 1, N = 14, O = 16]
(A*) 0.75 F (B) 0.15 F (C) 0.75 C (D) 0.125 C
3. A very thin copper plate is electro-plated with gold using gold chloride in HCl. The current was passed for 20 min.
and the increase in the weight of the plate was found to be 2g. [Au = 197]. The current passed was -
(A) 0.816 amp (B) 1.632 amp (C*) 2.448 amp (D) 3.264 amp
4. Acurrent of 4.0Ais passed for 5 hours through 1Lof 2M solution of nickel nitrate using two nickel electrode. The
molarity of the solution at the end of the electrolysis will be
(A) 1.5 M (B) 1.2 M (C) 2.5 M (D*) 2.0 M
Sol. The electrolysis of Ni(NO3
)2
in presence of Ni electrode will bring in following changes
Anode Ni 
 Ni2+
+ 2e–
Cathode Ni2+
+ 2e– 
 Ni
Equivalent of Ni2+
formed = Eq.of Ni2+
lost
Thus, there will be no equivalent change in concentration of Ni(NO3
)2
5. Electrolysis of a solution of HSO4

ions produces S2
O8

. Assuming 75% current efficiency, what current
should be employed to achieve a production rate of 1 mole of S2
O8

per hour ?
(A*)  71.5 amp (B) 35.7 amp (C) 142.96 amp (D) 285.93 amp
6. During an electrolysis of conc. H2
SO4
, perdisulphuric acid (H2
S2
O8
) and O2
form in equimolar amount. The
amount of H2
that will form simultaneously will be (2H2
SO4
 H2
S2
O8
+ 2H+
+ 2e–
)
(A*) thrice that of O2
in moles (B) twice that of O2
in moles
(C) equal to that of O2
in moles (D) half of that of O2
in moles
Sol. Anode 



















e
4
H
4
O
O
H
2
e
2
H
2
O
S
H
SO
H
2
2
2
8
2
2
4
2
Cathode {2H2
O  H2
+ 2OH–
– 2e–
} × 3.
Net : 2H2
SO4
+ 8H2
O  H2
S2
O8
+ O2
+ 3H2
+ 6H+
+ 6OH–
Hence ratio of 2
O
n and 2
H
n is 1 : 3.
7. You are given the following cell at 298 K with Eºcell
= 1.10 V
Zn(s) | Zn2+
(C1
) || Cu++
(C2
) | Cu(s)
where C1
and C2
are the concentration in mol/lit then which of the following figures correctly correlates Ecell
as a
function of concentrations..
x-axis : log
2
1
C
C
and y-axis : Ecell

2. (A) (B*) (C) (D)
8. Which solution is not a buffer solution ?
(A) NaCN (2 mole) + HCl (1 mole) in 5 L (B*) NaCN (1 mole) + HCl (1 mole) in 5L
(C) NH3 (2 mole) + HCl (1 mole) in 5 L (D) CH3COOH (2 mole) + KOH (1 mole) in 5L
Sol. (B) This is actually a solution of weak acid HCN with salt NaCl (having spectator ions only)
9. Which species has the lowest concentration in a solution prepared by mixing 0.1 mole each of HCN and
NaCN in 1L solution ? Ka (HCN) = 10–10.
(A) CN¯ (B) HCN (C*) H+ (D) OH¯
Sol. (C) [H+] = 10–10 M from Hendersen equation.
10. 0.1 M solution of 2 salts, NaX and NaY have pH 9 and 11 respectively. Both salts are of weak acid and strong
base, Which acis is stronger HX or HY ? Also calculate their ionisation constants.

3. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
51
SCQ
1. We have taken a saturated solution of AgBr. Ksp
of AgBr is 12 × 10–14
. If 10–7
mole of AgNO3
are added to
1 litre of this solution then the conductivity of this solution in terms of 10–7
Sm–1
units will be
[Given
º
)
Ag
( 
 = 4 × 10–3
Sm2
mol–1
º
)
Br
( 
 = 6 × 10–3
S m2
mol–1
,
º
)
NO
( 3

 = 5 × 10–3
Sm2
mol–1
]
(A*) 39 (B) 55 (C) 15 (D) 41
Sol. The solubility of AgBr in presence of 10–7
molar AgNO3
is 3 × 10–7
M.
Therefore [Br–
] = 3 × 10–4
m3
, [Ag+
] = 4 × 10–4
m3
and [NO3
–
] = 10–4
m3
Therefore total
= 
Br
+ 
Ag + 

3
NO = 39 Sm–1
2. At 298K the standard free energy of formation of H2O() is –237.20 kJ/mole while that of its ionisation into
H+ ion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be
(take F = 96500 C]
H2(g,1 bar) | H+ (1M) || OH¯ (1M) | O2 (g, 1bar)
(A*) 0.40 V (B) 0.81 V (C) 1.23 V (D) –0.40 V
Sol. Cell reaction
Cathode : H2O() +
2
1
O2(g) + 2e– 
 2OH¯(aq.)
Anode : H2(g) 
 2H+ (aq.) + 2e–
H2O() +
2
1
O2(g) + H2(g) 
 2H+(aq.) + 2OH¯(aq.)
Also we have
H2(g) +
2
1
O2(g) 
 H2O() G°f = –237.2 kJ/mole
H2O() 
 H+(aq.) + OH¯ (aq.) G° = 80 kJ/mole
Hence for cell reaction
G° = –77.20 kJ/mole
So, E° = –
nF
G

= 96500
x
2
77200
= 0.40 V
3. Which of following cell can produce more electrical work.
(A) pt, H2
| NH4
Cl || 0.1 M CH3
COOH | H2
, pt (B) pt, H2
| 0.1 M HCl || 0.1 M NaOH | H2
, pt
(C) pt, H2
| 0.1 M HCl || 0.1 M CH3
COOK | H2
, pt (D*) pt, H2
| 0.1 M CH3
COOK || 0.1 M HCl | H2
, pt
Sol. Ecell
= –
1
0591
.
0
log
c
a
]
H
[
]
H
[


For Ecell
to be highest [H+
]a
should be lower and [H+
]c
should be higher and that why anode compartment should
be more basic and cathodic compartment should be acidic.
4. Calculate the EMF of the cell at 298 K
Pt  H2
(1atm)  NaOH (xM), NaCl (xM)  AgCl (s) Ag
If
0
Ag
AgCl
Cl
E  = + 0.222 V

4. (A*) 1.048 V (B) – 0.04 V (C) – 0.604 V
(D) emf depends on x and cannot be determined unless value of x is given
Sol. anode :
2
1
H2

 H+
+ 1e–
cathode : AgCl + 1e– 
 Ag + Cl–
––––––––––––––––––––––––––––––––––
net :
2
1
H2
+ AgCl 
 


e
1
H+
+ Ag + Cl–
Ecell
= + 0.222 +
1
059
.
0
log
]
Cl
][
H
[
1


= + 0.222 + 0.059 log
]
Cl
[
]
10
[
]
OH
[
14 


= + 0.222 + 0.059 (14)
= + 1.048 volt
5. A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of copper was deposited
on the cathode. Calculate the current efficiency for the copper deposition. (Cu – 63.5)
(A*) 79% (B) 39.5 % (C) 63.25% (D) 63.5%
Sol. m (theoretical) =
96500
7200
1
.
0
5
.
63 

= 0.4738 g
 % efficiency =
4738
.
0
3745
.
0
× 100 = 79 %

5. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
52
SCQ
1. The emf of the cell Zn|ZnCl2
(0.05 M)|AgCl(s), Ag is 1.015 V at 298 K and the temperature coefficient of its emf
is – 4.92 × 10–4
K
V
. How many of the reaction thermodynamic parameters G, S and H are negative at 298
K?
(A) None of them (B) One of them (C)Two of them (D*)All of them
2. Number of electrons lost during electrolysis of 0.355 g of Cl¯ is -
(A) 0.01 (B*) 0.01 N0
(C) 0.02 N0
(D)
0 01
2 0
.
N
3. Asolution containing one mol per litre each of Cu(NO3
)2
, AgNO3
, Hg(NO3
)2
and Mg(NO3
)2
isbeing electrolysed by
using inert electrodes. The values of the standard oxidation potentials in volts are Ag/Ag+
= –0.8 V ; Hg/Hg2+
=
–0.79 V; Cu/Cu2+
= – 0.34 V ; Mg/Mg2+
= 2.37 V. The order in which metals will be formed at cathode will be -
(A) Ag, Cu, Hg, Mg (B) Ag, Hg, Cu, Mg (C*)Ag, Hg, Cu (D) Cu, Hg, Ag
Subjective
4. Find the thickness of the electro deposited silver if the surface area over which deposition occurred was 100 cm2
and a current of 0.2 A flowed for 1hr with the cathodic efficiency of 80%. Density of Ag = 10 g/cc (Ag = 108).
Sol. th × Area × density =
96500
3600
8
.
0
2
.
0
108 


;
th =
10
100
96500
3600
8
.
0
2
.
0
108





cm = 6.4 × 10–4 cm
SCQ
5. In the electrolysis of an aqueous potassium sulphate solution, the pH of the solution in the space near an
electrode increased. Which pole of the current source is the electrode connected to?
(A) The positive pole (B) Could be either pole
(C*) The negative pole (D) Cannot be determined
6. For a saturated solution of AgCl at 25°C, k = 3.4 × 10–6
ohm–1
cm–1
and that of H2
O () used is
2.02 × 10–6
ohm–1
cm–1
. m
for AgCl is 138 ohm–1
cm2
mol–1
then the solubility of AgCl in moles per liter will be-
(A*) 10–5
(B) 10–10
(C) 10–14
(D) 10–16
7. What must be the concentration of Ag+
in an aqueous solution containing Cu2+
= 1.0 M so that both the metals
can be deposited on the cathode simultaneously. Given that
0
Cu
Cu 2
E  = – 0.34 V and
0
Ag
Ag
E  = 0.812 V, T =
=
298 K
(A) nearly 10–19
M (B) 10–12
M (C*) 10–8
M (D) nearly 10–16
M
8. Electrolytic reduction of 6.15 g of nitrobenzene using a current effeciency of 40% will require which of the
following quantity of electricity. [C = 12, H = 1, N = 14, O = 16]
(A*) 0.75 F (B) 0.15 F (C) 0.75 C (D) 0.125 C

6. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
53
SCQ
1. The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79
V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is
(A) 0.5 mmol (B*) 1.0 mmol (C) 2.0 mmol (D) 2.5 mmol
Sol. AgCl + 1e– 
 Ag + Cl– E° = 0.2 V
Ag 
 Ag+ + 1e– E° = – 0.79 V
––––––––––––––––––––––––––––––––––––––––
AgCl 
 


e
1
Ag+ + Cl– E° = – 0.59 V
E° =
n
059
.
0
log K  – 0.59 =
1
059
.
0
log KSP
 KSP = 10–10
Now solubility of AgCl in 0.1 M AgNO3
S (S + 0.1) = 10–10  S = 10–9 mol/L
Hence 1 mole dissolves in 109 L solution
hence in 106 L amount that dissolves ins 1 m mol.
2. Calculate the cell EMF in mV for
Pt  H2
(1atm)  HCl (0.01 M)  AgCl(s) Ag(s) at 298 K
if Gf
° values are at 25°C
– 109.56
mol
kJ
for AgCl(s) and
– 130.79
mol
kJ
for (H+
+ Cl–
) (aq)
(A*) 456 mV (B) 654 mV (C) 546 mV (D) None of these
Sol. 0
reaction
cell
G
 = 2 (– 130.79) – 2 (– 109.56)
= – 42.46 kJ/mole
(for H2
+ 2AgCl 
 2Ag + 2H+
+ 2Cl–
)
 0
cell
E =
96500
2
42460



= + 0.220 V
Now Ecell
= + 0.220 +
2
059
.
0
log 4
)
01
.
0
(
1
= 0.456 V = 456 mV.
.
3. Adiponitrile is manufactured electrolytically from acrylonitrile
CH2
= CHCN 
 CN – (CH2
)4
– CN
How many kg of adiponitrile (molecular mass = 108) is produced in 9.65 hr using a current of 3750 Awith 80%
efficiency?
(A) 30 kg (B*) 58 kg (C) 60 kg (D) 80 kg
Sol. 2CH2
= CHCN + 2H+
+ 2e– 
 CN – (CH2
)4
CN
96500
3600
65
.
9
8
.
0
3750
10
2
108
m
3
–











 = 58.32 kg
Ans. 58 kg.

7. 4. To observe the effect of concentration on the conductivity, electrolytes of different nature are taken in two vessel
‘A’ and ‘B’. ‘A’ contains weak electrolyte e.g., NH4
OH and ‘B’ contains strong electrolyte e.g., NaCl. In both
container concentration of respective electrolyte is increased and conductivity observed :
(A) in ‘A’ conductivity increases, in ‘B’ conductivity decreases
(B) in ‘A’ conductivity decreases while in ‘B’ conductivity increases
(C) in both ‘A’ and ‘B’ conductivity increases
(D*) in both ‘A’ and ‘B’ conductivity decreases
Sol. Weak electrolyte is weakly ionizing substance, dilution promotes ionization thus conductivity.
For strong electrolyte, as concentration increases interionic attraction increases and conduction decreases.
5. A galvenic cell : Cu | Cu2+
|| Ag+
| Ag
initially contains 1 M Ag+
and 1 M Cu2+
ions. What will be of the cell potential after the passage of 9.65 A of
current for 1 h ?
(A) Existing cell operation speeds up and cell potential increases by 0.05 V
(B) Existing cell operation gets reversed and cell potential decreases by 0.05 V
(C*) Existing cell operation gets reversed and cell potential increases by 0.01 V
(D) Existing cell operation speeds up and cell potential decreases by 0.01 V
Sol. Cu + 2Ag+

 Cu2+
+ 2Ag
Ecell
= Eºcell
–
2
059
.
0
log 2
2
]
Ag
[
]
Cu
[


 Ecell
= Eºcell
On the passage of 9.65 A for 1 h (9.65 × 60 × 60 C) externally cell reaction is reversed, now Cu2+
will be
discharging (concentration of Cu2+
ion will decrease) and Ag will be ionizing (concentration of Ag+
will increase).
Final mol of Ag+
=
96500
60
60
65
.
9 

= 0.36 g-equivalent = 0.36 mol
and Final mol of Cu2+
=
96500
60
60
65
.
9 

= 0.36 g-equivalent = 0.18 mol
[Ag+
] = 1 + 0.36 = 1.36 M ; [Cu2+
] = 1 – 0.18 = 0.82 M
Hence, Ecell
= Eºcell
–
2
059
.
0
log 2
]
36
.
1
[
]
82
.
0
[
or Ecell
= Eºcell
+
2
059
.
0
log
]
82
.
0
[
]
36
.
1
[ 2
Ecell
= Eºcell
+ 0.010 V
6. In which of the following case, increase in concentration of ion cause increase in Ecell
?
(A) Pt(H2
) | H+
(aq) (B*) Ag, AgCl | Cl
(aq)
(C) Pt | quinhydrone | H+
(aq) (D) Ag | Ag+
(aq)
Sol. (A)
2
1
H2
(g) 
 H+
(aq) + e–
; Ecell
= Eºcell
– 0.0591 log [H+
]
(B) Ag(s) + Cl–
(aq) 
 AgCl(s) + e–
; Ecell
= Eºcell
– 0.0591 log 1 / [Cl–
]
(C) 


+ 2H+
+ 2e–
; Ecell
= Eºcell
– 0.0591 log [H+
]
(D) Ag(s) 
 Ag+
(aq) ; Ecell
= Eºcell
– 0.0591 log [Ag+
]

8. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
54
1. Giventhat Ni2+
| Ni = 0.25 V ; Cu2+
| Cu = 0.34 V
Ag+
| Ag = 0.80 V ; Zn2+
| Zn = – 0.76 V
Which of the following reactions under standard condition will not take place in the specified direction?
(A*) Ni2+
(aq) + Cu(s) 
 Ni(s) + Cu2+
(aq) (B) Cu(s) + 2Ag+
(aq) 
 Cu2+
(aq) + 2Ag(s)
(C*) Cu(s) + 2H+
(aq) 
 Cu2+
(aq) + H2
(g) (D) Zn(s) + 2H+
(aq) 
 Zn2+
(aq) + 3H2
(g)
Sol. Reaction (A) and (C) both have –ve emf and hence cannot take place in the specified direction.
2. In the equation ºm
=  + B C , the constant ‘B’ depends on :
(A) C1/2
(B*) stoichiometry of electrolyte
(C) resistance (D) conductivity
Sol. The value of B varies with different nature of electrolyte e.g., NaCl (Na+
Cl–
), MgO (Mg2+
, O2–
), MgCl2
(Mg2+
, 2Cl),
etc.
3. Which of the following is not a buffer solution ?
(A) 0.8 M H2
S + 0.8 M KHS (B) 2 M C6
H5
NH2
+ 2MC6
H5

NH3
Br–
(C) 3 M H2
CO3
+ 3 M KHCO3
(D*) 0.05 M KClO4
+ 0.05 M HClO4
Sol. HClO4
is a strong acid.
4. The emf of the cell
Ag | Ag+
(1 M) || I–
(1 M) | AgI | Ag is E
The solubility product of AgI can be expressed as :
(A) Ks
=
RT
303
.
2
nF
log E (B) ln K = nF 








E
T
E
(C) ln Ks
=
E
nF
(D*) log Ks
= RT
303
.
2
nFE
Sol. For the cell Ag | Ag+
|| I–
, AgI ; Ag
LHS electrode Ag 
 Ag+
+ e–
RHS electrode AgI + e– 
 Ag + I–
Cell reaction, AgI Ag+
+ I–
The equilibrium constant (K) = The solution product Ks
G = – nFE
Gº = – 2.303 RT log K
log K = RT
303
.
2
nFE
Comprehension # 1
(Read the following passage and answer the questions numbered 5 to 6. They have only one
correct option)
Metallic gold is frequently found in aluminosilicate rocks and it is finely dispersed among other
minerals. It may be extracted by treating the crushed rock with aerated sodium cyanide solution. During this
process metallic gold is slowly converted to [Au(CN)2
]–
which is soluble in water.
Au + CN

 [Au(CN)2
]–
After equilibrium has been reached, the aqueous phase is pumped off and metallic gold is recovered from it by
reacting the gold complex with zinc, which is converted to Zn(CN)4
I2–
Zn + [Au(CN2
)–

 [Zn(CN)4
]2–
+ Au
Gold in nature is frequently alloyed with silver which is also oxidised by aerated sodium cyanide solution.
[Zn(CN)4
]2–
+ 2e–

 Zn + 4CN
; Eº = – 1.26 V

9. [Au(CN)2
]–
+ e–

 Au + 2CN
; Eº = – 0.60 V
[Ag(CN)2
]–
+ e–

 Ag + 2CN
; Eº = – 0.31 V
5. Five hundred litres (500 L) of a solution of 0.0100 M in [Au(CN)2
]–
and 0.003 M in [Ag(CN)2
]
–
was evaporated to
one third of the original volume and was treated with zinc (40 g). Assuming that deviation from standard condi-
tions is unimportant in this case and all the concerned redox reaction go
essentially to completion. Hence, the concentrations of [Au(CN)2
]–
and of [Ag(CN)2
]–
after reaction are:
(A) Au(CN)2
]–
= 0.030 M ; [Ag(CN)2
]–
= 0.0030 M
(B) Au(CN)2
]–
= 0.0100 M ; [Ag(CN)2
]–
= 0.0090 M
(C*) Au(CN)2
]–
= 0.030 M ; [Ag(CN)2
]–
= 0.0020 M
(D) [Au(CN)2
]–
= [Ag(CN)2
]–
= 0.030 M
Sol.








]
)
CN
(
Ag
[
]
)
CN
(
Au
[
2
2
+ Zn 


 

duction
Re
Certainly according to spontaneity and availability substance will be reduced.
EºAg|Zn
= – 0.31 – (– 1.26) = 0.95 V
EºAu|Zn
= – 0.60 – (– 1.26) = 0.66 V
As EºAg|Zn
> EºAu|Zn
Hence, Ag complex will be reduced first :
Mole of Ag in 500 L = 500 × 0.0030 = 1.5 mol
Mole of Au in 500 L = 500 × 0.010 = 5.0 mol
Mole of Zn in 40 g =
38
.
65
40
= 0.61 mol
2[Ag(CN)2
]–
+ Zn 
 [Zn(CN)4
]2–
+ 2Ag
1 mol of zinc reacts with 2 mol of Ag(I) or Au(I), 0.61 mol of zinc reacts with 1.2 mol [Ag(CN)2
]–
Remaining [Ag(CN)2
]–
= 1.5 – 1.2 = 0.3 mol
[Ag(CN)2
]–
will not be reduced as zinc gets over.
Hence, concentration of [Ag(CN)2
]–
, reaction get over = 0.010 × 3 = 0.030 M
(due to evaporation to one-third)
Concentration of [Ag(CN)2
]–
when reaction gets over =
500
3
3
.
0 
= 0.002 M
6. [Au(CN)2
]–
is a very stable complex under certain conditions. The concentration of cyanide ion which is required
to keep 99 mol % of the gold in the form of the cyanide complex is :
(A) 2 × 10–28
M (B) 3 × 10–14
M (C) 5 × 10–28
M (D*) 5 × 10–14
M
Sol. Au+
+ 2CN–
[Au(CN)2
]–
99 mol% [Au(CN)2
]–
It means 


 )]
CN
(
Au
[
]
Au
[
)]
CN
(
Au
[
2
2
=
100
99
100[Au(CN)2
]–
= 99[Au+
] + 99[Au(CN)2
]–
[Au+
] =
99
)]
CN
(
Au
[ 2

– (I)
Kƒ
= 2
2
]
CN
[
]
Au
[
)]
CN
(
Au
[



Putting the [Au+
] from (I) in Kƒ
4 × 1028
= 2
]
CN
[
99

 [CN–
] = 5 × 10–14
M

10. 7. Match the following :
Column I Column II
(Salts) (pH Value)
(A) Potassium nitrite (p) pH =
2
1
(p 1
a
K + p 2
a
K )
(B)Anilinium chloride (q) pH =
2
1
(pKw
+ pKa
– pKb
)
(C) Sodium bicarbonate (r) pH =
2
1
(pKw
– pKb
– log C)
(D)Ammonium cyanide (s) pH =
2
1
(pKw
+ pKa
+ log C)
Ans. (A - s), (B - r) , (C - p) , (D - q)
Sol. Potassium nitrite (salt of weak acid and strong base) pH =
2
1
(pKw
+ pKa
+ log C)
Anilinium chloride (salt of strong acid and weak base) pH =
2
1
(pKw
– pKb
– log C)
Ammonium cyanide (salt of weak acid and weak base) pH =
2
1
(pKw
+ pKa
– pKb
)
Sodium bicarbonate (salt of diprotic acid) pH =
2
1
(p 1
a
K + p 2
a
K )

11. DAILY PRACTICE PROBLEMS
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1. In a face centred cubic arrangement of A and B atoms whose A atoms are at the corner of the unit cell and
B atoms at the face centres. One of the A atom is missing from one corner in unit cell. The simplest
formula of the compound is
(A) A7
B3
(B) AB3
(C*) A7
B24
(D) A2
B3
2. A cubic unit cell contains manganese ions at the corners and fluoride ions at the centre of each edge.
(a) What is the empirical formula
(b) What is the C.N. of the Mn ion ?
(c) Calculate the edge length of the unit cell if the radius of a Mn ion is 0.65Å and that of
F–
ion is 1.36Å.
Ans. (a) MnF3
(b) (c) 2.01
3. You have a face centred cubic lattice in which corner atoms are type A and face atoms are type B. In each
unit cell, one of the corner atoms is missing. What is the formula in terms of A and B?:
(A) A7
B12
(B*) A7
B24
(C) AB3
(D) A3
B2
4. Fraction of the total volume occupied by atoms in a simple cube is
(A*) /6 (B) 
3 /8 (C) 
2 /6 (D) /3
5. Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell
dimensions are a = 6.8 Å, b = 4.4 Å and c=7.2 Å. If the molar mass is 21.76, then the density of crystals
is :
(A*) 0.6708 g cm–2
(B) 1.6708 g cm–3
(C) 2.6708 g cm–3
(D) None of these.
6. The most unsymmetrical crystal system is:
(A) Cubic (B) Hexagonal (C*) Triclinic (D) Orthorhombic
7. A metal has face centered cubic arrangement. If length of the edge of the cell is x cm and M is its atomic
mass, then density will be equal to (N0
is Avogadro number):
(A) [(M)/(X3
× N0
)] g cm–3
(B) [(M × N0
) (X3
)] g cm–3
(C*) [(4M)/(X3
× N0
)] g cm–3
(D) [(M)/(4X3
× N0
)] gm cm–3
8. A metallic element has simple cubic arrangement. The number of unit cells in 100 g of this metal
(edge length = 288 pm, density = 7.2 g cm–3
) are 5.8 x 1023
. The total number of atoms in 100 g of the metal is:
(A) 5.8 x 1024
(B*) 5.8 x 1023
(C) 0.58 x 1023
(D) 58 x 1024

12. DAILY PRACTICE PROBLEMS
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1. The number of crystal systems known are :
(A*) 7 (B) 8 (C) 6 (D) 4
2. Tetragonal crystal system has the following unit cell dimensions:
(A) a = b = c and  =  =  = 90° (B*) a = b  c and  =  =  = 90°
(C) a  b  c and  =  =  = 90° (D) a = b  c and  =  = 90°,  = 120°
3. The lattice parameters of a given crystal are a = 5.62 Å, b = 7.41 Å and c = 9.48 Å. The three coordinate
axes are mutually perpendicular to each other. The crystal is :
(A) tetragonal (B*) orthorhombic (C) monoclinic (D) trigonal.
4. Three elements P, Q and R crystallize in a cubic solid lattice. The P atoms occupy the corners, Q atoms
the cube centres and R atoms the edges. The formula of the compound is:
(A) PQR (B) PQR2
(C*) PQR3
(D) PQ3
R.
5. Which of the following is not a crystal system ?
(A) Triclinic (B) Rhombohedral (C) Tetragonal (D*) Isomorphous.
6. An fcc lattice has lattice parameter a = 400 pm. Calculate the molar volume of the lattice including all the
empty space:
(A) 10.8 mL (B) 96 mL (C) 8.6 mL (D) 9.6 mL
7. The crystal system in which a  b  c and the angles  is
(A*) Triclinic (B) monoclinic (C) hexagonal (D) cubic
8. A body centred cubic arrangement is show :
O is the body centre ; A, B, C, D, .........., H are the corners. What is the magnitude of the angle AOB?
(A) 120° (B) 109° 28 (C) 104° 31 (D*) 70° 32

13. DAILY PRACTICE PROBLEMS
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1. The unit cell parameters of a rhombohedral crystal are
(A) a  b  c,  =  =  = 90° (B*) a = b = c,  =  =   90°
(C) a  b  c,      90° (D) a = b  c,  =  =  = 90°
2. A solid has a b.c.c. structure. If the distance of closest approach between the two atoms is 1.73 Å. The
edge belli length of the cell is ;
(A) 2 pm (B) )
2
/
3
( pm (C*) 200 pm (D) 142.2 pm
3. The compound AB crystallizes in cubic lattice in which both the elements have co-ordination number of
eight. The crystal class is :
(A*) simple cubic (B) face-centered cubic
(C) body-centered cubic (D) None of these.
4. If a metal has a bcc crystal structure, the coordination number is 8, because :
(A) each atom touches four atoms in the layer above it, four in the layer below it and none
in its own layer.
(B) each atom touches four atoms in the layer above it, four in the layer below it and one
in its own layer.
(C) two atoms touch four atoms in the layer above them, four in the layer below them,
and none in their own layer.
(D) each atom touches eight atoms in the layer above it, eight in the layer below it and
none in its own layer.
5. In a ccp structure, the :
(A) first and third layers are repeated
(B) first and fourth layers are repeated
(C) second and fourth layers are repeated
(D) first, third and sixth layers are repeated.
6. In a face centred cubic lattice the number of nearest neighbours for a given lattice point are
(A) 6 (B*) 8 (C) 12 (D) 14
7. How many 'nearest' and 'next nearest' neighbours respectively does potassium have in b.c.c. lattice
(A) 8, 8 (B*) 8, 6 (C) 6, 8 (D) 8, 2
8. The density of KBr is 2.75 gm/cc length of the unit cell is 654 pm. K = 38, Br = 80, then what is true about
the predicted nature of the solid.
(A*) Solid has F.C.C. structure with co-ordination number = 6
(B) Solid has simple cubic structure with co-ordination number = 4
(C) Solid has F.C.C. structure with co-ordination numbers-1
(D) None of these

14. DAILY PRACTICE PROBLEMS
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SCQ
1. CsBr has b.c.c. like structure with edge length 4.3 A. The shortest inter ionic distance in between Cs+
and
Br–
is :
(A*) 3.72 (B) 1.86 (C) 7.44 (D) 4.3
2. A metal crystallizes in a body centered cubic lattice (bcc) with the edge of the unit cell 5.2Å. The distance
between the next nearest neighbour is
(A) 10.4 Å (B*) 4.5 Å (C) 5.2Å (D) 9.0Å
3. Krypton crystallizes with four atoms per unit cell and unit cell is a cube. If the density of krypton is 3.19g/
ml,what is the edge length of the unit cell? What is the minimum distance between the two nearest
neighbours?
4. Potassium crystallizes in body centred cubic lattice with a unit cell length a = 5.2 Å
(A) What is the distance between nearest neighbours
(B) What is the distance between next nearest neighbours
(C) How many nearest neighbours does each K atom have
(D) How many next nearest neighbours does each K have
(E) What is calculated density of crystalline K.
5. A metal crystallizes in two cubic phases i.e., FCC and BCC whose unit cell lengths are 3.5Å and 3.0 Å
respectively. The ratio of their densities is :
(A) 3.12 (B) 2.04 (C) 1.30 (D) 0.72
Subjective
6. The coordinate no. of barium ion Ba2+
in Ba F2
is 8. What must be the C.N. of F–
ion.
7. If the ratio of co-ordination no. P to that of Q be Y : Z, then the formula of the solid is _____.
8. A mineral having formula AB2
crystallize in the cubic close packed lattice, with the A atoms occupying the
lattice points. What is the co-ordination no. of A atoms? of the B atoms? what fraction of tetrahedral sites
is occupied by B atoms.
Ans. 8, 4, 100%.

15. DAILY PRACTICE PROBLEMS
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SCQ
1. The numbers of tetrahedral and octahedral holes in a ccp array of 100 atoms are respectively
(A*) 200 and 100 (B) 100 and 200 (C) 200 and 200 (D) 100 and 100
2. The number of nearest neighbours and next near neighbours of a Na+
ion in a crystal of NaCl are respec-
tively -
(A) 6Na+
, 12Cl¯ (B) 6Cl¯ , 6Na+
(C) 12Cl¯, 6Na+
(D) 6Cl¯, 12Na+
3. In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral
and octahedral voids?
(A*) 0.543 (B) 0.732 (C) 0.414 (D) 0.637
Subjective
4. Show by drawing a diagram that in the NaCl lattice each ion occupies an octahedral void space
provided by Na
ions. How many ions surround each ion in the lattice?
Ans. 12
5. A simple cubic lattice consists of eight identical spheres of radius R in contact, placed at the corners of a cube.
What is the volume of the cubical box that will just enclose these eight spheres and what fraction of this volume
is actually occupied by the spheres?
Ans. 64R3, 52.36%
6. Copper has a face-centred cubic structure with a unit-cell edge length of 3.61Å. What is the size of the largest
atom which could fit into the interstices of the copper lattice without distorting it?
[Ans. 0.53Å]
7. A solid A+B– has NaCl type close packed structure. Compute the radius of the cation when the radius of anion
is 250 pm.
Can a cation C+ having a radius of 180 pm be accommodated into the tetrahedral site of the crystal A+B–? Give
reason for your answer
Sol. Since the solid A+B– has NaCl type close packed structure, it belongs to a system of coordination number 6. In
such case, the ratio of the cation to anion radii is given by


r
r
= 0.414
Since r– = 250 pm
 r+ = 0.414 × 250pm = 103.5 pm = Radius of the cation
For any tetrahedral site the ratio of cation to anion radii should be between 0.225 and 0.414
Now


r
r
= pm
250
pm
180
= 0.72
Since this ratio does not fall within the limit, the cation C+ having a radius of 180 pm cannot be slipped/
accommodated into the tetrahedral site of the crystal A+B–.

16. DAILY PRACTICE PROBLEMS
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Subjective
1. If 20 ml of 0.5 M Na2SO4 is mixed with 50 ml of 0.2 M H2SO4 & 30 ml of 0.4 M Al2(SO4)3 solution calculate.
[Na+], [H+], [SO4
2–], [Al3+]. Assuming 100% dissociation.
Sol. =
volume
moles
= 10 m. moles of Na2SO4
 20 m. moles of Na+
(P)  [Na+] =
100
20
= 0.2 M
(ii) [H+] = ?
10 m. moles H2SO4
20 m. moles H+
[H+] =
100
20
= 0.2 M
(iii) SO4
2– =
100
36
10
10 

=
100
56
= 0.56 M
(iv) Al3+ =
100
24
= 0.24 M
2. Calculate the osmotic pressure of a decinormal solution of cane sugar at 27°C.
Sol.
V
n
= molar concentration = 0.1 M  R = 0.082 litre atm K–1
mol–1
, T = 300 K
 we have P =
V
n
RT  = 0.1 x 0.082 x 300 = 2.46 atm
3. An aqueous solution contains 18 g of glucose (mol wt. = 180) per 0.5 L. Assuming the solution to be ideal,
calculate osmotic pressure at 27°C.
Sol. Mole of glucose =
180
18
=
10
1
, V = 0.5 liter
R = 0.082 , T = 300 K
we have , P =
V
n
RT
=
5
.
0
1
.
0
x 0.082 x 300 = 4.92 atm.
4. A solution of 1.73 g of ‘A’ in 100 cc of water is found to be isotonic with a 3.42% (weight/volume) solution
sucrose (C12
H22
O11
) Calculate molecular weight of A (C12
H22
O11
= 342)
Sol. We know that the isotonic solution have the same molar concentration (i.e moles.litre)
Let the molecular weight of A be M
moles of A = 1.73/M
 molar conc. of sucrose =
342
42
.
3
x
100
1000
= 0.1
M
3
.
17
= 0.1 ; M = 173.
5. At 12°C the osmotic pressure of a urea solution is 500 mm. The solution is diluted and the temperature is
raised to 27°C, when the osmotic pressure is found to be 100 mm. determine the extent of dilution.
Sol. Suppose V1
titres of the solution contains n moles of the solute at 12°C which was diluted to V2
litres at 27°C.
Thuswe have

17. 760
500
=
1
V
n
x 0.082 x 285 ...(i)
and
760
100
=
2
V
n
x 0.082 x 300
Dividing (1) by (2), we get
1
2
V
V
= 5.3.
SCQ
6. A sample of water has a hardness expressed as 80 ppm of Ca2+
. This sample is passed through an ion
exchange column and the Ca2+
is replaced by H+
. What is the pH of the water after it has been so treated?
[Atomic mass of Ca = 40]
(A) 3 (B) 2.7 (C) 5.4 (D*) 2.4
Sol. 106
ml water contains 80 gm of Ca2+
=
40
80
moles = 2 moles of Ca2+
= 2 × 2 moles of H+
ions
so 103
ml of H2
O will have = 4 × 10–3
moles of H+
ions
so pH = 3 – log 4 = 3 – 0.6 = 2.4.
Subjective
7. A membrane permeable only to water separates a 0.01 M solution of sucrose from a 0.001 M one. On which
solution must pressure be applied to bring the system into equilibrium? Find this pressure if the T = 300 K.
Ans. 0.22 atm.
Sol. Pressure will be applied to higher conc. side.
 = P1
– P2
= (C1
– C2
) × RT
= (10–2
– 10–3
) × 0.082 × 300 = 0.2214 atm.
8. The osmotic pressure of blood is 7 atm at 30°C. What is the molarity of the isotonic saline solution if the ‘i’ factor
for sodium chloride is taken to be 1.9?
Ans. 0.148 M.
Sol. For isotonic solution.
1
– 2
7 = 2 × M × RT.
= 1.9 × M × 0.82 × 303 = 0.148 M.
MTC
9. Column () Column ()
(A) 50 ml of 3M HCl + 150 ml of 1M FeCl3 (p) 1.85 m
(B) mole fraction of NaCl in aqueous solution
of NaCl is 0.1 then molality of the solution is (q) [Cl–
] = 3 M
(C) 10%(w/w) propanol (C3
H7
OH) solution has molality (r) [H+
] = 0.75 M
(D) 10.95% (w/v) HCl (s) 6.1 m
Ans. (A – q, r ; B – s ; C – p ; D – q)
Sol. (a) [Cl–
] =
200
3
1
150
3
50 



=
200
600
= 3 M
(b) molality = 1000
18
9
.
0
1
.
0


= 6.17 m
(c) Molality = 1000
90
60
10
 = 1.85 m
(d) Molarity of HCl =
100
5
.
36
95
.
10
× 1000 = 3 M

18. 10. Column I Column II
(A) 3.5 M solution of X (M = 100) specific gravity 1.35 (p) mole fraction of solute is 0.2
(B) 2 M 100 ml H2
SO4
is diluted to 500 ml (q) mass of the solute is 2.7 g
(C) 14.5 m solution of urea (N2
H4
CO) (r) molality = 3.5
(D) 50 ml of 0.1 M aq. solution of Glucose (180) (s) molarity 0.4
mixed with 50 ml of 0.2 M solution of Glucose
Ans. (A – r ; B – s ; C – p ; D – q)
Sol. (A) 3.5 M solution of X (M = 100) sp. gravity 1.35
Molality = )
350
1350
(
5
.
3

× 1000 = 3.5 m
Mole fraction of solute =
18
1000
5
.
3
5
.
3

= 0.06
(B) 2M, 100 ml is dilute to 500 ml
New molarity =
500
100
2
= 0.4 M
(C) 14.5 m solution of urea.
Xsolute
=
18
1000
5
.
14
5
.
14
 = 0.207
(D) 50 ml of 0.1 M aq. glucose sol. (180) mixed with 50 ml of 0.2 M sol. of glucose.
Total millimoles of glucose = 50 × 0.1 + 50 + 0.2
Mass of solute = 15 × 10–3
× 180 = 2.7 gm

19. DAILY PRACTICE PROBLEMS
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SCQ
1. Assuming the salts to be completely ionized in solution, which of the following has highest osmotic pressure.
(A) 1% CsCl w/w (B) 1% RbCl w/w (C) 1% KCl w/w (D*) 1% NaCl w/w
Sol. 1% wt. by wt.
1g of solute in 100 g of solution.
M
1
moles of solute in 100 g of solution.
Meghec mol. mass less no. of particles & less osmotic pressure.
2. Solution having osmotic pressure nearer to that of an equimolar solution of K4
[Fe(CN)6
] is:
(A) Na2
SO4
(B) BaCl2
(C*)Al2
(SO)3
(D)C12
H22
O11
3. Two aqueous solutions, one of the NaCl in water (A) and the other of C8
H15
O2
Na in water (B) are isotonic. If wA
and wB
are weight fractions of NaCl and C8
H15
O2
Na in solution A and B respectively, then (assuming that both
the salts dissociate completely) :
(A) wA
> wB
(B) wA
= wB
(C*) wA
< wB
(D) none of these
Sol. Two solution are isotonic Þ no. of moles of ions are same = no. of moles of molecule (as i = 2 for both)
 weight of B is Greatce than weight of A.
 WB
> WA
4. A sample of air is saturated with benzene (vapor pressure = 100 mm Hg at 298 K) at 298K, 750mm Hg pressure.
If it is isothermally compressed to one third of its initial volume, the final pressure of the system is
(A) 2250 torr (B) 2150 torr (C*) 2050 torr (D) 1950 torr
5. What weight of glucose dissolved in 100 grams of water will produce the same lowering of vapour pressure as
one gram of urea dissolved in 50 grams of water, at the same temperature?
(A) 3 gms (B) 5 gms (C*) 6 gms (D) 4 gms
Sol. Since same lowering in vapour pressure.
Xsolute of urea
= Xsolute of glucose
.
18
50
60
1
60
1

=
18
100
180
w
180
w

. 
18
50
60
1
=
18
100
180
w
(For dilute solution).
 w = 6 gm. of glucose.
6. Three solutions are prepared by adding 'w' gm of 'A' into 1kg of water, 'w' gm of 'B' into another 1 kg of water
and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed from these
solutions in sequence (A 
 B 
 C). The loss in weight of solution A was found to be 2 gm while
solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and
C is :
(A) MA : MB : MC = 4 : 3 : 5 (B) MA : MB : MC =
4
1
:
3
1
:
5
1
(C*) MC > MA > MB (D) MB > MA > MC
Sol. The loss in weight should be proportional to vapour pressure above that solution :
So, gm
2
P A
S 
gm
5
.
1
P B
S 
gm
5
.
2
P C
S 
So, maximum vapour pressure is above C solution hence, it is having minimum lowering and hence minimum
mole fraction (hence minimum number of moles of solute) So max. molar mass of substance.

20. MCQ
7. Which is the correct relation between osmotic pressure of 0.1 M NaCl solution & 0.1 M Na2
SO4
solution :
(A) the osmotic pressure of Na2
SO4
is less than NaCl solution
(B*) the osmotic pressure of Na2
SO4
is more than NaCl solution
(C*) the osmotic pressure of Na2
SO4
is 1.5 times that of NaCl solution
(D) the osmotic pressure of NaCl is 1.5 times that of Na2
SO4
solution
Subjective
8. A solution is prepared by dissolving 10 g of nonvolatile solute in 180 g of H2
O. If the relative lowering of vapour
pressure is 0.005, find the mol. wt of the solute.
Sol. Suppose the mol. wt. of the solute is M.
so moles of solute =
M
10
Moles of solvent (H2
O) =
18
180
= 10
mole fraction of solute =
10
M
/
10
M
/
10

=
M
1
1

we know, relative lowering of vapoure pressure = mol fraction of solute
0.005 =
M
1
1

so M = 199.
9. Match the following :
Column  column
(A) 120 g CH3
COOH in 1 Ltr solution
dsol
= 1.2 g / mL (p) M = 2
(B) 120 g glucose dissolved in 1 Ltr solution
(dsol
= 1.2 g/mL) (q) 10% w/w solution
(C) 2
2CONH
NH
X = 1/31 (aquous solution) (r) 12% w/v solution
(s) m = 1.85
(D) 19.6 % (w/v) H2
SO4
solution  (dsolution
= 1.2 g/mL)
Ans. [A – p,q,r,s] ; [B – q, r] ; [C – q, s] ; [D – p]
Sol. (A) 120 g CH3
COOH in 1L solution & dsol
= 1.2 g/ml
molarity =
1
60
120

= 2M
Molality =
)
120
–
1200
(
100
2
= 1.85 m
120 g CH3
COOH in 1000 Ml soltion  12 g in 100 ml of solution
 12% (w/v)
12 g in 100 × 1.2 g of solution
 10% w/w
(C) 2
2CONH
NH
X = 31
1 (aq. solution)
1 mole of urea in (30 × 18)g of solution.
molatity =
1000
18
30
1

=
18
30
1000

= 1.85 m
60 g of urea in (30 × 18) g of solution.
100 g of solution. mill have =
18
30
60

× 100 = 11% w/w
(D) 19.6 g of H2
SO4
in 100 ml.
196 g of H2
SO4
in 1000 ml  Molarity = 2 M
196 g of H2
SO4
in (100 × 1.2) g ol 1200 g of solution

21. molalty =
)
196
–
1200
(
1000
2
 2
10. Match the following :
Column-I Column-II
(A) 1 M glucose solution (p) 1 mol solute per litre solution
(B) 3 M urea solution (q) 180 g solute per litre solution
(C) 3 M CH3COOH solution (r) % w/v = 18% (solution)
(D) 1 M H2SO4 solution (s) % w/v = 9.8% (solution)
Ans. [A – p, q, r] ; [B – q, r] ; [C – q, r] ; [D – p, s]

22. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
62
1. 20g of a binary electrolyte (molecular weight = 100) are dissolved in 500 g of water. The freezing point of the
solution is – 0.74°C ; Kf
= 1.86 K moIality–1
, The degree of dissociation of electrolyte is
(A) 50% (B) 75% (C) 100% (D*) Zero
Sol. Tf
= 2Kf
m
0.74 = 2 × 1.36 × 0.4  2 = 0.9945  1
i = 1 +   1   0
2. 1M of glucose solution has a freezing point of 1.860
C. If 10 ml of 1M glucose is mixed with 30ml of 3M glucose
then the resultant solution will have a freezing point of
(A) 2.790
C (B*) 4.650
C (C) 5.580
C (D) 7.440
C
Sol. DTf
= Kf
xm
1.86 = 1.86 × m  m = 1
molelity =
)
180
–
d
1000
(
1000
1


= 1  1.18 g/ml
Mmix
=
40
3
30
1
10 


= 2.5 M
Tf
= – 4.65ºC
Tf
= – 4.65ºC
3. A solution of x moles of sucrose in 100 grams of water freezes at 0.20
C. As ice separates the freezing point
goes down to 0.250
C. How many grams of ice would have separated?
(A) 18 grams (B*) 20 grams (C) 25 grams (D) 23 grams
Sol. Tf
= Kf
× m
0.2 = 1.86 ×
100
1000
x 
 x = 0.01075 moles
0.25 = 1.86 ×
)
y
–
100
(
1000
01075
.
0 
 y = 20 gm of ice separate out.
4. 2.56g of sulfur in 100g of CS2
has depression in freezing point of 0.010
C.Kf
= 0.10
molal1
. Hence, the atomicity of
sulfur in CS2
is
(A) 2 (B) 4 (C) 6 (D*) 8
Sol. Tf
= Kf
m
10–2
= 0.1 × m  m = 0.1 m
molality =
100
M
1000
56
.
2


 M = 256
atomicity =
32
256
= 8
5. A solute’S’ undergoes a reversible trimerization when dissolved in a certain solvent. The boiling point
elevation of its 0.1 molal solution was found to be identical to the boiling point elevation in case of a 0.08
molal solution of a solute which neither undergoes association nor dissociation. To what percent had the
solute ‘S’ undergone trimerization?
(A*) 30% (B) 40% (C) 50% (D) 60%
Sol. 3S S3
1 0

23. 1–
3

 i = 1 –
3
2
Now 0.1 




 

3
2
1 = 0.08  = 0.3. 30% trimerization.
6. 1g of arsenic dissolved in 86 g of benzene brings down the freezing point to 5.31 °C from 5.50 °C. If Kf
of benzene
is 4.9
m
C

, the atomicity of the molecule is : (As – 75)
(A) 8 (B) 2 (C) 3 (D*) 4
Sol. Tf
= Kf
m
0.19 = 4.9 ×
1000
86
M
1
.
0.19 = 4.9 ×
86
M
1000

.
 M =
19
.
0
86
1000
9
.
4


= 300.
Atomicity =
75
300
= 4.
Comprehension # 1
Read the following comprehension carefully and answer the questions (7 to 10).
The main application of osmotic pressure measurement is in the determination of the molar mass of a substance
which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and
colloids. This isdue to the fact that even a very small concentration of the solution producesfairly large magnitude
of osmotic pressure. In the laboratory the concentrations usually employed
are of the order of 10–3
to 10–4
M. At concentration of 10–3
mol dm–3
, the magnitude of osmotic pressure at 300 K
is :
P = 10–3
× 0.082 × 300 = 0.0246 atm
or 0.0246 × 1.01325 × 105
= 2492.595 Pa
At this concentration, the values of other colligative properties such as boiling point elevation and depression in
freezing point are too small to be determined experimentally.
Further polymers have following two types of molar masses :
(A) Number average molar mass )
M
( n , which is given by


i
i
i
i
i
N
M
N
where Ni
is the number of molecules having molar mass Mi
.
(B) Mass average molar mass )
M
( m , which is given by 

i
i
i
i
2
i
i
M
N
M
N
Obviously the former isindependent of the individual characteristics of the molecules and gives equal weightage
to large and small molecules in the polymer sample. On the other hand later gives more weightage to the heavier
molecules. Infact with the help of a colligative property only one type of molar mass of the polymer can be
determined.
7. One gram each of polymerA (molar mass = 2000) and B(molar mass = 6000) is dissolved in water to form one

24. litre solution at 270
C. The osmotic pressure of this solution will be
(A*) 0.0164 atm (B) 1862 Pa (C) both are correct (D) none of these
Sol. P × 1 = 






6000
1
2000
1
× 0.082 × 300
 P = 0.0164 atm.
8. What will be the observed molecular weight of the above polymer sample generating osmotic pressure equal to
the answer in the previous question at 270
C :
(A) 4000 (B*) 3000 (C) 2500 (D) 1200
Sol. 0.0164 × 1 =
m
2
× 0.082 × 300
 m = 3000.
9. Answer of the previous question will correspond to
(A*) Number average molar mass of the polymer sample
(B) Mass average molar mass of the polymer sample
(C) For the given sample )
M
( n will be equal to )
M
( m
(D) It is impossible to determine )
M
( n by osmotic pressure measurements
Sol. n
M =
6000
/
4
1
1
= 3000
Since n
M is an average which gives equal weightage to large and small molecules and simply counts the
number of molecules, hence it isdetermined by the help of colligative properties– specially the osmotic pressure.
10. Which of the following solutions of NaOH in water will be able to generate osmotic pressure equal to the answer
of question number 1 at 270
C assuming NaOH to be completely ionised
(A) M/1000 (B) M/1500 (C) M/2000 (D*) M/3000
Sol. P =
3000
1
×0.082 × 300 = 0.0082 atm
But 'i' for NaOH = 2
 Pobserved
= 0.0082 × 2 = 0.0164 atm.

25. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
63
1. What is the mole ratio of benzene (PB
0
= 150 torr) and toluene (PT
0
= 50 torr) in vapour phase if the given solution
has a vapour pressure of 120 torr?
(A*) 7 : 1 (B) 7 : 3 (C) 8 : 1 (D) 7 : 8
Sol. P = PB
°XB
+ PT
° XT
120 = 150(XB
) + 50 (1 – XB
)
100 XB
= 70
XB
= 0.7
YB
=
P
P
X 0
B
B
=
120
150
x
7
.
0
= 0.075 1
7
Y
Y
T
B
 YT
= 1 – 0.875 = 0.125
2. LiquidsA and B form an ideal solution. At a certain temperature the total vapour pressure of a mixture of Aand
B is 400 mm. The mole fraction of Ain the liquid mixture in equilibrium with the vapour phase is 0.4. If the vapour
pressure ratio (PA
0
/PB
0
) for the pure liquids at this temperature is 1/6, what is the partial pressure of A in the
vapour phase? (A) 50mm (B) 60mm (C) 70mm (D*) 40mm
3. Which of the following are false for ideal solutions :
(A) Vmix
= 0 (B) Hmix
= 0 (C) Smix
> 0 (D*) Gmix
= 0
(E) Raoult's law is obeyed for entire concentration range and temperatures.
4. Mixture of volatile components A and B has total vapour pressure (in Torr) p = 254 – 119 xA
where xA
is mole
fraction of A in mixture. Hence
0
A
p and
0
B
p are (in Torr) (A) 254, 119 (B) 119, 254
(C*) 135, 254 (D) 119, 373
Sol. P = XA
PA
0
+ XB
PB
0
= (PA
0
– PB
0
)XA
+ PB
0
So PB
0
= 254
PA
0
– PB
0
= –119 PA
0
= 135
5. Which represents correct difference when non-volatile solute is present in an ideal solution ?
(I) ; (II)
Solvent
or Solution
(III)
Solvent
B.P.
Solution
(A*) I, II, III (B) I, III (C) II, III (D) I, III
Comprehension #
Read the following comprehension carefully and answer the questions (6 to 10).
A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the
entropy of the system. A liquid vaporises to form a more disordered gas. When a solute is present, there is
additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is

26. higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non volatile) lowers the
vapour pressure of a liquid, and hence a higher boiling point of the solution.
Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower the
temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.
Elevation of B.Pt. (Tb
) and depression of F.Pt. (Tf
) of a solution are the colligative properties which depend
only on the concentration of particles of the solute, not their identity. For dilute solutions, Tb
and Tf
are
proportional to the molality of the solute in the solution.
Tb
= Kb
m Kb
= Ebullioscopic constant =
vap
º
b
H
1000
M
RT
2

And Tf
= Kf
m Kf
= Cryoscopic constant =
fus
º
f
H
1000
M
RT
2

(M = molecular mass of the solvent)
The valuesof Kb
andKf
do dependonthepropertiesof the solvent. Forliquids, º
b
vap
T
H

isalmost constant.[Troutan’s
s
Rule, this constantfor most of the unassociated liquids(not having any strong bonding like Hydrogen bonding
in the liquid state) is equal to 90 J/mol.]
For solutes undergoing change of molecular state is solution (ionization or association), the observed T values
differ from the calculated ones using the above relations. In such situations, the relationships are modified as
Tb
= i Kb
m ; Tf
= i Kf
m
where i = Van't-Hoff factor, greater than unity for ionization and smaller than unity for association of the solute
molecules.
6. Depression of freezing point of which of the following solutions doesrepresent the cryoscopic constant of water?
(A) 6% by mass of urea is aqueous solution
(B) 100g of aqueous solution containing 18 g of glucose
(C*) 59 g of aqueous solution containing 9 g of glucose
(D) 1 M KCl solution in water.
Sol. Cryoscopic constant Kf
= Tf
of solution having unit molality of normal solutes
Molality of glucose solution in (c) =
180
)
9
59
(
1000
9



= 1
7. Dissolution of a non-volatile solute into a liquid leads to the -
(A) decrease of entropy
(B) increase in tendency of the liquid to freeze
(C) increases in tendency to pass into the vapour phase.
(D*) decrease in tendency of the liquid to freeze
Sol. Since the solution has greater entropy (disorder) than the pure liquid, so former has lesser tendency to freeze
i.e., the temperature has to be lowered to freeze the solution.
8. To aqueous solution of NaI, increasing amounts of solid Hg2
is added. The vapor pressure of the solution
(A) decreases to a constant value
(B*) increases to a constant value
(C) increases first and then decreases
(D) remains constant because Hg2
is sparingly soluble in water.
Sol. 2Na+
(aq) + 2–
(aq) + Hg2
(s) 
 Na+
(aq) + Hg4
2–
(aq)

27. The number of mole particle decreases from 4 (2Na+
+ 2–
) to 3 (2Na+
+ Hg4
2–
). Hence, the colligative property
will decrease ot, the vapour pressure will increase to a constant value until Na is completely consumed.
9. A liquid possessing which of the following characteristics will be most suitable for determining the molecular
mass of a compound by cryoscopic measurements?
(A) That having low freezing point and small enthalpy of freezing
(B*) That having high freezing point and small enthalpy of freezing
(C) That having high freezing point and small enthalpy of vaporisation
(D) That having large surface tension
Sol. Kf
=
fus
02
f
H
1000
M
RT

would be larger for larger value of T°f
and smaller value of enthalpy of fusion of the solid solvent.
10. A mixture of two immiscible liquids at a constant pressure of 1 atm boils at a temperature
(A) equal to the normal boiling point of more volatile liquid
(B) equal to the mean of the normal boiling points of the two liquids
(C) greater than the normal boiling point of either of the liquid
(D*) smaller than the normal boiling point of either of the liquid.
Sol. The vapour pressure of a mixture of two immiscible liquids is the sum of their vapour pressures in the pure
states, independent of their relative amounts. Hence, B.Pt. of the mixture will be less than that of either of the
liquids, remaining constant throughout.

28. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
64
1. In an ideal mixture of liquids A and B the mole fraction of A is 0.25. If the ratio of PA
0
to PB
0
is 7/3, how many
repeated distillations would be required as a “minimum” to obtain a small quantity of distillate which has a mole
fractionof A, better than 0.80?
(A) 4 (B) 2 (C*) 3 (D) 5
Sol. A = 0.25 B = 0.75
0
B
0
A
P
P
=
3
7
 0
B
P =
7
3 0
A
P
PT
= 0.25 × 0
A
P + 0.75 ×
7
3 0
A
P = 0
A
P 






28
9
4
1
yA
pT
= 0
A
P XA
 yA
= 






28
9
4
1
25
.
0
X
=













28
9
7
4
/
1
=
16
7
& yB
=
16
9
1
T
P =
16
7
× 0
A
P +
16
9
×
7
3 0
A
P = 0
A
P 







7
16
27
16
7
1
T
1
AP
y = 0
A
P 1
A
X  1
A
y =








7
16
27
49
16
7
=
)
27
49
(
16
7
16
7



=
76
9

11
T
P =
76
49
× 0
A
P +
76
27
×
7
3 0
A
P = 0
A
P 








76
7
81
7
49
11
A
y 11
T
P = 0
A
P 11
A
x  11
A
y = 









76
7
81
7
49
P
76
/
49
P
0
A
0
A
=
)
81
7
49
(
76
76
7
49




11
A
y = 0.8089  0.8
There will be 3 repeated distillation to get XA
= 0.8
2. To 100g of water some ethylene glycol ( |
OH
CH
OH
CH
2
2
) was added and the solution was colled to 100
C when 20g of ice
was separated. What was the amount of ethylene glycol added to water? What will be its boiling point? (Kf
for
water is 1.86 and Kb
is 0.513).
Sol. Tf
= Kf
m
10 = 13.86 ×
100
80
62
/
W
 W =
1000
86
.
1
62
80
10



= 26.67 gm
Tb
= 0.513 ×
86
.
1
10
= 2.76ºC
Tb
= 100 + Tb
= 100 + 2.76 = 102.76ºC
3. Two liquidsAand B form an ideal solution. The solution has a vapor pressure of 700 Torr at 80o
C. It is distilled till
2/3rd
of the solution is collected as condensate. The composition of the condensate is x'A
= 0.75 and that of the
residue is x''A
= 0.30. If the vapor pressure of the residue at 80o
C is 600 Torr, which of the following is/are true?

29. (A*) The composition of the original liquid was xA
= 0.6. (B*) PA
0
=
3
2500
Torr.
.
(C) The composition of the original liquid was xA
= 0.4. (D*) PB
0
= 500 Torr.
Sol. (A,B,D)
xA
PA
° + xB
PB
° = 700 ...(i)
xA
 PA
° + xB
 PB
° = 0.30 PA
° + 0.70 PB
° = 600 ...(ii)
if moles of A & B initially are x & y then
x = 0.75 ×
3
2
(x + y) + 0.30 ×
3
1
(x + y)
& xA
= y
x
x

or xB
= y
x
y

Solving gives
xA
= 0.6, xB
= 0.4 , PA
° =
3
2500
torr & PB
° = 500 torr.
4. How many grams of NaCl must be dissolved in 225 g of water to yield a solution having the same boiling point
as that containing 20 g glucose (M – 180) and 30 g sucrose (M – 342) in 225 g of water? (Na – 23, Cl – 35.5)
(A*) 5.8 (B) 11.6
(C) 12.3 (D) None of these
5. What is the degree of dissociation of trichloroacetic acid, if its 1.0 m solution has a freezing point of
– 2.53 °C? Kf
= 1.86
molal
K
Sol. Tf
= 2 Kf
m
2.53 = 2 × 1.86 × 1
2 = 1.36 = 1 + a.
 a = 0.36.
6. Non volatile compound 'X' dimerises according to Ist order kinetics (t1/2 = 10 min.) If 10 mol of X and 10 mol of
volatile solvent (P0 = 400 mm of Hg) are mixed initially. Calculate vapour pressure of solution after
20 minutes.
(A) 320 mm (B*) 246 mm (C) 200 mm (D) can not be predicted
Sol. 2X 
 X2
t = 0 10 mol –
t = 10 5 mol 2.5 mol
t = 20 2.5 mol (2.5 + 1.25) totalmol = 6.25 mol
Now Ps = P0 . Xsolvent = 400 ×
25
.
16
10
= 246.15.
7. The amount of benzene that will separate out (in grams) if a solution containing 7.32 g of triphenylmethane
in 1000 g of benzene is cooled to a temperature which is 0.2°C below the freezing point of benzene?
Kf = 5.12
mol
kg
K 
.
Sol. 0.2 = 5.12













1000
x
1000
244
32
.
7
 0.2 = 5.12 





 x
1000
30

x
1000
30

=
12
.
5
2
.
0

30.  1000 – x = 768  x = 232 g Ans.232.
Comprehension #
Read the following comprehension regarding Completely Immiscible Liquids : Steam
Distillation carefully and answer the questions (8 to 10).
It is probably true that no two liquids are absolutely insoluble in each other, but with certain pairs, eg.,
mercury and water and carbon disulfide and water, the mutual solubility is so small that the liquids may be
regarded as virtually immiscible. For systems of this type, each liquid exerts its own vapour pressure,
independent of the other, and the total vapour pressure is the sum of the separate vapour pressures of the
two components in the pure state at the given temperature. The composition of the vapour can be readily
calculated by assuming that the gas laws are obeyed; the number of moles of each constituent in the
vapour will then be proportional to its partial pressure, that is to say, to the vapour pressure of the substance
in the pure state. If 0
A
p and 0
B
p are the vapour pressures of the pure liquids A and B, respectively, at the
given temperature, and n’A and n’B are the numbers of moles of each present in the vapour, the total
pressure P and the same temperature is given by P = 0
,
B
0
A p
p
P 

(1)
and the composition of the vapour by 0
B
0
A
B
A
p
p
'
n
'
n
 (2)
To express the ratio of A to B in the vapour in terms of the actual weights wA and wB, the numbers of moles
must be multiplied by the respective molecular weights MA and MB ; hence,
B
0
B
A
0
A
B
B
A
A
B
A
M
p
M
p
M
'
n
M
'
n
w
w

 (3)
A system of two immiscible liquids will boil, that is, distill freely, when the total vapour pressure P is equal
to the atmospheric pressure. The boiling point of the mixture is thus lower than that of either constituent.
Further, since the total vapour pressure is independent of the relative amounts of the two liquids, the
boiling point, and hence the composition of the vapour and distillate, will remain constant as long as the
two layers are present.
The properties just described are utilized in the process of steam distillation, whereby a substance that is
immiscible, or almost immiscible, with water, and that has a relatively high boiling point, can be distilled
at a much lower temperature by passing steam through it. The same result should, theoretically, be
obtained by boiling a mixture of water and the particular immiscible substance, but by bubbling steam
through the latter the system is kept agitated, and equilibrium is attained between the vapour and the two
liquids. The mixture distills freely when the total pressure of the two components is equal to that of the
atmosphere.
It is seen that chlorobenzene, which has a normal boiling point of 132ºC, can be distilled with steam at a
temperature about 40º lower, the distillate containing over 70 percent of the organic compound. An
examination of the calculation shows that the high proportion by weight of chlorobenzene in the steam
distillate is due largely to the high molecular weight of this substance, viz., 112.5 as compared with that of
water. In addition this case is a particularly favorable one because chlorbenzene has a relatively high
vapour pressure in the region of 90º to 100ºC. In order that a liquid may be distilled efficiently in steam, it
should therefore be immiscible with water, it should have a high molecular weight, and its vapour pressure
should be appreciable in the vicinity of 100ºC. A liquid which is partially miscible with water, such as
aniline, may be effectively distilled in steam, provided the solubility is not very great. In calculating the
composition of the distillate, however, the pressures 0
A
p and 0
B
p in equation (3) would have to be replaced
by the actual partial pressures. Attention may be called to the fact that equation (3) can be employed to
determine the approximate molecular weight of a substance that is almost immiscible with water. This can
be done provided the composition of the steam distillate and the vapour pressures of the two components
are known.

31. 8. The hydrocarbon terpinene was found to distil freely in steam at a temperature of 95ºC, when the atmospheric
pressure was 744 mm; the vapour pressure of pure water at this temperature is 634 mm.
The distillate constained 55 per cent by weights of terpinene; calculate its molecular weight.
(A*) 127 (B) 121 (C) 132 (D) 120
Sol. If the terpinene is designated by A, and the water by B, it follows that at 95ºC, the boiling point of the
mixture, 0
B
p is 634 mm, and hence 0
A
p is equal to P – 0
B
p , i.e,
744 - 634 = 110 mm.
now using equation (3),
45
55
=
18
634
M
110 A


 MA = 127
9. Which of the following cannot be efficiently steam distilled ?
(A*) ethanol (B) chlorobenzene (C) aniline (D) o-nitrophenol
10. Following is the variation of vapour pressure with temperature for (1) water and (2) a substance to be
steam distilled. At what temperature will the mixture of water and that substance boil under a pressure of
740 mm Hg ?
t(ºC) pº pº
water substance
98 707.27 7.62
98.5 720.15 7.80
99.0 733.24 7.97
99.5 746.52 8.15
100.0 760.00 8.35
(A) 98.45ºC (B*) 98.95ºC (C) 99.25ºC (D) 99.75ºC
Sol. PtºC = C
º
t
0
water ,
P + C
º
t
0
.
subs ,
P
Pt = 98.5C = 720.15 + 7.8 = 727.95 mm of Hg
Pt = 99ºC = 733.24 + 7.97 = 741.21 mm of Hg
temp. at which Pt = 740 mm of Hg
= 98.5 +
)
95
.
727
–
21
.
741
(
)
95
.
727
–
740
(
× 0.5 = 98.95ºC

32. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
65
1. IUPAC name of compound
Cl
Br
is
(A) 3-Bromo-7-chloro-7-ethyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-3-methylnonane
(B*) 3-Bromo-7-chloro-5-(1,1-dimethylethyl)-7-ethyl-3-methyl-5-(2-methylpropyl)nonane
(C) 3-Bromo-7-chloro-7-ethyl-3-methyl-5-(1,1-dimethylethyl)-5-(2-methylpropyl)nonane
(D) 3-Bromo-5-(1,1-dimethylethyl)-5-(2-methylethyl)-7-chloro-7-ethyl-3-methylnonane
2. The correct structure/s of 4-(1-methylethyl)–5–(2-methylpropyl) octane is/are.
(A) (B)
(C*) (D*)
3. IUPAC name of the given structure is
(A) 3-Ethynyl-4-ethenyl-7-(2-methylbutyl)-8-methylnona-1,7-diene
(B*) 4-Ethenyl-3-ethynyl-5,8-dimethyl-7-propylnona-1,7-diene
(C) 3,4-Diethenyl-5,8-dimethyl-7-propylnona-7-ene-1-yne
(D) 4-Ethenyl-3-ethynyl-7-(2-methylbutyl)-8-methylnona-1,7-diene
4. IUPAC name of the compound
CH3
CH –
2 CH3
is
(A) 4-Ethyl-3-methylcyclopent-1-ene (B*) 3-Ethyl-4-methylcyclopent-1-ene
(C) 1-Ethyl-2-methylcyclopent-4-ene (D) 2-Ethyl-1-methylcyclopent-3-ene
5. IUPAC name of compound is
(A*) 3-Ethynylpenta-1, 4-diene (B) 3-Ethenylpent-1en-4-yne
(C) 3-Ethenylpent-4-en-1-yne (D) 4-Propynylpentadiene

33. 6. The correct IUPAC name of the following compound is :
(A) 5-Cyclopropylhept-2-en-1-oic acid (B*) 6-Cyclopropylhex-2-en-1-oic acid
(C) 5-Cyclopropylpent-1-en-carboxylic acid (D) 6-Cyclopropylpent-1-en-carboxylic acid
Subjective
7. Calculate molecular weight of the lowest alkane containing a sequence of 10
, 20
, 30
and 40
carbon
atoms.
Ans. CH3
– C(CH3
)2
– CH(CH3
) – CH2
– CH3
Mol Wt = 114

34. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
66
1. Correct IUPAC name of the compound CH –
3 CH – O – C
2
CH – CH – CH
2 3
CH2
CH3
O
is
(A) Ethyl pentanecarboxylate (B) Ethyl 3-pentanoate
(C) Ethoxy-3-pentanoate (D*) Ethyl 2-ethylbutanoate
2. Correct IUPAC name of the compound
O
O
O
Me
is
(A*) 2-Methylbut-2-enedioic anhydride (B) 3-Methylbut-2-enedioic anhydride
(C) 2-Methyl-1,4-diketobut-2-ene epoxy (D) 2-Methylcyclopentanoxy-1,4-dione
3. The correct IUPAC name of the compound is
(A) 2-Iodo-3-methyl-5-bromocyclopentanesulfonic acid
(B) 2-Bromo-4-methyl-5-iodocyclopentanesulfonic acid
(C) 5-Bromo-2-iodo-3-methylcyclopentanesulfonic acid
(D*) 5-Bromo-2-iodo-3-methylcyclopentanesulfonic acid
4. The IUPAC name of the compound H – C – N
O
CH3
CH – CH
2 3
is
(A*) N-ethyl, N-methyl, methanamide (B) N-methyl, N-ethyl methanamide
(C) N-ethyl, N-methyl formamide (D) N-ethyl, methyl methanamide
5. IUPAC name of the compound H – C – O – C – CH3
O O
is
(A*) Ethanoicmethanoic anhydride (B) Methanoicethanoic , anhydride
(C) Methanoic anhydride ethanoicanhydride (D) Formicacetic anhydride
6. Correct IUPAC name of the compound
(A) 4-(Ethyl methanolyoxy)phenylpropanoate
(B*) Ethyl 4-propanoyloxybenzenecarboxylate
(C) 4-(1-Oxo-2-oxabutyl)phenylpropanoate
(D) 1-(1-Oxo-2-oxabutyl)-4-(1-oxopropoxy)benzene

35. Sol. Tough
7. The IUPAC name of the following compound is
(A) 2-(Ethoxycarbonyl) benzalychloride
(B) Ethyl 2-(Chloroformyl) benzoate
(C) Ethyl 2-(Chloromethanoyl) benzoate
(D*) Ethyl 2-(Chlorocarbonyl) benzoate
Sol. Moderate
8. IUPAC name of the compounds
CH – C – H
2
O
CH – C – H
O
CH – C – H
2
O
are
(A*) 3-formyl, pentan (1,5)-dial (B) 1,2,3-formyl propane
(C) 3-methyl formyl, butan-(1,4)-dial (D) None of these

36. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
67
MCQ
1. The pair of compounds having the same general formula.
(A) and (B) and
H
H H
H
C = C = C
(C*) and HC  C – C  CH (D*) and
2. Which of the following molecules do not contain sp3 hybridised carbon atoms ?
(A) (B*) (C) (D*)
3. Which of the following compounds have same empirical formula.
(A) C6
H12
(B*) C2
H2
(C*) C6
H6
(D) C2
H6
Subjective
4. Calculate the molecular weight of the lowest hydrocarbon which contains sp & sp2
hybridised carbon atoms
only.
Ans. H2
C = C = CH2
M.W. = 40.
5. Identify the molecular weight of the compound ‘X’ containing carbon and hydrogen atoms only with 3 and 2
bonds in one molecule.
Sol. H – C  C – H, Mol wt. = 26
Matching
6. Match the following : (More than one option in column - II may match with single option in column-I).
Match the hybridization state of below listed carbon atoms.
CH2
= C = CH – CH2
– C  C – CH2
– NH2
Column - I
Carbon atoms
Column - II
Hybridization state
(A) C1 (P) sp
(B) C2 (Q) sp
2
(C) C5 (R) sp
3
(D) C6 (S) dsp
2
Sol. A  R ; B  P ; C  Q ; D  P
7. Match the following :
(p) Saturated compound
(q) Heterocyclic compound
(r) Unsaturated compound
(s) Hydrocarbon
(A)
(B)
(C)
(D)
O
I Compounds II Class of compounds
OH
Sol. A 
 p, s , B 
 p, q , C 
 r, s , D 
 p

37. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
1. 1-(1-Chloropropyl)-4-(2-chloropropyl)cyclohexane and 1-(2-Chloropropyl)-4-(3-chloropropyl)cyclohexane are
(A) Chain isomers (B*) Position isomers (C) Optical isomers (D) Identical compounds
Sol. and(,oa) are position isomers .
2. and are
(A) Position isomers (B) Chain isomers
(C) Functional isomers (D*) Metamers
Sol. All isomeric esters are metamers.
3. How many structural isomers of tertiary amines corresponding to molecular formula C6H15Nare possible.
(2 Marks, 2 min)
(A) 4 (B) 5 (C) 6 (D*) 7
MCQ
4. Consider the following structures (3 Marks, 3 min)
I
Cl
–
CH
–
CH
–
CH
–
CH 2
2
2
3 CH – C – CH
3 3
CH3
Cl
II
Cl
III
– CH – Cl
2
IV
(A*) I & II are chain isomers (B) I & III are functional isomers
(C) II & IV are functional isomers (D*) III & IV are chain isomers
5. Which of the following pair represents the correct relationship (3 Marks, 3 min)
I II Relationship
(A)
NH2
OH
Cl
NH2
OH
Cl
Positional Isomers
(B*) Chain Isomer
(C*)
NH – C H
3 7 H C – N – C H
3 2 5
Functional Isomers
(D*) CH –CH –CH –C–OCH
3 2 2 3
O
CH –CH –C–OCH –CH
3 2 2 3
O
Metamer Isomers
68

38. Matching
6. Match the relationship between the compounds of column-I with column-II. (5 Marks, 6 min)
Column-I Column-II
(A) Pentanetetraene and Penta-1,4-diyne
(B) Benzene and Toluene
(C) Cyclohexane and 1,2,3-Trimethylcylopropane
(D) Methylpropanoate and Ethylethanoate
(P) Chain isomer
(Q) Functional isomer
(R) Metamer
(S) Homologues
Ans. Q
A  S
B  C - P, D - R
7. STATEMENT -1
4-Methylphenol and phenylmethanol are functional isomers. (2 Marks, 2 min)
STATEMENT -2
Isomeric alcohols and phenols have different chemical properties and therefore they are functional isomers.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 isNOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Ans. A

39. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
1. An organic compound has molecular formula C9H18. Its all carbon atoms are sp3 hybridised and its all hydrogen
atoms are identical. Its structural formula can be :
(A) (B*) (C) (D)
2. Which of the following hydrocarbons give same product on hydrogenation.
(A*) 2-Methyl hex-1-ene & 2-Methyl hex-3-ene
(B) 3-Ethyl hex-1-en-4-yne & 2-Methylhept-2-en-4-yne
(C) 3-Ethylcycloprop-1-ene & 1,2-Dimethylcycloprop-1-ene
(D*) 2-Methylbut-2-ene & 3-Methylbut-1-ene
3. Which of the following pair represents functional and chain isomer respectively of
CH – CH – OH
2
2
(A*)
CH3
CH3
OH
&
CH2 – OH
CH3
(B)
CH2 3
– O – CH
&
CH2 3
– CH
OH
(C*)
CH2 3
– O – CH
&
CH3
CH –
2 OH
(D)
CH3
CH3
OH
&
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
CH3
CH3
OH
4. The possible structures of molecular formula C4
H6
O2
(A)
O O
O O
O O
O O
O O
O O
O O
O O
O O
O O
O O
O O
(B*)
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
(C*)
O
O
(D*)
OH
HO
5. Match the following:-
Column I Column II
Compound Degree of unsaturation
(A) (P) 2
(B)
H C – CH – CH – C = CH
3 2 3
CH3
(Q) 5
(C) (R) 4
69

40. (D) (S) 3
Sol. (A)  (S), (B)  (R), (C)  (Q), (D)  (P)
6. What is the lowest molecular weight of a saturated cyclic hydrocarbon which has all its carbon atoms different.
(5 Marks, 5 min)
Ans. Mol. Wt. 84
CH – CH
3 2 CH3

41. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
70
SCQ
1. Which of the following alkanes contains primary, secondary, tertiary and quaternary carbon atoms together
(A) (CH3
)3
CH (B) (C2
H5
)3
CH (C*) (CH3
)3
CCH2
CH(CH3
)2
(D) (CH3
)4
C
2. IUPAC name of sec-butyl is
(A) 2-Methylpropyl (B*) 1-Methylpropyl (C) 1,1-Dimethylethyl (D) 2-Methylethyl
3. In the following compound bond x and y are made by the overlap of which type of hybridized orbital respectively
(A) sp2
and sp2
, sp2
and sp (B*) sp3
and sp2
, sp2
and sp
(C) sp2
and sp , sp3
and sp2
(D) sp3
and sp2
, sp and sp
4. The following hydrocarbon group
3
2
2
3
CH
|
—
CH
—
CH
—
C
—
H
|
CH
is named as –
(A) 1, 1-Dimethylpropyl (B) 3, 3-Dimethylpropyl (C*) 3-Methylbutyl (D) 2-Methylbutyl
5. Which of the following has longest chain of carbon
(A)
3
2
3
2
3
3
2
CH
—
CH
|
CH
—
CH
—
CH
—
CH
—
CH
|
CH
—
CH
(B)
(C*) (D)
3
2
3
2
2
2
3
3
CH
—
CH
|
CH
—
CH
—
CH
—
CH
—
CH
—
CH
—
CH
|
CH
MCQ
6. Neohexane contains
(A*) Four 1º carbon atoms and two 2º hydrogen (B*) Twelve 1º hydrogen but no 3º carbon
(C) Two 2º carbon & one 4º carbon (D) One 4º carbon and two 3º hydrogen

42. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
71
SCQ
1. 1-(1-Chloropropyl)-4-(2-chloropropyl)cyclohexane and 1-(2-Chloropropyl)-4-(3-chloropropyl)cyclohexane are
(A) Chain isomers (B*) Position isomers (C) Optical isomers (D) Identical compounds
Sol. and are position isomers .
2. and are
(A) Position isomers (B) Chain isomers
(C) Functional isomers (D*) Metamers
Sol. All isomeric esters are metamers.
3. How many structural isomers of tertiary amines corresponding to molecular formula C6H15Nare possible.
(A) 4 (B) 5 (C) 6 (D*) 7
MCQ
4. Consider the following structures
I
Cl
–
CH
–
CH
–
CH
–
CH 2
2
2
3 CH – C – CH
3 3
CH3
Cl
II
Cl
III
– CH – Cl
2
IV
(A*) I & II are chain isomers (B) I & III are functional isomers
(C) II & IV are functional isomers (D*) III & IV are chain isomers
5. Which of the following pair represents the correct relationship
I II Relationship
(A)
NH2
OH
Cl
NH2
OH
Cl
Positional Isomers
(B*) Chain Isomer
(C*)
NH – C H
3 7 H C – N – C H
3 2 5
Functional Isomers
(D*) CH –CH –CH –C–OCH
3 2 2 3
O
CH –CH –C–OCH –CH
3 2 2 3
O
Metamer Isomers

43. Matching
6. Match the relationship between the compounds of column-I with column-II.
Column-I Column-II
(A) Pentanetetraene and Penta-1,4-diyne
(B) Benzene and Toluene
(C) Cyclohexane and 1,2,3-Trimethylcylopropane
(D) Methylpropanoate and Ethylethanoate
(P) Chain isomer
(Q) Functional isomer
(R) Metamer
(S) Homologues
Ans. Q
A  S
B  C - P, D - R

44. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
72
1. Ahydrocarbon (R) has six membered ring in which there is no unsaturation. Two alkyl groups are attached to the
ring adjacent to each other. One group has 3 carbon atoms with branching at 1st
carbon atom of chain and
another has 4 carbon atoms. The larger alkyl group has main chain of three carbon atoms of which second
carbon is substituted. Correct IUPAC name of compound (R) is
(A) 1-(1-Methylethyl)-2-(1-methylpropyl)cyclohexane
(B) 1-(2-Methylethyl)-2-(1-methylpropyl)cyclohexane
(C*) 1-(1-Methylethyl)-2-(2-methylpropyl)cyclohexane
(D) 1-(1-Methylethyl)-2-butylcyclohexane
Sol. Moderate
1-(1-Methylethyl)-2-(2-methylpropyl)cyclohexane
2. STATEMENT -1 : 4-Methylphenol and phenylmethanol are functional isomers.
STATEMENT -2 : Isomeric alcohols and phenols have different chemical properties and therefore they are
functional isomers.
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 isNOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
3. Which of the following are functional isomers of methyl ethanoate
(A*) CH3
– CH2
– COOH (B*)
CH
|
OH
C
CH3
H
||
O
(C*) CH – O – CH – C – H
3 2
||
O
(D*) CH – C – CH
2 3
||
O
|
OH
4. Match items in List I with items in List II

45. Ans. A 
 s B 
 r C 
 q D 
 p
Small Write up - 1 (5 to 7)
A hydrocarbon ‘X’ has 2 geometrical isomers. It has only one  bond in its molecule along with 11 
bonds.
5. Which of the following can be the structural isomer of the hydrocarbon ‘X’.
)
(
CH
CH
CH
CH 3
2
2
I



)
(
CH
|
CH
C
CH
3
2
3
II


(A) III, IV (B*) I, II (C) II, IV (D) I, III
6. The ring chain functional isomer of hydrocarbon ‘X’ is / are
(A) I, II (B) II, III (C) III, IV (D*) I, IV
7. The hydrocarbon ‘X’ does not have any
(A) Position isomer (B) Chain isomer (C) Geometrical isomer (D*) Metamer

46. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
73
1. Among the following amino acids the R-enantiomer is represented by :
(A) (B*) (C) (D)
2. In the given compounds I and II are related as [2 Marks, 2 min]
(A) Identical (B) Enantiomers (C*) Diastereomers (D) Structural isomers
3. The stereochemical formula of diastereomer ‘Y’ of optically active compound ‘X’ is
X = 2, 3-Dihydroxybutanedioic acid. [2 Marks]
(A) (B*) (C) (D)
4. The correct relation between (A), (B),(C) are.
H
HO COOH
OH
H COOC
3
H
(I)
COOH
HO
COOCH3
H
HO
H
(II)
Identical Diastereomer Enantiomer
(A) I, II II, III
(B) I, II II, III
(C) I, II II, III
(D) II, III I, III
Ans. (B)
Sol. Moderate
(i)

47. (ii)
(iii)
I and II : – Diastereomer
II and III : – Enantiomer
5. Match the relationship and properties of the compounds of different pairs given in column I. Only one
option of 'I' matches with one option of II and III. [5 Marks]
Ans. A – M – P B – J – Q C – K – R D – L – S
6. STATEMENT - 1 : All physical and chemical properties of enantiomers are identical .
STATEMENT - 2 : Enantiomers are optically active mirror image isomers.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D*) Statement-1 is False, Statement-2 is True

48. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
74
1. Select the correct statement
(A*) A racemic mixture can be distinguished from a meso or an achiral compound by optical resolution.
(B*) d and  stereoisomer are enantiomer.
(C*) A chiral compound can have no assymetric carbon.
(D*) There is no obvious correlation that exists between the configurations (R or S) of enantiomers and the
direction [(+) or (–)] in which they rotate plane polarised light.
2. STATEMENT 1 : Compoundshaving only one chiral centre can have both enantiomer and optical diastereomer.
STATEMENT 2 : Optical diastereomer may or may not have chirality.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D*) Statement-1 is False, Statement-2 is True
Ans. (D)
3. STATEMENT-1 : The following compounds are optically inactive meso compounds [2 Marks]
STATEMENT-2 : The meso compounds do not have any chiral (asymmetric) C* atom so have optical rotation
zero.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Ans. C
4. (+)-Mandelic acid has a specific rotation of + 158º. What would be the observed specific rotation of each of the
following mixture ? [Mark 2+ 1 +2]
(a) 25% (–)-mandelic acid and 75% (+)-mandelic acid
(b) 50% (–)-mandelic acid and 50% (+)-mandelic acid
(c) 75% (–)-mandelic acid and 25% (+)-mandelic acid.
Sol. Enantiomeric excess =



d
–
d
Optical purity =
pure
]
[
observed
]
[


(a) 50% (+) mandelic acid [] =
100
50
158 

= 79º
(b) Racemic mixture, so [] = 0
(c) 50% (–) mandelic acid [] =
100
50
158 

= – 79º

49. 5. Matching (Topic : Optical)
(A) (p) Geometrical Isomers
(B) (q) Optical isomers
(C) (r) Compound contaning plane of symmetry
(D) (s) Compound contain center of symmetry
(A)  p,q (B)  p,q (C)  p,q (D)  p,r
6. STATEMENT 1 : For the given compound the two isomers of configurationsTrans - (R) and Trans - (S) are
enantiomers
STATEMENT 2 : The two isomers are mirror image stereoisomers
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Ans. (A)

50. DAILY PRACTICE PROBLEMS
Subject : Chemistry Date of Distribution : DPP No. Class : XII Batch :
FACULTY
COPY
75
1. Match the following [5 Marks]
Ans. (A) – r (B) – q (C) – p (D) – q
Sol.
2. C6
H4
= is ozonolysed in presence of Zn/H2
O to give a molecule which on reaction with H2
NOH
(excess) gives compound (M). How many stereoisomers of (M) are possible. Draw them.

51. Ans.
It has three geometrical centres with symmetry
 six geometrical isomers.
3. Specify the configuration of (commonly known as valine) k
(A) D
(B*) L
(C) R
(D) Cannot be specified because it is not an -amino acid
Sol. (Easy) It can be written as (S)  (L)
4. Which one among the following is not diastereomeric pair.
(A*) I and III (B) I and II (C) II and III (D) I and IV
5. Which of the following pairs can be separated by fractional distillation. (3 Marks, 3 min)
(A*) H
Me
Me
H
O
NH
NH and H
Me
H
Me
O
NH
NH
(B)
H
H
OH
OH
COOH
COOH
and
HOOC
HO
H
COOH
OH
H

52. (C*) CH –C–O–C H
3 3 7
O
and CH –O–C–C H
3 2 5
O
(D)
H
COOH
NH2
CH3
CH3
H
and
H
COOH
NH2
CH3
CH3
H
6. Identify which of the the given substance is/are chiral :
(A*) (B)
(C*) (D)