1. DPP No.1
Max. Time : 29 Total Time : 29 min.
Single choice Objective ('–1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21]
Subjective Questions ('–1' negative marking) Q.8 (4 marks 4 min.) [4, 4]
Match the Following (no negative marking) Q.9 (4 marks 4 min.) [4, 4]
1. 84 g of Iron (Fe) is reacted with sufficient amount of steam to produced 44.8 L, H2
gas at S.T.P. according to
the following reaction, a Fe + b H2
O 
 c Fe3
O4
+ dH2
. The stoichiometric coefficients of the reaction is
(At. wt., Fe = 56, O = 16, H = 1)
(1) 4, 3, 1, 4 (2*) 3, 4, 1, 4 (3) 1, 4, 2, 3 (4) none of these
Sol.
a
Fe
of
Mole
=
d
H
of
Mole 2
a
/
56
84
=
d
4
.
22
/
8
.
44
d
2
a
5
.
1
 
4
3
d
a

So, Ans. B.
2. Which statement is wrong-
(1*) Oxidation number of oxygen is +1 in peroxides
(2) Oxidation number of oxygen is +2 in oxygen difluoride
(3) Oxidation number of oxygen is 
1
2
in superoxides
(4) Oxidation number of oxygen is –2 in most of is compound
Sol. Oxidation number of oxygen is (–1) in peroxide.
3. Which of the following is not a redox reaction?
(1) Mg + N2
Mg3
N2
(2) MnO4
–
+ C2
O4
2–
 Mn2+
+ CO2
(3) CuSO4
+ Kl  Cu2
2
+ 2
+ K2
SO4
(4*) AgCl + NH3
[Ag(NH3
)2
] Cl
Sol. Those reaction in which oxidation number of any element do not change not a redox reaction.
AgCl + NH3

 [Ag(NH3
)2
]Cl.
4. In the reaction, 2S2O3
2–
+ I2  S4O6
2–
+ 2I
–
, the eq. wt. of S4O6
–2
is equal to its -
(1) Mol. wt. (2*) Mol. wt./2 (3) 2 × mol. wt. (4) Mol. wt./6.
Sol. I2 + 2S2O3
2–

 S4O6
2–
+ 2I
–
.
–
2
6
4O
S
E =
t
tan
reduc
or
oxidant
of
molecule
per
by
loss
or
gain
electron
of
.
no
.
wt
molecular
=
2
M
.
2S2O3
2–

 S4O6
2–
+ 2e
–
.
5. In the following change -
3Fe + 4H2O  Fe3O4 + 4H2. If the atomic weight of iron is 56, then its equivalent weight will be -
(1) 42 (2*) 21 (3) 63 (4) 84
Sol. 3
0
Fe + 4H2O 

3
/
8
4
3O
Fe

+ 4H2
3Fe + 4H2O 
 Fe3O4 + 8H
+
+ 8e
–
V.F. of Fe =
3
8
.
EFe =
.
F
.
V
mass
Atomic
=
3
/
8
56
= 21.
Subject : Physical/Inorg.Chemistry Date : DPP No. 11 Class : XIII Course :
DAILY PRACTICE PROBLEMS (DPP)
FACULTY
COPY

2. 6. How many millilitres of a 9 N H2
SO4
solution will be required to neutralize completely 20 mL of a 3.6 N
NaOH solution?
(1) 18.0 mL (2*) 8.0 mL (3) 16.0 mL (4) 80.0 mL
Sol. NNaOH
= 3.6 ; VNaOH
= 20 ; 4
2SO
H
N = 9 ; 4
2SO
H
V = ?
4
2SO
H
V =
9
20
6
.
3 
= 8 mL.
7. Calculate the normality of an NaOH solution, 21.5 mL of which is required to convert 0.240 g of NaH2
PO4
in a solution to monohydrogen phosphate.
(1) 1.093 N (2*) 0.093 N (3) 0.048 N (4) 0.93 N
Sol. Equivalent of NaOH = Equivalent of NaH2
PO4
21.5 × 10–3
× N =
120
24
.
0
× 1  N = 0.093
8. Identify the oxidant and the reductant in the following reactions :
(a) KMnO4
+ KCl + H2
SO4

 MnSO4
+ K2
SO4
+ H2
O + Cl2
(b) FeCl2
+ H2
O2
+ HCl 
 FeCl3
+ H2
O
Ans. (a) 4
)
7
(
O
Mn
K

+
)
1
(–
Cl
K + H2
SO4 
 4
)
2
(
SO
Mn

+ K2
SO4
+ H2
O +
)
0
(
2
Cl .
4
)
7
(
O
Mn
K

(oxidant ) 
 4
)
2
(
SO
Mn

(reduction half).
)
1
(–
Cl
K (reductant) 

)
0
(
2
Cl (oxidant half).
(b) 2
)
2
(
Cl
Fe

+
)
1
(–
2
2O
H + HCl 
 3
)
3
(
Cl
Fe

+
)
2
(–
2 O
H (oxidation half)
2
)
2
(
Cl
Fe

(reductant) 
 3
)
3
(
Cl
Fe

(oxidation half).
)
1
(–
2
2O
H (oxidant) 
 H2
O2–
(reduction half).
9. Match the following :
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+
]=0.75M
(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 =
90
60
1000
10


= 1.85 m.
(D) Molarity of HCl =
100
5
.
36
95
.
10
× 1000 = 3 M

3. DPP No.2
Max. Marks : 30 Total Time : 30 min.
Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18]
Multiple choice objective ('–1' negative marking) Q.7 to Q.8 (4 marks 4 min.) [8, 8]
Subjective Questions ('–1' negative marking) Q.9 (4 marks 4 min.) [4, 4]
1. How many millilitres of 0.150 M H2
SO4
(sulphuric acid) are required to react with 1.68 g of sodium hydrogen
carbonate, (NaHCO3
), according to the following equation ?
H2
SO4
(aq) + 2NaHCO3
(aq) 
 Na2
SO4
(aq) + 2H2
O() + 2CO2
(g)
(1) 55.55 mL (2*) 66.66 mL (3) 25.55 mL (4) 40.88 mL
Sol. Moles of H2
SO4
=
2
NaHCO
of
moles 3
1000
MV
= wt
mol
wt
×
2
1
V =
84
68
.
1
×
2
1000
×
15
.
0
1
= 66.66 ml
2. A sample of H2
SO4
(density 1.787 g mL–1
) is labelled as 86% by weight. What is molarity of acid ? What
volume of acid has to be used to make 1 litre of 0.2 M H2
SO4
?
(1*) 15.68 M, 12.75 m (2) 10.70 M, 12.75 m (3) 15.68 M, 10.75 m (4) 25.22 M, 20.55 m
Sol. M =
98
787
.
1
86
× 10 = 15.68 M
M1
V1
= M2
V2
15.68 × V1
= 0.2 × 1000
V1
= 12.75 ml
3. What is the normality of the H2
SO4
solution, 18.6 mL of which neutralizes 30.0 mL of a 1.55 N KOH
solution?
(1) 5.0 N (2) 1.25 N (3*) 2.5 N (4) 3.5 N
Sol. VNaOH
= 18.6 ; NNaOH
= ? ; NKOH
= 1.55 ; VKOH
= 30
NNaOH
=
6
.
18
30
55
.
1 
= 2.5 N.
4. In the equation , SnCl2 + 2HgCl2  Hg2Cl2 + SnCl4
The equivalent weight of stannous chloride (molecular weight = 190) will be -
(1) 190 (2*) 95 (3) 47.5 (4) 154.5
Sol. SnCl2 + 2HgCl2 
 Hg2Cl2 + SnCl4
2
SnCl
E =
t
tan
reduc
or
oxidant
of
molecule
per
by
loss
or
gain
electron
of
.
no
.
wt
molecular
=
2
190
= 95.
Sn
2+

 Sn
4+
+ 2e
–
.
5. Four different identical vessels at same temperature contains one mole each of C2
H6
, CO2
, C2
and H2
S
at pressures P1
, P2
, P3
and P4
respectively. The value of van der Waals constant ‘a‘ for C2
H6
, CO2
, Cl2
and H2
S is 5.562, 3.640, 6.579 and 4.490 atm.L2
.mol–2
respectively. Then
(1*) P3
< P1
< P4
< P2
(2) P1
< P3
< P2
< P4
(3) P2
< P4 < P1
< P3
(4) P1
= P2
= P3
= P4
6. Under critical states of a gas for one mol of a gas, compressibility factor is -
(1*)
8
3
(2)
3
8
(3) 1 (4)
4
1
Subject : Physical/Inorg.Chemistry Date : DPP No. 12 Class : XIII Course :
DAILY PRACTICE PROBLEMS (DPP)

4. 7. Critical temperature for a particular gasis– 177°C then for which of the following case value of compressibility
factor of the gas may be more than unity.
(1) at 0°C and 0.01 atm (2*) at 0°C and 2000 atm
(3*) at 60°C and 0.01 atm (4*) at 60°C and 10 atm
Sol. at very high Pressure Z = 1+
RT
Pb
Z > 1
for particular realgas above boyle temp Z > 1.
8. The vander waal gas constant ‘a’ is given by
(1)
3
1
VC
(2*) 2
C
C V
P
3 (3)
C
C
P
RT
8
1
(4*)
64
27
C
2
C
2
P
T
R
9. Balance the following redox equations and identify the valency factor (n-factor) for different compunds or
molecule involved in the reactions at reactant and product side.
(i) K2
Cr2
O7
+ H2
O2
+ H2
SO4

 K2
SO4
+ Cr2
(SO4
)3
+ H2
O + O2
Ans. K2
Cr2
O7
+ 3H2
O2
+ 4H2
SO4

 K2
SO4
+ Cr2
(SO4
)3
+ 7H2
O + 3O2
.
Sol. Mass Balance and Charge Balance :
Remove the spectator ion — 2K+
, SO4
2–
.
Cr2
O7
2–
+ H2
O2
+ 2H+

 2Cr3+
+ H2
O + O2
.
Oxidation Half :
H2
O2

 O2
+ 2H+
+ 2e–
.
Reduction Half :
Cr2
O7
2–
+ 14H+
+ 6e–

 2Cr3+
+ 7H2
O.
Total loss electrons = total gain electrons.
3H2
O2
+ Cr2
O7
2–
+ 8H+

 2Cr3+
+ 7H2
O + 3O2
.
Add the spectator ion — 2K+
, SO4
2–
.
3H2
O2
+ K2
Cr2
O7
+ 4H2
SO4

 Cr2
(SO4
)3
+ K2
SO4
+ 7H2
O + 3O2
.
(ii) Zn + NaNO3
+ NaOH 
 Na2
ZnO2
+ H2
O + NH3
Ans. 4Zn + NaNO3
+ 7NaOH = 4Na2
ZnO2
+ 2H2
O + NH3
.
Sol. Mass Balance and Charge Balance :
Remove the spectator ion — Na+
Zn + NO3
–
+ OH–

 ZnO2
2–
+ H2
O + NH3
.
Oxidation Half :
Zn + 4OH–

 ZnO2
2–
+ 2H2
O + 2e–
.
Reduction Half :
NO3
–
+ 6H2
O + 8e–

 NH3
+ 9OH–
.
Total loss electrons = total gain electrons.
4Zn + 7OH–
+ NO3
–

 4ZnO2
2–
+ 2H2
O + NH3
.
Add the spectator ion — Na+
.
4Zn + 7NaOH + NaNO3

 4Na2
ZnO2
+ 2H2
O + NH3
.