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PAGE NO. # 1
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
CHEMISTRY
DAILY PRACTICE PROBLEMS
D P P
COURSE NAME : UMANG (UP) DATE : 29.07.2013 to 03.08.2013 DPP NO. 39
TARGET
JEE (ADVANCED) : 2014
* Marked Questions are having more than one correct option.
DPP No. # 39
1. Maganese achieves its maximum oxidation state in which of these compounds :
(A) MnO2
(B) Mn3
O4
(C) KMnO4
(D) K2
MnO4
2. Oxidation number of underlined elements in N2
O5
, SO3
2–
, NH4
+
are :
(A) + 5, + 2, – 3 (B) + 6, – 2, + 3 (C) + 6, + 2, – 3 (D) + 5, + 4, – 3
3. Phosphorous has the oxidation state of + 3 in :
(A) Phosphorus acid (H3
PO3
) (B) Orthophosphoric acid (H3
PO4
)
(C) Metaphosphoric acid (HPO3
) (D) Pyrophosphoric acid (H4
P2
O7
)
4. Which statement is wrong :
(A) Oxidation number of oxygen is + 1 in peroxides
(B) Oxidation number of oxygen is + 2 in oxygen difluroide
(C) Oxidation number of oxygen is
1
2
 in superoxides
(D) Oxidation number of oxygen is – 2 in most of its compounds
5. The oxidation number and covalency of sulphur in the sulphur molecule (S8
) are respectively :
(A) 0 and 2 (B) + 6 and 8 (C) 0 and 8 (D) + 6 and 2
6. The oxidation number of oxygen in F2
O is :
(A) + 1 (B) – 2 (C) + 2 (D) – 3
7. Calculate individual and average oxidation number (if required) of the marked element and also draw the
structure of the following compounds or molecules :
(1) Na2
S2
O3
(2) Na2
S4
O6
(3) H2
SO5
(4) H2
S2
O8
(5) H2
S2
O7
(6) S8
(7) HNO4
(8) C3
O2
(9) OsO4
(10) PH3
(11) CrO4
2–
(12) Cr2
O7
2–
(13) CrO2
Cl2
(14) CrO5
(15) Na2
HPO4
(16) FeS2
(17) C6
H12
O6
(18) NH4
NO3
8. Calculate individual and average oxidation number (if required) of the marked element and also draw the
structure of the following compounds or molecules :
(1) FNO3
(2) CaOCl2
(3) XeO3
F2
(4) LiAlH4
(5) Na3
AlF6
(6) P4
(7) O3
(8) I (IO3
)3
(9) Fe3
O4
(10) CsI3
(11) KO3
(12) O2
F2
(13) H2
SiF6
(14) H3
PO3
(15) POCl3
(16) Si(OH)4
(17) Mg2
C3
(18) CaC2
(19) Be2
C
PAGE NO. # 2
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
ANSWER KEY
DPP No. # 39
1. (C) 2. (D) 3. (A) 4. (A) 5. (A) 6. (C)
7. (1) + 2 (6, – 2) (2) + 5/2 (5, 5, 0, 0) (3) + 6
(4) + 6 (+ 6, + 6) (5) + 6 (+ 6, + 6) (6) 0
(7) + 5 (8) 4/3 (+ 2, + 2, 0) (9) + 8 (10) – 3 (11) + 6
(12) + 6 (+ 6, + 6) (13) + 6 (14) + 6
(15) + 5 (16) + 2 (17) 0 (18) – 3, + 5
8. (1) + 5 (2) 0 (– 1, + 1) (3) + 8
(4) + 3 (5) + 3 (6) 0 (0, 0, 0, 0) (7) 0 (0, – 1, + 1)
(8) + 3, + 5 (9)
8
( 2, 3)
3
  (10)
1
3

(11)
1
3
 (12) + 1 (+ 1, + 1) (13) + 4
(14) + 3 (15) + 5 (16) + 4
(17)
4
(0, 4, 0)
3
  (18) – 1 (– 1, – 1) (19) – 4
PAGE NO. # 1
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
CHEMISTRY
DAILY PRACTICE PROBLEMS
D P P
COURSE NAME : UMANG (UP) DATE : 05.08.2013 to 10.08.2013 DPP NO. 40 TO 42
TARGET
JEE (ADVANCED) : 2014
* Marked Questions are having more than one correct option.
DPP No. # 40
1. Identify the reactants, if any, oxidized or reduced.
(a) NaH + H2
O  NaOH + H2
(b) 2FeCl3
+ H2
S  2FeCl2
+ S + 2HCl
(c) 3Mg + N2
 Mg3
N2
(d) AgCN + CN–
 Ag(CN)2
–
(e) SnCl2
+ Br2
 SnCl2
.Br2
(f) AlCl3
+ K  Al + 3KCl
(g) 2CuSO4
+ 4KI 
 Cu2
I2
+ K2
SO4
+ I2
(h) NH4
NO2
 N2
+ 2H2
O
(i) CaC2
+ 2H2
O  Ca(OH)2
+ C2
H2
(j) SnS2
+ S–2
 SnS3
–2
(k) 3O2
 2O3
2. Select the oxidants and reductants in the given reactions .
(a) Ag(NH3
)2
+
+ 2H+
 Ag+
+ 2NH4
+
(b) KIO3
+ 5KI + 6HCl  3I2
+ 6KCl + 3H2
O
(c) 2CuSO4
+ SO2
+ 2KBr + 2H2
O  2CuBr + 2H2
SO4
+ K2
SO4
(d) C2
H4
+ Br2
 C2
H4
Br2
(e) 2I–
+ H2
O2
 2OH–
+ I2
(f) Si + 2KOH + H2
O  K2
SiO3
+ 2H2
(g) 2Cu+2
+ 4I–
 Cu2
I2
+ I2
3. Balance the following equation using desired medium –
(a) C2
H5
OH + Cr2
O7
2–
+ H+
 Cr3+
+ C2
H4
O + H2
O
(b) Sn(OH)3
–
+ Bi(OH)3
+ OH–
 Sn(OH)6
2–
+ Bi
(c) IO3
–
+ N2
H4
+ HCl  N2
+ ICl2
–
+ H2
O
(d) Hg2
Cl2
+ NH3
 Hg + HgNH2
Cl + NH4
Cl
(e) Zn + NO3
–
+ H+
 Zn2+
+ NH4
+
+ H2
O
(f) I2
+ NO3
–
+ H+
 IO3
–
+ NO2
+ H2
O
(g) MnO4
–
+ SO2
2–
+ H2
O  MnO2
+ SO4
2–
+ OH–
(h) H2
O2
+ ClO2
+ OH–
 ClO2
–
+ O2
+ H2
O
(i) ClO–
+ CrO2
–
+ OH–
 Cl–
+ CrO4
2–
+ H2
O
(j) Cl2
+ KOH  KClO3
+ KCl + H2
O
PAGE NO. # 2
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
DPP No. # 41
Balance the following equation using desired medium
1. H2
O2
+ KMnO4
 MnO2
+ KOH + O2
+ H2
O
2. HNO2
+ KMnO4
+ H2
SO4
 HNO3
+ KMnO4
+ K2
SO4
+ H2
O
3. N2
H4
+ AgNO3
+ KOH  N2
+ Ag + KNO3
+ H2
O
4. P + HNO3
 HPO3
+ NO + H2
O
5. K2
Cr2
O7
+ HCl  KCl + CrCl3
+ H2
O + Cl2
6. MnO4
–
+ C2
O4
2–
+ H+
 CO2
+ Mn2+
+ H2
O
7. Cr2
O7
2–
+ C2
O4
2–
+ H+
 Cr3+
+ CO2
+ H2
O
8. KMnO4
+ H2
S + H2
SO4
 KHSO4
+ MnSO4
+ S + H2
O
9. Cu(NH3
)4
Cl2
+ KCN + H2
O  K2
Cu(CN)3
+ NH3
+ KCNO + NH4
Cl + KCl
10. Ag + KCN + H2
O + O2
 KAg(CN)2
+ KOH
11. Fe3
O4
+ MnO4
–
+ H2
O  Fe2
O3
+ MnO2
+ OH–
12. C2
H5
OH + MnO4
–
+ OH–
 C2
H3
O–
+ MnO2
+ H2
O
13. Cr2
O7
2–
+ SO3
2–
+ H+
 Cr3+
+ SO4
2–
+ H2
O
14. ClO3
–
+ SO2
+ H+
 ClO2
+ HSO4
–
15. Mn2+
+ S2
O8
2–
+ H2
O  MnO4
–
+ HSO4
–
+ H+
DPP No. # 42
* Equivalent weight of any substance is the weight in gram which combines with 1 gram of hydrogen ,
8 grams of oxygen, 35.5 grams of Cl and 108 grams of Ag.
* Equivalent weight =
factorn''
weightMolecular
1. The equivalent weight of a metal is 36. What weight of the metal would give 9.322 gm of its chloride?
2. On heating 5 × 10–3
equivalent of CaCl2
.xH2
O, 0.18 gm of water is obtained, what is the value of x ?
3. The equivalent weight of a metal is double that of oxygen. How many times is the weight of its oxide
greater than the weight of the metal ?
4. 0.224 gm of a metal when dissolved in dilute acid liberates 285 cc of hydrogen, measured at 17ºC
and 78 cm of mercury pressure. Find the equivalent weight of the metal.
5. 1.82 g of a metal required 32.5 mL of N HCl to dissolve it. What is equivalent weight of metal ?
6. The equivalent weight of a metal is 30. What volume of chlorine would be liberated at 27ºC and 750
mm pressure by 0.6 gm of the metal ? [Aqueous tension at 27ºC = 26.6 mm]
7. The molecular formula of an oxide of an element X is Xm
On
what is the equivalent weight of themetal?
[Given atomic weight of X = a]
8. 12 g of an element combines with 32 g of oxygen. What is the equivalent weight of the element if the
equivalent weight of oxygen is 8 ?
9. Equivalent weight of sulphur in SCl2
is 16, what is the equivalent weight of S in S2
Cl2
?
[S = 32 , Cl = 35.5]
10. 1.60 gm of a metal were dissolved in HNO3
to prepare its nitrate. The nitrate was strongly heated to
give 2 g oxide. Calculate equivalent weight of metal.
11. From 3.36 gm of Fe, 4.80 gm of iron oxide is obtained. What is the equivalent weight of Fe?
12. 1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate equivalent weight of metal.
13. An element forms two oxides, the weight ratio composition in them, A : O = x : y in the first oxide and
y : x in the second oxide. If the equivalent weight of A in the first oxide is 10.33, Calculate the equivalent
weight of A in the second oxide.
PAGE NO. # 3
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
ANSWER KEY
DPP No. # 40
1. Rxn. No. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)
Oxidized : NaH, H2
S, Mg, none, SnCl2
, K, KI, NH4
+
, none, none none.
Reduced : H2
O, FeCl3
, N2
none, Br2
, AlCl3
, CusO4
, NO2
–
, none, none, none
2. (a) not (b) KIO3
-oxidant, I–
reductant
(c) CuSO4
-oxidant, SO2
reductant (d) Br2
-oxidant, C2
H4
reductant
(e) H2
O2
-oxidant, I–
reductant (f) KOH-oxidant, Si reductant
(g) Cu++
-oxidant, I–
reductant
3. (a) 3C2
H5
OH + Cr2
O7
2–
+ 8H+
 2Cr3+
+ 3C2
H4
O + 7H2
O
(b) 3Sn(OH)3
–
+ 2Bi(OH)3
+ 3OH–
 3Sn(OH)6
2–
+ 2Bi
(c) IO3
–
+ N2
H4
+ 2HCl  N2
+ ICl2
–
+ 3H2
O
(d) 2Hg2
Cl2
+ 4NH3
 2Hg + 2HgNH2
Cl + 2NH4
Cl
(e) 4Zn + NO3
–
+ 10H+
 4Zn2+
+ NH4
+
+ 3H2
O
(f) I2
+ 10NO3
–
+8H+
 2IO3
–
+ 10NO2
+ 2H2
O
(g) 2MnO4
–
+ 3SO2
2–
+ H2
O  2MnO2
+ 3SO4
2–
+ 2OH–
(h) H2
O2
+ 2ClO2
+ 2OH–
 2ClO2
–
+ O2
+ 2H2
O
(i) 3ClO–
+ 2CrO2
–
+ 2OH–
 3Cl–
+ 2CrO4
2–
+ H2
O
(j) 3Cl2
+ 5KOH  KClO3
+ 5KCl + 3H2
O
DPP No. # 41
1. 3H2
O2
+ 2KMnO4
 2MnO2
+ 2KOH +3O2
+ 2H2
O
2. 5HNO2
+ 2KMnO4
+ 3H2
SO4
 5HNO3
+ 2KMnO4
+ K2
SO4
+ 3H2
O
3. N2
H4
+ 4AgNO3
+ 4KOH  N2
+ 4Ag + 4KNO3
+ 4H2
O
4. 3P + 5HNO3
 3HPO3
+ 5NO + H2
O
5. K2
Cr2
O7
+ 8HCl  2KCl + 2CrCl3
+ 7H2
O + 3Cl2
6. 2MnO4
–
+ 5C2
O4
2–
+ 16H+
 10CO2
+ 2Mn2+
+ 8H2
O
7. Cr2
O7
2–
+ 3C2
O4
2–
+ 14H+
 2Cr3+
+ 6CO2
+ 7H2
O
8. 2KMnO4
+ 5H2
S + 4H2
SO4
 2KHSO4
+ 2MnSO4
+ 5S + 8H2
O
9. 2Cu(NH3
)4
Cl2
+ 7KCN + H2
O K2
Cu(CN)3
+ 6NH3
+ KCNO + 2NH4
Cl + 2KCl
10. 4Ag + 8KCN + 2H2
O + O2
 4KAg(CN)2
+ 4KOH
11. 6Fe3
O4
+ 2MnO4
–
+ H2
O  9Fe2
O3
+ 2MnO2
+ 2OH–
12. 3C2
H5
OH + 2MnO4
–
+ OH–
 3C2
H3
O–
+ 2MnO2
+ 5H2
O
13. Cr2
O7
2–
+ 3SO3
2–
+ 8H+
 2Cr3+
+ 3SO4
2–
+ 4H2
O
14. 2ClO3
–
+ SO2
+ H+
 2ClO2
+ HSO4
–
15. 2Mn2+
+ 5S2
O8
2–
+ 8H2
O  2MnO4
–
+ 10HSO4
–
+ 6H
DPP No. # 42
1. 4.6935 gm 2. x = 4 3. 1.5 4. 9.105 5. 56
6. 0.2584 lit 7. m × a/2n 8. 3 9. 32 10. 32
11. 18.67 12. 38 13. 6.2
PAGE NO. # 1
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
CHEMISTRY
DAILY PRACTICE PROBLEMS
D P P
COURSE NAME : UMANG (UP) DATE : 12.08.2013 to 17.08.2013 DPP NO. 43 & 44
TARGET
JEE (ADVANCED) : 2014
* Marked Questions are having more than one correct option.
DPP No. # 43
1. White phosphorus reacts caustic soda, the products are PH3
and NaH2
PO2
. The reaction is an example
of :
(A) Oxidation (B) Reduction (C) Disproportionation (D) Neutralisation
2. The reaction 3 ClO–
(aq) ClO3
–
(aq) + 2Cl–
(aq) is an example of :
(A) Oxidation (B) Reduction (C) Disproportionation (D) Decomposition
3. Which of the following changes does not involve either oxidation or reduction :
(A) VO2+
 V2
O3
(B) Na  Na+
(C) Zn2+
Zn (D) CrO4
2–
Cr2
O7
2–
4. Which of the following is not a redox reaction?
(A) Mg + N2
 Mg3
N2
(B) MnO4
–
+ C2
O4
2–
 Mn2+
+ CO2
(C) CuSO4
+ KI 2CuI + I2
+ K2
SO4
(D) AgCl + NH3
[Ag(NH3
)2
]Cl
5.* Which of the following can show disporoportionation reaction :
(A) ClO4
–
(B) Cl–
(C) ClO2
–
(D) ClO3
–
6. 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
(c) Cu + HNO3
(dil) Cu (NO3
)2
+ H2
O + NO
(d) Na2
HAsO3
+ KBrO3
+ HCl NaCl + KBr + H3
AsO4
(e) I2
+ Na2
S2
O3
Na2
S4
O6
+ NaI
7. Balance the following redox equations and determine the valency factor (n-factor) for oxidant and
reductant involved in the reactions :
(i) K2
Cr2
O7
+ H2
O2
+ H2
SO4
K2
SO4
+ Cr2
(SO4
)3
+ H2
O + O2
(ii) Zn + NaNO3
+ NaOH Na2
ZnO2
+ H2
O + NH3
(iii) Al  [Al(OH)4
]–
+ H2
(basic)
(iv) Cu3
P + Cr2
O7
2–
Cu2+
+ H3
PO4
+ Cr3+
+ H2
O (acidic)
(v) ClO3
–
+ Fe2+
+ H+
Cl–
+ Fe3+
+ H2
O
(vi) N2
O4
+ BrO3
–
NO3
–
+ Br–
(vii) S2
O3
2–
+ Sb2
O5
SbO + H2
SO3
(viii) Cr2
O7
2–
+ I–
+ H+
Cr3+
+ I2
+ H2
O
(ix) IO4
–
+ I–
+ H+
I2
+ H2
O
PAGE NO. # 2
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
DPP No. # 44
1. In the reaction 2S2
O3
2–
+ I2
S4
O6
2–
+ 2I–
, the equivalent wt. of S4
O6
2–
is equal to :
(A) Mol wt. (B) Mol. wt./2 (C) 2 × Mol. wt. (D) Mol. wt./6
2. Equivalent weight of NH3
in the change N  NH3
is :
(A) 17/6 (B) 17 (C) 17/2 (D) 17/3
3. The molecular weight of the compounds (a) Na2
SO4
, (b) Na3
PO4
. 12H2
O and (c) Ca3
(PO4
)2
respectively
are X, Y and Z. The correct set of their equivalent weights will be :
(A) (a)
X
2
(b)
Y
3
(c)
Z
6
(B) (a) x (b)
Y
3
(c)
Z
3
(C) (a)
X
2
(b) Y (c)
Z
3
(D) (a) X (b) Y (c) Z
4. In the following reaction :
3Fe + 4H2
O Fe3
O4
+ 4H2
, if the atomic weight of iron is 56, then its equivalent weight will be :
(A) 42 (B) 21 (C) 63 (D) 84
5. When one mole NO3
–
is converted into 1 mole NO2
, 0.5 mole N2
and 0.5 mole N2
O respectively, it
accepts x, y and z mole of electrons. Then, x, y and z are respectively :
(A) 1, 5, 4 (B) 1, 2, 3 (C) 2, 1, 3 (D) 2, 3, 4
6. In the equation, SnCl2
+ 2HgCl2
Hg2
Cl2
+ SnCl4
:
The equivalent weights of Stannous chloride (molecular weight = 190) will be :
(A) 190 (B) 95 (C) 47.5 (D) 154.5
7. In the reaction
2CuSO4
+ 4KI 2CuI + I2
+ 2K2
SO4
the equilvalent weight of CuSO4
will be :
(A) 79.75 (B) 159.5 (C) 329 (D) None of these
8. In the following half-reaction, hydrazine is oxidized to N2
.
N2
H4
+ OH–
N2
+ H2
O + e–
The equivalent weight of N2
H4
(hydrazine) is :
(A) 8 (B) 16 (C) 32 (D) 64
PAGE NO. # 3
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
ANSWER KEY
DPP No. # 43
1. (C) 2. (C) 3. (D) 4. (D) 5.* (CD)
6. (a) (+ 7) (– 1) (+ 2) (0)
KMnO4
+ KCl + H2
SO4
 MnSO4
+ K2
SO4
+ H2
O + Cl2
(+ 7) (+ 2)
KMnO4
(oxidant)  MnSO4
(reduction half)
(– 1) (0)
KCl (reducant)  Cl2
(oxidant half)
(b) (+ 2) (– 1) (+ 3) (–2)
FeCl2
+ H2
O2
+ HCl  FeCl3
+ H2
O (oxidant half)
(+ 2) (+ 3)
FeCl2
(reductant)  FeCl3
(oxidation half)
(– 1)
H2
O2
(oxidant)  H2
O2–
(reduction half)
(c) (0) (+5) 2+ 2+
Cu + HNO3
(dil)  Cu (NO3
)2
+ H2
O + NO
(0) 2+
Cu (reductant)  Cu (NO3
)2
(oxidation half)
+5 +2
HNO3
(oxidant)  NO (reduction half)
(d) +3 + 5 – 1 + 5
Na2
HAsO3
+ KBrO3
+ HCl  NaCl + KBr + H3
AsO4
+3 +5
Na2
HAsO3
(reductant)  H3
AsO4
(oxidation half)
+5 – 1
KBrO3
(oxidant)  KBr
(e) 0 + 2 + 2.5 –1
I2
+ Na2
S2
O3
 Na2
S4
O6
+ NaI
0 –1
I2
(oxidant)  NaI (reduction half)
+2 +2.5
Na2
S2
O3
(reducant)  Na2
S4
O6
(oxidant half)
7. (i) K2
Cr2
O7
+ 3H2
O2
+ 4H2
SO4
K2
SO4
+ Cr2
(SO4
)3
+ 7H2
O + 3O2
v.f. = 6 v.f. = 2
(ii) 4Zn + NaNO3
+ 7NaOH = 4Na2
ZnO2
+ 2H2
O + NH3
v.f. = 2 v.f. = 8
(iii) 2OH–
+ 6H2
O + 2Al  3H2
+ 2(Al(OH)4
]–
v.f. = 3
(iv) 6Cu3
P + 11Cr2
O7
2–
+ 124H+
18Cu2+
+ 6H3
PO4
+ 22Cr3+
+ 53H2
O
v.f. = 11 v.f. = 6
(v) 6H+
+ ClO3
–
+ 6Fe2+
Cl–
+ 6Fe3+
+ 3H2
O
v.f. = 6 v.f. = 1
(vi) 3N2
O4
+ BrO3
–
+ 3H2
O 6NO3
–
+ Br–
+ 6H+
v.f. = 2 v.f. = 6
(vii) 3S2
O3
2–
+ 2Sb2
O5
+ 6H+
+ 3H2
O 4SbO + 6H2
SO3
v.f. = 4 v.f. = 6
(viii) Cr2
O7
2–
+ 6I–
+ 14H+
2Cr3+
+ 3I2
+ 7H2
O
v.f. = 6 v.f. = 1
(ix) IO4
–
+ 7I–
+ 8H+
4I2
+ 4H2
O
v.f. = 7 v.f. = 1
DPP No. # 44
1. (B) 2. (D) 3. (A) 4. (B) 5. (A) 6. (B) 7. (B)
8. (A)
PAGE NO. # 1
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
CHEMISTRY
DAILY PRACTICE PROBLEMS
D P P
COURSE NAME : UMANG (UP) DATE : 26.08.2013 to 31.08.2013 DPP NO. 45 & 47
TARGET
JEE (ADVANCED) : 2014
* Marked Questions are having more than one correct option.
DPP No. # 45
1. Suppose a compound contains atoms A, B and C. The oxidation number of A is +2, B is +5 and C is
– 2. The possible formula of the compound would be :
(A) ABC2
(B) A2
(BC3
)2
(C) A3
(BC4
)2
(D) A3
(B4
C)2
2. 50 mL of 5.6% KOH (w/v) is added to 50 mL of a 5.6% HCI (w/v) solution. The resulting solution will
be
(A) neutral (B) alkaline (C) strongly alkaline (D) acidic
3. Number of moles of electrons taken up when 1 mole of NO3

ions is reduced to 1 mole of NH2
OH is :
(A) 2 (B) 4 (C) 5 (D) 6
4. As2
O3
is oxidised to H3
AsO4
by KMnO4
in acidic medium. Volume of 0.02M KMnO4
required for this
purpose by 1mmol of As2
O3
will be :
(A) 10 mL (B) 20 mL (C) 40 mL (D) 80 mL
5. A 25-mL HCl solution containing 3.65 g HCl/L is neutralized by 50 mL an NaOH solution. Again, 25 mL
of the same NaOH solution neutralized by 50 mL of an H2
SO4
solution of unknown strength.The
normality of the H2
SO4
solution is
(A) 0.25 N (B) 0.025 N (C) 0.05 N (D) 0.50 N
6. The total volume of 0.1 M KMnO4
solution that are needed to oxidize 144 mg of ferrous oxalate and 152 mg
of ferrous sulphate in a mixture in acidic medium is :
(A) 5 mL (B) 2 mL (C) 8 mL (D) None of these
7. The equivalent weight of a metal is double than of oxygen. How many times is the weight of it’s oxide
greater than the weight of the metal ?
(A) 1.5 (B) 2 (C) 3 (D) 4
8. How much mass of K2
C2
O4
is require for complete oxidation in titration with 0.2 M, KMnO4
. The
volume of KMnO4
used in acidic medium is 500 ml and calculate the volume of CO2
obtained at S.T.P.
in the reaction.
9. The percent yield for the following reaction carried out in carbon tetrachloride (CCI4
) solution is 80%
Br2
+ CI2
 2BrCI
(a) What amount of BrCI would be formed from the reaction of 0.025 mol Br2
and 0.025 mol CI2
?
(b) What amoumt of Br2
is left unchanged?
10. 24 ml of a solution of Fe2+
ions was titrated with a solution of the oxidizing agent Cr2
O7
2–
. 40 ml of 0.02 M
K2
Cr2
O7
solution was required.What is the molarity of the Fe2+
solution.
PAGE NO. # 2
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
DPP No. # 46
1. 25 mL of 2N HCl, 50 mL of 4N HNO3
and x mL of 2M H2
SO4
are mixture together and the total volume is
made up to 1 L after dilution. 50 mL of this acid mixture completely reacted with 25 mL of a 1N Na2
CO3
solution. The value of x is :
(A) 250 mL (B) 62.5 mL (C) 100 mL (D) None of these
2. One mole of H2
SO4
will exactly neutralize
(A) 4 moles of an ammonia solution (B) 1 mole of Ba(OH)2
(C) 3 moles of Al (OH)3
(D) 0.5 mole of Ba(OH)2
3. When ferrous oxalate is titrated against K2
Cr2
O7
, meq of Fe2+
, C2
O4
2
and Cr2
O7
2
in this redox reac-
tion are x, y and z respectively. Then
(A) x = y (B) x + y = z (C) x + 2y = z (D) 2x + 6y = 6z
4. What volume of 2N K2
Cr2
O7
solution is required to oxidise 0.81 g of H2
S in acid medium?
(A) 47.8 ml (B) 23.8 ml (C) 40 ml (D) 72 ml
5. The number of moles of KMnO4
that will be needed to react with one mole of sulphite ion in acid
solution is
(A) 4/5 (B) 3/5 (C) 1 (D) 2/5
6. 4 mole of a mixture of Mohr’s salt and Fe2
(SO4
)3
requires 500 mL 1M K2
Cr2
O7
for complete oxidation in
acidic medium. The mole % of the Mohr’s salt in the mixture is :
(A) 25 (B) 50 (C) 60 (D) 75
7. Asolution of Na2
S2
O3
is standardized iodometrically against 0.167 g of KBrO3
. This process requires 50 mL
of the Na2
S2
O3
solution. What is the normality of the Na2
S2
O3
?
(A) 0.2 N (B) 0.12 N (C) 0.72 N (D) 0.02 N
8. In an iodometric estimation, the following reaction occur
2Cu2+
+ 4I–
 Cu2
I2
; I2
+ 2Na2
S2
O3  2NaI + Na2
S4
O6
0.12 mole of CuSO4
was added to excess of KI solution and the liberated iodine required 120 mL of hypo.
The molarity of hypo solution was :
(A) 2 (B) 0.20 (C) 0.1 (D) 1.0
9. It requires 20 ml of 1 M X3+ to titrate 10 ml of 2 M Sn2+ to Sn4+. What is the oxidation state of the 'X'
in the product.
10. 25 ml of a solution of H2S (it convert into S) was titrated with a solution of the oxidizing agent Cr2O7
2–. 100
ml of 0.01 M K2Cr2O7 solution was required. What is the molarity of the H2S solution ?
11. How many ml of 0.3M K2Cr2O7 (acidic) is required for complete oxidation of 10 ml of 0.1 M FeC2O4
solution.
12. Calculate the mass of oxalic acid which can be oxidized by 100ml of MnO4
– solution, 10ml of same solution
of MnO4
– is capable of oxidizing 50ml of 1 N I– to I2.
13. Butane, C4
H10
, burns with the oxygen in air to give carbon dioxide and water.
What is the amount (in moles) of carbon dioxide produced from 0.15 mol C4
H10
?
C4
H10
(g) + O2
(g)  CO2
(g) + H2
O(g) (not balanced)
DPP No. # 47
1. A 10-volume H2
O2
solution is equal to
(A) 3% (w/v) H2
O2
(B) 30 g/L H2
O2
(C) 1.76 N (D) all of these
2. A 5.0-mL solution of H2
O2
liberates 0.508 g of iodine from an acidified Kl solution. The volume strength
of the H2
O2
solution at STP is approximately :
(A) 4.00 (B) 4.5 (C) 6.05 (D) 5.5
PAGE NO. # 3
ETOOS ACADEMY Pvt. Ltd
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
3. What volume of O2
measured at standard conditions will be formed by the action of 100 ml of 0.5 N
KMnO4
on hydrogen peroxide in an acid solution? The skeleton equation for the reaction is
KMnO4
+ H2
SO4
+ H2
O2
 K2
SO4
+ MnSO4
+ H2
O + O2
(A) 0.12 l (B) 0.28 l (C) 0.56 l (D) 1.12 l
4. 100 ml of 0.1M NaAl(OH)2
CO3
is neutralised by 0.25 N HCl to form NaCl, AlCl3
and CO2
. Volume of
HCl required is
(A) 10 mL (B) 40 mL (C) 100mL (D) 160 mL
5. 100 mL of 0.1N I2
oxidizes Na2
S2
O3
in 50 ml solution to Na2
S4
O6
. The normality of this hypo solution
against KMnO4
(which oxidizes it to Na2
SO4
) would be
(A) 0.1 (B) 0.2 (C) 1.0 (D) 1.6.
6. Which of the following has two unpaired electrons ?
(A) Si, Mg (B) S, Mg (C) S, Si (D) S, Fe.
7. Total number of electrons contained in all the p-orbitals of bromine is:
(A) 5 (B) 17 (C) 19 (C) 23
8. The number of unpaired electrons in Mn2+
is:
(A) 5 (B) 3 (C) 4 (D) 1
9.* 1M 100 ml NaCl is mixed with 3M 100 ml HCl solution and 1M 200 ml CaCl2
solution. Find the ratio of the
concentration of cation and anion.
(A) The ratio of concentration of cation and anion = 3/4M
(B) The ratio of concentration of cation and anion = 2 M
(C) [Cl–
] = 2 M (D) [Cl–
] = 3/2 M
Subjective
10. The reaction Cl2
(g) + S2
O3
2-
 SO4
2-
+ Cl-
is to be carried out in basic medium. Starting with 0.15 mol
of Cl2
, 0.01 mol S2
O3
2-
and 0.3 mol of OH-
, how many moles of OH-
will be left in solution after the
reaction is complete. Assume no other reaction occurs.
11. A piece of brass weighing 304.8 mg was dissolved and prepared for iodometric titration. Excess K was
added and the liberated iodine required 30.0 ml of 0.08 M hypo. Find the % of Cu in the brass.
ANSWER KEY : DPP No. # 45
1. (C) 2. (D) 3. (D) 4. (C) 5. (B) 6. (C) 7. (A)
8. 41.5 gm K2
CO4
; 11.2 lt CO2
.
9. (a) Br2
+ CI2
 2BrCI
1 mol 1mol 2mol
0.025 0.025 0.050 mol
(b) Theoretical yield of BrCI = 0.050 mol
actual yield is =
100
8005.0 
= 0.04 mole
Br2
left unreacted = 0.025 – 0.02 = 0.005 mol.
10. 0.2 M.
DPP No. # 46
1. (B) 2. (B) 3. (B) 4. (B) 5. (D) 6. (D) 7. (B)
8. (D) 9. + 1 10. 0.12 M. 11. 1.67 ml. 12. 22.5 g 13. 0.60 mol CO2
DPP No. # 47
1. (D) 2. (B) 3. (C) 4. (D) 5. (D) 6. (C) 7. (B)
8. (A) 9.* (AC) 10. 0.2 moles OH-
. 11. 50%

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  • 1. PAGE NO. # 1 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 CHEMISTRY DAILY PRACTICE PROBLEMS D P P COURSE NAME : UMANG (UP) DATE : 29.07.2013 to 03.08.2013 DPP NO. 39 TARGET JEE (ADVANCED) : 2014 * Marked Questions are having more than one correct option. DPP No. # 39 1. Maganese achieves its maximum oxidation state in which of these compounds : (A) MnO2 (B) Mn3 O4 (C) KMnO4 (D) K2 MnO4 2. Oxidation number of underlined elements in N2 O5 , SO3 2– , NH4 + are : (A) + 5, + 2, – 3 (B) + 6, – 2, + 3 (C) + 6, + 2, – 3 (D) + 5, + 4, – 3 3. Phosphorous has the oxidation state of + 3 in : (A) Phosphorus acid (H3 PO3 ) (B) Orthophosphoric acid (H3 PO4 ) (C) Metaphosphoric acid (HPO3 ) (D) Pyrophosphoric acid (H4 P2 O7 ) 4. Which statement is wrong : (A) Oxidation number of oxygen is + 1 in peroxides (B) Oxidation number of oxygen is + 2 in oxygen difluroide (C) Oxidation number of oxygen is 1 2  in superoxides (D) Oxidation number of oxygen is – 2 in most of its compounds 5. The oxidation number and covalency of sulphur in the sulphur molecule (S8 ) are respectively : (A) 0 and 2 (B) + 6 and 8 (C) 0 and 8 (D) + 6 and 2 6. The oxidation number of oxygen in F2 O is : (A) + 1 (B) – 2 (C) + 2 (D) – 3 7. Calculate individual and average oxidation number (if required) of the marked element and also draw the structure of the following compounds or molecules : (1) Na2 S2 O3 (2) Na2 S4 O6 (3) H2 SO5 (4) H2 S2 O8 (5) H2 S2 O7 (6) S8 (7) HNO4 (8) C3 O2 (9) OsO4 (10) PH3 (11) CrO4 2– (12) Cr2 O7 2– (13) CrO2 Cl2 (14) CrO5 (15) Na2 HPO4 (16) FeS2 (17) C6 H12 O6 (18) NH4 NO3 8. Calculate individual and average oxidation number (if required) of the marked element and also draw the structure of the following compounds or molecules : (1) FNO3 (2) CaOCl2 (3) XeO3 F2 (4) LiAlH4 (5) Na3 AlF6 (6) P4 (7) O3 (8) I (IO3 )3 (9) Fe3 O4 (10) CsI3 (11) KO3 (12) O2 F2 (13) H2 SiF6 (14) H3 PO3 (15) POCl3 (16) Si(OH)4 (17) Mg2 C3 (18) CaC2 (19) Be2 C
  • 2. PAGE NO. # 2 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 ANSWER KEY DPP No. # 39 1. (C) 2. (D) 3. (A) 4. (A) 5. (A) 6. (C) 7. (1) + 2 (6, – 2) (2) + 5/2 (5, 5, 0, 0) (3) + 6 (4) + 6 (+ 6, + 6) (5) + 6 (+ 6, + 6) (6) 0 (7) + 5 (8) 4/3 (+ 2, + 2, 0) (9) + 8 (10) – 3 (11) + 6 (12) + 6 (+ 6, + 6) (13) + 6 (14) + 6 (15) + 5 (16) + 2 (17) 0 (18) – 3, + 5 8. (1) + 5 (2) 0 (– 1, + 1) (3) + 8 (4) + 3 (5) + 3 (6) 0 (0, 0, 0, 0) (7) 0 (0, – 1, + 1) (8) + 3, + 5 (9) 8 ( 2, 3) 3   (10) 1 3  (11) 1 3  (12) + 1 (+ 1, + 1) (13) + 4 (14) + 3 (15) + 5 (16) + 4 (17) 4 (0, 4, 0) 3   (18) – 1 (– 1, – 1) (19) – 4
  • 3. PAGE NO. # 1 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 CHEMISTRY DAILY PRACTICE PROBLEMS D P P COURSE NAME : UMANG (UP) DATE : 05.08.2013 to 10.08.2013 DPP NO. 40 TO 42 TARGET JEE (ADVANCED) : 2014 * Marked Questions are having more than one correct option. DPP No. # 40 1. Identify the reactants, if any, oxidized or reduced. (a) NaH + H2 O  NaOH + H2 (b) 2FeCl3 + H2 S  2FeCl2 + S + 2HCl (c) 3Mg + N2  Mg3 N2 (d) AgCN + CN–  Ag(CN)2 – (e) SnCl2 + Br2  SnCl2 .Br2 (f) AlCl3 + K  Al + 3KCl (g) 2CuSO4 + 4KI   Cu2 I2 + K2 SO4 + I2 (h) NH4 NO2  N2 + 2H2 O (i) CaC2 + 2H2 O  Ca(OH)2 + C2 H2 (j) SnS2 + S–2  SnS3 –2 (k) 3O2  2O3 2. Select the oxidants and reductants in the given reactions . (a) Ag(NH3 )2 + + 2H+  Ag+ + 2NH4 + (b) KIO3 + 5KI + 6HCl  3I2 + 6KCl + 3H2 O (c) 2CuSO4 + SO2 + 2KBr + 2H2 O  2CuBr + 2H2 SO4 + K2 SO4 (d) C2 H4 + Br2  C2 H4 Br2 (e) 2I– + H2 O2  2OH– + I2 (f) Si + 2KOH + H2 O  K2 SiO3 + 2H2 (g) 2Cu+2 + 4I–  Cu2 I2 + I2 3. Balance the following equation using desired medium – (a) C2 H5 OH + Cr2 O7 2– + H+  Cr3+ + C2 H4 O + H2 O (b) Sn(OH)3 – + Bi(OH)3 + OH–  Sn(OH)6 2– + Bi (c) IO3 – + N2 H4 + HCl  N2 + ICl2 – + H2 O (d) Hg2 Cl2 + NH3  Hg + HgNH2 Cl + NH4 Cl (e) Zn + NO3 – + H+  Zn2+ + NH4 + + H2 O (f) I2 + NO3 – + H+  IO3 – + NO2 + H2 O (g) MnO4 – + SO2 2– + H2 O  MnO2 + SO4 2– + OH– (h) H2 O2 + ClO2 + OH–  ClO2 – + O2 + H2 O (i) ClO– + CrO2 – + OH–  Cl– + CrO4 2– + H2 O (j) Cl2 + KOH  KClO3 + KCl + H2 O
  • 4. PAGE NO. # 2 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 DPP No. # 41 Balance the following equation using desired medium 1. H2 O2 + KMnO4  MnO2 + KOH + O2 + H2 O 2. HNO2 + KMnO4 + H2 SO4  HNO3 + KMnO4 + K2 SO4 + H2 O 3. N2 H4 + AgNO3 + KOH  N2 + Ag + KNO3 + H2 O 4. P + HNO3  HPO3 + NO + H2 O 5. K2 Cr2 O7 + HCl  KCl + CrCl3 + H2 O + Cl2 6. MnO4 – + C2 O4 2– + H+  CO2 + Mn2+ + H2 O 7. Cr2 O7 2– + C2 O4 2– + H+  Cr3+ + CO2 + H2 O 8. KMnO4 + H2 S + H2 SO4  KHSO4 + MnSO4 + S + H2 O 9. Cu(NH3 )4 Cl2 + KCN + H2 O  K2 Cu(CN)3 + NH3 + KCNO + NH4 Cl + KCl 10. Ag + KCN + H2 O + O2  KAg(CN)2 + KOH 11. Fe3 O4 + MnO4 – + H2 O  Fe2 O3 + MnO2 + OH– 12. C2 H5 OH + MnO4 – + OH–  C2 H3 O– + MnO2 + H2 O 13. Cr2 O7 2– + SO3 2– + H+  Cr3+ + SO4 2– + H2 O 14. ClO3 – + SO2 + H+  ClO2 + HSO4 – 15. Mn2+ + S2 O8 2– + H2 O  MnO4 – + HSO4 – + H+ DPP No. # 42 * Equivalent weight of any substance is the weight in gram which combines with 1 gram of hydrogen , 8 grams of oxygen, 35.5 grams of Cl and 108 grams of Ag. * Equivalent weight = factorn'' weightMolecular 1. The equivalent weight of a metal is 36. What weight of the metal would give 9.322 gm of its chloride? 2. On heating 5 × 10–3 equivalent of CaCl2 .xH2 O, 0.18 gm of water is obtained, what is the value of x ? 3. The equivalent weight of a metal is double that of oxygen. How many times is the weight of its oxide greater than the weight of the metal ? 4. 0.224 gm of a metal when dissolved in dilute acid liberates 285 cc of hydrogen, measured at 17ºC and 78 cm of mercury pressure. Find the equivalent weight of the metal. 5. 1.82 g of a metal required 32.5 mL of N HCl to dissolve it. What is equivalent weight of metal ? 6. The equivalent weight of a metal is 30. What volume of chlorine would be liberated at 27ºC and 750 mm pressure by 0.6 gm of the metal ? [Aqueous tension at 27ºC = 26.6 mm] 7. The molecular formula of an oxide of an element X is Xm On what is the equivalent weight of themetal? [Given atomic weight of X = a] 8. 12 g of an element combines with 32 g of oxygen. What is the equivalent weight of the element if the equivalent weight of oxygen is 8 ? 9. Equivalent weight of sulphur in SCl2 is 16, what is the equivalent weight of S in S2 Cl2 ? [S = 32 , Cl = 35.5] 10. 1.60 gm of a metal were dissolved in HNO3 to prepare its nitrate. The nitrate was strongly heated to give 2 g oxide. Calculate equivalent weight of metal. 11. From 3.36 gm of Fe, 4.80 gm of iron oxide is obtained. What is the equivalent weight of Fe? 12. 1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate equivalent weight of metal. 13. An element forms two oxides, the weight ratio composition in them, A : O = x : y in the first oxide and y : x in the second oxide. If the equivalent weight of A in the first oxide is 10.33, Calculate the equivalent weight of A in the second oxide.
  • 5. PAGE NO. # 3 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 ANSWER KEY DPP No. # 40 1. Rxn. No. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) Oxidized : NaH, H2 S, Mg, none, SnCl2 , K, KI, NH4 + , none, none none. Reduced : H2 O, FeCl3 , N2 none, Br2 , AlCl3 , CusO4 , NO2 – , none, none, none 2. (a) not (b) KIO3 -oxidant, I– reductant (c) CuSO4 -oxidant, SO2 reductant (d) Br2 -oxidant, C2 H4 reductant (e) H2 O2 -oxidant, I– reductant (f) KOH-oxidant, Si reductant (g) Cu++ -oxidant, I– reductant 3. (a) 3C2 H5 OH + Cr2 O7 2– + 8H+  2Cr3+ + 3C2 H4 O + 7H2 O (b) 3Sn(OH)3 – + 2Bi(OH)3 + 3OH–  3Sn(OH)6 2– + 2Bi (c) IO3 – + N2 H4 + 2HCl  N2 + ICl2 – + 3H2 O (d) 2Hg2 Cl2 + 4NH3  2Hg + 2HgNH2 Cl + 2NH4 Cl (e) 4Zn + NO3 – + 10H+  4Zn2+ + NH4 + + 3H2 O (f) I2 + 10NO3 – +8H+  2IO3 – + 10NO2 + 2H2 O (g) 2MnO4 – + 3SO2 2– + H2 O  2MnO2 + 3SO4 2– + 2OH– (h) H2 O2 + 2ClO2 + 2OH–  2ClO2 – + O2 + 2H2 O (i) 3ClO– + 2CrO2 – + 2OH–  3Cl– + 2CrO4 2– + H2 O (j) 3Cl2 + 5KOH  KClO3 + 5KCl + 3H2 O DPP No. # 41 1. 3H2 O2 + 2KMnO4  2MnO2 + 2KOH +3O2 + 2H2 O 2. 5HNO2 + 2KMnO4 + 3H2 SO4  5HNO3 + 2KMnO4 + K2 SO4 + 3H2 O 3. N2 H4 + 4AgNO3 + 4KOH  N2 + 4Ag + 4KNO3 + 4H2 O 4. 3P + 5HNO3  3HPO3 + 5NO + H2 O 5. K2 Cr2 O7 + 8HCl  2KCl + 2CrCl3 + 7H2 O + 3Cl2 6. 2MnO4 – + 5C2 O4 2– + 16H+  10CO2 + 2Mn2+ + 8H2 O 7. Cr2 O7 2– + 3C2 O4 2– + 14H+  2Cr3+ + 6CO2 + 7H2 O 8. 2KMnO4 + 5H2 S + 4H2 SO4  2KHSO4 + 2MnSO4 + 5S + 8H2 O 9. 2Cu(NH3 )4 Cl2 + 7KCN + H2 O K2 Cu(CN)3 + 6NH3 + KCNO + 2NH4 Cl + 2KCl 10. 4Ag + 8KCN + 2H2 O + O2  4KAg(CN)2 + 4KOH 11. 6Fe3 O4 + 2MnO4 – + H2 O  9Fe2 O3 + 2MnO2 + 2OH– 12. 3C2 H5 OH + 2MnO4 – + OH–  3C2 H3 O– + 2MnO2 + 5H2 O 13. Cr2 O7 2– + 3SO3 2– + 8H+  2Cr3+ + 3SO4 2– + 4H2 O 14. 2ClO3 – + SO2 + H+  2ClO2 + HSO4 – 15. 2Mn2+ + 5S2 O8 2– + 8H2 O  2MnO4 – + 10HSO4 – + 6H DPP No. # 42 1. 4.6935 gm 2. x = 4 3. 1.5 4. 9.105 5. 56 6. 0.2584 lit 7. m × a/2n 8. 3 9. 32 10. 32 11. 18.67 12. 38 13. 6.2
  • 6. PAGE NO. # 1 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 CHEMISTRY DAILY PRACTICE PROBLEMS D P P COURSE NAME : UMANG (UP) DATE : 12.08.2013 to 17.08.2013 DPP NO. 43 & 44 TARGET JEE (ADVANCED) : 2014 * Marked Questions are having more than one correct option. DPP No. # 43 1. White phosphorus reacts caustic soda, the products are PH3 and NaH2 PO2 . The reaction is an example of : (A) Oxidation (B) Reduction (C) Disproportionation (D) Neutralisation 2. The reaction 3 ClO– (aq) ClO3 – (aq) + 2Cl– (aq) is an example of : (A) Oxidation (B) Reduction (C) Disproportionation (D) Decomposition 3. Which of the following changes does not involve either oxidation or reduction : (A) VO2+  V2 O3 (B) Na  Na+ (C) Zn2+ Zn (D) CrO4 2– Cr2 O7 2– 4. Which of the following is not a redox reaction? (A) Mg + N2  Mg3 N2 (B) MnO4 – + C2 O4 2–  Mn2+ + CO2 (C) CuSO4 + KI 2CuI + I2 + K2 SO4 (D) AgCl + NH3 [Ag(NH3 )2 ]Cl 5.* Which of the following can show disporoportionation reaction : (A) ClO4 – (B) Cl– (C) ClO2 – (D) ClO3 – 6. 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 (c) Cu + HNO3 (dil) Cu (NO3 )2 + H2 O + NO (d) Na2 HAsO3 + KBrO3 + HCl NaCl + KBr + H3 AsO4 (e) I2 + Na2 S2 O3 Na2 S4 O6 + NaI 7. Balance the following redox equations and determine the valency factor (n-factor) for oxidant and reductant involved in the reactions : (i) K2 Cr2 O7 + H2 O2 + H2 SO4 K2 SO4 + Cr2 (SO4 )3 + H2 O + O2 (ii) Zn + NaNO3 + NaOH Na2 ZnO2 + H2 O + NH3 (iii) Al  [Al(OH)4 ]– + H2 (basic) (iv) Cu3 P + Cr2 O7 2– Cu2+ + H3 PO4 + Cr3+ + H2 O (acidic) (v) ClO3 – + Fe2+ + H+ Cl– + Fe3+ + H2 O (vi) N2 O4 + BrO3 – NO3 – + Br– (vii) S2 O3 2– + Sb2 O5 SbO + H2 SO3 (viii) Cr2 O7 2– + I– + H+ Cr3+ + I2 + H2 O (ix) IO4 – + I– + H+ I2 + H2 O
  • 7. PAGE NO. # 2 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 DPP No. # 44 1. In the reaction 2S2 O3 2– + I2 S4 O6 2– + 2I– , the equivalent wt. of S4 O6 2– is equal to : (A) Mol wt. (B) Mol. wt./2 (C) 2 × Mol. wt. (D) Mol. wt./6 2. Equivalent weight of NH3 in the change N  NH3 is : (A) 17/6 (B) 17 (C) 17/2 (D) 17/3 3. The molecular weight of the compounds (a) Na2 SO4 , (b) Na3 PO4 . 12H2 O and (c) Ca3 (PO4 )2 respectively are X, Y and Z. The correct set of their equivalent weights will be : (A) (a) X 2 (b) Y 3 (c) Z 6 (B) (a) x (b) Y 3 (c) Z 3 (C) (a) X 2 (b) Y (c) Z 3 (D) (a) X (b) Y (c) Z 4. In the following reaction : 3Fe + 4H2 O Fe3 O4 + 4H2 , if the atomic weight of iron is 56, then its equivalent weight will be : (A) 42 (B) 21 (C) 63 (D) 84 5. When one mole NO3 – is converted into 1 mole NO2 , 0.5 mole N2 and 0.5 mole N2 O respectively, it accepts x, y and z mole of electrons. Then, x, y and z are respectively : (A) 1, 5, 4 (B) 1, 2, 3 (C) 2, 1, 3 (D) 2, 3, 4 6. In the equation, SnCl2 + 2HgCl2 Hg2 Cl2 + SnCl4 : The equivalent weights of Stannous chloride (molecular weight = 190) will be : (A) 190 (B) 95 (C) 47.5 (D) 154.5 7. In the reaction 2CuSO4 + 4KI 2CuI + I2 + 2K2 SO4 the equilvalent weight of CuSO4 will be : (A) 79.75 (B) 159.5 (C) 329 (D) None of these 8. In the following half-reaction, hydrazine is oxidized to N2 . N2 H4 + OH– N2 + H2 O + e– The equivalent weight of N2 H4 (hydrazine) is : (A) 8 (B) 16 (C) 32 (D) 64
  • 8. PAGE NO. # 3 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 ANSWER KEY DPP No. # 43 1. (C) 2. (C) 3. (D) 4. (D) 5.* (CD) 6. (a) (+ 7) (– 1) (+ 2) (0) KMnO4 + KCl + H2 SO4  MnSO4 + K2 SO4 + H2 O + Cl2 (+ 7) (+ 2) KMnO4 (oxidant)  MnSO4 (reduction half) (– 1) (0) KCl (reducant)  Cl2 (oxidant half) (b) (+ 2) (– 1) (+ 3) (–2) FeCl2 + H2 O2 + HCl  FeCl3 + H2 O (oxidant half) (+ 2) (+ 3) FeCl2 (reductant)  FeCl3 (oxidation half) (– 1) H2 O2 (oxidant)  H2 O2– (reduction half) (c) (0) (+5) 2+ 2+ Cu + HNO3 (dil)  Cu (NO3 )2 + H2 O + NO (0) 2+ Cu (reductant)  Cu (NO3 )2 (oxidation half) +5 +2 HNO3 (oxidant)  NO (reduction half) (d) +3 + 5 – 1 + 5 Na2 HAsO3 + KBrO3 + HCl  NaCl + KBr + H3 AsO4 +3 +5 Na2 HAsO3 (reductant)  H3 AsO4 (oxidation half) +5 – 1 KBrO3 (oxidant)  KBr (e) 0 + 2 + 2.5 –1 I2 + Na2 S2 O3  Na2 S4 O6 + NaI 0 –1 I2 (oxidant)  NaI (reduction half) +2 +2.5 Na2 S2 O3 (reducant)  Na2 S4 O6 (oxidant half) 7. (i) K2 Cr2 O7 + 3H2 O2 + 4H2 SO4 K2 SO4 + Cr2 (SO4 )3 + 7H2 O + 3O2 v.f. = 6 v.f. = 2 (ii) 4Zn + NaNO3 + 7NaOH = 4Na2 ZnO2 + 2H2 O + NH3 v.f. = 2 v.f. = 8 (iii) 2OH– + 6H2 O + 2Al  3H2 + 2(Al(OH)4 ]– v.f. = 3 (iv) 6Cu3 P + 11Cr2 O7 2– + 124H+ 18Cu2+ + 6H3 PO4 + 22Cr3+ + 53H2 O v.f. = 11 v.f. = 6 (v) 6H+ + ClO3 – + 6Fe2+ Cl– + 6Fe3+ + 3H2 O v.f. = 6 v.f. = 1 (vi) 3N2 O4 + BrO3 – + 3H2 O 6NO3 – + Br– + 6H+ v.f. = 2 v.f. = 6 (vii) 3S2 O3 2– + 2Sb2 O5 + 6H+ + 3H2 O 4SbO + 6H2 SO3 v.f. = 4 v.f. = 6 (viii) Cr2 O7 2– + 6I– + 14H+ 2Cr3+ + 3I2 + 7H2 O v.f. = 6 v.f. = 1 (ix) IO4 – + 7I– + 8H+ 4I2 + 4H2 O v.f. = 7 v.f. = 1 DPP No. # 44 1. (B) 2. (D) 3. (A) 4. (B) 5. (A) 6. (B) 7. (B) 8. (A)
  • 9. PAGE NO. # 1 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 CHEMISTRY DAILY PRACTICE PROBLEMS D P P COURSE NAME : UMANG (UP) DATE : 26.08.2013 to 31.08.2013 DPP NO. 45 & 47 TARGET JEE (ADVANCED) : 2014 * Marked Questions are having more than one correct option. DPP No. # 45 1. Suppose a compound contains atoms A, B and C. The oxidation number of A is +2, B is +5 and C is – 2. The possible formula of the compound would be : (A) ABC2 (B) A2 (BC3 )2 (C) A3 (BC4 )2 (D) A3 (B4 C)2 2. 50 mL of 5.6% KOH (w/v) is added to 50 mL of a 5.6% HCI (w/v) solution. The resulting solution will be (A) neutral (B) alkaline (C) strongly alkaline (D) acidic 3. Number of moles of electrons taken up when 1 mole of NO3  ions is reduced to 1 mole of NH2 OH is : (A) 2 (B) 4 (C) 5 (D) 6 4. As2 O3 is oxidised to H3 AsO4 by KMnO4 in acidic medium. Volume of 0.02M KMnO4 required for this purpose by 1mmol of As2 O3 will be : (A) 10 mL (B) 20 mL (C) 40 mL (D) 80 mL 5. A 25-mL HCl solution containing 3.65 g HCl/L is neutralized by 50 mL an NaOH solution. Again, 25 mL of the same NaOH solution neutralized by 50 mL of an H2 SO4 solution of unknown strength.The normality of the H2 SO4 solution is (A) 0.25 N (B) 0.025 N (C) 0.05 N (D) 0.50 N 6. The total volume of 0.1 M KMnO4 solution that are needed to oxidize 144 mg of ferrous oxalate and 152 mg of ferrous sulphate in a mixture in acidic medium is : (A) 5 mL (B) 2 mL (C) 8 mL (D) None of these 7. The equivalent weight of a metal is double than of oxygen. How many times is the weight of it’s oxide greater than the weight of the metal ? (A) 1.5 (B) 2 (C) 3 (D) 4 8. How much mass of K2 C2 O4 is require for complete oxidation in titration with 0.2 M, KMnO4 . The volume of KMnO4 used in acidic medium is 500 ml and calculate the volume of CO2 obtained at S.T.P. in the reaction. 9. The percent yield for the following reaction carried out in carbon tetrachloride (CCI4 ) solution is 80% Br2 + CI2  2BrCI (a) What amount of BrCI would be formed from the reaction of 0.025 mol Br2 and 0.025 mol CI2 ? (b) What amoumt of Br2 is left unchanged? 10. 24 ml of a solution of Fe2+ ions was titrated with a solution of the oxidizing agent Cr2 O7 2– . 40 ml of 0.02 M K2 Cr2 O7 solution was required.What is the molarity of the Fe2+ solution.
  • 10. PAGE NO. # 2 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 DPP No. # 46 1. 25 mL of 2N HCl, 50 mL of 4N HNO3 and x mL of 2M H2 SO4 are mixture together and the total volume is made up to 1 L after dilution. 50 mL of this acid mixture completely reacted with 25 mL of a 1N Na2 CO3 solution. The value of x is : (A) 250 mL (B) 62.5 mL (C) 100 mL (D) None of these 2. One mole of H2 SO4 will exactly neutralize (A) 4 moles of an ammonia solution (B) 1 mole of Ba(OH)2 (C) 3 moles of Al (OH)3 (D) 0.5 mole of Ba(OH)2 3. When ferrous oxalate is titrated against K2 Cr2 O7 , meq of Fe2+ , C2 O4 2 and Cr2 O7 2 in this redox reac- tion are x, y and z respectively. Then (A) x = y (B) x + y = z (C) x + 2y = z (D) 2x + 6y = 6z 4. What volume of 2N K2 Cr2 O7 solution is required to oxidise 0.81 g of H2 S in acid medium? (A) 47.8 ml (B) 23.8 ml (C) 40 ml (D) 72 ml 5. The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acid solution is (A) 4/5 (B) 3/5 (C) 1 (D) 2/5 6. 4 mole of a mixture of Mohr’s salt and Fe2 (SO4 )3 requires 500 mL 1M K2 Cr2 O7 for complete oxidation in acidic medium. The mole % of the Mohr’s salt in the mixture is : (A) 25 (B) 50 (C) 60 (D) 75 7. Asolution of Na2 S2 O3 is standardized iodometrically against 0.167 g of KBrO3 . This process requires 50 mL of the Na2 S2 O3 solution. What is the normality of the Na2 S2 O3 ? (A) 0.2 N (B) 0.12 N (C) 0.72 N (D) 0.02 N 8. In an iodometric estimation, the following reaction occur 2Cu2+ + 4I–  Cu2 I2 ; I2 + 2Na2 S2 O3  2NaI + Na2 S4 O6 0.12 mole of CuSO4 was added to excess of KI solution and the liberated iodine required 120 mL of hypo. The molarity of hypo solution was : (A) 2 (B) 0.20 (C) 0.1 (D) 1.0 9. It requires 20 ml of 1 M X3+ to titrate 10 ml of 2 M Sn2+ to Sn4+. What is the oxidation state of the 'X' in the product. 10. 25 ml of a solution of H2S (it convert into S) was titrated with a solution of the oxidizing agent Cr2O7 2–. 100 ml of 0.01 M K2Cr2O7 solution was required. What is the molarity of the H2S solution ? 11. How many ml of 0.3M K2Cr2O7 (acidic) is required for complete oxidation of 10 ml of 0.1 M FeC2O4 solution. 12. Calculate the mass of oxalic acid which can be oxidized by 100ml of MnO4 – solution, 10ml of same solution of MnO4 – is capable of oxidizing 50ml of 1 N I– to I2. 13. Butane, C4 H10 , burns with the oxygen in air to give carbon dioxide and water. What is the amount (in moles) of carbon dioxide produced from 0.15 mol C4 H10 ? C4 H10 (g) + O2 (g)  CO2 (g) + H2 O(g) (not balanced) DPP No. # 47 1. A 10-volume H2 O2 solution is equal to (A) 3% (w/v) H2 O2 (B) 30 g/L H2 O2 (C) 1.76 N (D) all of these 2. A 5.0-mL solution of H2 O2 liberates 0.508 g of iodine from an acidified Kl solution. The volume strength of the H2 O2 solution at STP is approximately : (A) 4.00 (B) 4.5 (C) 6.05 (D) 5.5
  • 11. PAGE NO. # 3 ETOOS ACADEMY Pvt. Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303 3. What volume of O2 measured at standard conditions will be formed by the action of 100 ml of 0.5 N KMnO4 on hydrogen peroxide in an acid solution? The skeleton equation for the reaction is KMnO4 + H2 SO4 + H2 O2  K2 SO4 + MnSO4 + H2 O + O2 (A) 0.12 l (B) 0.28 l (C) 0.56 l (D) 1.12 l 4. 100 ml of 0.1M NaAl(OH)2 CO3 is neutralised by 0.25 N HCl to form NaCl, AlCl3 and CO2 . Volume of HCl required is (A) 10 mL (B) 40 mL (C) 100mL (D) 160 mL 5. 100 mL of 0.1N I2 oxidizes Na2 S2 O3 in 50 ml solution to Na2 S4 O6 . The normality of this hypo solution against KMnO4 (which oxidizes it to Na2 SO4 ) would be (A) 0.1 (B) 0.2 (C) 1.0 (D) 1.6. 6. Which of the following has two unpaired electrons ? (A) Si, Mg (B) S, Mg (C) S, Si (D) S, Fe. 7. Total number of electrons contained in all the p-orbitals of bromine is: (A) 5 (B) 17 (C) 19 (C) 23 8. The number of unpaired electrons in Mn2+ is: (A) 5 (B) 3 (C) 4 (D) 1 9.* 1M 100 ml NaCl is mixed with 3M 100 ml HCl solution and 1M 200 ml CaCl2 solution. Find the ratio of the concentration of cation and anion. (A) The ratio of concentration of cation and anion = 3/4M (B) The ratio of concentration of cation and anion = 2 M (C) [Cl– ] = 2 M (D) [Cl– ] = 3/2 M Subjective 10. The reaction Cl2 (g) + S2 O3 2-  SO4 2- + Cl- is to be carried out in basic medium. Starting with 0.15 mol of Cl2 , 0.01 mol S2 O3 2- and 0.3 mol of OH- , how many moles of OH- will be left in solution after the reaction is complete. Assume no other reaction occurs. 11. A piece of brass weighing 304.8 mg was dissolved and prepared for iodometric titration. Excess K was added and the liberated iodine required 30.0 ml of 0.08 M hypo. Find the % of Cu in the brass. ANSWER KEY : DPP No. # 45 1. (C) 2. (D) 3. (D) 4. (C) 5. (B) 6. (C) 7. (A) 8. 41.5 gm K2 CO4 ; 11.2 lt CO2 . 9. (a) Br2 + CI2  2BrCI 1 mol 1mol 2mol 0.025 0.025 0.050 mol (b) Theoretical yield of BrCI = 0.050 mol actual yield is = 100 8005.0  = 0.04 mole Br2 left unreacted = 0.025 – 0.02 = 0.005 mol. 10. 0.2 M. DPP No. # 46 1. (B) 2. (B) 3. (B) 4. (B) 5. (D) 6. (D) 7. (B) 8. (D) 9. + 1 10. 0.12 M. 11. 1.67 ml. 12. 22.5 g 13. 0.60 mol CO2 DPP No. # 47 1. (D) 2. (B) 3. (C) 4. (D) 5. (D) 6. (C) 7. (B) 8. (A) 9.* (AC) 10. 0.2 moles OH- . 11. 50%