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CS– 1
STOICHIOMETRY
C1
C2
In this chapter we will discuss the calculations based on chemical equations. It has been classified into two
parts :
1.
Mole Concept
2.
Equivalent Concept
MOLE CONCEPT :
In mole concept we deal with different types of relations like weight-weight, weight-volume, or
volume-volume relationship between reactants or products of the reaction.
Mole concept is based on balanced chemical chemical reaction. Some basic definitions used in mole
concept are as follows :
Limiting Reagent : A reagent which is consumed completely during the chemical reaction.
Number of moles of a substance(n) 
weight of substance
atomic or molecular weight
Also, Number of moles of a substance(n) 
Given number of molecules
Avogadro number
In gas phase reaction number of moles of a gas (n) =
1.
2.
C3
PV
,
RT
At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure.
In aq. solution
n = MV [M - molarity, V - volume of solution]
Practice Problems :
Chlorine can be produced by reacting H2SO4 acid with a mixture of MnO2 and NaCl. The reactions
follows the equation : 2NaCl + MnO2 + 3H2SO4  2NaHSO4 + MnSO4 + Cl2 + H2O what volume of
chlorine at STP can be produced from 100 g of NaCl ? (At. wt. Na = 23, Cl = 35.5)
(a)
19.15 lt
(b)
30 lt
(c)
29 lt
(d)
5 lt
A solution contains 5 g of KOH was poured into a solution containing 6.8 g of AlCl3, find the mass of
precipitate formed [At. wt. : H-1, Al-27, Cl-35.5, K-39]
(a)
2.3 g
(b)
23 g
(c)
32 g
(d)
0.32 g
[Answers : (1) a (2) a]
DIFFERENT WAYS OF EXPRESSING THE CONCENTRATION TERMS :
Important Definitions :
mass percent 
mass of solute
 100
mass of solution
Molarity(M) 
No. of moles of solute
, unit of molarity are mol/lit., M or molar..
Vol. of solution in L
Normality ( N ) 
Molality (m ) 
No. of gramequiva lents of solute , unit of normality are g-eq./lit., N or normal.
Vol. of solution in L
No. of moles of solute
, unit of molality are mol/kg, m or molal.
wt . of solvent
Mole fraction( x A ) 
ppm 
nA .
n A  nB
mass of solute
 10 6
mass of solution
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CS – 2
No. of gram equivalents of solute(neq ) 
Equivalent weight 
Weight of solute
Equivalent weight
Molecular weight (or ) Atomic weight (or ) Ionic weight
n factor
The relation between different concentration terms :
1.
neq = nmol × n-factor
2.
neq = Normality × Volume (L)
3.
Number of moles(nmol) = Molarity × Volume (L)
4.
Normality = Molarity × n-factor
5.
M
10xd
M
6.
m
7.
m
x B  1000
(1  x B )M A
M  1000
1000d  MM 
(d  density of solution in g/ml, M  molar mass of solute, xB and xA are mole fraction of solute and
solvent respectively, MA  molar mass of solvent)
Calculation of ‘n’ Factor for Different Compounds :
1.
2.
Acids : n = basicity
H3PO4 n = 3
H3PO3 n = 2
H3PO2 n = 1
H3BO3 n = 1
Bases : n = acidity of base
e.g. Ammonia and all amines are monoacidic bases,
NaOH(n = 1), Na2CO3(aq) n = 2, NaHCO3(n = 1)
3.
Salt : (Which does not undergo redox reactions)
n factor = Total cationic or anionic charge, e.g. Na3PO4 n = 3, Ba3(PO4)2 n = 6
4.
C4
Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Or
number of electron lost or gained from one mole of the compound.
EQUIVALENT CONCEPT
It is based on law of equivalence which is explained as follows :
Law of chemical equivalents : In a chemical reaction the equivalents of all the species (reactants or
products) are equal to each other provided none of these compounds is in excess.
N1V1 = N2V2 (when normalities and volumes are given).
If the number of equivalence of both the reactants are different then reactant with the lesser number of
equivalence will be the limiting reagent.
Application of equivalent concept : It is used in acid base titration, back titration and double titration,
similarly in redox titration. Equivalent concepts can be used on all reaction whether they are balanced or
not balanced but mole concept is used in solving the problems when the reactions are balanced.
Basic principles of tirations :
In voltmetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react with
a known volume of a standard solution slowly. A chemical reaction takes place between the solute of an
unknown substance and the solute of the standard solution. The completion of the reaction is indicated by
the end point of the reaction, which is observed by the colour change either due to the indicator or due to the
solute itself. Whether the reactions during the analysis are either between an acid and or base or between
O.A. and R.A., the law of equivalence is used at end point.
Following are the different important points regarding this process :
(i)
In case of acid base titration at the equivalence point
(neq)acid = (neq)base
(ii)
In case of redox titration
(neq)oxidant = (neq)reductant
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(iii)
If a given volume of solution is diluted then number of moles or number of equivalence of
solute remains same but molarity or normality of the solution decreases.
(iv)
If a mixture contains more than one acids and is allowed to react completely with the base then
at the equivalence point, (neq) acid1 + (neq) acid2 + ... = (neq) base
(v)
Similarly if a mixture contains more than one oxidising agents then at equivalence point,
(neq) O.A1 + (neq) O.A2 +... = (neq) reducing agent.
(vi)
If it is a difficute to solve the problem through equivalence concept then use the mole concept.
Back titration :
This is a method in which a substance is taken in excess and some part of its has to react with another
substance and the remaining part has to be titrated against standard reagent.
Double titration :
This is a titration of specific compound using different indicators. Let us consider a solid mixture of NaOH,
Na2CO3 and inert impurities.
When the solution containing NaOH and Na2CO3 is titrated using phenolphalein indicator following reaction takes place at the phenolphthalein end point –
NaOH + HCl  NaCl + H2O
Na2CO3 + HCl  NaHCO3 + H2O
Here, eq. of NaOH 
1
eq. of Na 2 CO 3  eq. of HCl
2
(n  2)
When methyl orange is used, Na2CO3 is converted into NaCl + CO2 + H2O
Hence, eq. of NaOH + eq. of Na 2 CO 3 = eq. of HCl
(n 2)
TITRATION OF MIXTURE OF BASES WITH TWO INDICATORS
Every indicator has a working range
Indicator
pH range
Behaving as
Phenolphthalein
8 — 10
weak organic acid
Methyl orange
3 — 4.4
weak organic base
Thus methyl orange with lower pH range can indicate complete neutralisation of all types
of bases. Extent of reaction of different bases with acid (HCl) using these two indicators
summarised below
Phenolphthalein
Methyl Orange
NaOH
100% reaction is indicated
100 % reaction is indicated
NaOH + HCl  NaCl + H2O
NaOH + HCl  NaCl + H2O
Na2CO3
50% reaction upto NaHCO3
100% reaction is indicated
stage is indicated
Na2CO3 + 2HCl  2NaCl + H2O
Na2CO3 + HCl  NaHCO3 + NaCl
+ CO2
NaHCO3
No reaction is indicated
NaHCO3 + HCl  NaCl + H2O +
CO2
100% reaction is indicated
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CS – 4
Electron exchanged Eq. wt.
or change in O.N.
Species
Changed to
Reactions
1.
MnO4— (O.A.)
Mn2+ in acidic
medium
MnO4— + 8H+ + 5e—  Mn2+ +
4H2O
5
E
2.
MnO4— (O.A.)
MnO2 in basic
medium
MnO4— + 3e— + 2H2O  MnO2 +
4OH—
3
E
3.
MnO4— (O.A.)
MnO42— in
MnO4— + e— + 2H2O  MnO22—
neutral medium
1
E
4.
Cr2O72—(O.A.)
Cr3+ in acidic
medium
Cr2O72— + 14H+ + 6e—  2Cr3+ +
7H2O
6
E
5.
MnO2(O.A.)
Mn2+ in acidic
medium
MnO2 + 4H+ + 2e—  Mn2+ +
2H2O
2
E
6.
Cl2(O.A.)
(in bleaching
powder)
CuSO4 (O.A.)
(in iodometric
titration)
S2O32— (R.A.)
Cl—
Cl2 + 2e—  2Cl—
2
E
Cu+
Cu2+ + e–  Cu+
1
E
S4O62—
2S2O32—  S4O62— + 2e—
7.
8.
+
5
—
H2O2(O.A.)
H2 O
H2O2 + 2H + 2e  2H2O
2
10.
H2O2(R.A.)
O2
2
11.
Fe2+ (R.A.)
Fe3+
H2O2  O2 + 2H+ + 2e—
(O.N. of oxygen in H2O2 is (–1)
per atom)
Fe2+  Fe3+ + e—
M
2
M
E
2
E
E
1
Estimation of
Reaction
Relation between O.A. and R.A.
I2
I2 + 2Na2S2O3  2NaI + Na2S4O62—
I2 + 2S2O32—  2I— + S4O62—
I2  2I—  2Na2S2O3
Eq. wt. (Na2S2O3) = E 
2.
3.
4.
5.
CuSO4
CaOCl2
MnO2
IO3—
2CuSO4  I2  2Na2S2O3
CaOCl2 + H2O  Ca(OH)2 + Cl2
Cl2 + 2KI  2KCl + I2
Cl2 + 2I—  2Cl— + I2
CaOCl2  Cl2  I2  2I—  2Na2S2O3
IO3— + 5I— + 6H+  3I2 + 3H2O
Eq. wt. of CuSO4 =
M
1
M
2
Eq. wt. of CaOCl2 =
MnO2  Cl2  I2  2I—  2Na2S2O3
Eq. wt. of MnO2 =
M
2
IO3—  3I2  6I  6Na2S2O3
Eq. wt. IO3— =
Einstein Classes,
M
1
M
1
2CuSO4 + 4KI  Cu2I2 + 2K2SO4 + I2
or 2Cu2+ + 4I—  Cu2I2 + I2
white ppt.

MnO2 + 4HCl (conc.) 
MnCl2 +

Cl2 + 2H2O
Cl2 + 2KI  2KCl + I2
or MnO2 + 4H+ + 2Cl—  Mn2+ + 2H2O
+ Cl2
Cl2 + 2I—  I2 + 2Cl—
M
1
2 (for two molecules) E  M  M
9.
1.
M
5
M
3
M
1
M
6
M
2
M
2
M
6
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CS– 5
6.
—
+
—
H2O2 + 2I + 2H  I2 + 2H2O
H2 O 2
H2O2  I2  2I  2Na2S2O3
Eq. wt. H2O2 =
7.
Cl2 + 2I—  2Cl— + I2
Cl2
M
2
Cl2  I2  2I—  2Na2S2O3
M
2
Eq. wt. of Cl2 =
8.
O3 + 6I— + 6H+  3I2 + 3H2O + O2
O3
O3  3I2
Eq. wt. of O3 =
ClO—
9.
ClO— + 2I— + 2H+  H2O + Cl— + I2
M
2
ClO—  I2  2I  2Na2S2O3
Eq. wt. of ClO— =
Cr2O72—
10.
Cr2O72— + 14H+ + 6I—  3I2 + 2Cr3+ +
7H2O
M
2
Cr2O72—  3I2  6I—
Eq. wt. of Cr2O72— =
M
6
Practice Problems :
1.
[Na+] in a solution prepared by mixing 30.00 mL of 0.12 M NaCl with 70 mL of 0.15 M Na2SO4 is
(a)
2.
(c)
0.210 M
(d)
0.246 M
(b)
MnO2
(c)
MnO4–
(d)
MnO42–
6.20 g
(b)
7.75 g
(c)
10.5 g
(d)
21.0 g
–
When BrO ion reacts with Br ion in acid solution Br2 is liberated. The equivalent weight of KBrO3
in this reaction is
M/8
(b)
M/3
(c)
M/5
(d)
M/6
The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous
oxalate in acidic solution is
(a)
6.
Mn2O3
–
3
(a)
5.
0.141 M
The anion nitrate can be converted into ammonium ion. The equivalent mass of NO3– ion in this
reaction would be
(a)
4.
(b)
The equivalent mass of MnSO4 is half of its molar mass when it is converted to
(a)
3.
0.135 M
3/5
(b)
2/5
(c)
4/5
(d)
1
5 ml of N-HCl, 20 ml of N/2-H2SO4 and 30 ml of N/3 – HNO3 are mixed together and the volume
made to 1 litre.
(i)
The normality of the resulting solution is
(a)
N/5
(b)
N/10
(c)
N/20
(d)
N/40
(d)
2.5 g
(d)
7.
(ii)
The wt. of pure NaOH required to neutralize the above solution is
(a)
10 g
(b)
2g
(c)
1g
0.7 g of a sample of Na2CO3.xH2O were dissolved in water and the volume was made to 100 ml, 20 ml
of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x is
(a)
7
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3
(c)
2
(d)
5
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8.
–
4
100 mL of 1 M KMnO4 oxidised 100 mL of H2O2 in acidic medium (when MnO is reduced to Mn2+);
volume of same KMnO4 required to oxidise 100 mL of H2O2 in basic medium (when MnO4–. is
reduced to MnO2) will be
(a)
9.
500
mL
3
(c)
300
mL
3
(d)
100 mL
0.2 g
(b)
0.4 g
(c)
0.6 g
(d)
1.0 g
3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 mL of 2 M KMnO4 solution is acidic
medium. Hence mol fraction of FeSO4 in the mixture is
(a)
11.
(b)
100 mL of a mixture of NaOH and Na2SO4 is neutralised by 100 mL of 0.5 M H2SO4. Hence amount
of NaOH in 100 mL mixture is
(a)
10.
100
mL
3
1
3
(b)
2
3
(c)
2
5
(d)
3
5
5.3 g of M2CO3 is dissolved in 150 mL of 1 N HCl. Unused acid required 100 mL of 0.5 N NaOH.
Hence equivalent weight of M is
(a)
23
(b)
12
(c)
24
(d)
13
[Answers : (1) d (2) b (3) b (4) c (5) a (6) (i) d (ii) c (7) c (8) b (9) b (10) a (11) a]
C5
VOLUME STRENGTH OF H2O2
x volume of H2O2 means x litre of O2 is liberated by 1 volume of H2O2 on decomposition
2H 2 O 2 
 2H 2 O  O 2
68 gm
22.4 lit at STP
Volume strength of H2O2 solution = N × 5.6......
(where N is the normality of the H2O2
solution
Practice Problems :
1.
(a)
Calculate the strength of ‘20 V’ of H2O2 in terms of :
(i)
(b)
2.
normality (ii) grams per litre (iii) molarity and (iv) percentage
Calculate the volume strength of 2.0 N H2O2 solution.
In a 50 ml solution of H2O2 an excess of KI and dilute H2SO4 were added. The I2 so liberated required
20 ml of 0.1 N Na2S2O3 for complete reaction. Calculate the strength of H2O2 in grams per litre.
[Answers : (a) (i) 3.58 N (ii) 60.86 g/lit. (iii) 1.79 M (iv) 6.086% (W/V) (b) 11.2 V (2) 0.68 g/litre]
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SINGLE CORRECT CHOICE TYPE
1.
2.
1 g of the carbonate of a metal was dissolved in
25 ml of N-HCl. The resulting liquid required 5 ml
of N-NaOH for neutralization. The eq. wt. of the
metal carbonate is
8.
6.
Molality of 18 M H2SO4 (d = 1.8 gmL–1) is
30
(a)
36 mol kg–1
(b)
200 mol kg–1
(c)
20
(d)
None
(c)
500 mol kg–1
(d)
18 mol kg–1
If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na3PO4,
the maximum amount of Ba 3(PO4)2 that can be
formed is
(a)
0.70 mol
(b)
0.50 mol
(c)
0.20 mol
(d)
0.10 mol
9.
When one gram mol of KMnO4 reacts with HCl,
the volume of chlorine liberated at NTP will be
(a)
11.2 litres
(b)
22.4 litres
(c)
44.8 litres
(d)
56.0 litres
34 g of hydrogen peroxide is present in 1120 ml of
solution. This solution is called
(a)
10 vol solution (b)
20 vol solution
(c)
30 vol solution (d)
32 vol solution
To prepare a solution that is 0.50 M KCl starting
with 100 mL of 0.40 M KCl
(a)
add 0.75 g KCl
(b)
add 20 mL of water
(c)
add 0.10 mol KCl
(d)
evaporate 10 mL water
In hot alkaline solution, Br2 disproportionates to
Br– and BrO3–
13.
3Br2 + 6OH–  5Br– + BrO3– + 3H2O
hence equivalent weight of Br 2 is (molecular
weight = M)
(a)
(c)
7.
0.08 M KCl and 0.01 M HCl
(b)
12.
5.
(d)
50
11.
4.
0.08 M KCl and 0.01 M KOH
(a)
10.
3.
(c)
M
6
(b)
3M
5
(d)
M
5
14.
5M
5
When 80 mL of 0.20 M HCl is mixed with 120 mL
of 0.15 M KOH, the resultant solution is the same
as a solution of
15.
1 g equiv. of a substance is the weight of that amount
of a substance which is equivalent to
(a)
0.25 mol of O2
(b)
0.50 mol of O2
(c)
1 mol of O2
(d)
8 mol of O2
The molality of a H2SO4 solution is 9. The weight of
the solute in 1 kg H2SO4 solution is
(a)
900.0 g
(b)
468.65 g
(c)
882.0 g
(d)
9.0 g
The density of 1 M solution of NaCl is 1.0585 g/mL.
The molality of the solution is
(a)
1.0585
(b)
1.00
(c)
0.10
(d)
0.0585
Which is false about H3PO2
(a)
it is tribasic acid
(b)
one mole is neutralised by 0.5 mol
Ca(OH)2
(c)
NaH2PO2 is normal salt
(d)
it disproportionates to H3PO3 and PH3 on
heating.
When KMnO 4 acts as an oxidising agent and
ultimately forms [MnO4], MnO2, Mn2O3, Mn2+ then
the number of electrons transferred in each case
respectively is
(a)
3, 5, 7, 1
(b)
1, 5, 3, 7
(c)
4, 3, 1, 5
(d)
1, 3, 4, 5
With increase of temperature, which of these
changes ?
(a)
mole fraction
(b)
Fraction of solute present in water
(c)
molality
(d)
weight fraction of solute
In a compound C, H and N atoms are present in
9:1:3. 5 by weight of compound is 108. Molecular
formula of compound is
(a)
0.16 M KCl and 0.02 M HCl
(a)
C2H 6N2
(b)
C 3H 4N
(b)
0.08 M KCl
(c)
C9H12N3
(d)
C6H 8N2
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16.
17.
8 g of sulphur are burnt to form SO2 which is
oxidised by Cl2 water. The solution is treated with
BaCl2 solution. The amount of BaSO4 precipitated
is
(a)
1 mol
(b)
0.5 mol
(c)
0.24 mol
(d)
0.25 mol
(c)
19.
20.
21.
22.
24.
One mole of a mixture of CO and CO2 requires
exactly 20 gram of NaOH in solution for complete
conversion of all the CO2 into Na2CO3. How many
grams of NaOH would it require for conversion into
Na2CO3 if the mixture (one mole) is completely
oxidised to CO2
(a)
18.
23.
60 grams
(b)
40 grams
(d)
80 grams
20 grams
One gram of a mixture of Na2CO3 and NaHCO3
consumes y gram equivalents of HCl for complete
neutralisation. One gram of the mixture is strongly
heated, the cooled and the residue treated with HCl.
How many gram equivalents of HCl would be
required for complete neutralization ?
(a)
2 y gram equivalent
(b)
y gram equivalents
(c)
3y/4 gram equivalents
(d)
3y/2 gram equivalents
25.
To neutralise completely 20 mL of 0.1 M aqueous
solution of phosphorus acid (H3PO3), the volume
of 0.1 M aqueous KOH solution required is
(a)
10 mL
(b)
20 mL
(c)
40 mL
(d)
60 mL
Excess of KI reacts with CuSO4 solution and then
Na 2S 2O 3 solution is added to it. Which of the
statements is incorrect for this reaction ?
(a)
Cu2I2 formed
(b)
CuI2 is formed
(c)
Na2S2O3 is oxidised
(d)
evolved I2 is reduced
Two solutions of a substance (non electrolyte) are
mixed in the following manner 480 ml of 1.5 M first
solution and 520 mL of 1.2 M second solution. What
is the molarity of the final mixture ?
(a)
1.20 M
(b)
1.50 M
(c)
1.344 M
(d)
2.70 M
If equal volumes of 1 M KMnO4 and 1 M K2Cr2O7
solutions are allowed to oxidise Fe (II) to Fe(III),
then Fe (II) oxidised will be
(a)
more by KMnO4
(b)
more by K2Cr2O7
(c)
equal in both cases
(d)
none of these
0.5 g of fuming H2S2O7(Oleum) is diluted with
water. This solution is completely neutralized by
26.7 ml of 0.4 N NaOH. The percentage of free SO3
in sample is
(a)
30.6%
(b)
40.6%
(c)
20.6%
(d)
50%
One mole of N2H4 loses 10 mol of electrons to form
a new compound Y. Assuming that all the nitrogen
appears in the new compound. What is the
oxidation state of nitrogen in Y.
(a)
–1
(b)
–3
(c)
+3
(d)
+5
25 ml of a solution of barium hydroxide on
titration with a 0.1 molar solution of hydrochloric
acid gave a titre value of 35 ml. The molarity of
barium hydroxide solution was
(a)
0.35
(b)
0.07
(c)
0.14
(d)
0.28
Einstein Classes,
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
1.
a
11.
b
2.
d
12.
a
3.
d
13.
d
4.
a
14.
b
5.
a
15.
d
6.
c
16.
d
7.
c
17.
a
8.
c
18.
b
9.
a
19.
b
10.
b
20.
c
21.
22.
23.
24.
25.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
c
b
c
b
c
CS– 9
EXCERCISE BASED ON NEW PATTERN
1.
2.
COMPREHENSION TYPE
Comprehension-1
A 2.0 g sample of a mixture containing sodium
carbonate, sodium bicarbonate and sodium
sulphate is gently heated till the evolution of CO2
ceases. The volume of CO2 at 750 mm Hg pressure
and at 298 K is measured to be 123.9 ml. A 1.5 g of
the same sample requires 150 ml of M/10 HCl for
complete neutralisation.
The percentage composition of sodium carbonate
in the mixture is
(a)
42%
(b)
18%
(c)
53%
(d)
none
The percentage composition of sodium
bicarbonate in the mixture is
(a)
42%
(b)
18%
(c)
53%
(d)
none
7.
8.
(A)
(B)
(C)
3.
4.
5.
6.
The amount of HCl used for complete
neutralization of 2g of sample is
(a)
0.73
(b)
0.42
(c)
0.53
(d)
0.18
Comprehension-2
Reducing sugars are sometimes characterized by a
number RCu, which is defined as the number of mg
of copper reduced by 1 g of the sugar, in which the
half-reaction for the copper is
Cu2+ + OH—  Cu2O + H2O
(unbalanced)
It is sometimes more convenient to determine the
reducing power of a carbohydrate by an indirect
method. In this method 43.2 mg of the
carbohydrate was oxidized by an excess of
K3Fe(CN)6. The Fe(CN)64— formed in this reaction
required 5.29 cm3 of 0.0345 N Ce (SO 4) 2 for
reoxidation to Fe(CN)63— [the normality of the
cerium(IV) sulfate solution is given with respect to
the reduction of Ce4+ to Ce3+].
The number of milimole of Ce(SO 4) 2 used for
re-oxidation to Fe(CN)63– is
(a)
0.183
(b)
0.0915
(c)
1.83
(d)
9.14
43.2 mg of sugar was reduced by
(a)
11.01 mg Cu2+ (b)
11.6 mg Cu2+
2+
(c)
1.101 mg Cu
(d)
1.16 mg Cu2+
The RCu value for the sample is
(a)
269
(b)
2.69
(c)
26.9
(d)
0.269
Comprehension-3
An acid solution of a KReO4 sample containing
26.83 mg of combined rhenium was reduced by
passage through a column of granulated zinc. The
effluent, including was washing from the column,
was then titrated with 0.1000 N KMnO4; 11.45 mL
of the standard permanganate was required for the
reoxidation of all the rhenium to the perrhenate ion,
Einstein Classes,
(D)
(A)
(B)
(C)
(D)
1.
2.
ReO 4 —. Assuming that rhenium was the only
element reduced.
Number of equivalence of KMnO4 used
(a)
1.145
(b)
1.145 × 10–3
–3
(c)
11.45 × 10
(d)
3.25 × 10–3
What is the oxidation state to which rhenium was
reduced by the zinc column ?
(a)
0
(b)
–1
(c)
–2
(d)
–3
MATRIX-MATCH TYPE
Matching-1
Column - A
0.1M of MnO4–
in acidic medium
0.6 mol of KMnO4
in acidic medium
Molarity of pure water
(density of water = 1g/ml)
0.083 molar Cr2O72–
in acidic medium
Column - B
(P)
oxidised 0.25M
C2O42–
(Q)
oxidised 0.5M
Fe2+
(R)
oxidised
0.166M
FeC2O4
(S)
oxidises 1 mol
of Fe(C2O4)2
(T)
5.55 × 10
Matching-2
Column-A
Column-B
Total no. of electrons in (P)
3.01 × 1021
1.6 g methane are
No. of sulphate ions
(Q)
6.02 × 1023
present in 50mL of 0.1M
H2SO4 solution are
500cm3 of 0.2 molar NaCl (R)
6.02 × 1022
3
is added to 100cm of
0.5 molar AgNO3. Thus no.
of ions of AgCl formed are
The no. of of Fe2+ ions
(S)
3.01 × 1022
formed when excess of
iron is treated with 5 ml
of 0.04 N HCl
(T)
6.02 × 1020
MULTIPLE CORRECT CHOICE TYPE
Assuming complete dissociation of H 2 SO 4 as
(H2SO4  2H+ + SO42–), the number of sulphate ions
present in 50 ml of 0.1 M H2SO4 solution are
(a)
5 × 10–3 mol
(b)
3.01 × 1021
23
(c)
5 × 10
(d)
3.01 × 10–3 mol
In the following redox reaction 2MnO4– + 10Cl– +
16H+  2Mn2+ 5Cl2 + 8H2O. Pick up the correct
statements.
(a)
MnO4– is reduced
(b)
Cl– is oxidising agent
(c)
MnO4– is an oxidising agent
(d)
Cl– is reduced
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 10
3.
4.
5.
Which of the following represents redox reaction
(s) ?
(a)
Cu + Cu2+  2Cu+
(b)
MnO4– + Mn2+ + OH–  MnO2 + H2O
(c)
Cr2O72– + 2OH–  2CrO42– + H2O
(d)
2CrO42– + 2H+  Cr2O72– + H2O
Which of the following are redox reactions ?
(a)
Zn + 2HCl  ZnCl2 + H2
(b)
Al(OH)3 + 3HCl  AlCl3 + 3H2O
(c)
Disproportionation of Cu+ ions in a given
solution
(d)
Ag+(aq.) + I–(aq.)  Ag I (s)
Which of the following is disproportionation ?
(a)
2Cu+  Cu + Cu2+
(b)
3Cl2 + 6OH–  ClO3– + 5Cl– + 3H2O
(c)
2H2S + 8O2  2H2O + 3S
1
Cl  NaCl
2 2
Oxidation number of C is zero in
(a)
CHCl3
(b)
CH2Cl2
(c)
C6H12O6
(d)
CO
Among the species given below which can act as
oxidising as well as reducing agent ?
(a)
SO2
(b)
SO3
(c)
H2 O 2
(d)
H 2S
Which of the following changes involve oxidation ?
(a)
change of Zn to ZnSO4 by reaction with
H2SO4
(b)
change of Cl2 to chloride ion
(c)
change of H2S to S
(d)
change of sodium sulphite to sodium
sulphate.
(d)
6.
7.
8.
Na +
1.
2.
3.
4.
5.
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
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
STATEMENT-1 : 22.4 L of ethane at N.T.P.
contains one mole of hydrogen molecules.
STATEMENT-2 : One mole of hydrogen
molecules at N.T.P. occupies 22.4 L of volume.
STATEMENT-1 : The masses of oxygen which
combine with fixed mass of nitrogen in N2O, NO,
N2O3 bears a simple multiple ratio.
STATEMENT-2 : The combination according to
law of multiple proportions.
STATEMENT-1 : 8.075 × 10–2 kg of Glauber’s salt
is dissolved in water to obtain 1 dm3 of solution of
density 1077.2 kg m–3 . The molarity of the
resultant solution is 0.25 M .
STATEMENT-2 : The volume in mL of 0.5 M
H2SO4 needed to dissolve 0.5 g of copper (II)
carbonate is 24.3 mL
STATEMENT-1 : Basicity of an acid can change
in different reactions.
STATEMENT-2 : As the equivalent weight always
remains same.
STATEMENT-1 : If equal volume of C M KMnO4
and C M K2Cr2O7 solutions are allowed to oxidised
Fe2+ to Fe3+ in acidic medium, then Fe2+ oxidised
by KMnO 4 is more then K 2 Cr 2 O 7 of same
concentrations.
STATEMENT-2 : Number of moles of electrons
gained by 1 mole of K2Cr2O7 is more than the 1
mole of KMnO4
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
c
2.
a
7.
b
8.
b
3.
a
4.
b
5.
b
6.
a
[A-P, Q, R ; B-S ; C-T ; D-P]
2.
[A-Q ; B-P ; C-S ; D-R]
3.
a, b
4.
a, c
5.
a, b
6.
b, c
3.
B
4.
C
5.
D
MATRIX-MATCH TYPE
1.
MULTIPLE CORRECT CHOICE TYPE
1.
a, b
2.
a, c
7.
a, c
8.
a, c, d
ASSERTION-REASON TYPE
1.
D
2.
Einstein Classes,
A
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS– 11
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
2.68 × 10–3 mol of a solution containing an ion An+
require 1.61 × 10–3 mol of MnO4– for the oxidation
of An+ to AO3– in acid medium. What is the value of
n?
How much 1.00 M HCl should be mixed with what
volume of 0.250 M HCl in order to prepare 1.00 L
of 0.500 M HCl ?
Calculate the final concentration of HNO 3 if
0.20 mol HNO3 is added to a beaker containing
2.0 L of 1.1 M HNO3 and enough pure water is
added to give a final volume of 3.0 L.
If 40.00 mL of 1.600 M HCl and 60.00 mL of
2.000 M NaOH are mixed, what are the molar
concentrations of Na + , Cl – , and OH – in the
resulting solution ? Assume a total volume of
100.00 mL.
Calculate the molarity of the original H 3PO 4
solution if 20.0 mL of H3PO4 solution is required to
completely neutralize 40.0 mL of 0.0500 M Ba(OH)2
solution.
What volume of 96.0% H2SO4 solution (density 1.83
g/mL) is required to prepare 2.00 L of 3.00 M H2SO4
solution ?
How many mL of 0.5000 M KMnO4 solution will
react completely with 20.00 g of K 2C 2O 4.H 2O
according to the following equation ?
16H+ + 2MnO4– + 5C2O42–  10CO2 + 2Mn2+ + 8H2O
When 50.00 mL of a nitric acid solution was titrated
with 0.334 M NaOH, it requires 42.80 mL of the
base to achieve the equivalence point. What is the
molarity of the nitric acid solution ? What mass of
HNO3 was dissolved in 90.00 mL of solution ?
The acidic substance in vinegar is acetic acid
CH3COOH. When 6.00 g of a certain vinegar was
titrated with 0.100 M NaOH, 40.11 mL of base had
to be added to reach the equivalence point. What
percent by mass of this sample of vinegar is acetic
acid ?
Calculate the percent of BaO in 29.0 g of a mixture
of BaO and CaO which just reacts with 100.8 mL
of 6.00 M HCl. BaO + 2HCl  BaCl2 + H2O ;
CaO + 2HCl  CaCl2 + H2O
Calculate the normality of each of the following
solutions : (a) 7.88 g of HNO3 per L solution (b)
26.5 g of Na2CO3 per L solution (if acidified to form
CO2).
What volumes of 12.0 N and 3.00 N HCl must be
mixed to give 1.00 L of 6.00 N HCl ?
One gram of a mixture of CaCO3 and MgCO3 gives
240 ml of CO2 at N.T.P. Calculate the percentage
composition of mixture (Ca = 40, Mg = 24, C = 12,
O = 16).
Einstein Classes,
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
(a)
What volume of 5.00 N H2SO4 is required
to neutralize a solution containing
2.50 g NaOH ?
(b)
How many g pure H2SO4 are required ?
A 0.250 g sample of a solid acid was dissolved in
water and exactly neutralized by 40.0 mL of 0.125
N base. What is the equivalent weight of the acid ?
Exactly 50.0 mL of Na2CO3 solution is equivalent
to 56.3 mL of 0.102 N HCl in an acid-base
neutralization. How many g CaCO 3 would be
precipitated if an excess of CaCl2 solution were
added to 100 mL of this Na2CO3 solution ?
How many cm3 of concentrated sulfuric acid, of
density 1.84 g/cm3 and containing 98.0% H2SO4 by
weight, should be taken to make 1.00 L of normal
solution. Assume complete ionisation.
A 40.8 mL sample of an acid is equivalent to
50.0 mL of Na2CO3 solution, 25.0 mL of which is
equivalent to 23.8 mL of a 0.102 N HCl. What is the
normality of the first acid ?
Given the unbalanced equation
KMnO4 + KI + (H)2SO4  (K)2SO4 + MnSO4 + I2 +
H2 O
(a)
How many g KMnO4 are needed to make
500 mL 0.250 N solution ?
(b)
How many g KI are needed to make
25.0 mL 0.360 N solution ?
A solution contains 4 g of Na2CO 3 and NaCl in
250 ml. 25 ml of this solution required 50 ml of
N/10 HCl for complete neutralization. Calculate %
composition of mixture.
The density of a 2.0 M solution of acetic acid
(MW = 60) in water is 1.02g/mL. Calculate the mole
fraction of acetic acid.
The density of a 2.03 M solution of acetic acid in
water is 1.017 g/mL. Calculate the molality of the
solution.
Calculate the (a) molar concentration and (b)
molality of a sulfuric acid solution of density
1.198 g/cm3, containing 27.0% H2SO4 by weight.
A solution contains 57.5 mL ethyl alcohol (C2H5OH)
and 600 mL benzene (C6H6). How many g alcohol
are in 1000 g benzene ? What is the molality of the
solution ? Density of C2H5OH is 0.800 g/mL; of
C6H6, 0.900 g/mL.
One gram of an alloy of aluminium and
magnesium when treated with excess of dil. HCl
froms magnesium chloride, aluminium chloride and
hydrogen. The evolved hydrogen collected over
mercury at 0 0 C has a volume of 1.20 litres at
0.92 atm. pressure. Calculate the composition of the
alloy. [H = 1, Mg = 24, Al = 27].
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 12
26.
27.
28.
29.
30.
A gas mixture of 3.0 litres of propane and butane
on complete combustion at 250C produced 10 litres
of CO 2. Find out the composition of the gas
mixture.
In an Industrial process for producing acetic acid,
oxygen gas is bubbled into acetaldehyde CH3CHO,
containing manganese (II) acetate (catalyst) under
pressure at 600C.
2CH3CHO(l) + O2(g)  2CH3COOH(l)
In a laboratory test of this reaction, 20.0 g CH3CHO
and 10.0 g O2 were put into a reaction vessel. (a)
How many grams of acetic acid can be produced
by this reaction from these amounts of reactants ?
(b) How many grams of the excess reactant
remaining after the reaction is complete ?
The hourly energy requirements of an astronaut
can be satisfied by the energy released when
34 grams of sucrose are “burned” in his body. How
many grams of oxygen would he needed to be
carried in spaces capsule to meet his requirement
for one day ?
1.84 g of a mixture of CaCO3 and MgCO3 are heated
strongly till no further loss of weight takes place.
The residue weighs 0.96 g. Find the percentage
composition of the mixture.
(Mg = 24, Ca = 40, C = 12, O = 16)
1.0 g of a mixture of Potassium chloride and
Potassium iodide dissolved in water and
precipitated with Silver nitrate, gave 1.618 g of
silver halides. Calculate the percentage of each in
the mixture (K = 39, Cl = 35.5, Ag = 108, I = 127).
31.
32.
33.
20% surface sites have adsorbed N2. On heating N2
gas evolved from sites and were collected at
0.001 atm and 298 K in a container of volume is
2.46 cm 3 . Density of surface sites is
6.023 × 1014/cm2 and surface area is 1000 cm2, find
out the number of surface sites occupied per
molecule of N2.
A plant virus is found to consist of uniform
cylindrical particles of 150 Å in diameter and
5000 Å long. The specific volume of the virus is
0.75 cm3/g. If the virus is considered to be a single
particle, find its molar mass.
(a)
Calculate the amount of calcium oxide
required when it reacts with 852 g of
P4O10.
(b)
34.
35.
6CaO  P4O 10 
 2Ca 3 ( PO 4 ) 2
[At. wt. : Ca = 40, P = 31, O = 16]
How many milliliters of 0.5 M H2SO4 are
needed to dissolve 0.5 g of copper (II)
carbonate ?
CuCO 3  H 2 SO 4 
 CuSO 4  H 2 O  CO 2 
1.5 gm of an impure sample of (NH4)2SO4 was boiled
with excess of Caustic soda solution in a Kjeldahl’s
flask and the ammonia evolved was passed into 200
ml of semi-normal H2SO4 solution. Thepartially
neutralized acid was made to 500 ml with distilled
water. 25ml of this diluted acid required 40.8 ml of
decinormal Caustic soda for complete
neutralization. Calculate the percentage purity of
Ammonium sulphate.
A sample of a metal (M) carbonate was neutralized
by 10 ml of 0.1 N-hydrochloric acid and the
resulting chloride gave 0.0517 g of phosphate
[M3(PO4)2]. Calculate the eq. wt. of M. (the formula
of Phosphoric acid is H3PO4). Determine the atomic
weight of M ?
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
One litre of a mixture of O2 and O3 at STP was
allowed to react with excess of acidified solution of
KI. The iodine liberated require 40mL of M/10
sodium thiosulfate solution for titration. What is
the mass percentage of ozone in mixture.
3.
A mixture of FeO and Fe3O4 when heated in air to a
constant weight gains 5 % in its weight. Find the
composition of initial mixture. (Fe = 56, O = 16)
4.
A 10.0 g sample of “gas liquor” is boiled with an
excess of NaOH, and the resulting ammonia is
KI
passed into 60 cm3 if 0.90 N H2SO4. Exactly 10.0
2
O 3  I 2  O 2  O
cm3 of 0.40 N NaOH is required to neutralize the
What volume of 3.00 M HNO3 can react completely
excess sulfuric acid (not neutralized by the NH3).
with 15.0 g of a brass (90.0% Cu, 10.0% Zn)
Determine the percent ammonia in the “gas liquor”
according to the following equations ?
examined.
Cu + 4H+(aq) + 2NO3–(aq)  2NO2(g) + Cu2+ + 2H2O
5.
A 10.0 mL portion of (NH4)2SO4 solution was treated
4Zn + 10H+(aq) + NO3–(aq)  NH2+ + 4Zn2+ + 3H2O
with excess NaOH. The NH 3 gas evolved was
What volume of NO2 gas at 250C and 1.00 atm
absorbed in 50.00 mL of 0.1000 N HCl. To
pressure would be produced ?
neutralize the remaining HCl, 21.50 mL of 0.0980
N NaOH was required. What is the molar
concentration of the (NH4) 2SO 4 ? How many g
(NH4)2 SO4 are in 1 L solution ?
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS– 13
6.
7.
8.
9.
10.
11.
12.
13.
A solution contained the mixture of Na2CO3 and
NaCl. 25 ml of the solution required 20.4 ml of 0.095
N-HCl to convert the carbonate into the chloride.
25 ml of the later solution (containing chloride
converted from carbonate) required 38.76 ml of
N/10-AgNO 3 to ppt, the chloride completely.
Calculate the strength of Na2CO3 and NaCl in one
litre of the original solution.
10 g of a mixture of Cu2S and CuS was treated with
200 mL of 0.75 M MnO 4 – in acidic solution
producting SO2, Cu2+ and Mn2+ . SO2 was bubbled
off and excess of MnO4– was titrated with 175 mL
of 1.00 M Fe2+. Calculate the CuS in the mixture.
Determine the volume of dilute nitric acid (density
1.11 g/mL, 19.0% HNO3 by weight) that can be
prepared by diluting with water 50 mL of the
concentrated acid (density 1.42 g/mL, 69.8% HNO3
by weight). Calculate the molar concentrations and
molalities of the concentrated and dilute acids.
A 1.00 g sample of KClO3 was heated under such
conditions that a part of it decomposed according
to the equation
(i) 2KClO3 = 2KCl + 3O2
and the remaining underwent change according to
the equation
(ii) 4KClO3 = 2KClO4 + KCl
If the amount of oxygen evolved was 146.8 ml at
S.T.P., calculate percentage by weight of KClO4 in
the residue (K = 39.1, Cl = 35.5)
To a 25 ml H 2 O 2 solution, excess of acidified
solution of potassium iodide was added. The iodine
liberated required 20 ml of 0.3 N sodium
thiosulphate solution. Calculate the volume strength
of H2O2 solution.
A solution contains Na2CO3 and NaHCO3. 20 cm3
of this solution requires 5.0 cm3 of 0.1 M H2SO4
solution for neutralization using phenophthalein as
the indicator. Methyl orange is then added and
further 5.0 cm3 of 0.2 M H2SO4 was required.
Calculate the masses of Na2CO3 and NaHCO3 in
1 L of this solution.
A 1.2 g of a mixture containing H2C2O4.2H2O and
KHC2O4.H 2O and impurities of a neutral salt,
consumed 18.9 ml of 0.5 N NaOH for
neutralization. On titrating with KMnO4 solution
0.4 g of the same substance needed 21.55 ml of 0.25
N KMnO4. Calculate the percentage composition
of the substance.
A mixture of FeO and Fe 2 O 3 is reacted with
acidified KMnO4 solution having a concentration
of 0.2278 M, 100 ml of which was used. The
solution then was treated with Zn dust which
converted the Fe3+ of the solution to Fe2+ . The Fe2+
required 1000 ml of 0.13 M K2Cr2O7 solution. Find
the mass % of FeO and Fe2O3.
Einstein Classes,
14.
15.
16.
17.
18.
19.
20.
10 g of a mixture of anhydrous nitrates of two metals A and B were heated to a constant weight and
gave 5.531 g of a mixture of the corresponding oxides. The equivalent weights of A and B are 103.6
and 31.8 respectively. What was the percentage of
A in the mixture ? (N = 14, O = 16).
Equal weights of mercury and iodine are allowed
to react completely to form a mixture of mercurous
and mercuric iodides. Calculate the ratio of the
weights of mercurous and mercuric iodides formed.
(I = 127, Hg = 201).
5 ml of 8 N nitric acid, 4.8 ml of 5 N HCl acid and a
certain volume of 17 M sulphuric acid are mixed
together and made up to 2 litre. Thirty ml of this
acid mixture exactly neutralise 42.9 ml of sodium
carbonate solution containing one gram of Na2CO3.
10H2O in 100 ml of water. Calculate the mass in
gram of the sulphate ions in solution.
200 ml of a solution of mixture of NaOH and
Na2CO3 was first titrated with phenolphthalein and
N/10 HCl. 17.5 ml of HCl was required for the end
point. After this methyl orange was added and 2.5
ml of same HCl was again required for next end
point. Find out amounts of NaOH and Na2CO3 in
mixture.
50 ml of solution, containing 1g each of Na2CO3,
NaHCO3 and NaOH, was titrated with N HCl.
What will be the titre readings if
(a)
only phenolphthalien is used as
indicator ?
(b)
only methyl orange is used as indicator
from the very beginning ?
(c)
methyl orange is added after the first end
point with phenolpthalein ?
A mixture of calcium carbonate and sodium
chloride weighing 3.20 g was added to 100 ml of
1.02 N HCl. After the reaction had ceased the
liquid was filtered, the residue washed and the
filtrate was made up to 200 ml. 20 ml of this dilute
solution required 25 ml of N/5 NaOH for
neutralization. Calculate the % of CaCO3 in the
mixture. (Ca = 40).
A solution contains a mixture of sulphuric acid and
oxalic acid, 25 ml of the solution require 35.5 ml of
N/10-NaOH for neutralization and 23.45 ml of
N/10 - KMnO4 for oxidation. Calculate (a) the
Normality of the solution with regard to sulphuric
acid and oxalic acid (b) the number of g of each of
these substances present in one litre of the solution.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 14
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
4.
6.
9.
11.
13.
14.
15.
18.
20.
22.
23.
n=2
2.
333 mL of 1m HCl
0.56 M OH–, 0.640 Cl–, 1.2MNa+
335 mL
7.
86.9 mL
4.01 %
10.
65.5 % BaO
(a)
0.1251 N
(b)
0.5 N
CaCO3 = 62.5 %, MgCO3 = 37.5%
(a)
12.5 mL
(b)
3.07 gm
50g/eq
16.
0.575 gm
0.119 N
19.
(a)
3.95 gm
Na2CO3 = 66.25%, NaCl = 33.75%
2.27 m
(a)
3.30 m
(b)
3.78 m
3.
5.
8.
0.80 m
0.0667 M
1.62 gm, 0.286 M
12.
.33 L, .667 L
17.
(b)
21.
27.2 cm3
1.49 gm
0.038
24.
0.54 kg C6H6, 1.85 m
25.
26.
Al% = 54.87%, Mg% = 45.13%
66.66% propane, 33.33 % butane
27.
(a)
(b)
28.
29.
916.2 gm
CaCO3 = 54.35%, MgCO3 = 45.65 %
30.
31.
34.
2
80.96
KCl = 39.6 %
KI = 60.4%
(a) 1008 g (b) 8.097 mL
32.
35.
70.96 × 106 g/mol
20.03, 40.06
33.
27.24 gm
2.73 gm
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
6.57 %
3.
% of FeO = 20.25%
% of Fe3O4 = 79.75 %
2.
VHNO3  302L, VNO 2  10.4L
4.
8.5%
5.
0.145 M, 19.1 g/L
6.
Na2CO3 = 4.109 gm, NaCl = 4.535 gm
7.
CuS = 57.4 %
8.
15.7 M, 3.35 M, 36.8 m, 3.73 m
9.
49.83%
10.
1.344
12.
KHC2O4.H2O = 80.9 %, H2C2O4.2H2O = 14.7 %
13.
% FeO = 13.34 %, % Fe2O3 = 86.66 %
15.
0.513 : 1
11.
Na2CO3 = 5.3 gm, NaHCO3 = 4.2 gm
16.
14.
6.528 gm
17.
32.28
WNaOH = 0.06gm
WNa 2CO3  0.0265 gm
18.
(a)
19.
81.25%
34.4 mL
(b)
55.8 mL
(c)
21.3 mL
20.
H2SO4 = 0.0482N, 2.36 g/L
H2C2O4 = 0.0938N, 4.225 g/L
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111