Download Stoichiometry Notes

CS – 1
S TO I C H I O M E T RY
Syllabus :
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The concept of atoms and molecules; Dalton’s
atomic theory; Mole concept; Chemical formulae;
Balanced chemical equations; Calculations (based
on mole concept) involving common oxidationreduction, neutralization, and displacement
reactions; Concentration in terms of mole fraction,
molarity, molality and normality.
CONCEPTS
C1
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
C2A 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.
Some basic formulas and definitions used in mole concept are as follows :
Mole is the unit representing 6.022 × 1023 numbers (atoms, molecules or ions). A mole also represents gram
molecular mass or gram atomic mass.
Masses of atoms, molecules and nuclear particles are expressed in a specially defined unit, called atomic
mass unit (u or amu). One amu =
1
6.02  10 23
g / atom .
Molar mass of a substance is the mass of 6.02 × 1023 particles present in it.
An empirical formula represents the simplest whole number ratio of various atoms present in a
compound, whereas the molecular formula shows the exact number of different types of atoms present in a
molecule of a compound.
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) =
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RT
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CS – 2
At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure. Mole concept is based
on balanced chemical reaction. If amounts of both reactants are given then find the limiting reagent. It is
that reagent which is consumed completely in a irreversible chemical reaction.
Class Discussion Problems :
1.
Hydrogen reacts with nitrogen to produce ammonia according to this equation :
3H2(g) + N2(g)  2NH3(g)
Determine how much ammonia would be produced when 200.0 g of hydrogen react.
2.
If 20.0 g of CaCO3 is treated with 20.0 g of HCl, how many grams of CO2 can be generated according
to the following equation :
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
Given : Molar mass of (CaCO3 = 100 g mol–1, HCl = 36.5 g mol–1, CO2 = 44.0 g mol–1)
Zinc and hydrochloric acid react according to the reaction :
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3.
Zn(s) + 2HCl (aq)  ZnCl2(aq) + H2(g)
If 0.30 mol Zn are added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are
produced ?
4.
A compound contains 4.07% hydrogen, 22.47% carbon and 71.65% chlorine. Its molar mass is
98.96 g. What are its empirical and molecular formulas ? [At. wt. : H = 1, C = 12, Cl = 35.5]
5.
Find the number of atom in 1g of 12C.
Home Work Problems :
In a reaction A + B2  AB2. Identify the limiting reagent, if any, in the following reaction mixtures.
6.
7.
(i)
300 atoms of A + 200 molecules of B
(ii)
2 mol A + 3 mol B
(iii)
100 atoms of A + 100 molecules of B
(iv)
5 mol A + 2.5 mol B
(v)
2.5 mol A + 5 mol B
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following
chemical equation : N2(g) + 3H2(g)  2NH3(g). [At. wt. : N = 14, H = 1]
8.
(i)
Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g
of dihydrogen.
(ii)
Will any of the two reactants remain unreacted ?
(iii)
If yes, which one and what would be the mass ?
Which one of the following will have largest number of atoms ?
(i)
1 g Au(s)
(ii)
1 g Na(s)
(iii)
1 g Li(s)
(iv)
1 g of Cl2(g)
[At. Wt. Au = 197, Na = 23, Li = 7, Cl = 35.5]
[Answers : (1) 1132.2 g (2) 8.80 g of CO2 (3) 0.26 mol (4) CH2Cl, C2H4Cl2 (5) 5.016 × 1022 (6) (i) B is the
limiting reagent (ii) A is the limiting reagent (iii) No limiting reagent (iv) B is the limiting reagent
(v) A is limiting reagent (7) (i) 2428.57 g NH3 (ii) H2 will remain unreacted (iii) 571.4 g (8) 1 g of Li has
the largest number of atoms]
C3
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
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CS – 3
Normality ( N ) 
No. of gramequiva lents of solute , unit of normality are g-eq./lit., N or normal.
Vol. of solution in L
No. of gram equivalents of solute(neq ) 
Equivalent weight 
Molality (m ) 
Weight of solute
Equivalent weight
Molecular weight (or ) Atomic weight (or ) Ionic weight
n factor
No. of moles of solute
, unit of molality are mol/kg, m or molal.
wt . of solvent
nA .
n A  nB
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Mole fraction( x A ) 
ppm 
mass of solute
 10 6
mass of solution
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.
Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Or
number of electron lost or gained from one mole of the compound.
Class Discussion Problems :
1.
If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M
solution ?
2.
A sample of drinking water was found to be severely contaminated with chloroform, CHCl3
supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
3.
(i)
Express this in per cent by mass.
(ii)
Determine the molality of chloroform in the water sample.
Commercially available concentrated hydrochloric acid contains 38% HCl by mass.
(a)
What is the molarity of this solution ? The density is 1.19 g mL–1.
(b)
What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCl ?
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CS – 4
4.
500 mL of 0.250 M Na2SO4 solution is added to an aqueous solution of 15.00 g BaCl2 resulting in the
formation of the white precipitate of insoluble BaSO4. How many moles and how many grams of
BaSO4 are formed ? (At. wt. Ba = 137, S = 32, Na = 23, O = 16, Cl = 35.5)
5.
Calculate the volume of 1.00 mol L–1 aqueous sodium hydroxide that is neutralised by 200 mL of
2.00 mol L–1 aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralisation
reaction is : NaOH(aq) + HCl (aq)  NaCl (aq) + H2O(l)
6.
What volume of 0.30 M Na2SO4 solution is required to prepare 2.0 L of a solution 0.40 M in Na+ ?
Home Work Problems :
7.
What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to
make a final volume up to 2 L ?
8.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction
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CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O (l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl ?
The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate the molality of the solution.
9.
[Answers : (1) 25.22 mL (2) (i) 1.5 × 10–3% (ii) 1.255 × 10–4 m (3) (a) 12.38 M (b) 807.7 mL
(4) 0.072 mol BaCl2, 16.80 g BaSO4 (5) 400 mL, 23.40 g (6) 1.3 L (7) 0.0292 mol L–1 (8) 0.94 g CaCO3
(9) 2.79 m]
C4
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 volumetric 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 oxidising agent and reducing agent., 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
(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 difficult to solve the problem through equivalence concept then use the mole concept.
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CS – 5
Back titration :
This is a method in which a substance is taken in excess and some part of it 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 phenolphthalein indicator following
reaction takes place at the phenolphthalein end point –
NaOH + HCl  NaCl + H2O
Na2CO3 + HCl  NaHCO3 + H2O
1
eq. of Na 2 CO 3  eq. of HCl
2
(n  2)
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Here, eq. of NaOH 
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
Redox titration :
Class Discussion Problems :
1.
What is the normality of 0.300 M H3PO3 when it undergoes the following reaction ?
H3PO3 + 2OH–  HPO32– + 2H2O
2.
What volumes of 12.0 N and 3.00 N HCl must be mixed to give 1.00 L of 6.00 N HCl ?
3.
Determine the normality of an H3PO4 solution, 40.0 mL of which neutralized 120 mL of 0.531 N
NaOH.
4.
A 48.4 mL sample of HCl solution requires 1.240 g of pure CaCO3 for complete neutralization.
Calculate the normality of the acid. (Molecular wt. of CaCO3 = 100 g/mol)
5.
In standardizing HCl, 22.5 mL was required to neutralize 25.0 mL of 0.100 N Na2CO3 solution. What
is the normality of the HCl solution ? How much water must be added to 200 mL of it to make it
0.100 N ? Neglect volume changes.
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CS – 6
6.
Given the unbalanced equation
KMnO4 + KI + H2SO4  K2SO4 + MnSO4 + I2 + H2O
(a) How many g KMnO4 are needed to make 500 mL of 0.250 N solution ? (b) How many g KI are
needed to make 25.0 mL of 0.360 N solution ?
(At. wt. K = 39, Mn = 55, O = 16, I = 127)
7.
Determine the mass of KMnO4 required to make 80.0 mL of N/8 KMnO4 when the latter acts as an
oxidizing agent in acid solution and Mn2+ is a product of the reaction.
Home Work Problems :
8.
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).
[At. wt. C = 12, N = 14, O = 16, Na = 23]
(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 ? [Molecular wt. of NaOH = 40, H2SO4 = 98]
10.
Exactly 50.0 mL of a solution of Na2CO3 was titrated with 65.8 mL of 3.00 N HCl.
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9.
If the density of the Na2CO3 solution is 1.25 g/mL, what percent Na2CO3 by weight does it contain ?
(At. wt. Na = 23, C = 12, O = 16)
11.
Calculate the mass of oxalic acid, H2C2O4, which can be oxidized to CO2 by 100.0 mL of an MnO4–
solution 10.0 mL of which is capable of oxidizing 50.0 mL of 1.00 N I– to I2.
(At. wt. H = 1, C = 12, O = 16)
12.
Find the equivalent weight of KMnO4 in the reaction Mn2+ + MnO4– + H 2O  MnO 2 + H +
(unbalanced). How many g MnSO4 is oxidized by 1.25 g KMnO4 ? (molar mass MnSO4 = 151 g)
[Answers : (1) 0.600 N (2) 0.333 L of conc. solution, 0.667 L of dil. solution (3) 1.59 N (4) 0.512 N
(5) 0.111 N, 22 mL (6) (a) 3.95 g (b) 1.49 g KI (7) 0.316 g (8) 0.1251 N, 0.500 N (9) (a) 12.5 mL
(b) 3.07 g (10) 16.8% Na2CO3 (11) 22.5 g (12) 1.79 g MnSO4]
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
Class Discussion Problems :
1.
Calculate the strength of ‘20 V’ of H2O2 in terms of normality
2.
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 : (1) 3.58 N (2) 0.68 g/litre]
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CS – 7
Electron exchanged Eq. wt.
or change in O.N.
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
neutral medium
MnO4— + e— + 2H2O  MnO22—
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.
Cl—
Cl2 + 2e—  2Cl—
2
E
Cu+
Cu2+ + e–  Cu+
1
E
8.
Cl2(O.A.)
(in bleaching
powder)
CuSO4 (O.A.)
(in iodometric
titration)
S2O32— (R.A.)
S4O62—
2S2O32—  S4O62— + 2e—
9.
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—
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Species
7.
1.
2 (for two molecules) E 
3.
E
1
Reaction
Relation between O.A. and R.A.
I2
I2 + 2Na2S2O3  2NaI + Na2S4O62—
I2 + 2S2O32—  2I— + S4O62—
I2  2I—  2Na2S2O3
4.
5.
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

MnO2 + 4HCl (conc.) 
MnCl2 +

Cl2 + 2H2O
Cl2 + 2KI  2KCl + I2
or MnO2 + 4H+ + 2Cl—  Mn2+ + 2H2O
+ Cl2
Cl2 + 2I—  I2 + 2Cl—
IO3— + 5I— + 6H+  3I2 + 3H2O
Eq. wt. of CuSO4 =
M
1
M
1
2CuSO4 + 4KI  Cu2I2 + 2K2SO4 + I2
or 2Cu2+ + 4I—  Cu2I2 + I2
white ppt.
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
2
M
E
2
Estimation of
CuSO4
M
1
E
Eq. wt. (Na2S2O3) = E 
2.
M
5
M
3
M
1
M
6
M
2
M
2
M
6
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18
Ph. : 9312629035, 8527112111, E-mail [email protected],
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CS – 8
6.
H2 O 2
H2O2 + 2I— + 2H+  I2 + 2H2O
H2O2  I2  2I—  2Na2S2O3
Eq. wt. H2O2 =
7.
Cl2
Cl2 + 2I—  2Cl— + I2
M
2
Cl2  I2  2I—  2Na2S2O3
M
2
Eq. wt. of Cl2 =
8.
O3
O3 + 6I— + 6H+  3I2 + 3H2O + O2
O3  3I2
M
2
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Eq. wt. of O3 =
9.
ClO—
ClO— + 2I— + 2H+  H2O + Cl— + I2
ClO—  I2  2I  2Na2S2O3
Eq. wt. of ClO— =
10.
Cr2O72—
Cr2O72— + 14H+ + 6I—  3I2 + 2Cr3+ +
7H2O
Cr2O72—  3I2  6I—
Eq. wt. of Cr2O72— =
11.
Fe2+
Fe2+  Fe3+ + e—
MnO + 8H+ + 5e—  Mn2+ +
4H2O
M
2
M
6
5Fe2+ = MnO4—
—
4
Eq. Wt. Fe2+ =
M
1
Eq. Wt. MnO4— =
12.
Fe2+
Cr2O
13.
C2O42—
2—
7
Fe2+  Fe3+ + e—
+ 14H+ + 6e—  2Cr3+ +
7H2O
C2O42—  2CO2 + 2e—
MnO4— + 8H+ + 5e  Mn2+ +
4H2O
M
5
6Fe2+  Cr2O72—
Eq. Wt. Cr2O72— =
M
6
5C2O42—  2MnO4—
Eq. wt. C2O42— =
M
2
Eq. wt. MnO4— = M/5
14.
H2 O 2
H2O2  2H+ + O2 + 2e—
5e– + MnO4—  Mn2+
5H2O2  2MnO4—
Eq. wt. H2O2 =
M
2
M
5
M
Eq. wt. As2O3 =
4
Eq. wt. MnO4— =
15.
As2O3
Einstein Classes,
As2O3 + 5H2O  2AsO43— + 10H+
+ 4e—
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CS – 9
EXERCISE
TYPE – 1 (SINGLE CORRECT CHOICE TYPE)
1.
(a)
50
(b)
30
(c)
20
(d)
None
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
11.
The normality of an H3PO4 solution 40.0 mL of
which neutralized 120 mL of 0.531 N NaOH is
(a)
1.00 N
(b)
1.59 N
(c)
3.00 N
(d)
5.02 N
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
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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
10.
3.
4.
5.
6.
7.
8.
(a)
0.70 mol
(b)
0.50 mol
(c)
0.20 mol
(d)
0.10 mol
12.
The equivalent mass of MnSO4 is half of its molar
mass when it is converted to
(a)
Mn2O3
(c)
–
(b)
MnO2
23
(b)
12
(c)
24
(d)
13
(a)
11.2 litres
(b)
22.4 litres
(c)
44.8 litres
(d)
56.0 litres
(a)
2.5
(b)
1
(c)
0.5
(d)
0.25
(d)
MnO
2–
4
When one gram mol of KMnO4 reacts with HCl,
the volume of chlorine liberated at NTP will be
34 g of hydrogen peroxide is present in 1120 ml of
solution. This solution is called
The number of moles of KMnO4 that will be needed
to react completely with one mole of ferrous
oxalate in acidic solution is
A solution contains Na2CO3 and NaHCO3. 10 ml of
the solution required 2.5 ml of 0.1 M H2SO4 for
neutralisation using phenolphthalein as indicator.
Methyl orange in then added when a further 2.5 ml
of 0.2 M H 2SO 4 was required. The amount of
Na2CO3 and NaHCO3 in 1 litre of the solution is
(a)
3/5
(b)
2/5
(a)
5.3 and 4.2 g
(b)
3.3 g and 6.2 g
(c)
4/5
(d)
1
(c)
4.2 g and 5.3 g
(d)
6.2 g and 3.3 g
(a)
10 vol solution (b)
20 vol solution
(c)
30 vol solution (d)
32 vol solution
14.
(a)
0.135 M
(b)
0.141 M
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
(c)
0.210 M
(d)
0.246 M
(a)
30.6%
(b)
40.6%
(c)
20.6%
(d)
50%
[Na+] in a solution prepared by mixing 30.00 mL of
0.12 M NaCl with 70 mL of 0.15 M Na2SO4 is
10 mL of H2O2 solution
15.
(volume strength = x)
N
required 10 mL of
MnO4– solution in acidic
0.56
16.
medium. Hence x is
9.
(a)
Hydrogen peroxide in aqueous solution decomposes
on warming to give oxygen according to the
equation 2H2O 2(aq.)  2H2O(l) + O2(g) under
conditions where one mole of gas occupies 24 dm3,
100 cm3 of X M solution of H2O2 produces 3 dm3 of
O2 X is thus
MnO4
13.
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)
0.56
(b)
5.6
(c)
0.1
(d)
10.0
Molality of 18 M H2SO4 (d = 1.8 gmL–1) is
(a)
36 mol kg–1
(b)
200 mol kg–1
(c)
500 mol kg–1
(d)
18 mol kg–1
Einstein Classes,
17.
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
(b)
3
(c)
2
(d)
5
100 mL of 1 M KMnO4 oxidised 100 mL of H2O2 in
acidic medium (when MnO4– 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
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CS – 10
(a)
(c)
(b)
300
mL
3
(d)
500
mL
3
The no. of of Fe2+ ions
(S)
3.01 × 1022
(T)
6.02 × 1019
formed when excess of
iron is treated with 5 ml
of 0.04 N HCl
100 mL
For a given mixture of NaHCO 3 and Na 2CO 3,
volume of given HCl required is x mL with
phenolphthalein indicator and further y mL
required with methyl orange indicator. Hence
volume of HCl for complete reaction of NaHCO3 is
(a)
2x
(b)
(c)
y
(d)
TYPE-4
(MULTIPLE CORRECT CHOICE TYPE)
1.
Assuming complete dissociation of H 2 SO 4 as
(H2SO4  2H+ + SO42–), the number of sulphate ions
present in 50 ml of 0.1 N H2SO4 solution are
(a)
x
2
15.05 × 1020
(b)
1.505 × 1021
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18.
100
mL
3
(D)
19.
2.
(y – x)
1 mol of ferric oxalate is oxidised by x mol of MnO4–
and also 1 mol of ferrous oxalate is oxidised by y
mol of MnO4– in acidic medium. The ratio
20.
(c)
(a)
2:1
(b)
1:2
(c)
3:1
(d)
1:3
x
is
y
3.
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)
1
3
(b)
2
3
(c)
2
5
(d)
3
5
4.
TYPE-2 (COMPREHENSION TYPE)
5.
1.
2.
3.
TYPE-3 (MATRIX-MATCH TYPE)
(A)
Column-A
Column-B
Total no. of electrons in
(P)
3.01 × 1021
(Q)
6.02 × 1023
No. of sulphate ions
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
Which of the following is disproportionation ?
(a)
2Cu+  Cu + Cu2+
(b)
3Cl2 + 6OH–  ClO3– + 5Cl– + 3H2O
(c)
2H2S + 8O2  2H2O + 3S
(d)
Na +
1
Cl  NaCl
2 2
Oxidation number of C is zero in
(a)
CHCl3
(b)
CH2Cl2
(c)
C6H12O 6
(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
(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
H2SO4 solution are
500cm3 of 0.2 molar NaCl (R)
3.01 × 10–3 mol
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.
present in 50mL of 0.1M
(C)
(d)
TYPE-5 (Assertion-Reason Type)
1.6 g methane are
(B)
5 × 1023
6.02 × 1022
is added to 100cm3 of
0.5 molar AgNO3. Thus no.
of ions of AgCl formed are
1.
STATEMENT-1 : 22.4 L of ethane at N.T.P.
contains one mole of hydrogen molecules.
STATEMENT-2 : One mole of hydrogen
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CS – 11
molecules at N.T.P. occupies 22.4 L of volume.
2.
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.
3.
6.
A 10.0 mL portion of (NH4)2SO4 solution was treated
with excess NaOH. The NH 3 gas evolved was
absorbed in 50.00 mL of 0.1000 N HCl. To
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 ?
7.
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.
8.
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.
9.
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.
10.
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).
STATEMENT-1 : 8.075 × 10–2 kg of Glauber’s salt
is dissolved in water to obtain 1 dm3 of solution .
The molarity of the resultant solution is 0.0025 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 (At. Wt of Cu = 63.5)
STATEMENT-1 : Basicity of an acid can change in
different reactions.
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4.
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 ?
STATEMENT-2 : As the equivalent weight always
remains same.
5.
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
TYPE – 6 (SUBJECTIVE TYPE PROBLEMS)
1.
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.
2.
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 ?
3.
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).
4.
A solution contains 4 g of Na2CO3 and NaCl in
250 ml. 25 ml of this solution required 50 ml of
N/10 HCl for complete neutralization. Calculate %
composition of mixture.
5.
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
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ANSWERS
TYPE – 1
a
d
b
d
a
a
d
d
c
b
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
c
a
a
a
c
c
b
d
a
a
8.
(a)
27.24 gm
(b)
0.145 M, 19.1 g/L
KHC2O4.H2O = 80.9 %, H2C2O4.2H2O =
14.7 %
% FeO = 13.34 %, % Fe2O3 = 86.66 %
9.
10.
6.528 gm
81.25%
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1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
5.
6.
7.
CS – 12
2.73 gm
TYPE – 2
1.
2.
3.
TYPE – 3
A-R ; B-P ; C-S ; D-T
TYPE – 4
1.
2.
3.
4.
5.
a, b
a, c
a, b
b, c
a, c
TYPE – 5
1.
2.
3.
4.
5.
D
A
B
C
D
TYPE – 6
1.
2.
3.
4.
0.56 M OH–, 0.640 Cl–, 1.2MNa+
4.01 %
CaCO3 = 62.5 %, MgCO3 = 37.5%
Na2CO3 = 66.25%, NaCl = 33.75%
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18
Ph. : 9312629035, 8527112111, E-mail [email protected],
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