This document provides an overview of key chemistry concepts for exams. It defines matter and its three states. Compounds have a constant composition while mixtures have a variable composition. It also defines valence and provides formulas for common cations and anions. Additional concepts covered include the mole, atomic and molecular weights, gas laws, empirical and molecular formulas, and several important reaction types. Concentration terms like density, specific gravity, and percentages are also defined.

1. 1
SOME BASIC CONCEPTS OF CHEMISTRY
KEY CONCEPTS FOR JEE (MAIN) - NEET
1. Matter
Matter may be defined as anything which has mass
and occupies space. e.g. water, air, milk, salt, sand,
oxygen etc.
Matter may be classified into three states namely
solid, liquid and gas. At macroscopic level matter
can be classified as mixture or pure substance as :
Mixture Pure Substance
Matter
Heterogeneous Homogeneous
mixture
Element Compound
2. Valancy and formula of some
radicals (Cations and anions)
(a) CATIONS :
VALANCY – 1
Ammonium NH4
+
Sodium Na+
Potassium K+
Rubidium Rb+
Cesium Cs+
Silver Ag+
Copper (Cuprous) Cu+
Gold (Aurous) Au+
VALANCY – 2
Magnesium Mg2+
Calcium Ca2+
Stroncium Sr2+
Barium Ba2+
Zinc Zn2+
Cadmium Cd2+
Nickel Ni2+
Copper (Cupric) Cu2+
Mercury (Mercuric) Hg2+
Lead (Plumbus) Pb2+
Tin (Stannous) Sn2+
Iron (Ferrous) Fe2+
VALANCY – 3
Iron (Ferric) Fe3+
Aluminium Al3+
Chromium Cr3+
Gold (Auric) Au3+
(b) ANIONS :
VALANCY – 1
Hydroxide OH¯
Nitrate NO3¯
Nitrite NO2¯
Permanganate MnO4¯
Bisulphite HSO3¯
Bisulphate HSO4¯
Bicarbonate
(Hydrogen carbonate)
HCO3¯
Dihydrogen phosphate H2PO4¯
Perchlorate ClO4¯
Chlorate ClO3¯
Chlorite ClO2
–
Hypochlorite ClO¯
Iodate IO3¯
Periodate IO4¯
Meta aluminate AlO2¯
Meta borate BO2¯
Cyanide [ CN ]–
••••CN–
Isocyanide [–
N 

C ]••••NC–
Cyanate [–
O–CN]••••
••
••CNO–

2. 2
Isocyanate [O=C=N –
]
••
••
Thiocyanate SCN–
Formate HCOO–
Perhydroxyl ion HOO–
Hypophosphites H2PO2
–
Benzoate C6H5COO–
Salicylates C6H4(OH)COO–
Acetate CH3COO–
Metaphosphate PO3
–
VALANCY – 2
Carbonate CO3
2–
Sulphate SO4
2–
Sulphite SO3
2–
Sulphide S2–
Thiosulphate S2O3
2–
Tetrathionate S4O6
2–
Oxalate C2O4
2–
Silicate SiO3
2–
Hydrogen phosphate HPO4
2–
Manganate MnO4
2–
Chromate CrO4
2–
Dichromate Cr2O7
2–
Zincate ZnO2
2–
Stannate SnO3
2–
Hexaflurosilicate
(or silicofluorides)
SiF6
2–
Tartrates C4H4O6
2–
Phosphite HPO3
2–
Chromate CrO4
2–
Pyroborate B4O7
2–
Dithionite S2O4
2–
Peroxodisulphate S2O8
2–
Silicate SiO3
2–
Succinate C4H4O4
2–
VALANCY – 3
Hexacyano ferrate (III) or
Ferricynide
[Fe(CN)6]3–
Phosphate PO4
3–
Borate (orthoborate) BO3
3–
Arsenate AsO4
3–
Arsenite AsO3
3–
Nitride N3–
Phosphide P3–
VALANCY – 4
Pyrophosphate P2O7
4–
Hexacyano ferrate (II) or
Ferrocyanide
[Fe(CN)6]4–
3. Formula of simple compounds
Sr.
no.
Name of
Compound
Symbols with
valancy
Formula
1. Calcium chloride Ca2 Cl1 CaCl2
2.
Magnesium
sulphate
Mg2 SO4
2
Mg2(SO4)2 ~
MgSO4
(Simple
Ratio)
3. Stannic Sulphide Sn4
S2
SnS2
4.
Potassium
perchlorate
1
4
1
ClOK KClO4
5. Sodium Zincate Na1 2
2ZnO Na2ZnO2
6.
Magnesium
bicarbonate
1
3
2
HCOMg Mg(HCO3)2
7.
Sodium
carbonate
2
3
1
CONa Na2CO3
8.
Ammonium
Oxalate
2
42
1
4 OCNH (NH4)2C2O4
9.
Sodium
thiosulphate
2
32
1
OSNa Na2S2O3
10.
Potassium
permanganate
1
4
1
MnOK KMnO4
11. Sodium Iodate 1
3
1
IONa NaIO3
12.
Sodium
periodate
1
4
1
IONa NaIO4
4. Laws of chemical combination
(i) Law of conservation of mass - [Lavoisier, 1744]
Matter is neither created nor destroyed in the
course of chemical reaction although it may
change from one form to other

3. 3
(ii) Law of definite proportion [Proust, 1799]
The composition of a compound always
remains constant (i.e. the ratio of weights of
different elements in a compound) no matter by
whatever method, it is prepared or obtained
from different sources.
(iii) Law of multiple proportion [John Dalton, 1804]
According to this law, when two elements A
and B combine to form more than one chemical
compound then different weights of A, which
combine with a fixed weight of B, are in a
proportion of simple whole number.
(iv) Law of reciprocal proportions [Ritche, 1792-94]
If two elements A and B combine separately
with third element C to form two different
compounds and if A and B also combine
together to form a compound then they do so in
a ratio of their masses equal or multiple or
submultiples of ratio of their masses which
combine with a definite mass of C.
(v) The law of Gaseous volume : [Gay Lussac 1808]
According to this law, when gases combine,
they do so in volume which bear a simple ratio
to each other and also to the product formed
provided all volumes are measured under
similar conditions.
5. Concept of mole
(i) Definition : One mole is amount of a substance
that contains as many particles or entities as
there are atoms in exactly 12 gram of the
carbon (12
C – isotope).
(ii) 1 mole  collection of 6.022 × 1023
particles or
entities.
(iii) 1 mole atoms  1 gram-atom  gram atomic mass
(iv) 1 mole molecules  1 gram-molecule  gram
molecular mass
(v) 1 mole ions  1 gram-ion  gram ionic mass
6. Atomic weight and
atomic mass unit (amu)
(i) The atomic weight (or atomic mass) of an
element may be defined as the average relative
weight (or mass) of an atom of the element
with respect to the (1/12)th
mass of an atom of
carbon (mass number 12)
Thus, atomic weight =
12
)12.nomass(CofatomanofWeight
elementtheofatomanofWeight

(ii) If we express atomic weight in grams, it
becomes gram atomic weight (symbol gm-atom).
(iii) 1 gm-atom of any element contain NA number
of atoms.
(iv) The atomic mass unit (amu or u) is defined as
the (1/12)th
of the mass of single carbon atom
of mass number 12.
Thus, 1 amu or u = 1.667 × 10–24 gm
= 1.667 × 10–27 kg.
7. Molecular weight and formula weight
(i) Molecular weight is defined as the weight of a
molecule of a substance compared to the
(1/12)th
of the mass of a carbon atom (mass
number = 12).
(ii) In ionic compounds, as for example, NaCl,
CaCl2, etc. there are no existence of molecules.
For ionic compounds, instead of "molecular
weight" we use a new term known as "formula
weight". "Formula weight" is defined as the
total weights of atoms present in the formula of
the compound.
8. The average atomic mass and average
molecular mass
(i) Average atomic mass : Let us consider, an
element X, is available in the earth as isotopes
of 21 a
n
a
n X,X ,……, na
n X , the percentage
abundance of the given isotopes in earth are
x1 , x2, ........, xn respectively.

100
xa......xaxa
X
ofmassatomicaveragethe
nn2211 

(ii) Average molecular mass : Let us consider, in
a container,
n1 moles of substance X1 (mol. wt M1) present
n2 moles of substance X2 (mol. wt M2) present
........................................................................

4. 4
nn moles of substance Xn (mol. wt Mn) present
hence, the total number of moles of substance
present in the container = n1 + n2 + ....... + nn
Total mass of the substance present in the
container = n1M1 + n2M2 + ……… + nnMn






 nj
1j
j
nj
1j
jj
avg
n
Mn
M
9. Use of mole, GAM and GMM
(i) GAM = 1 gram-atom
= 1 mole-atoms
(GAM  gram atomic mass)
(ii) GMM = 1 gram-molecule
= 1 mole molecules
(GMM  gram molecular mass)
(iii) Molar mass : Mass of one mole of particles or
entities of a substance is known as molar mass
of a substance.
(iv) No. of moles =
massMolar
massGiven
=
M
w
(When mass of substance is given)
(v) No. of moles =
AN
particlesof.noGiven
AN
N

(When no. of particles of substance are given)
Here NA = Avogadro's No. = 6.022 × 1023
(vi) At STP : Number of moles (for ideal gas)
=
711.22
)litresin(gasofVolume
According to IUPAC recommendations STP refers
273.15 K (or 0ºC) temperature and 1 bar pressure.
In old books STP refers 273.15 K and 1 atm
pressure. Volume of 1 mole ideal gas at STP is
considered as 22.4 L.
10. conversion of volume of gases into
mass(i) For Ideal Gases : PV = nRT (where n
indicates number of moles of gas and R is
universal gas constant)
(ii) As we know, n =
M
w
, where w is the mass of
gaseous substance).
RT
M
w
PV  or
RT
PVM
w 
Therefore if we know, pressure, volume,
temperature and molecular weight of gas, we
can calculate its mass.
(iii) Value of R = 8.314 J/mol-K
= 0.0821 atm-L/mol-K
= 2 cal/mol-K (approx)
= 1/12 bar-L/mol-K
(iv) Density of gas may be calculated as
d =
RT
PM
V
w

11. Empirical formula and molecular
formula
(i) Empirical formula (simplest formula) : The
empirical formula of a compound reflects the
simple ratio of atoms present in the formula
units of the compound.
(ii) Molecular formula : The molecular formula is
the actual number of atoms of the constituent
elements that comprise a molecule of the
substance.
Molecular formula = (Empirical formula)n
Here n = 1, 2, 3.......
12. Some Important Reactions
(i) Decomposition Reaction :
CaCO3(s) 
CaO(s) + CO2(g)
MgCO3(s) 
MgO(s) + CO2(g)
SrCO3(s) 
SrO(s) + CO2(g)
2NaHCO3 
Na2CO3 + H2O + CO2
2KHCO3 
K2CO3 + H2O + CO2
2 HI 
H2 + I2
2 NH3 
N2 + 3H2
Carbonates of Ist group elements i.e. Na, K, Rb, Cs
do not decompose on heating.
(ii) Displacement Reactions :

5. 5
Zn(s) + CuSO4  ZnSO4 + Cu
Fe + CuSO4  FeSO4 + Cu
Pb + CuCl2  PbCl2 + Cu
Zn + H2SO4  ZnSO4 + H2
(iii) Double Displacement Reactions :
NaCl + AgNO3  AgCl + NaNO3
white precipitate
NaBr + AgNO3  AgBr  + NaNO3
Yellow
NaI + AgNO3  AgI  + NaNO3
Yellow
Na2SO4 + BaCl2  BaSO4  + 2NaCl
white
Similarly sulphides as HgS (Black), PbS (Black),
Bi2S3 (Black), CuS (Black), CdS (Yellow), As2S3
(Yellow), Sb2S3 (orange), SnS (Brown),
SnS2(Yellow) give precipitate.
Carbonates of 2nd group elements also give
precipitate.
(iv) Neutralisation Reactions :
Reaction between acids (contain replaceable H+
ion) and bases (containing replaceable OH¯
ion) is known as neutralisation reaction
Examples :
NaOH + HCl  NaCl + H2O
2 KOH + H2SO4  K2SO4 + 2H2O
13. Eudiometry
(i) Different solutions used for absorbing gases:
Sr. No. Gas(es) Solution (or solvent)
1. CO2, SO2, Cl2 KOH or NaOH
(aq. solution)
2. O2…….. Alkaline Pyrogallol
3. CO……. Ammonical Cu2Cl2
4. O3…….. Mineral turpentine oil
5. NH3 & HCl…. Water
6. Water (vapour) Silica gel or
anhydrous CaCl2
On cooling if volume of gaseous mixture decreases
then this is because of condensation of H2O(V).
14. Concentration terms
(i) Density ()
=
cetansubstheofvolume
ceantsubstheofMass
In c.g.s. and MKS units, density is expressed in
gm/cm3 or gm/ml and kg/m3 respectively.
(ii) Relative density
=
cetansubsreferenceofDensity
cetansubsanyofDensity
(iii) Specific gravity
=
Cº4atOHofDensity
cetansubsanyofDensity
2
(iv) Weight by weight percentage (% w/w) or
percentage by weight
= 100
solutionofweight
soluteofweight

(v) Weight by volume percentage (%w/v) or
percentage by volume
= 100
solutionofvolume
soluteofweight

(vi) volume by volume percentage (%v/v) or
percentage by strength
= 100
solutionofvolume
soluteofvolume

(vii)mole percentage (% mol/mol) or percentage
by mole
= 100
solventofMolessoluteofMoles
)soluteofMoles(


Do remember, for the calculation of strength
(% w/w, %w/v etc) the solute must be completely
dissolved into the solution, otherwise, the given
terminologies will be invalid.
If, anything is not specified, % strength generally
means % by mass.
(viii) Parts per million (PPM)

6. 6
= 6
10
solutionincompoundsallofpartsof.noTotal
soluteofpartsof.No

PPM is generally expressed as w/w (mass to mass)
PPM can also be expressed as w/v (mass to volume)
or V/V (volume to volume)
(ix) gram per litre (gm/lit): It is the amount of
solute in gm dissolved in 1 litre (1000 ml) of
solution.
(x) Formality
=
)litresin(solutionofvolume
soluteofunitsformulaofmolesofNumber
(xi) Molality =
kginsolventofmass
soluteofmolesofNumber
(xii) Molarity =
)litresin(solutionofVolume
soluteofmolesof.No
Molarity =
1M
dx10 
;
Here x = %
w
w
of solute
d = density of solution in gm/mL
M1 = molar mass of solute
(xiii) Mole fraction :
Xsolute =
solventsolute
solute
nn
n

Xsolvent =
solventsolute
solvent
nn
n

Xsolute + Xsolvent = 1
15. Relation between concentration terms
(i) m =
21
1
M)X1(
X1000

(ii) m =
1MMd1000
M1000

(iii)
2111
1
M)X1(MX
Xd1000
M


Here m = molality
M = molarity
d = density of solution in gm/mL
X1 = mole fraction of solute
M1 = molar mass of solute
M2 = molar mass of solvent
(iv) PPM = % 





ionconcentrat
w
w
× 104
(v) Gram per litre = M × M1
(vi) Gram per litre = 10 × 





v
w
%
(vii) % d
W
w
%
v
w







All these above relations 16 (i – vii) are applicable
only to binary solutions.
16. Vapour density
=
pressureand.tempsameatgasHofDensity
pressure&.tempsameatvapourofDensity
2
Vapour density =
2
1
× molecular mass
17. % Yield of reaction
amount of a definite product
actually produced in a reaction
Maximum possible amount of the
same product which can be produced
× 100=
18. Oleum (H2SO4 + SO3)
% of free SO3 =
18
80)100x( 
Here x = strength of oleum sample in percentage.
(x is always greater than 100)

7. SOLVED EXAMPLES FOR JEE(MAIN) - NEET
Ex.1 An atom of element 'X' is 1.02 times heavier
than that of an atom of 'Y'. An atom of 'Y' is
0.1809 times heavier than that of an atom of
oxygen. What is the atomic weight of 'X' ?
(A) 2.952 (B) 5.314 (C) 4.4 (D) 2.6
Sol.(A) Atomic wt of
X =
Yatomanofwt
Xofatomanofwt
×
Oatomanofwt
16
1
Yofatomanofwt

= 1.02 × 16 × 0.1809
= 2.952
Ex.2 A gas container contain 35 gm nitrogen gas, 64
gm O2
gas and 0.112 lit of CH4
at STP, what is
average molecular weight of the gas mixture?
(A) 32 (B) 30.44 (C) 28 (D) 16
Sol.(B) 35 gm nitrogen =
28
35
= 1.25 mole
64 gm oxygen =
32
64
= 2 mole
0.112 lit CH4 at STP =
4.22
112.0
= 0.005 mole
nj = 1.25 + 2 + 0.005 = 3.255
 Mavg =
255.3
16005.03222825.1 
=
255.3
08.06435 
=
255.3
08.99
= 30.44
Ex.3 A porous catalyst has an internal surface area
of 800 m2 per cm3 of bulk material. Fifty
percent of the bulk volume consists of the pores
(holes), while the other 50% of the volume is
made of solid substance. Assume that the pores
are all cylindrical tubules of uniform diameter d
and length  and that the measured surface area
is the total area of the curved surfaces of the
tubules. What is the diameter of the pores ?
Sol. Basis = 1 cm3 of catalyst
 total volume of pores = 0.5 × 1 = 0.5 cm3.
now consider, n number of total pores
present
 the volume of each pore =
4
1
d2

4
1
d2× n = 0.5 cm3.
surface area of total pores
= 800 m2 = 8 × 106 cm3 = n × d



dn
nd
4
1 2


=
4
d
= 26
3
cm108
cm5.0

or d = 6
108
45.0


cm = 0.25 × 10–6 cm
= 25 × 10–8 cm = 25 Å
 the diameter of pore = 25 Å
Ex.4 In a victor Meyer determination of the relative
molecular mass of benzene, the heating vessel
was maintained at 120ºC. A mass of 0.1528 gm
of benzene was used and the volume of
displaced air collected over water at 15ºC, was
48 cm3. The barometric pressure was 743 mm
mercury. Calculate the relative molecular mass
of benzene. The vapour pressure of water at
15ºC = 13 mm Hg.
Sol. Actual pressure of displaced air = 743 – 13
= 730 mm =
760
730
atm
15 ºC = 15 + 273
= 288 K
V = 48 cm3
= 48 × 10–3 lit.
W = 0.1528

8.  M =
PV
WRT
=
048.0
760
730
28882.01528.0


= 78.26 Hence, the molecular weight of
benzene = 78.26
Ex.5 Which of the following data illustrates the law
of conservation of mass ?
(A) 56 gm of CO reacts with 32 gm of oxygen
produce 44 gm of CO2
(B) 1.70 gm of AgNO3
reacts with 100 ml. of
0.1M HCl to produce 1.435 gm of AgCl
and 0.63 gm of HNO3
(C) 12 gm of C is heated in vaccum and on
cooling there is no change in mass
(D) None of the above
Sol.(B)
Ex.6 If a mixture containing 3 moles of hydrogen
and 1 mole of nitrogen is converted completely
into ammonia the ratio of initial and final
volumes under the same temperature and
pressure would be -
(A) 3 : 1 (B) 1 : 3 (C) 2 : 1 (D) 1 :2
Sol.(C) Initial volumes  4 at the definite temp & press
Final volumes  2 at the same temp & press.
The ratio of initial and final volumes
= 4 : 2 = 2 : 1
Ex.7 Atomic mass of H, O, C and B are 1.008,
15.9956, 12 and 10.81 respectively. Which
element you do select to define the atomic
weight of x (given mass of an atom of x = 25
amu; mass number of H, O, C and B are 1, 16,
12 and 11 respectively).
Sol. In H scale, the atomic mass of x
=
008.1
1
1
25

= 24.80
In O scale, the atomic mass of
x =
9956.15
16
1
25

= 25.00062
In C scale, the atomic mass of
x =
12
12
1
25

= 25
In B scale, the atomic mass of
x =
81.10
11
1
25

= 25.43
Since in the carbon scale, the atomic weight of
x is whole number (having no fractional part
and equals to 25), therefore we do select C
scale.
Ex.8 There are 10 gm of mixture of NaCl and NaBr.
If the amount of sodium is 25% of the weight
of total mixture, calculate the amount of NaCl
and NaBr present in the mixture. (Given,
atomic weights of Na, Cl and Br are 23, 35.5
and 80 respectively).
Sol. Let in the given mixture NaCl present = x gm.
in the given mixture NaBr present = (10 – x) gm.
The formula weight of NaCl = 23 + 35.5 = 58.5
and formula weight of NaBr = 23 + 80 = 103
the total amount of Na present in the mixture
=
5.58
23
x +
103
23
(10 – x) = 0.25 × 10 = 2.5; on
solving x = 1.5734 and 10 – x = 8.4266
the amount of NaCl and NaBr present in the
given mixture are 1.5734 gm and 8.4266 gm
respectively.
Ex.9 A sample of potato starch was ground in a ball
mill to give a starch like molecule of lower
molecular weight. The product analyzed
0.086% Phosphorus. What is the minimum
molecular weight of starch?
(Given, atomic weight of P = 31).

9. Sol. In one molecule of starch minimum one P atom
must be present.
0.086 gm. of P present in 100 gm of starch
31gm. of P present in
=
086.0
100
× 31 of starch
= 36046.51 gm
the minimum mol. wt. of starch = 3.6046 × 104
Ex.10 60 gm of sucrose (C12H22O11) and 90 gm of
glucose (C6H12O6) are dissolved in 1000 ml of
water. The specific gravity of the resulting
solution is 1.1 gm/ml. What is the percentage
by moles of sucrose present in the solution ?
Sol. The molecular weight of sucrose is
= 12 × 12 + 22 × 1 + 16 × 11 = 342
The molecular weight of glucose
= 12 × 6 + 12 × 1 + 6 × 16 = 180
Now, total weight of resulting solution
= 1.1 × 1000 = 1100 gm.
wt of pure water present in 1000 ml of
solution = 1100 – 90 – 60 = 950 gm.
moles of water present in solution =
18
950
 moles of sucrose present in solution =
342
60
 moles of glucose present in solution =
180
90
 mole % of sucrose in solution
=
180
90
342
60
18
950
342
60

× 100
=
5.00754.0527777
1001754.0


=
4531.53
1001754.0 
= 0.328
Ex.11 5000 ml of 0.6 (M) H2SO4 acid is mixed with 2
lit of 98% (w/v) of H2SO4 (sp. gravity 1.84). If
the specific gravity of the resulting mixture 1.4,
what is the molarity strength of the mixture?
(Given that the specific gravity of 0.6 (M)
H2SO4 is 1.02 and also consider there is no loss
of mass due to mixing).
Sol. It has been given that there is no loss of mass
due to mixing of two acid solution.
 mass of 5 lit of 0.6 (M) H2SO4 + mass 2 lit
of 98% (w/v) H2SO4 = mass of the resulting
solution.
mass of the resulting solution
= 5000 × 1.02 + 2000 × 1.84
= 5100 + 3680 = 8780 gm.
Hence, the volume of the resulting mixture
=
4.1
8780
= 6271.43 ml
The moles of H2SO4 present in first kind of
acid solution = 5000 × 0.6 × 10–3 = 3 moles.
The amount of H2SO4 present in 2 lit of second
kind of acid solution =
98
2980
= 20 moles.
total moles of H2SO4 present in the resulting
acid solution = 3 + 20 = 23 moles.
the concentration of resulting acid solution
=
43.6271
23
× 100 = 3.6674 (M)
Ex.12 What is the volume of free SO3 obtained from
100 gm of oleum (considered as solution of
SO3 in H2SO4) that is labelled with "109%
H2SO4" at the pressure of 1.5 atm and 400 K ?
(Given 1 mole of H2O combine with 1 mole of
SO3)

10. Sol. "109% H2SO4", means on dilution of 100 gm of
oleum 109 gm of pure H2SO4 is obtained, when
are the free SO3 would combine with water to
form H2SO4.
9 gm of H2O will combine are the free SO3
present in 100 gm of the oleum to give a total
of 109 gm of pure H2SO4.
9 gm. of H2O =
18
9
= 0.5 mole of H2O.
Since, 1 mole H2O combines with 1 mole of
SO3. (SO3 + H2O = H2SO4)
0.5 mole H2O combines with 0.5 mole of SO3.
free SO3 present in 100 gm of oleum = 0.5 mole
volume of free SO3 =
P
nRT
=
5.1
400082.05.0 
= 10.933 lit
Ex.13 A 1.00 gm sample of KClO3 was heated under
such conditions that a part of it decomposed
according to the equation
(i) 2 KClO3 = 2 KCl + 3 O2
and the remaining underwent change
according to the equation
(ii) 4 KClO3 = 3 KClO4 + KCl If the amount of
oxygen involved was 146.8 ml at STP,
calculate percentage by weight of KClO4 in
the residue. (Given at wt of K and Cl are
39 & 35.5 respectively).
Sol. 146.8 ml of O2 at STP
=
22400
328.146 
gm of oxygen
Let us consider, x gm of KClO3 is decomposed
as equation (i). Therefore, the amount of KClO3
decomposed as per equation (ii)
= (1 – x) gm.
As per equation (i) from,
2 × 122.5 gm of KClO3 oxygen obtained
= 6 × 16 gm.
from x gm of KClO3 oxygen obtained
=
5.1222
x166


gm =
5.122
x48

5.122
x48
=
22400
328.146 
or x = 0.5352 gm
1 – x = 1 – 0.5352 = 0.4698 gm.
In the residue KCl and KClO4 will be present,
so, now we are to calculate the amount of KCl
and KClO4 present in the mixture. As per
equation (i) from x gm KClO3, KCl obtained
=
5.122
x5.74
gm.
As per equation (ii) from (1 – x) gm KClO3 ,
KClO4 obtained =
5.1224
)x–1(5.1383


gm.
As per equation (ii) from (1 – x) gm KClO3,
KCl obtained =
5.1224
)x–1(5.74

gm.
Total mass of the residue
=
5.122
x5.74
+
5.1224
)x–1(5.1383


+
5.1224
)x–1(5.74

Putting 1 – x = 0.4648,
The total mass of the residue
= 0.3941 + 0.07067 + 0.3255 = 0.79027 gm.
Putting 1 – x = 0.4648, mass of KClO4
= 0.3941 gm.
mass of KClO4 present in the residue
=
79027.0
3941.0
× 100 = 49.87 %
Ex.14 A natural gas sample contain 84% (by volume)
CH4, 10% C2H6, 3% C3H8 and 3% N2. If a
series of catalytic reaction could be used for
converting all carbon atoms of the gas into
butadiene, C4H6, with 80% efficiency, how
much butadiene could be prepared from 1000
lit of natural gas at 10 atm pressure and 400 K
temperature ?

11. Sol. We know that volume percent = mole percent.
Avg. molecular wt of natural gas, Mavg =
i
ii
n
Mn


We have considered 100 mole of natural gas as
basic.
Mavg =
100
28344330101684 
= 18.6
(molecular wt of CH4, C2H6 C3H8 and N2 are
16, 30 44 and 28 respectively)
Now, press P = 10 atm, volume, 103 lit
temperature T = 400 K and
R = 0.082 lit atm /k/mole
weight of given natural gas,
w =
RT
PVM
=
400082.0
6.181010 3


= 5670.7 gm.
=
6.18
7.5670
mole
In 100 moles of mixture, moles of C present
= 84 × 1 + 10 × 2 + 3 × 3 = 113 moles
moles of C present in 5670.7 gm of natural
gas =
100
113
×
6.18
7.5670
from 4 moles of C, C4H6
obtained = 1 mole
from
100
113
×
6.18
7.5670
of C, C4H6 obtained
=
4
1
×
100
113
×
6.18
7.5670
mole
Since, efficiency of the process is 80%,
therefore the moles C4H6 actually produced
= 0.8 ×
4
1
×
100
113
×
6.18
7.5670
mole
= 0.8 ×
4
1
×
100
113
×
6.18
7.5670
= 68.902 mole
the amount of C4H6 obtained = 68.902 × 54
gm = 3720.75 gm.
Ex.15 (a) What mass of P4O10 will be obtained from
the reaction of 1.33 gm of P4 and 5.07 gm of
O2?
(b) What mass of P4O6 will be obtained from
the reaction of 4.07 gm P4 and 2.01 gm of
O2 ? (Given atomic weight of P = 31)
Sol. (a) 1.33 gm P4 =
431
33.1

moles of P4
5.07 gm O2 =
32
07.5
moles of O2
P4 + 5 O2 = P4 O10
From 5 moles of O2, moles of P4O10
obtained = 1 mole
 from
32
07.5
moles of O2 moles of P4O14
obtained =
5
1
×
32
07.5
= 0.03169 moles.
from 1 moles of P4, moles of P4 O10
obtained = 1 moles
from
431
33.1

moles of P4, moles of P4O10
obtained =
431
33.1

= 0.010726
 P4 is the limiting reactant
amount of P4 O10 obtained
= 0.010726 moles
= 0.010726 × 284 (mol. wt of P4O10 = 284)
= 3.046 gm
(b) 4.07 gm P4 =
124
07.4
moles of P4
2.01 gm O2 =
32
01.2
moles of O2
P4 + 3 O2 = P4O6
from 1 mole of P4, P4O6 obtained = 1 mole
  from
124
07.4
moles, P4, P4 O6 obtained
=
124
07.4
moles = 0.03282 moles
  from moles of O2, P4O6 obtained
=
4
1
×
32
01.2
= 0.02094 moles
Hence, O2 is limiting reactant,

12. the amount of P4O6 obtained = 0.02094
moles
= 0.02094 × 220 (mol wt of P4O6 = 220)
= 4.6068 gm
Ex.16 15 ml of an oxide of nitrogen was taken in an
eudiometer tube and mixed with hydrogen till
the volume was 42 ml. On sparking, the
resulting mixture occupied 27 ml. To the
mixture 10 ml of oxygen was added and on
explosion again the volume shift to 19 ml. Find
the formula of the oxide of nitrogen that was
originally admitted. Both explosions lead to the
formation of water.
Sol. Let the formula of the oxide of nitrogen is NxOy
the reaction of this oxide with hydrogen is as
follows-
Nx Oy + y H2 =
2
x
N2 (g) + y H2O ()
1 vol y vol
2
x
vol y vol (as it was liquid)
15 ml 15 y ml 15
2
x
ml. (as it was liquid its
volume is negligible)
Now, volume of H2 introduced = 42 – 15 = 27 ml.
and volume of H2 required to react = 15 y ml.
volume of H2 remaining = (27 – 15 y) ml.
After the reaction, remaining gases are
N2 and H2
15
2
x
+ (27 – 15 y) = 27
or
2
x
= y or x = 2y …….(1)
To oxidize the remaining hydrogen O2 has been
introduced,
H2 (g) +
2
1
O2 (g) = H2O ()
(27 – 15y) ml 





2
y15–27
ml. (27 – 15 y) ml.
(as it was liquid)
after the complete reaction of H2, the
unreacted O2 left = 10 – 





2
y15–27
ml.
And N2 is also present in the residual gas
mixture.

2
x15
+ 10 – 





2
y15–27
= 19 or
2
7–yy15x15 
= 19 or x + y = 3 ……. (2)
on solving, x = 2 and y = 1
the formula of oxide of nitrogen is = N2O
Ex.17 A sample of coal gas contains 50% H2, 30%
CH4, 14% CO and 6% C2H4. 100 ml of coal gas
is mixed with 150 ml of O2 and the mixture is
exploded. What will be the volume and
composition of mixture when cooled to original
conditions ?
Sol. In 100 ml of given coal gas contain H2 = 50 ml,
CH4 = 30 ml, CO = 14 ml C2H4 = 6 ml
H2 (g) +
2
1
O2 (g) = H2O ()
50 ml 25 ml 50 ml
(as it is liquid, its volume is negligible)
CH4 (g) + 2 O2 (g) = CO2 (g) + 2H2O ()
30 ml 60 ml 30 ml 60 ml (as it is liquid)
CO (g) +
2
1
O2 (g) = CO2 (g)
14 m l7 ml 14 ml
C2H4 (g) + 3O2 (g) = 2CO2 (g) + 2H2O ()
6 ml 18 ml 12 ml 12 ml
Total volume of O2 consumed
= 25 + 60 + 7 + 18 = 110 ml
volume of O2 remaining = 150 – 110 = 40 ml
total volume of CO2 produced = 30 + 14 + 12
= 56 ml
Hence % by volume of CO2 =
96
56
× 100
= 58.334

13. Hence % by volume of O2 =
96
40
× 100
= 41.666
total volume of product gas = 40 + 56 = 96
ml and it contain 58.334 % CO2 and 41.666%
O2 by volume
Ex.18 Nine volumes of a mixture of gaseous organic
compound A and just sufficient amount of
oxygen required for complete combustion
yielded on burning four volumes of CO2, six
volumes of H2O vapour and two volumes of
N2, all being measured at the same temperature
and pressure. If compound A contained C, H
and N only, what is the molecular formula of A ?
Sol. This problem requires a direct application of
Avogadro's principle under the same conditions
of temperature and pressure, equal volumes of
gas contain the same number of molecules.
since in one molecule of N2 two N atoms are
present, in one molecule of H2O two H atoms
are present, in 1 molecule of CO2, one C atom
is present.
C : H : N = volume of CO2 : 2 × volume of
H2O : 2 × volume of N2 = 4 : 2 × 6 : 2 × 2
= 2 : 6 : 2
Thus, 1 molecule of gaseous organic compound
contains 2 carbon atoms, 6 H atoms and 2 N
atoms, therefore its molecular formula is
C2H6N2.
Ex.19 On heating 60 ml of a mixture of equal
volumes of chlorine and its gaseous oxide and
cooling to atmospheric temperature, the
resulting gas measured 75 ml. Treatment of this
resulting gas mixture with caustic soda
(absorbs chlorine) solution resulted in a
contracts to 15 ml. Assuming that all
measurements were made at the same
temperature and pressure, deduce the formula
of oxide. (Consider due to heating entire
chlorine oxide is decomposed).
Sol. Let the formula of chlorine oxide is Clx Oy
volume of Cl2 = 30 ml.
and volume of Clx Oy= 30 ml
when the gas mixture is heated only ClxOy will
decomposed as,
Clx Oy =
2
x
Cl2 +
2
y
O2
30 ml
2
x30
ml
2
y30
ml
30 +
2
x30
+
2
y30
= 75 or 15 (x + y) = 45
or x + y = 3 .......... (1)
In the mixture, total volume of Cl2 gas present
=
2
x30
+ 30 = 75 – 15 = 60 or 15 x = 30
or x = 2 y = 1
formula of chlorine oxide is Cl2O.
Ex.20 Sodium chlorate, NaClO3 can be prepared by
the following series of reactions
2 KMnO4 + 16HCl  2 KCl + 2 MnCl2 + 8 H2O + 5Cl2
6 Cl2 + 6 Ca (OH)2  Ca (ClO3)2 + 5 CaCl2 + 6H2O
Ca (ClO3)2 + Na2SO4  CaSO4 + 2 NaClO3
what mass of NaClO3 can be prepared from
100 ml of concentrated HCl [density 1.18
gm/ml & 36% HCl (w/w)] ? Assume all other
substances are present in excess amounts.
Sol. Each of the given equations are balanced, our
aim is to produce NaClO3 from HCl. HCl
involved into the Ist step and NaClO3 is
produced at the IIIrd step so, to corelate the
amount of these two substances, we are to add
the three given equation as
2 KMnO4 + 16 HCl 
2 KCl + 2 MnCl2 + 8 H2O + 5Cl2] × 6
6 Cl2+6 Ca (OH)2 
Ca (ClO3)2 + 5 CaCl2 + 6 H2O] × 5
Ca (ClO3)2 + Na2 SO4 
CaSO4 + 2 NaClO3] × 5

14. ———————————————————
12 KMnO4 + 96 HCl + 30 Ca(OH)2 + 5 Na2
SO4 = 12 KCl + 12 MnCl2 + 78 H2O + 25
CaCl2 + 5 CaSO4 + 10 NaClO3.
Given, all reactants are excess except for HCl.
Mass of 100 ml of HCl = 100 × 1.18 = 118 gm.
the mass of pure HCl = 118 × 0.36 gm
= moles of HCl [mol wt of HCl = 36.5]
As per the net equation, from 96 moles of HCl,
NaClO3 produced = 10 moles
1 moles of HCl, NaClO3 produced
=
96
10
moles
5.36
36.0118
moles of HCl, NaClO3 produced
=
96
10
×
5.36
36.0118
moles
=
96
10
×
5.36
36.0118
× 106.5 gm.
[mol. wt of NaClO3 = 106.5]
= 12.91 gm.
Hence the amount of NaClO3 produced = 12.91 gm.
Ex.21 Chlorine gas can be produced in the laboratory
by the reaction
K2Cr2O7 + 14HCl = 2KCl + 2CrCl3+ 7H2O +3Cl2
If 68 gm sample that is 96% K2Cr2O7 is
allowed to react with 320 ml of HCl solution
having density of 1.15 gm/ml and containing
30% by weight HCl, what mass of Cl2 is
generated ? (Given atomic mass of K and Cr
are 39 and 52 respectively).
Sol. Mol. wt of K2Cr2O7 = 39 × 2 + 52 × 2 + 16 × 7
78 + 104 + 112 = 294
In 68 gm of given sample, amount of pure
K2Cr2O7 present = 68 × 0.96 = 65.28 gm.
= 0.222 moles
If, 0.222 moles of K2Cr2 O7 completely reacted,
moles of Cl2 formed = 3 × 0.222 = 0.666
The mass of given HCl = 320 × 1.15 gm.
The amount of pure HCl = 320 × 1.15 × 0.3
gm =
5.36
3.015.1320 
moles
from
5.36
3.015.1320 
moles of HCl, moles of
Cl2 produced =
14
3
×
5.36
3.015.1320 
= 0.6481
HCl is the limiting reactant, hence amount of
Cl2 produced = 0.6481 moles
= 0.6481 × 2 × 35.5
= 46.015 gm.
Ex.22 A sample of NaCl contains NaNO3 as an
impurity, 250 ml of its solution were prepared
by dissolving 1.25 gm of the sample. 25 ml of
this solution required 17.75 ml. of (M/10)
AgNO3 solution. Calculate the composition of
the solution in gm/litre.
Sol. Let in 1.25 gm of impure NaCl sample, NaCl
present = x gm
in the given sample NaNO3 present
= (1.25 – x) gm.
250 ml. of resulting solution contain x gm of
NaCl
25 ml of resulting solution contain =
10
x
gm
of NaCl =
10
x
×
5.58
1
formula wt of NaCl
formula wt of NaCl = 23 + 35.5 = 58.5)
17.75 ml of (M/10) AgNO3 = 17.75 ×
10
1
× 10–3
moles of AgNO3
= 1.775 × 10–3 moles of AgNO3
AgNO3 + NaCl = AgCl + NaNO3.
1 moles of AgNO3 reacts with 1 formula wt of
NaCl.
1.775 × 10–3 moles of AgNO3 reacts with
1.775 × 10–3 formula wt. of NaCl

10
x
×
5.58
1
= 1.775 × 10–3

15. or x = 1.775 × 10 × 58.5 × 10–3 = 1.0384 gm.
in 1.25 gm of sample NaNO3 present
= 1.25 – 1.0384 = 0.2116 gm.
concentration of NaCl in the resulting
solution
=
250
0384.1
× 1000 = 4.1536 gm/lit. Ans.
concentration of NaNO3 in the resulting
solution
=
250
2116.0
× 1000 = 0.2116 × 4
= 0.8464 gm/lit.
Ex.23 A mixture of HCOOH and H2C2O4 is heated
with conc H2SO4. The gas produced is
collected and on treating with KOH solution
the volume of the gas decreases by
6
1
th.
Calculate molar ratio of two acids in original
mixture.
Sol. Let we have a mole of HCOOH & b mole of
H2C2O4.
HCOOH

  42SOH.Conc
CO + H2
O
a mole a mole
H2C2O4

  42SOH.Conc
CO2 + CO + H2O
b mole b mole b mole
Total mole of gases obtained = a + 2b
When this gaseous mixture will be passed
through KOH solution, CO2
will be absorbed
by which volume of gas will decrease. Hence,
from question
b =
6
1
(a + 2b) 
b
a
=
1
4
Ex.24 A piece of aluminium weighing 2.7 g is treated
with 75.0 ml of H2
SO4
(sp. gravity 1.18
containing 24.7 percent H2
SO4
by weight).
After the metal is completely dissolved, the
solution diluted to 400 ml. Calculate the
molarity of free H2
SO4
in the resulting solution.
Sol. Molarity of H2
SO4
solution taken
=
10098
10007.24


× 1.18 = 2.974 M
Now, mole of Al taken =
27
7.2
= 0.1
and mole of H2SO4 taken =
1000
974.275
= 0.223
2 Al + 3 H2SO4  Al2 (SO4)3 + 3 H2
2 mole 3 mole
 0.1 mole 0.1 ×
2
3
= 0.15 mole
(Here aluminium is limiting reagent)
Hence, mole of H2SO4 remained = 0.223 – 0.15
= 0.073
But now the volume of solution is made to be
400 ml and hence molarity of free H2SO4 in the
resulting solution =
400
073.0
× 1000
= 0.1825 M