FOREWORD

Dear Students,

Welcome aboard to the BSEB coaching program for NEET (UG).

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Anand Kishor, IAS seem insurmountable at times, but always remember that it's through
Chairman overcoming these challenges that you grow stronger and more resilient. Your
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the path to success and persistence is the vehicle you arrive in.

Wishing you all the success and fulfilment on your academic journey.
S.NO. TOPIC NAME PAGE NO.

1. Introduction .............................................................................................................................. 01
2. Classification of Matter ............................................................................................................ 01
3. Laws Of Chemical Combination ............................................................................................... 02
4. Atomic terms ........................................................................................................................... 06
5. Natural Abundance of Isotopes of Elements ............................................................................ 07
6. Atomic Mass and Molecular Mass .......................................................................................... 10
7. Mole Concept .......................................................................................................................... 12
8. Concepts related to Density .................................................................................................... 17
9. Relative Density....................................................................................................................... 18
10. Vapour Density ........................................................................................................................ 19
11. Average Molecular mass of Mixture or Observed Molecular Weight ........................................ 19
12. Percentage Composition ......................................................................................................... 20
13. Experimental methods for determination of elements in organic compound ............................ 21
14. Empirical Formula and Molecular Formula .............................................................................. 23
15. Stoichiometry .......................................................................................................................... 28
16. Percentage Purity.................................................................................................................... 34
17. Methods of expressing the concentration of a solution ............................................................ 41
18. Some Experimental Methods .................................................................................................. 53
19. Assignments
Exercise-1 ............................................................................................................................... 57
Exercise-2 ............................................................................................................................... 59
Exercise-3 ............................................................................................................................... 67
Exercise-4 ............................................................................................................................... 73
Exercise-5 ............................................................................................................................... 75
20. Answer Key ........................................................................................................................... 79
STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

Chemistry is the science of substances, their properties, structures and their transformation. It deals with
the study of material objects. In order to achieve correct results, one has to rely upon the various skills
connected with the measurements of quantities during a physical or chemical change. The degree of
accuracy is closely linked with precision of the measuring instrument as well as on the skill of the person
engaged in measurement.

Anything which has mass and occupies space is called matter. Everything around us, for Example:,
house, tree, water, air, and all living beings etc., are composed of matter. Matter can exist in three
physical states viz. solid, liquid and gas.

This classification of matter is based upon chemical composition of various substances. According to this
matter can be further divided into two types, pure substance and mixture. Mixtures are also of two types,

homogenous mixtures and heterogeneous mixtures.

Only one type of More than one type of

substance present substance present
Matter

Pure Substance Mixture

One type of atoms More than one Uniform Irregular

type of atoms composition composition

Element Compound Homogeneous Heterogeneous

Elements are the fundamental materials of which all matter is composed. whose smallest unit is known
as atom. It is composed of one type of atoms Total 118 elements are known till date of which metals are
92 nonmetals are 20 and metalloids are 6.

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A non-elemental pure substance is called a compound . Compounds are chemical substances made up of
two or more elements that are chemically bound together in a fixed ratio are linked by chemical bonds
formed due to chemical reaction. The resulting molecule is a electrically neutral particle of constant
continuous composition.

Mixtures are the aggregate of more than one type of pure substance whose chemical identity remains
maintained even in mixtures. Their constituent ratio may vary unlike compound.
For Example: sugar + water = sugar syrup, Gun-powder 75 % KNO3 10% sulphur + 15% carbon
There are two types of mixture (a) homogeneous (b) heterogeneous
(a) Homogeneous mixtures are those whose composition for each part remains constant.
For Example: aqueous and gaseous solution etc.
(b) Heterogeneous mixtures are those whose composition may vary for each and every part.
For Example: soil, concrete mixtures etc.

In order to understand the composition of the compounds, it is necessary to have a theory which accounts
for both qualitative and quantitative observations during chemical changes. Observations of chemical
reactions were most significant in the development of a satisfactory theory of the nature of matter. These
observations of chemical reactions are summarized in certain statements known as laws of chemical
combination.

The law was proposed by Lavoisier :In a chemical change total mass remains conserved i.e. mass before
the reaction is always equal to mass after the reaction.

1
H2 + O  H2O (l)
2 2
2 grams 16 grams 18 grams
mass before the reaction = 2 + 16 = 18 gm
mass after the reaction = 18 gm

Example 1:
A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two
substances react, releasing carbon dioxide gas to the atmosphere. After reaction, the contents of the
reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction?
Solution:
The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final mass
of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due to the
mass of released carbon dioxide gas.
Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm

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The law was proposed by Proust : All chemical compounds are found to have constant composition irrespective
of their method of prepration or sources. In H2O, Hydrogen & oxygen combine in 2 : 1 molar ratio and 1:8
mass ratio, this ratio remains constant whether it is Tap water, river water or seawater or produced by any
chemical reaction.It can be explained by the following example.

Example 2:
The following are results of analysis of two samples of the same or two different compounds of phosphorus
and chlorine. From these results, decide whether the two samples are from the same or different com-
pounds. Also state the law, which will be obeyed by the given samples.

Amount P Amount Cl
Compound A 1.156 gm 3.971 gm
Compound B 1.542 gm 5.297 gm
The mass ratio of phosphorus and chlorine in compound A, mP : m Cl = 1.156:3.971 = 0.2911:1.000
The mass ratio of phosphorus and chlorine in compound B, mP : m Cl = 1.542:5.297 = 0.2911:1.000
As the mass ratio is same, both the compounds are same and the samples obey the law of definite
proportion.

The law was proposed by Dalton : When one element combines with the other element to form two or more
different compounds, the mass of one element, which combines with a constant mass of the other bear a
simple ratio to one another.
Example 3:
Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that these
figures shows the law of multiple proportion.
First oxide Second oxide
Carbon 42.9 % 27.3 %
Oxygen 57.1 % 72.7%
Solution:
Given In the first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon.

42.9
1 part of oxygen will combine with part of carbon = 0.751
57.1
Similarly in 2nd oxide

27.3
1 part of oxygen will combine with part of carbon = 0.376
72.7
The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1
This is a simple whole no ratio this means above data shows the law of multiple proportion.

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Example 4:
Two oxide samples of lead were heated in the current of hydrogen and were reduced to the metallic lead.
The following data were obtained
(i) Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm
(ii) Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm.
Show that the data illustrates the law of multiple proportion.
Solution:
When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the loss
in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the hydrogen.
Therefore,
the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm.

mPb 3.21 13.375

The mass ratio of lead and oxygen, r1 = m  0.244  1.000
O

and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm.

mPb 1.067 6.669

The mas ratio of lead and oxygen, r2 = m  0.16  1.000
O

Now, r1 : r2 = 13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple
proportion.

The law was proposed by Richter : When two elements combine seperately with definite mass of a third
element, then the ratio of their masses in which they do so is either the same or some whole number
multiple of the ratio in which they combine with each other.
This law can be understood easily with the help of the following Examples.
Example 5:
In the set (H2O , SO2. H2S.) explain law of reciprocal proportions

At. Mass 16

O
O

SO
2
H

H S
H2S
At. Mass 1 At. Mass 32

Hydrogen combine with oxygen to form H2O whereas sulphur combines with it to form SO2. Hydrogen and
sulphur can also combine together to form H2S. The formation of these compounds is shown in fig.
In H2O, the ratio of masses of H and O is 2 : 16.
In SO2, the ratio of masses of S and O is 32 : 32.
Therefore, the ratio of masses of H and S which combines with a fixed mass of oxygen (say 32 parts)
will be 4 : 32 i.e. 1 : 8 ...(i)

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When H and S combine together, they form H2S in which the ratio of masses of H and S is
2 : 32 i.e., 1 : 16 ...(ii)

1 1
The two ratios (i) and (ii) are related to each other as : or 2 : 1
8 16

i.e., they are whole number multiples of each other.

Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole number multiple
of the ratio in which H and S combine together.
Example 6:
Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27 % carbon
and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89% oxygen, by mass. Show
that the data illustrates the law of reciprocal proportion.
Solution:
Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third
element. Now, let the fixed mass of carbon = 1 gm. Then,

25 1
the mass of hydrogen combined with 1 gm carbon in methane =  gm
75 3

72.73 8
and the mass of oxygen combined with 1 gm carbon in carbon dioxide =  gm
27.27 3

1 8 1
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r 1 = : 
3 3 8

11.11 1
Now, the mass ratio of hydrogen and oxygen in water, r2 = =
88.89 8
As r1 and r2 are same , the data is according to the law of reciprocal proportion.

When two or more gases react with one another, their volumes bear simple whole number ratio with one
another and to the volume of products (if they are also gases) provided all volumes are measured under
identical conditions of temperature and pressure.

H 2(g)  2HCl(g) 1: 1: 2

 Cl2(g) 
1 voume 1 volume 2 volumes

i.e. one volume of hydrogen reacts with one volume of chlorine to form two volumes of HCl gas. i.e. the
ratio by volume which gases bears is 1:1:2 which is a simple whole number ratio.
Similarly other Example: is:
2SO 2  O2  2SO3 2 : 1: 2
2 volume 1 volume 2 volume

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Example 7:
2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces
7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure and
temperature. Show that the data illustrates Gay Lussac’s law of volume combination.

Vhydrocarbon : Voxygen : Vcarbon dioxide : Vwater vapour

= 2.5 : 12.5 : 7.5 : 10.0
= 1: 5 : 3 : 4 (simple ratio)
Hence, the data is according to the law of volume combination.
Example 8:
In the reaction, N2  3H2 
 2NH3 , the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3 : 2.
These figures illustrate the law of:
(A) constant proportions (B) Gay-Lussac
(C) multiple proportions (D) reciprocal proportions
Solution:
The above ratio of 1 : 3 : 2 illustrates the Gay-Lussac law of combining volume.
Hence (B) is correct

(a) Atomic Number (Z)

The atomic number of an element is the number of protons contained in the nucleus of the atom of that
element.
(b) Nucleons
Protons and neutrons are present in a nucleus. So, these particles are collectively known as nucleons.
(c) Mass Number (A)
The total number of protons and neutrons present in the nucleus is called the mass number of the
element.
(d) Neutrons
The number of neutrons present in nucleus is calculated as A-Z.
(e) Nuclides
Various species of atoms in general. A nuclide has specific value of atomic number and mass number.
IUPAC notation of an atom (nuclide)
Let X be the symbol of the element. Its atomic number be Z and mass number be A. Then the element
can be represented as zXA.

A MASS NUMBER

SYMBOL OF ELEMENT

ATOMIC NUMBER Z

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Example 9:

Find number of Protons, Electrons, Nucleons, Mass number, Neutrons of 11Na23, 17 Cl35 & 26 Fe56
Solution :
Number of Protons Number of Electrons Number of Nucleons Mass number Number of Neutrons
Nuclides
=Z =Z =A =A = Z-A
11Na23 11 11 23 23 12
35
17 Cl 17 17 35 35 18
26 Fe56 26 26 56 56 30

(f) Isobars:
Atoms of different elements which have same mass number but different atomic numbers are called
isobars.

18
Ar40, 19K40, & 20 Ca40 are isobars. All have same mass number 40 but their atomic numbers are different.
(g) Isotones:
Atoms of different elements containing same number of neutrons in their nucleus are called isotones.

6
C14, 7N15 and 8O16 are isotones as all contain 8 neutrons (A-Z).

6
C14 (14-6=8), 7N15 (15-7=8) and 8O16 (16-8=8)

(e) Isotopes:
Atoms of the same element having same atomic number but different mass numbers are called isotopes.
(i) Hydrogen has three isotopes
1 – protium, 1H2 – Deuterium, 1H3 – Tritium
1H

(ii) Chlorine has two isotopes

35 , 37
17Cl 17Cl
(iii) Oxygen has three isotopes
16, 17, 18
8O 8O 8O
Isotopes have different number of neutrons.

Note: (a) Isotopes show different physical properties.

(b) Isotopes shows similiar chemical properties.

Isotopes of an element are found in nature always together in fixed ratio or percentage known as natural
abundance

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Elements are found in different isotopic forms (atoms of same elements having different atomic mass), so
the atomic mass of any element is the average of all the isotopic mass within a given sample. Example: In
all sample of chlorine collected from nature, 17Cl35 is 75% and 17Cl37 is 25%
we use atomic weight of chlorine 35.5. which is average of atomic mass of its two isotopes
It is calculated as given below :

Average atomic mass  % abundance  atomic mass

100

natural abundence for 35 37 is

17Cl is 75% and for 17Cl 25%

75  35  25  37
So, Aveage Atomic mass Cl =  3 5 .5
100

Example 10:
Use the date given in the following table to calculate the molar mass of naturally occurring argon.
Isotope Isotopic molar mass Abundance
Ar
36
35.96755 g mol -1
7.1%
Ar
38
37.96272 g mol-1 16.3%
Ar
40
39.9624 g mol -1
76.6%

Solution:

35.96755  7.1  37.96272  16.3  39.9624  76.6

Molar mass of Ar  = 39.352 g mol-1
100
Example 11:
Carbon occurs in nature as a mixture of carbon 12 and carbon 13. The average atomic mass of carbon is
12.011. What is the percentage abundance of carbon 12 in nature?
Solution:

Average atomic mass  % abundance  atomic mass

100

x  12  (100  x)13
i.e. 12.011 
100

1201.1 = 12x + 1300 – 13x

x = 1300 – 1201.1 = 98.9%

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1. Which of the following example of a Homogeneous mixture.

(A) Water + Alcohol (B) Water + Sand (C) Water + Oil (D) None of these
2. Which of the following is not an element
(A) Diamond (B) Ozone (C) sulphur (D) silica
3. Which mixture is called as solution.
(A) Hetrogeneous mixture (B) Homogeneous mixture (C) Both (A) and (B) (D) None of these
4. Which of the following is a compound
(A) graphite (B) producer gas (C) cement (D) marble
5. Which of the following statements is/are true :
(A) An element of a substance contains only one kind of atoms.
(B) A compound can be decomposed into its components.
(C) All homogeneous mixtures are called as solutions.
(D) All of these
6. Which one of the following is not a mixture :
(A) Tap water (B) Distilled water (C) Salt in water (D) Oil in water
7. Which one of the follwing forms part of seven basic SI units :
(A) Joule (B) Candela (C) Newton (D) Pascal
8. What mass of silver nitrate (AgNO3) will react with 5.85 g of sodium chloride (NaCl) to produce 14.35 g of
silver chloride (AgCl) and 8.5 g of sodium nitrate (NaNO3), if the law of conservation of mass is true?
(A) 17 g (B) 34 g (C) 68 g (D) 51 g
9. Irrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass
of hydrogen. This explaines the
(A) law of definite proportions (B) law of multiple proportions
(C) law of reciprocal proportions (D) law of conservation of mass
10. Among the following pairs of compounds, the one that illustrates the law of multiple proportions is
(A) NH3 andNCl3 (B) H2S andSO2 (C) CS2 andFeSO4 (D) CuO andCu2 O
11. Law of reciprocal proportion can be verified by following compounds -
(A) CO2, CH4, H2O (B) N2O, N2O3, N2O5 (C) NaCl, Na2CO3, NaOH (D) H2O, HCl, NaCl
12. Oxygen combines with two isotopes of carbon C and 14C to form two sample of carbon dioxide. The data
12

illustrates
(A) Law of conservation of mass (B) Law of multiple proportions
(C) Law of reciprocal proportions (D) None of these
13. The law of conservation of mass holds good for all of the following except
(A) All chemical reactions (B) Nuclear reactions
(C) Endothermic reactions (D) Exothermic reactions
14. A chemical equation is balanced according to the law of
(A) Multiple proportions (B) Constant proportions (C) Reciprocal proportions (D) Conservation of mass
15. The atomic weight of Cu is 63.6. There are only two naturally occurring isotopes of copper, 63Cu and 65Cu.
The natural abundance of the 63Cu isotope is approximately
(A) 20% (B) 70% (C) 30% (D) 80%

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1
The quantity of the mass of an atom of carbon-12 is known as the atomic mass unit and is abbreviated
12
as amu. The masses of atoms, molecules & ions are generally expressed in amu.
It is also known as unified unit (u)

1
It is defined as number of times a given atom is heavier than chosen standard 1 amu i.e.mass of of an
12
atom of carbon-12 .
It is unitless.
Example: Na = 23 , Ca = 40 ,Fe = 56 ,O = 16 etc

It is defined as number of times a given molecules is heavier in comparison to the mass of chosen
1
standard mass of an atom of carbon-12 i.e 1 amu.
12
It is also unitless.
It can be also obtained by adding relative atomic mass of all the atoms present in molecule
Example: H2O = 18 , CO2 = 44 , H2SO4 = 98 ,Na2CO3 = 106 etc.

The atomic mass of an element is actully mass of one atom of the element

AtomicMass or Weight
Relative atomic mass of an element =
1
Mass of th part of mass of C - 12 atom
12

1
Atomic mass = Relative atomic mass X Mass of th part of mass of C - 12 atom
12
As Na has RAM equal to 23
so atomic mass or weight of Na = 23x1 amu = 23 amu

The molecular mass is actully mass of one molecule

Molecular mass or weight

Relative molecular mass of an element  1
Mass of th part of mass of C  12 atom
12

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1
Molecular mass = Relative molecular mass X Mass of th part of mass of C - 12 atom
12
As H2O has RMM equal to 18
Thus, Molecular mass or weight of H2O = 18 x 1 amu =18 amu

It is the number of atoms of carbon persent in 12 gm carbon.

It is found as 6.023 × 1023 atom.
It is represented by NA.

Atomic mass of carbon is 12 amu

Mass of 6.023×1023 (Avagadro Number) carbon atom is 12×6.023×1023 amu
By defination of Avagadro number mass of carbon atoms is 12 gram
12×6.023×1023 amu = 12 gram
6.023 × 1023 amu = 1 gram

1
1 am u  g  1.66  10  24 g
6.023  10 23

Gram atomic mass of element is equal to its atomic mass reported in grams it is also the mass of specified
number of atoms (equal to Avogadro number) of that element. One gram atomic mass of any element
contains the same number of atoms of that element as there are carbon atoms in exactly 12 gm of
carbon-12.
1 atom of carbon = 12 amu
6.023 × 1023 atoms of carbon = 12 × 6.023 × 1023 amu = 12 g = gram atomic weight
1 atom of sodium = 23 amu
6.023 × 1023 atoms of sodium = 23 × 6.023 × 1023 amu = 23 g = gram atomic weight
1 atom of oxygen = 16 amu
6.023 × 1023 atoms of oxygen = 16 × 6.023 × 1023 amu = 16 g = gram atomic weight

Gram molecular mass of molecule is equal to its molecular mass reported in grams it is also the mass of
specified number of molecules (equal to Avogadro number) of that molecule. One gram molecular mass of
any element contains the same number of molecules of that molecule as there are carbon atoms in exactly
12 gm of carbon-12.
1 molecule of water = 18 amu
6.023 × 1023 molecules of water = 18 × 6.023 × 1023 amu = 18 g = gram molecular weight
1 molecule of CO2 = 23 amu
6.023 × 1023 molecules of CO2 = 44 × 6.023 × 1023 amu = 44 gm = gram molecular weight
1 molecule of H2SO4 = 98 amu
6.023 × 1023 molecules of H2SO4 = 98 × 6.023 × 1023 amu = 98 gm = gram molecular weight

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Definition of mole : One mole is the collection of that many entities as there are number of atoms
exactly in 12 g of C –12 isotope.
or 1 mole = collection of 6.023× 1023 species
(species may be atoms or molecules or ions or any other countable objects)
1 Mole = 6.023 × 1023 Particles.
1 Mole atom = 6.023 × 1023 atoms.
1 Mole molecule = 6.023 × 1023 Molecules
1 Mole Electron = 6.023 × 1023 Electrons.
Also 6.023 × 1023 = NA = Avogadro's No.

Note: 1 mole of atoms is also known as 1 g-atom

1 mole of molecules is also known as 1 g-molecule and
1 mole of ions is also known as 1 g-ion

(A) When number of molecules or atoms or ions given

no. of atoms  N 
number of moles of atoms = Avogadro ' s number n = 
 NA 

no. of molecules  N 
number of moles of molecules= Avogadro's number n = 
 NA 

no. of ions  N 
number of moles of ions = Avogadro 's number  n = N 
 A 

(B) When weight of molecules or atoms or ions given , then

Mass of one mole atoms is called as gram atomic mass or molar mass

weight of atoms in grams  weight 

number of moles of atoms = n  
gram atomic mass  molar mass 

Mass of one mole molecules is called as gram molecular mass mass or molar mass

weight of molecules in grams  weight 

number of moles of molecules=  n= 
gram molecular mass  molar mass 

Mass of one mole ions is called as gram ionic mass or molar mass

weight of ions in grams  n = weight 

number of moles of ions =  
gram ionic mass  molar mass 

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(C) Mole calculation when matter in gaseous form ;
Lets study few properties of gases to understand the calculation
(i) Volume
Definition : Volume of gas is nothing but volume of the container in which it is present.
Relation between different units of volume
1 m3 = 103 dm3 = 103 litre =106 cm3 = 106 ml = 109 mm3.
(ii) Temperature
Definition : Degree of hotness or coldness of a body is measured by temperature
It is expressed in
(I) Celcius scale, (II) Kelvin scale,and (III) Fahrenheit scale
Interconversion relations

C K  273 F  32
 
100 100 180
C–Celcius scale,K–Kelvin scale, F–Fahrenheit scale

Note: In all the calulation in chemistry, temperature must be expressed in kelvin scale.
i.e.,tC  273  TK

(iii) Pressure
Definition : Force acting per unit area

F
P Units in SI pascal i.e. Newton/m 2 (1N/m 2 = 1Pa)
A
Units of pressure generally used is bar, atmosphere(atm) ,torr or height of Hg
1 bar =105 Pa or N/m 2
1 atm = 76 cm of Hg
= 760 mm of Hg
= 760 torr
= 1.01325×105 Pa or N/m 2
= 101.325 kPa
= 1.01325 bar
= 14.7 lb/In2 (Psi)
Ideal gas equation PV = nRT
Where P = Pressure of gas, V = Volume of gas
n = Number of moles of gas T = Temperature(Kelvin)
–1 –1
R = Gas constant = 0.082 atm Ltr K mole
–1 –1
= 0.083 bar Ltr K mole
3 –1 –1
= 8.314 Pa meter K mole

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Standard conditions
(i) NTP condition : NTP stands for normal temperature & pressure.
At NTP, pressure is equal to 1 atm and temperature is equal to 273 K.
Under these conditions, the volume of 1 mol of gas i.e molar volume is 22.4 litre

Volum e of gas at NTP in Liters  n = VNTP 

number of moles of at NTP =  22.4 
22.4 

(ii) STP condition :STP for standard temperature pressure.

At STP pressure is equal to 1 bar and temperature is equal to 273.15 K.
Under these conditions, the volume of a gas i.e molar volume is 22.7 liters

Volume of gas at STP in Liters  n = VSTP 

number of moles of at STP =  
22.7  22.7 

Relation between to P,V & T to convert any given state of a gaseous system to NTP or
STP :

P1V1 P2 V2

T1 T2

Where P1, V1, T1 are pressure, volume and temperature at given state of gaseous system and
P2, V2, T2 are pressure, volume and temperature at NTP or STP state of gaseous system

Example 12:
Number of moles in 46 grams of sodium. (Na=23)
Solution:

g 46
moles   2
M0 23

Example 13:
Number of moles in 54 grams of vapours phosphorus (P4 ) (P=31)
(A) 4.355 (B) 43.55 (C) 0.4355 (D) 0.4355 × 1023
Solution:

g 54
moles    0.4355 Hence (C) is correct option.
M0 124

Example 14:
Calculate the mass of sodium which contains same number of atoms as are present in 4 g of calcium.
Atomic mass of sodium and calcium are 23 and 40 respectively.

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Solution:
40 g of calcium contains = 6.023 × 1023 atoms

6.023  10 23 atoms
4 g of calcium contain =  4 atoms  6.023  1022 atoms
40

Now, 6.023 × 1023 atoms of sodium = 23g

23
 6.023 × 1023 atoms of sodium will have mass =  6.023  1022  2.3g
6.023  1023

Example 15:
Calculate the number of moles in each of the following:
(i) Number of moles in 49 g of H2SO4
(ii) 44.8 liters of carbon dioxide at NTP
(iii) Number of moles in 25 g of CaCO3
(iv) 9.0 grams of aluminium.
Solution:
(i) 1 mole Of H2SO4 = 98g

49 1
Moles =   0.5
98 2

(Mol. mass= 2 × 1 + 96 = 98)

(ii) 1 mole of CO2 = 22.4 litters at NTP i.e. 22.4 liters of CO2 at NTP = 1 mole

1
 44.8 litters of CO2 at NTP =  44.8 = 2 moles CO2
22.4

(iii) Number of moles in 25 gm of CaCO3

25
moles   0.25
100

(Mol. Mass = 40 + 12 + 16 × 3 = 100)

(iv) 1 mole of Al = 27 g of Al  Atomic mass of aluminium  27g

i.e. 27 g of aluminium = 1 mole of Al

1
9 g of aluminium =  9 = 0.33 mole of Al
27

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1. 1 amu is equal to

1 1
(A) of O16 (B) o f C12 (C) an atom of H (D) 1.66  10 23 kg
16 12
2. The weight of one atom of Uranium is 238 amu. Its actual weight is .... gm.
(A) 1.43 × 1026 (B) 3.94 × 10–22 (C) 6.99 × 10–23 (D) None of these
3. The number of atoms in 0.004 g of magnesium will be
(A) 4  1020 (B) 8  1020 (C) 1020 (D) 6.02  1020
4. If the mass of an electron is 9 × 10–28 grams, weight of one mole of electrons is
(A) 9 × 10–28 gm (B) 6 × 10–28 (C) 1.008 gm (D) 0.00054 gm
5. The number of milli moles present in 1.0 gram of water
(A) 1.0 (B) 18 (C) 55.55 (D) 100
6. The actual weight of a molecule of water is
(A) 18 g (B) 2.99 × 10–23 g
(C) Both (A) & (B) are correct (D) None of these
7. The number of gram molecules of oxygen in 6.02 × 1024 CO molecules is –
(A) 10 g-molecules (B) 5 g-molecules (C) 1 g-molecules (D) 0.5 g-molecules
8. What is the number of molecules in 16 grams of sulphur dioxide (SO2 ) ?
(A) 3 × 1023 (B) 2.5×1023 (C) 1.5×1023 (D) 4.5×1023
9. The volume of a gas in discharge tube is 1.12 × 10–7 ml. at NTP. Then the number of molecule of gas in the
tube is -
(A) 3.01 × 104 (B) 3.01 × 1015 (C) 3.01 × 1012 (D) 3.01 × 1016
12
10. Sum of number of protons, electrons and neutrons in 12gm of 6 C is :-

(A) 1.8 (B) 12.044 × 1023 (C) 1.084 × 1025 (D) 10.84 × 1023
11. The charge on 1 gram ions of Al3+ is : (NA = Avogadro number, e = charge on one electron)

1 1 1
(A) N e coulomb (B) × NAe coulomb (C) × NAe coulomb (D) 3 × NAe coulomb
27 A 3 9

12. The no. of electrons present in one mole of Azide ion are N3  


(A) 21N (B) 20N (C) 22N (D) 43N

13. 4.5 moles of oxygen atoms are present in
(A) 1.5 moles of SrCO3 (B) 1 mole of SrCO3
(C) 2.5 moles of SrCO3 (D) 2.25 moles of SrCO3
14. The gas having same number of molecules as 16g. of oxygen is
(A) 16g. of O3 (B) 16g. of SO3 (C) 48g. of SO3 (D) 1g of hydrogen
15. The weight of gaseous mixture containing 6.02 × 10 molecules of nitrogen and 3.01 × 1023 molecules of
23

sulphur dioxide
(A) 46g (B) 92g (C) 60g (D) 30g

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16. If the weight of 5.6 litres of a gas at N.T.P is 11 grams, the gas is

(A) Phosphine PH3  (B) Phosgene  COCl2 

(C) Nitric oxide (NO) (D) Nitrous oxide N2 O 

17. 8 gms of O2 has the same number of molecules as

(A) 7 gm CO (B) 14 gm of CO (C) 22 gm of CO2 (D) 44 gms of CO2
18. Maximum number of atoms are present in
(A) 11.35 lit. of SO2 at STP (C) 22.7 lit. of Helium at STP
(C) 2.0 gms. of hydrogen (D) 11.2 litres of methane at NTP
19. Which of the following has highest mass
(A) 50 gms of iron (B) 5 moles of nitrogen
(C) 1 gm atom of silver (D) 5 × 1023 atoms of carbon
20. Maximum number of electrons are present in
(A) 2.27 litre of SO2 at S.T.P. (B) 0.2 moles of NH3
(C) 1.5 g moles of oxygen (D) 2 mole atoms of sulphur
21. Which of the following contains least number of molecules
(A) 11 gms of CO2 (B) 32 gm of SO2
(C) 2 gms of hydrogen (D) 4 gms of helium
22. 10 grams of each O2, N2 and Cl2 are kept in three bottles. The correct order of arrangment of bottles
containing decreasing number of molecules.
(A) O2, N2, Cl2 (B) Cl2, N2, O2 (C) Cl2, O2, N2 (D) N2, O2, Cl2
23. The number of Cl and Ca ions in 222g. of CaCl2 are
– +2

(A) 4NA, 2NA (B) 2NA, 4NA (C) 1NA, 2NA (D) 2NA, 1NA
24. If 0.25 g of a substance when vaporized displaced 50 cm 3 of air at STP. The gram molecular mass of the
substance will be

(A) 50 g (B) 100 g (C) 112 g (D) 127.5 g

It is of two type.
1. Absolute density
2. Relative density
For liquid and solids
mass
(i) Absolute density = unit for absolute density are g/cc or g/L or Kg/m 3
volume

relation between units Kg/m 3 = g/L and also Kg/m 3 =1000 g/cc

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density of the subs tance

(ii) specific gravity = density of water at 4C

If a material has density 1.8 g/mL then its specfic gravity is 1.8
For simplification, we can conclude that the density and specific gravity of any substance is numerically
same, but density has a definite unit, but specific gravity has no unit. (dimension less)

mass mass
For gases : Absolute density = d = 
volume V

mass mass
PV = nRT so PV  RT also P(Mol.wt)  RT
mol.wt V

PM
PM = dRT d Molecular weight is shown as M
RT

Mol wt Mol wt
dNTP  g/L dSTP  g/L
22.4 22.7

It is the density of a substance with respect to any other substance.

Relative density of A with respect to B

PM A
density of A Mol.wt A
RD   RT 
density of B PMB Mol.wt B
RT

Example 16:
What is the density of SO2 with respect to CH4 under similar conditions?
Solution :

PMSO 2
density of SO 2 64
Re lative density   RT  4
density of CH4 PM CH4 16
RT
Example 17:
Find the density of CO2(g) with respect to N2O(g) under similar conditions.
Solution :
density of CO2 44
Relative density   1
density of N2O 44

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Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature
and pressure.

dgas PMgas RT
Vapour density  
dH2 PMH2 RT

Mgas Mgas
V.D. = M = Mgas = 2 x V.D.
H2 2

Example 18:
What is V.D. of oxygen gas
Solution :

32
V.D = = 16
2
Example 19:
7.5 litre of the particular gas at N.T.P. weigh 16 gram. What is the vapour density of gas?
Solution :
7.5 litre = 16 gram
7.5 16
moles =  M = 48 gram
22.4 M
48
V.D. = 24
2

When container contains more than one gas then molecular mass of mixture is termed as mean
molecular mass (Molecular mass of mixture) which is average mass of one mole of mixture in the
container.

Total mass of all the gases present in the container

Average molecular mass = Total no. of moles of all gases present in the container

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Example 20:
The molar composition of polluted air is as follows :
Gas molar mass mole percentage composition
Oxygen 32 16%
Nitrogen 28 80%
Carbon dioxide 44 03%
Sulphur dioxide 64 01%
What is the average molecular weight of the given polluted air ?
Solution :

jn

n M j j j n

Mavg =
j1
jn Here n j = 100

j1
nj
j1

16 x 32  80 x 28  44 x 3  64 x 1 512  2240  132  64 2948

 Mavg = = = = 29.48
100 100 100

Here we are going to find out the percentage of each element in the compound by knowing the molecular
formula of compound.
We know that according to law of definite proportions any sample of a pure compound always possess
constant ratio with their combining elements.
Example 21:
Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources. i.e. 1
mole of ammonia always contains 1 mol of N and 3 mole of H. In other words 17 gm of NH3 always contains
14 gm of N and 3 gm of H. Now find out % of each element in the compound.

Mass of N in 1 mol NH3

Mass % of N in NH3 =  100 = 14 gm × 100 = 82.35 %
Mass of 1 mol of NH3 17

Mass of H is 1 mol NH3 3

Mass % of H in NH3 = Mass of 1 mol e of NH  100 = × 100 = 17.65 %
3 17
Example 22:
What is the percentage of calcium and oxygen in calcium carbonate (CaCO 3) ?
Solution :

40 48
% Ca   100  40% %O  100  48%
100 100

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Example 23:
A compound of sodium contains 11.5% sodium then find the minimum molar mass of the compound.
Solution :

1 23
11.5   100  M  200
M

A sample of organic compound on qualitative analysis (with elements present) was found to contain
C,S,H,N,P,O,Cl,Br & I. The following experiment were performed. Calculate % composition of each element.

These two elements are always estimated together by Liebig’s combustion method. A weighted amount of
the compound is heated strongly with excess copper oxide in an atmosphere of air or oxygen. The
constituents hydrogen and carbon are thus oxidised to water and carbon dioxide, which are collected
separately and weighed. The percentage of carbon and hydrogen in the compound can be calculated as
given below.

12 Weight of CO 2
% of Carbon = ×  100
44 Weight of sample

2 Weight of H2O
% of Hydrogen = × ×100
18 Weight of sample

A known mass of the organic substance containing halogen is heated with fuming nitric acid along with
few crystals of silver nitrate in a sealed tube. The silver halide is formed which is separated, washed dried
and weighed. From the mass of silver halide obtained the percentage of halogen is calculated.

35.5 Weight of AgCl

% of Chlorine = × ×100
143.5 Weight of sample

80 Weight of AgBr
% of Bromine = ×  100
188 Weight of sample

127 Weight of AgI

% of Iodine = ×  100
235 Weight of sample

Sulphur is also estimated by the carius method. In this case, the organic compound is heated only with
nitric acid. Sulphur present in the compound is thus oxidised to sulphuric acid which is treated with barium

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chloride to precipitate barium sulphate. The precipitate of barium sulphate is washed, dried and weighed.
From the weighted mass of barium sulphate obtained, the percentage of sulphur is calculated.

32 Weight of BaSO4
% of Sulphur = ×  100
233 Weight of sample

Phosphorus is estimated like sulphur, an organic compound that is heated with fuming nitric acid. Phos-
phorus in the compound is thus oxidised to phosphoric acid which is precipitated by adding a magnesia
mixture. The precipitate of magnesium ammonium phosphate is ignited to obtain magnesium pyrophos-
phate.
2MgNH4PO4 (Magnesium ammonium phosphate)  Mg2P2O7 + 2NH3 + H2O
Magnesium pyrophosphate is weighed and the percentage of phosphorus calculation is

62 Weight of Mg2P2O7
% of Phosphorus = ×  100
222 Weight of sample

(i) Duma’s Method

This method can be applied to organic compounds containing nitrogen. The method is based on the
principle that if an organic compound containing nitrogen is heated with copper oxide, free nitrogen and
oxides of nitrogen are formed along with other products (carbon dioxide, water vapour etc). The oxides of
nitrogen are reduced to free nitrogen on passing over heated copper and the whole of nitrogen is collected
over KOH solution. The volume of nitrogen collected is measured and from this, the percentage of ‘N’
present in the compound is calculated.

28 Volume of nitrogen gas at NTP

% of Nitrogen = × ×100
22400 Weight of sample

(ii) Kjeldahl’s Method

W g sample was reacted with concentrated H2 SO 4 to cause conversion of all N2 present in the compound.
Ammonium sulphate (NH4 )2 SO4 solution was then added with excess of NaOH and all NH3 librated was
reacted with H2 SO 4 solution. The residual acid was reacted with NaOH solution to complete naturalization
of residual acid. Calculate % composition of N2 from above data ?
Solution :
conc. H SO NaOH (for neutralisation of excess H SO )
Organic Compound 
2
M1V1
4
(NH4 )2 SO 4 
M2 V2
2 4

 NH3

2NH3  H2SO 4  (NH4 )2 SO 4 and residual acid H2 SO 4  2 NaOH  Na2SO4  2H2O

moles of ammonia can be calculated by solving stoichiometry of above reaction

moles of NH3
% of Nitrogen = 14×  100
Weight of sample

% of oxygen in Organic Compound= 100 - [sum of % of all other element]

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Example 24:
In the estimation of sulphur by carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of
barium sulphate. Find the percentage of sulphur in the given compound.
Solution :

Mass of BaSO4  32 100 0.668  32  100

Percentage of sulphur in the compound = × Mass of compound =
233 233  0.468
Percentage of sulphur in the compound = 19.6%

Example 25:
0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in carius estimation. Calculate the
percentage of chlorine present in the compound.
Solution :

35.5  0.5740
Percentage of chlorine = × 100 = 37.6%
143.5  0.3780
Example 26:
0.2585g of an organic compound containing iodine was heated with excess of strong nitric acid and silver
nitrate in Carius tube. The precipitate of silver iodide was filtered, washed and dried. Its weight was found
to be 0.3894g. Calculate the percentage of iodine in the compound.
Solution :
Mass of organic compound = 0.2585g
Mass of AgI formed = 0.3894g

0.3894  127
Mass of I present in 0.2585g of organic compound = g
235

0.3894  127  100

Percentage of I = = 81.4%
235  0.2585

Empirical formula : Formula depicting constituent atom in their simplest ratio.

Molecular formula : Formula depicting actual number of atoms in one molecule of the compound

Relation between Molecular formula &Empirical formula

Molecular formula = Empirical formula × n

Molecular mass
n=
Empirical Formula mass

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Example 27:
Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene and
benzene are 26 and 78 respectively. Deduce their molecular formulae.
Solution :
For acetylene
 Empirical Formula is CH
Step-1
The empirical formula of the compound is CH  Empirical formula mass= (1 × 12) + 1 = 13.
Molecular mass = 26
Step-2
To calculate the value of ‘n’

Molecular mass 26
n = Empirical formula mass = =2
13

Step-3
To calculate the molecular formula of the compound.
Molecular formula = n × (Empirical formula of the compound)
= 2 × CH = C2 H2
Thus the molecular formula is C2 H2
Similarly for benzene
To calculate the value of ‘n’

Molecular mass 78
n = Empirical formula mass = =6
13

thus the molecular formula is 6 × CH = C6H6

To calculate the empirical formula of the compound. we must have either percenage of each element or
grams of each element in a given sample of compound. Let’s study the given example

Example 28:
An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition.
C = 40.684% ; H = 5.085% and O = 54.228%
The molecular weight of the compound is 118. Calculate the molecular formula of the compound.

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Solution:
Step-1
To calculate the empirical formula of the compound.

Element Symbol Percentage At. mass Relative no. Simplest Simplest whole
Percentage
of element of element of atoms = atomic ratio no. atomic ratio
At. mass

Carbon C 40.687 12 40.687 3.390

= 3.390 =1 2
12 3.389
5.085 5.085
Hydrogen H 5.085 1 = 5.085 =1.5 3
1 3.389
54.228 3.389
16 = 3.389 =1 2
Oxygen O 54.228 16 3.389

 Empirical Formula is C2 H3 O2
Step-2 To calculate the empirical formula mass.
The empirical formula of the compound is C2 H3 O2 .
 Empirical formula mass
= (2 × 12) + (3 × 1) + (2 × 16) = 59.
Step-3 To calculate the value of ‘n’

Molecular mass 118

n = Empirical formula mass = =2
59

Step-4 To calculate the molecular formula of the salt.

Molecular formula = n × (Empirical formula) = 2 × C2 H3 O2 = C4 H6 O4
Thus the molecular formula is C4 H6 O4.
Example 29:
An oxide of nitrogen gave the following precentage composition :
N = 25.94
and O = 74.06
Calculate the empirical formula of the compound.
Solution :
N2O5
Example 30:
Hydroquinone, used as a photographic developer, is 65.4%C, 5.5% H, and 29.1% O, by mass. What is the
empirical formula of hydroquinone ?
Solution :
C3H3O

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Example 31:
An organometallic compound on analysis gave the following results :
C = 64.4%, H = 5.5% and Fe = 29.9%. Determine its empirical formula.
Solution :
The given data is tabulated as follows. Since, the sum of mass % is 99.8, hence there is negligible oxygen
in the given compound.

Simplest whole
Element Mass % Atomic mass Atomic ratio Simplest atomic ratio
number ratio

C 64.4 12 64.4/12 = 5.37 5.37/0.53= 10.1 10

H 5.5 1 5.5/1 = 5.5 5.5/0.53= 10.4 10

Fe 29.9 56 29.9/56 = 0.53 0.53/0.53=1 1

Thus, the empirical formula of the compound is C10H10Fe.

1. The density of a gas at N.T.P. is 1.43 gram per litre. The molecular weight of the gas is
(A) 28 (B) 30 (C) 32 (D) 35
2. A mixture of O 2 and a gas with molecular mass 80 is present in the molar ratio a : b. This mixture has
on average molar mass of 40. What would be the average molar mass if the gases are mixed in the
ratio of b : a ?
(A) 36 (B) 72 (C) 82 (D) 18
3. If a pure compound is composed of X2 Y3 molecules and consists of 60% X by weight, what is the atomic
weight of Y in terms of atomic weight of X?

9 4 3 2
(A) AX (B) AX (C) AX (D) AX
4 9 2 3
4. A gas is found to have the formula (CO)x. It's VD is 70 the value of x must be:-
(A) 7 (B) 4 (C) 5 (D) 6
5. Molecular weight = vapour density ×2, is valid for
(A) metals (B) non metals (C) solids (D) gases
6. Density of ozone relative to methane under the same conditions of temperature & pressure is
(A) 1 (B) 3 (C) 1.5 (D) 2.5
7. The percentage composition of nitrogen in urea is about [Molecular formula of urea is CO(NH2)2]
(A) 38.4 (B) 46.6 (C) 59.1 (D) 61.3
8. Caffine has molecular weight of 194. It contains 28.9% by mass of nitrogen.Then the number of nitrogen
atoms of nitrogen in one molecule is
(A) 2 (B) 3 (C) 4 (D) 5

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9. The haemoglobin from the red blood corpuscles of most mammals contains 0.3 % of iron by weight. . The
number of iron atoms in each molecule of haemoglobin is 4. Find the molecular weight of haemoglobin
(atomic weight of iron = 56) :
(A) 22456.34 (B) 56783.23 (C) 74666.67 (D) 67500.5
10. A carbonic compound contains 40% carbon, 13.3% hydrogen and 46.7% nitrogen. What is the empirical
formula?

(A) CH2N (B) C2H4N (C) CH4N (D) CH2N3

11. A monobasic acid with molar mass 63 containing nitrogen, hydrogen and oxygen only has the percentage
composition by weight as: N = 22.22%, H = 1.59%. Determine the molecular formula of the acid.

(A) HNO 2 (B) HNO3 (C) HNO 4 (D) None of these

12. Which of the following compounds has same empirical formula as that of glucose
(A) CH3CHO (B) CH3COOH (C) CH3OH (D) C2H6
13. A gas is found to contain 2.34 g of Nitrogen and 5.34 g of oxygen. Simplest formula of the compound is
(A) N2O (B) NO (C) N2O3 (D) NO2
14. 2.2 g of a compound of phosphorous and sulphur has 1.24 gms of 'P' in it. Its emperial formula is
(A) P2S3 (B) P3S2 (C) P3S4 (D) P4S3
15. On analysis, a certain compound was found to contain iodine and oxygen in the mass ratio of 254:80.
The formula of the compound is : (At mass of I = 127, O = 16)
(A) I2O3 (B) I2O7 (C) I2O (D) I2O5
16. The number of atoms of Cr and O are 4.8 × 10 and 9.6 × 10 respectively. Its empirical formula is
10 10

(A) Cr2O3 (B) CrO2 (C) CrO3 (D) None

17. 14 g of element X combine with 16 g of oxygen. On the basis of this information, which one of the following
is correct ?
(A) The element X could have an atomic weight of 7 and its oxide formula XO
(B) The element X could have an atomic weight of 14 and its oxide the formula X 2O
(C) The element X could have an atomic weight of 7 and its oxide is X2O
(D) The element X could have an atomic weight of 14 and its oxide is XO2
18. The simplest formula of a compound containing 50% of element X (atomic mass 10) and 50% of element
Y (atomic mass 20) is
(A) XY (B) X2Y (C) XY3 (D) X2Y2
19. In a compound C, H, N atoms are present in 9 : 1: 3.5 by weight. Molecular weight of compound is 108.
Its molecular formula is
(A) C2H6N2 (B) C3H4N (C) C6H8N2 (D) C9H12N3
20. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The
percentage of bromine in the compound is : (Atomic mass of Ag = 108 & Br = 80 )
(A) 36 (B) 48 (C) 60 (D) 24

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21. On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of
water. Determine the percentage composition of carbon and hydrogen in the compound.
(A) 3.64% (B) 4.58% (C) 6.98% (D) 5.60%
22. In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 44.8mL of dry nitrogen
gas collected at 273K temperature and 760mm pressure. Calculate the percentage composition of nitrogen
in the compound.
(A) 18.67% (B) 14.50% (C) 16.88% (D) 15.60%
23. In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out
the percentage of bromine in the compound.
(A) 22.11% (B) 34.04% (C) 46.14% (D) 38.12%
24. In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the
percentage of sulphur in the compound?
(A) 42.10% (B) 54.04% (C) 48.14% (D) 56.12%

Stoichiometry is the calculations of the quantities of reactants and products involved in a chemical reaction.
Following methods can be used for solving problems.
(a) Mole Method (For Balanced reaction)
(b) POAC method (Balancing not required)
(c) Equivalent concept (we will study in mole concept II)

Lets consider the following balanced equation

2A  3B  4C  5D
The number written before atoms/molcule in a balanced chemical equation are known as stoichiometric
co-efficient. In mole concept method it is required to take data in terms of mole. Calculations are done
using mole only and then the mole is converted into required data

2A  3B  4C  5D
x mole ? ? ?

Suppose x mole of A is given and it is asked to calculate the moles of other reactants and products involved
in the reaction.
Divide the balancing co-efficient of each substance by the co-efficient of given substance (A) then multiply
with the given mole (x) of (A)

2A  3B  4C  5D
3 4 5
x mole x mole x mole x mole
2 2 2

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Similar study here

7P + 2Q → 3R + 5S
10 mol ? ? ?
2 3 5
When taken x10 mol x10 mol x10 mol
7 7 7
Required Formed Formed

Example 32:
Calculate the amount of oxygen required for complete combution for 39 gm benzene and also calculate
amount of products formed ?

2C6H6  15O2  12CO2  6H2O

(Data in gm) 39 gm ? ? ?
39
(Data in mole)  0.5 mole
78
15 12 6
0.5 mole  0.5  0.5mole  0.5mole
2 2 2
 3.75 mole 3 mole 1.5 mole
 3.75  32 gm 3  44  132gm 1.5  18gm
 3.75  22.4L(at NTP) 3  22.4L (at NTP) 1.5  6.023  10 23
 3.75  6.023  1023 molecules molecules

Example 33:
By heating 10g of CaCO 3.What is the weight of CaO & CO2 obtained in this reaction?
Solution :

CaCO3 
 CaO  CO2
10g
10 10 10
mole mole mole
100 100 100

10
Mole of CaCO3  mole
100

10 10
Mole of CaO  mole g of CaO   56  5  6 g
100 100

10 10
Mole of CO2  mole g of CO2   44  4  4 g
100 100

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Example 34:
Calculate the volume of hydrogen liberated at 27°C and 760 mm pressure by treating
1.2g of magnesium with excess of hydrochloric acid.
Solution :

Mg  2HCl  MgCl2  H2

1.2g
1.2
mole
24

1.2
Mole of Mg  mole
24

1.2
Mole of H2  mole
24

1.2 0821 300

Vol of H2   = 1.23 litre.
24 1atm

Limiting Reagent : The reactant which is totally consumed first during the course of reaction and when it
is consumed reaction stops.
Lets consider the balanced equation :

2A  3B  4C  5D  6E
a mole b mole c mole 0 0
To determine the limiting reagent divide the given mole with the balancing co-efficient

A B C
a b c
2 3 4
The smallest of them is the limiting reagent other are excessive reagent

b
Suppose   is smallest in these then B is called the limiting reagent, A and C are excessive reagent and
3
these are not consumed fully during reaction,
All calculation of consumed or formed should be done with moles of limiting reagent only like

2A  3B  4C  5D  6E
Initially given
a mole b mole c mole 0 0

Limiting reagent is B which is fully consumed but the amount of A consumed is 2 b mole , the mole of C
3

4 5 6
consumed is b mole and mole D and E formed are b mole & b mole
3 3 3

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Finally present

2A  3B  4C  5D  6D

 2   4  5 6
 a  b  mole 0mole  a  b  mole b mole b mole
 3   3  3 3
unreacted unreacted formed formed

It is based on the concept that atoms can neither be created nor be destroyed during chemical reaction,
only recombination of atoms take place. Atoms are conserved,moles of atoms shall also be conserved in a
chemical reaction.This principle is applicable even when we don’t get the idea of balanced chemical equa-
tion on problem. The strategy here will be around a particular atom. We focus on an atom and conserve it
in that reaction principle can under stand by the following Example:.
Consider the decomposition of KClO3(s)  KCl (s) + O2(g) (unbalanced chemical reaction)
Apply the principle of atom conservation (POAC) for K atoms.
Moles of K atoms in reactant = moles of K atoms in products
or moles of K atoms in KClO3 = moles of K atom in KCl.
Now, since 1molecule of KClO3 contains 1 atom of K
or 1 mole of KClO3 contains 1 mole of K, similarly, 1 mole of KCl contains 1 mole of K
Thus, moles of K atoms in KClO3 = 1 × moles of KClO3
and moles of K atoms in KCl = 1 × moles of KCl.
 moles of KClO3 = moles of KCl
Again, applying the principle of atom conservation for O atoms,
moles of O in KClO3 = 3 × moles of KClO3
moles of O in O2 = 2 × moles of O2
 3 × moles of KClO3 = 2 × moles of O2

1. In a gaseous reaction
aA + bB  cC + dD,
which statement is wrong ?
(A) a litre of A combines with b litre of B to give C and D
(B) a mole of A combines with b moles of B to give C and D
(C) a gm of A combines with b gm of B to give C and D
(D) a molecules of A combines with b molecules of B to give C and D
2. Calculate the mass of iron, which will be converted, into its oxide (Fe3O4) by the action of 18g of steam on it.
(A) 21 g (B) 42 g (C) 64 g (D) 51 g

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3. Calculate the volume of CO 2 formed at NTP by heating 4.2 g of NaHCO 3.

 2NaHCO3  Na2CO3  H2O  CO2 

(A) 0.28L (B) 0.112L (C) 1.12L (D) 0.56L
4. Calculate the amount of CaCO3 required to be heated in order to collect 1.135 litre of CO 2 at STP.
(A) 2g (B) 3g (C) 4g (D) 5g
5. Calculate the volume of O 2 required for complete combustion of 1 litre CH4 gas at NTP.
(A) 2L (B) 4L (C) 6L (D) 8L
6. When sodium bicarbonate is heated 1.806 x 10 molecules of water is obtained. Then find the volume of
24

CO2(g) obtained at NTP and amount of NaHCO3 needed for this reaction.

2NaHCO3 (s) 
 Na2CO3 (s)  H2O(g)  CO 2 (g)

(A) 67.2 L, 504 g (B) 22.4 L, 252 g (C) 67.2 L, 252 g (D) 22.4 L, 504 g
7. How many mole of Zn(FeS2) can be made from 2 mole zinc, 3 mole iron and 5 mole sulphur.

 Zn  Fe  2S  ZnFeS2 
(A) 2 mole (B) 3 mole (C) 4 mole (D) 5 mole

8. A chemist wants to prepare diborane by the reaction : 6LiH  8BF3  6LiBF4  B 2H6

If he starts with 2.0 moles each of LiH & BF3. How many moles of B2H6 can be prepared.
(A) 0.25 (B) 0.5 (C) 2.0 (D) 2.5
9. If 0.5 moles of BaCl2 is mixed with 0.2 moles of Na3PO4 the maximum number of moles of Ba3(PO4)2 that
can be formed is
3BaCl2 + 2Na3 PO4  Ba3 (PO4)2 + 6NaCl
(A) 0.7 (B) 0.5 (C) 0.3 (D) 0.1
10. The moles of O2 required for reacting with 6.8 g of ammonia.
(4NH3 + 5O2  4NO + 6H2O) is
(A) 5 (B) 2.5 (C) 1(D) 0.5
11. Carbon reacts with chlorine to form CCl4. 36 g of carbon was mixed with 142 g of Cl2. Calculate mass of
CCl4 produced and the remaining mass of reactant.
(A) 24 g, 72 g (B) 24 g, 154 g (C) 12 g, 72 g (D) 12 g, 154 g
12. Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen :

KO2  s   H2 O  l 
 KOH  s   O2  g

If a reaction vessel contains 0.158 mol KO2 and 0.10 mol H2O, how many moles of O2 can be produced
(A) 0.01185 (B) 0.1185 (C) 1.185 (D) 11.85

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actual yield
The percentage yield of product  theoretical yield  100

Example 35:
Al and KClO3 react together to form Al2O3 according to
80%
2KClO3  2KCl  3O 2

90%
4Al  3O2  2Al2 O3
4 mol of KClO3 on reaction with excess of Al forms how many moles of Al2O3.
Solution :

80 90
mole of Al2O3 = 4    2.88 mol
100 100
Percentage purity of sample:
A given sample may not be 100% pure it may contain some part of impurity and it is assumed that
impuirity will not be participitate in the chemical reaction.

Wreacted
% purity = Wgiven × 100

B
A
C
When a substance reacts simultaneously to give two different products a parallel or competitive reaction is
said to take place.
Example 36:
3 B + 4C
A
5 D + 2E
If initially 10 moles of A were taken and after the completion of reaction (100% of A reacted) it was found 20
moles of D was formed. Find the number of moles of B, C and E formed.
Solution :

A  3B + 4C
x 3x 4x
A  5D + 2E x + y = 10 5y = 20 x=6&y=4
y 5y 2y

B = 3x = 18 C = 4x = 24 E = 2y = 8

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When a reaction occur in two or more steps such that one of the product of 1st step is reactant in next
step.

A 
 B 
C
Example 37:
NH3 is obtained in the following steps

(i) Ca  2C 
 CaC2

(ii) CaC2  N2 
 CaCN2  C

(iii) CaCN2  3H2O 

 2NH3  CaCO3

To obtain 2 mol NH3, Calcium required is

Solution :

n(Ca)  n(CaC2 ) ... (i)

n(CaC2 )  n(CaCN2 ) .... (ii)

2× n(CaCN2) = 1 × mole NH3 .... (iii)

from (i), (ii) and (iii), n(Ca)  1 mol

Many samples of chemicals are not pure. We can define percent purity as

massof mass of pure compound in the impure sample

 100
total mass of impure sample

If an impure sample of a chemical of known percent purity is used in a chemical reaction, the percent purity
has to be used in stoichiometric calculations. Conversely, the percent purity of an impure sample of a
chemical of unknown percent purity can be determined by reaction with a pure compound as in an acid-
base titration. Percent purity can also be determined, in theory, by measuring the amount of product
obtained from a reaction. This latter approach, however, assumes a 100% yield of the product.

Example 38:
On thermal decomposition of 600 g of limestone 44.8 L of CO2 is released at STP. The percentage purity of
limestone is
Solution:
CaCO3(s) 
 CaO(s) + CO2(g)

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44.8
ncu2   2moles
22.4

 nCaCO3  2moles  WCaCO3  nCaCO3  MCaCO3

= 2 × 100 = 200 g

200
 % purity of CaCO3  100 = 33.33 %
600

Example 39:
5g of MnO2 on reaction with HCl forms 1.12 L of Cl2 gas at S.T.P. What is the percentage purity of MnO 2 ?
Solution:

1.12
nCl2 =  0.05moles
22.4

MnO2  4HCl 
 MnCl2  Cl2  2H2O

 1 mole of MnO2 yield 1 mole of Cl2

 nMnO2  0.05mole  WMnO2  4.35g

4.35
%purity of MnO2   100  87%
5

1. A power company burns approximately 474 tons of coal per day to produce electricity. If the sulphur content
of the coal is 1.30% by weight, how many tons SO 2 are dumped into the atmosphere each day ?
(A) 12.3 (B) 18.6 (C) 24.6 (D) 36.8
2. The percent loss in weight after complete decomposition of a pure sample of potassium chlorate
KClO3  s  
 KCl  s   O2  g is

(A) 19.18 (B) 29.18 (C) 39.18 (D) 49.18

3. 5 g of CaCO 3 when heated the CO2 liberated was found 1 litre at STP. Calculate the percentage
purity of the CaCO 3 sample. CaCO3  CaO  CO2 

(A) 88.1% (B) 8.81% (C) 44.1% (D) 4.41%

4. When 5 g mixture of NaHCO3 and Na2CO3 was heated, 560 ml of CO2 was collected at NTP. Calculate
percentage of composition of the mixture.  2NaHCO3  Na2CO3  H2O  CO2 
(A) 16%NaHCO3, 84%Na 2CO3 (B) 76%NaHCO 3, 24%Na 2CO 3

(C) 84%NaHCO3, 16% Na 2 CO3 (D) 24%NaHCO3, 76%Na2 CO3

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5. Sulphur trioxide may be prepared by the following two reactions :

S8  8O 2  g   8SO2  g  2SO 2  g   O2  g  

 2SO3  g 

How many grams of SO3 will be produced from 1 mol of S8 ?

(A) 320 g (B) 640 g (C) 1120 g (D) 1280 g
6. NH3 is obtained in the following steps

(i) Ca  2C 
 CaC2 (ii) CaC2  N2 
 CaCN2  C

(iii) CaCN2  3H2O 

 2NH3  CaCO3

Moles of Calcium required , to obtain 2 mol NH3 is

(A) 2 mol (B) 1 mol (C) 1.5 mol (D) 2.5 mol
7. NaOH is formed according to reaction

1
2Na  O2 
 Na2O Na2 O  H2 O 
 2NaOH
2
To form 4 g of NaOH mass of Na required is
(A) 4.6 g (B) 4.0 g (C) 2.3 g (D) 0.23 g
8. An alloy of aluminium, and copper was treated with aqueous HCl. The aluminium dissolved according
3
to the reaction : Al + 3H+  Al3+ + H,
2 2
but the copper remained as pure metal. A 0.5g sample of the alloy gave 560 cc of H 2 measured at 273
K and 1 atm pressure. What is the weight percentage of Al in the alloy ?
(A) 85.3% (B) 75.3% (C) 65.3% (D) 90%
9. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of this ore would have
to be processed in order to obtain 1.00g of pure silver. (Ag = 108, S = 32)  Ag2 S  2Ag  S

(A) 74.6g (B) 85.7g (C) 107.9g (D) 134.0g

10. 9 moles of “D” and 14 moles of E are allowed to react in a closed vessel according to given reactions.
Calculate number of moles of B formed in the end of reaction, if 4 moles of G are present in reaction vessel.
(Percentage yield of reaction is mentioned in the reaction)

Step-1 3D  4E 
80%
 5C  A

Step-2 3C  5G 
50%
 6B  F
(A) 2.4 (B) 30 (C) 4.8 (D) 1
11. 21.6 g of silver coin is dissolved in HNO3. When NaCl is added to this solution, all silver is precipitated as
AgCl. The weight of AgCl is found to be 14.35 g then % of silver in coin is :

Ag  HNO3  AgNO3
AgNO3  NaCl  AgCl  Na 2CO3

(A) 50% (B) 75% (C) 100% (D) 15%

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12. Al and KClO3 react together to form Al2O3 according to
50%
2KClO3  2KCl  3O 2

20%
4Al  3O2  2Al2 O3

4 mol of KClO3 on reaction with excess of Al forms how many moles of Al2O3.
(A) 1.82 (B) 0.8 (C) 2 (D) 0.4
13. Percentage yield of NH3 in the following reaction is 80%.

NH2CONH2  2NaOH  Na2CO3  2NH3

The mass of NH3 formed when 6 g NH2CONH2 reacts with 8 g of NaOH is

(A) 2.72 g (B) 3.4 g (C) 4.25 g (D) 11.2 g

2B + 3C
14. A
4D + 5E

If initially 10 moles of A were taken and after the completion of reaction (100% of A reacted) it was found 16
moles of D was formed. Find the number of moles of B, C and E formed.
(A) B = 20mol, C = 18mol and E = 12 mol (B) B = 12mol, C = 18mol and E = 20 mol
(C) B = 18mol, C = 12mol and E = 20 mol (D) B = 12mol, C = 20mol and E = 18 mol
15. The chief ore of Zn is the sulphide, ZnS. The ore is concentrated by froth floatation process and then heated
in air to convert ZnS to ZnO.

2 ZnS  3O 2 
80%
 2 ZnO  2SO 2

ZnO  H2SO 4 

100%
 ZnSO 4  H2 O

2 ZnSO 4  2H2 O 

80%
 2 Zn  2H2SO 4  O2
The number of moles of ZnS required for producing 2 moles of Zn will be :
(A) 3.125 (B) 2 (C) 2.125 (D) 4

The volumetric analysis of gaseous reaction using eudiometric tube called Eudiometry or “Volume analysis
of gas” In eudiometric tube all the measurement of volume is done at constant pressure and temperature
and given gaseous reaction at least two components are gases.
The edudiometric relationship amongst gases, when they react with one another, is governed by two laws,
they are illustrated as, Gay-Lussac law and Avogadro’s law.
(1) Gay-Lussac Law:
According to Gay Lussac’s law, the volumes of gaseous reactants reacted and the volumes of gaseous
products formed, are always bear a simple ratio at same temperature and pressure.

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(2) Avogadro’s Law:
According Avogadro law A samples of different gases which contain the same number of molecules occupy
the same volume at the same temperature and pressure. This law is also known as Avogadro’s hypothesis
lets us consider following
From experiment it was found that 10 L of N2 require 30 L of H2 to produce 20 L NH3.

N2 + 3H2  2NH3

 volume ratio = 1 : 3 : 2 [Stoichiometric coefficient]

Since the reaction is carried out at constant T and P, then, volume of gaseous mixture is directly proportional
to number of moles of gases and change in volume is proportional to change in number of moles of gases.

V n
 V  n
The expression can be used if pressure of individual species used [at constant T and V]

Pn
 P  n

Hence, for

N2  3H2  2NH3
1 mole 3mole 2mole
1 atm of N2 3 atmof N2 2 atm of NH3

So,

Vn [at constant T and P]

Pn [at constant T and V]

Analysis of gaseous substances in eudiometric tubes

Gaseous reactions are studied in this tube. After the completion of reaction, the reaction is brought back
to original temperature.
To estimate the volume of each gas, certain solvents are used.
Solvent Gases absorbed
1. KOH CO2, SO2, Cl2
2. CuSO4/CaCl2 H2O (g)
3. H2O NH3, HCl
4. Terpentine oil O3
5. Alk. Pyrogallol O2
6. Ammonical Cu2Cl2 CO
We will consider only the volume of gases and not of solid and liquid as they have very less volume
compared to gas. Hence, the volume of solids and liquids is neglected with respect to gases.
(i) Reaction analysis

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Example 40:
20 mL of CO was mixed with 50 mL of oxygen and the mixture was exploded. On cooling, the resulting
mixture was shaken with KOH. Find the volume of the gas that is left.
Solution :
2CO + O2  2CO2
20ml 10ml 20ml
for 20 mL of CO the volume of O2 reacted will be 10 mL. Thus, the volume of O2 remaining unreacted is
40 mL. Further, as KOH absorbs CO2 produced in the reaction, the only gas left is O2, the volume of which
is 40 mL
(ii) Molcular formula determination
Example 41:
10 mL of alkane on complete combustion gives 30 mL of CO2. Calculate molecular formula ?

 y y
C xHy   x   O2  xCO2  H2O
 4 2
1 mole gives x mole
1 mL give x mL
10mL 10  xmL  30mL given
and x3

 y y
10mL 10  x   ml 10x mL 10  mL
 2 2

Since, molecular formula of alkane CxH2x 2  CxHy

So, y = 8
Therefore, molecular formula = C3H8
(iii) Mixture analysis
Example 42:
50 mL of a mixture of CO and CH4 was exploded with 85 mL of O2. The volume of CO2 produced was 50 mL.
Calculate the percentage composition of the gaseous mixture if all volumes are measured under the same
conditions, and the given volume of O2 is just sufficient for the combustion of 50 mL of the mixture of CO and
CH4.
Solution:
Assume volume of CH4 = x mL and volume of CO = y mL.
x + y = 50 ..............................(i)

2CO + O2  2CO2

x x/2 x

CH4  2O2  CO2  2H2 O

y 2y y
x/2 + 2y = 85 ............................(ii)

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on solving the volume of CH4 = x=40 mL and volume of CO = y = (50 – 40) = 10 mL.

40
 percentage of CH4 in the mixture = × 100 = 80% and percentage of CO = 20%.
50

1. What volume of oxygen gas (O2) measured at 0ºC and 1 atm, is needed to burn completely 1 L of propane
gas (C3H8) measured under the same conditions?
(A) 10 L (B) 7 L (C) 6L (D) 5 L
2. Calculate the volume of CO2 produced by the combustion of 40 mL of acetone(CH3COCH3) in the presence
of excess of oxygen.
(A) 120 mL. (B) 100 mL (C) 80 mL (D) 60 mL
3. If a mixture containing 12 litres of chlorine is exploded in an eudiometer tube, what will be the composition
of the resulting mixture by volume ?
(A) HCl = 22.4 litre, H2 = 0.8 litre. (B) HCl = 22.4 litre, H2 = 0.4 litre.
(C) HCl = 11.2 litre, H2 = 0.4 litre. (D) HCl = 11.2 litre, H2 = 0.8 litre.
4. 500 mL of a hydrocarbon gas, burnt in excess of oxygen, yields 2500 mL of CO2 and 3 litres of water vapour
all volumes being measured at the same tempearature and pressure.What is the formula of the
hydrocarbon ?
(A) C3H8 (B) C4H10 (C) C5H12 (D) C6H12
5. When 0.05 litre of a mixture of hydrogen and oxygen was exploded, 0.005 litre of oxygen remined. Calclate
the initial composition of the mixture in per cent by volume.
(A) O2 : 40.0% , H2 : 60% (B) O2 : 20.0% , H2 : 30%
(C) O2 : 60.0% , H2 : 40% (D) O2 : 30.0% , H2 : 20%
6. 12 mL of a gaseous hydrocarbon was exploded with 50 mL of oxygen. The volume measured after explo-
sion was 32 mL. After treatment with KOH the volume diminished to 8 mL. Determine the fomula of the
hydrocarbon.
(A) C3H8 (B) C2H6 (C) C4H8 (D) C5H12
7. The explosion of a mixture consisiting of one volume of a gas being studied and one volume of H2 yielded
one volume of water and one volume of N2 , the volumes being measured under indentical conditions. Find
the formula of the gas being studied.
(A) N2O (B) NO2 (C) NO (D) N2O3
8. 5 mL of a gas containing C and H was mixed with an excess of oxygen (30 mL) and the mixture exploded
by means of an electric spark. After the explosion the volume of the mixed gases remaining was 25 mL. On
adding a concentrated solution of KOH, the volume further diminished to 15 mL, the residual gas being pure
oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas.
(A) C4H8 (B) C3H6 (C) C5H10 (D) C2H4 .
9. 20 mL of a gas containing H and S was heated with tin. When the reaction was over, there was no change
in volume. The residual gas was hydrogen. If the molecular weight of the gas is 34, calculate the molecular
formula.
(A) H2S2 (B) HS (C) H2S (D) NONE

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10. When a certain quantity of oxygen was ozonised in a suitable apparatus, the volume decreased by 4 mL.
On addition of turpentine the volume further decreased by 8 mL. All volumes were measured at the same
temperature and pressure. From these data, establish the formula of ozone.
(A) O2 + O (B) O3 (C) O2 (D) O
11. 20 mL of a mixture of C2H2 and CO was exploded with 30 mL of oxygen. The gases after the reaction had
a volume of 34 mL. On treatment with KOH, 8 mL of oxygen remained. Calculate the composition of the
mixture.
(A) C2H2 : 8 mL ; CO : 12 mL (B) C2H2 : 6 mL ; CO : 14 mL
(C) C2H2 : 6 mL ; CO : 12 mL (D) C2H2 : 8 mL ; CO : 14 mL
12. On passing 25 mL of a gaseous mixture of N2 and NO over heated copper, 20 mL of the gas remained.
Calculate the percentage of each gas in the mixture.
(A) N2 : 40% ; NO : 80% (B) N2 : 40% ; NO : 60%
(C) N2 : 60% ; NO : 40% (D) N2 : 30% ; NO : 60%
13. 38 mL of a mixture of CO and H2 was exploded with 31 mL of O2 . The volume after the explosion was 29 mL
which reduced to 12 mL when shaken with KOH. Find the percentage of CO and H 2 in the mixture.
(A) CO = 44.7% ; H2 = 55.3%. (B) CO = 44.7% ; H2 = 55.3%
(C) CO = 44.7% ; H2 = 55.3% (D) CO = 44.7% ; H2 = 55.3%
14. A mixture of 20 mL of CO, CH4 and N2 was burnt in an exces of oxygen, resulting in reduction of 13 mL of
volume. The residual gas was treated with KOH solution when there was a further reduction of 14 mL in
volume. Calculate the volumes of CO and CH4 in the given mixture.
(A) 12 mL , 6 mL (B) 10 mL , 4 mL (C) 5 mL , 3 mL (D) 8 mL , 6 mL
15. When 100 mL of an O2 & O3 mixture was passed through turpentine, there was reduction of volume by 20
mL. If 100 mL of such a mixture is heated, what will be the increase in volume ?
(A) 10 mL (B) 6 mL (C) 8 mL (D) 12 mL

The concentration of a solution can be expressed in a number of ways. The important methods are:
(a) Strength of solution : It is defined as amount of solute in grams present in one litre solution

weight of solute (gram) w

S = =
volume of solution in litre V in (l )

(b) Mass percent or percentage by mass (% w/w):

Mass of solute
%(w/w) Mass percentage of solute =  100
Mass of solution

Mass of solute  100 Mass of solute  100

= =
Mass of solute  Mass of solvent Volume of solution  Density of solution

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(c) Percent mass by volume (%w/v):

Mass of solute
%(w/v) =  100
Volume of soltuion
(d) Percent volume by volume(% v/v)

Volume of solute
%(v/v) =  100
Volume of soltuion

Sl.
No.
Percentage concentration Physical meaning

1. 20%(w/w) NaCl Solution in water In 100g Solution NaCl = 20g; H2O=80 gm

2. 30%(w/v) NaCl Solution in water In 100 gm Solution NaCl = 30g; H2O=(100×dsolution–30)

3. 25%(v/v) Alcohol Solution in water In 100 ml Solution 25 ml alcohol and 75 ml water.

Note: %w/v = % w/w × dsolution

Example 43:
20 cm3 of an alcohol is dissolved in 80 cm 3 of water. Calculate the percentage of alcohol in solution.
Solution :
Volume of alcohol = 20 cm 3
Volume of water = 80 cm3

20
 percentage of alcohol =  100 = 20.
20  80
(e) Parts per million (ppm) : when solute is very less

Mass of solute
ppm by mass = × 106
Mass of solution

Massof solute
  106
Massof Solvent

(f) Parts per billion (ppb) : when solute is very very less

Mass of solute
ppm by mass = × 109
Mass of solution

Massof solute
  109
Massof Solvent

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Example 44:

One litre of sea water weighs 1030 g and contains about 6  10–3g of dissolved O2. Calculate the
concentration of dissolved oxygen in ppm.
Solution :

Mass of O2 in mg = 6  10–3g  103 mg/g = 6mg

Mass of O2 in mg 6 6  1000
ppm of O2 in 1030 g sea water = Mass of sea water in kg  (1030 /1000) kg   5.8 ppm
1030

(g) Mole fraction :

Let n moles of solute (A) and N moles of solvent (B) be present in a solution.

n N
Mole fraction of solute = = XA , Mole fraction of solvent = = XB
N n N n

In binary solution, XA + XB = 1

Mole fraction is independent of temperature of the solution.

XA n
Note: 1. Ratio of mole fraction is equal to ratio of their moles X  N
B

2. If total mole of mixture is assumed as one then mole fraction can be taken as mole.

(h) Molality

No. of moles of solute

Molality (m) = weight (in kg) of solvent

Let wA grams of the solute of molecular mass mA be present in wB grams of the solvent, then

wA
Molality (m) = m  w  1000
A B

Note : Molality is the most convenient method to express the concentration because it involves the mass of
liquids rather than their volumes. It is also independent of the variation in temperature.
Example 45:
225 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in terms of
molality. (Mol. wt. of urea = 60)
Solution :
Mass of urea = 5 gm
Molecular mass of urea = 60

5
Number of moles of urea = = 0.083
60
Mass of solvent = (255 – 5) = 250 gm

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Number of moles of solute

 Molality of the solution =  1000
Mass of solvent in gram

0.083
=  1000 = 0.332
250
(i) Molarity (Molar concentration)

No. of moles of solute

Molarity (M) =
Volume (in litre) of solution

wA
Molarity of the solution = m  V  1000
A mL

Note : On increasing temperature Molarity decreases.

Number of moles of solute = Molarity × volume (in litre)
Number of millimoles of solute = Molarity × volume (in mL)

Dilution :
A solution can be diluted by adding solvent. During addition of solvent number of millimoles of solute
remain constant.
M1V1 = M2V2
Millimoles before dilution Millimoles After dilution
Molarity of mixing :
M1V1 + M2V2 + M3V3 = MR(V1 + V2 + V3)
MR = resultant molarity
Example 46 :
Calculate the molarity of pure water.
Solution:
Molecular weight of water = 18gm; density of water = 1gm/ml
If volume of water is 1L then its mass is 1000gm.

1000
Number of moles =
18

1000
Molarity of water = = 55.55 M
18

Example 47:
149 gm of potassium chloride (KCl) is dissolved in 10 L of an aqueous solution. Determine the molarity of
the solution (K = 39, Cl = 35.5)
Solution :
Molecular mass of KCl = 39 + 35.5 = 74.5 gm

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149 gm
 Moles of KCl = 74 .5 gm = 2

2
 Molarity of the solution = = 0.2 M
10
Example 48:
Calculate the resultant molarity of following :
(a) 200 ml 1M HCl + 300 ml water
(b) 1500 ml 1M HCl + 18.25 g HCl
(c) 200 ml 1M HCl + 100 ml 0.5 M HCl
Solution:
(a) 200 ml 1M HCl + 300 ml water
M1V1 = M2V2

M2 = 200  1
 0.4M
(200  300)

(b) 1500 ml 1M HCl + 18.25 g HCl

18.25
Millimole of HCl = 1500×1 +  1000  2000
36.5

2000
Molarity =  1.33M
1500
(c) 200 ml 1M HCl + 100 ml 0.5 M HCl
M1V1 +M2V2 = MfVf

200  1  100  0.5 250

Mf    0.833
300 300
Example 49:
Calculate the molarity of H+ ion in the resulting solution when 200 ml 1M HCl is mixed with 200 ml 1M
H2SO4
Solution :
For HCl M1 = 1 M V1 = 200 mL
For H2SO4 M2 = 1 M V2 = 200 mL
nHCl = MHCl x VHCl (L) = 1 x 0.2 =0.2mole

HCl  H+ + Cl– nH = 0.2 (from HCl)

nH2SO 4  MH2SO4  VH2SO4 = 1 × 0.2 = 0.2 mole

H2SO4  2H+ + SO4–2 nH = 0.4

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from H2SO4

Total H+ = nH (from HCl) + nH (from H2SO4) = 0.2 + 0.4 = 0.6
Total volume = 200 + 200 = 400 mL = 0.4 L MR = Resultant molarity

nH 0.6
=  =1.5 Ans.
Vsolution 0.4

Example 50 :
What are the final concentration of all ions when the following solutions are mixed.
50 ml of 0.12 M Fe(NO3)3 + 100 ml of 0.1 M FeCl3 + 100 ml of 0.26 M Mg(NO3)2

1 18  52 70
[NO3–] = =  = 0.28
2 250 250

100  0.1 3
[Cl–] = =0.12 M
250

100  0.26  1
[Mg++] = =0.104 M
250

50  0.12  1  100  0.1  1

[Fe3+] = = 0.064 M
250
Example 51 :
CaCO3 reacts with aq. HCl to give CaCl2 and CO2 according to reaction
CaCO3 (s)  2HCl(aq)  CaCl2  CO2  H2O

How much mass of CaCO3 is required to react completly with 100 ml of 0.5 M HCl
Solution :
millimole of HCl = 100 × 0.5 = 50

2 mole of HCl reacts  1 moles CaCO3

1
1 mole of HCl reacts  moles CaCO3
2
1
50 mmole of HCl reacts  × 50 = 25 mmole of CaCO3
2
25 25
mole of CaCO3 = mass of CaCO3 = × 100 = 2.5 gm.
1000 1000
Example 52 :
Na2CO3 + 2HCl  2NaCl + CO2 + H2O

(a) moles of NaCl formed when 10.6 gm of Na2CO3 is mixed with 100 ml of 0.5 M HCl solution.
(b) Calculate the concentration of each ion in the solution after the reaction
(c) Volume of CO2 liberated at N.T.P.

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Solution :
molar mass of Na2CO3 = 106 gm

10.6
Na2CO3 = × 1000 = 100 m mole
106
HCl = 100 × 0.5 = 50 milli mole
Na2CO3 + 2HCl  2NaCl + CO2 + H2O

1
limiting reagent  HCl  1 mole HCl reacts = mole of Na2CO3
2
50 mmole HCl reacts = 25 mmole of Na2CO3
remaining Na2CO3 = 100 – 25 = 75 mmole

50
(a) 2mole HCl gives = 2 mole of NaCl moles of NaCl =50 mmole= = 0.05
1000

25
(c) volume of CO2 =  22.4 = 0.556 L
1000

50  150 200
[Na+] =  =2M
100 100

50
[Cl–] = = 0.5 M
100

75
[CO32–] = = 0.75 M
100
(i) Formality :
Since molecular weight of ionic solids is not determined accurately as they do not exist as molecular
units and therefore molecular weight of ionic solid is often referred as formula weight and molarity as
formality.

wt. of solute
Formality = i.e., molarity
Formula wt.  V(in l )

(a) Relation between mole fraction and Molality :

X A  1000 wA  1000
XB  mB = m = wB  mA

(b) Relation between molarity and % solute (W/V):

let it contains x% (W/V) solute by mass.

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mass of solute in gram x  10

Number of moles of solute in 1 litre = grams molecular mass of solute = M
s o lu te

%(W/V)×10
M=
mA

(c) Relation between molarity and % solute by mass :

Let d = density of solution in g/mL and let it contains x% (w/w) solute by mass.

mass of solute in gram x × d ×10

Number of moles of solute in 1 litre = grams molecular mass of solute = M
solute

x × d ×10
M= mA

Example 53:
The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Calculate the molarity
of the solution.
Solution :
In solving such numericals , the following formula can be applied:

% strength of solution  density of solution 10

Molarity =
Mol. mass

13  1.09  10
M= = 1.445 M
98

Strengthof solution(gL1 )
(d) Molarity 
Molecular weight

(e) Relationship between molality and molarity :

1000 × M
Molality (m) = 1000 × d - M × M
solute
Example 54:
Density for 2 M CH3COOH solution is (1.2 g/mL)
Calculate
(1) Molality
(2) mole fraction of CH3COOH acid
(3) % (w/w) after the solution
(4) % (w/v) for the solution

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Solution:
Let volume & solution = 1 Litre

n solute
Molarity = v =2
solution

n solute = 2 (CH3COOH)

mass of solute
molar mass of solute = 2

mass of solute = 2 × 60 = 120 g.

mass of solution = volume of solution × density of solution
= 1200 g
= mass of solute + mass of solvent
 mass of solvent = 1200 – 120 = 1080 g = 1.08 kg

1080
moles of solvent = = 60 mole
18

n solute 2
(1) Molality =  = 1.85 M Ans.
w solvent (kg) 1.08

nCH3CCOH 2
(2) X CH3COOH =  = 0.0322
nCH3CCOH  nH2O 2  60

wt. of solute (gm) 120

(3) % w/w = wt. of solution (gm) × 100 = ×100 = 10%
1200

wt. of solute (gm) 120

(4) % w/v = volume. of solution (mL ) × 100 = ×100 = 12%
1000
Example 55:
10% w/w urea solution is 1.2 g/mL calculate
(1) % w/v
(2) molarity
(3) molality
(4) mole fraction of urea
Solution:
% w/w = 10%
It means
10 g urea is present in 100 gm of solution
Let 100 gm of solution is taken
msolution = 100 gm

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m solution 100 gm
vsolution = Density  1.1g / mL = 91 mL

= 0.91 L

Murea = 10 gm
murea 10 1
nurea = M.M.  
urea 60 6
msolution = msolute + msolvent= 100
n H2O 90
msolvent =  = 5 mole
MM 18

wt. of urea 10
(1) % (w/v) = volume. of solution(mL ) × 100 =  100 = 10.99%
91

nurea 1/ 6
(2) molarity = v 
solution 0 .91 = 1.83 M

nurea 1/ 6
(3) molality = w 
solution 0.9 = 1.85 M
(4) mole fraction of urea

nurea 1/ 6 1/ 6
Xurea = n =  = 1/31 Ans.
urea  n H2O 1/ 6  5 31/ 6

The concentration of H2O2 is usually represented in terms of volume. If a sample of H2O2 is labeled as ‘x
volume’, it means that 1 volume of H2O2 solution gives ‘x volumes’ of O2 gas at STP on complete
decomposition.
Consider the decomposition of H2O2 as

2H2O2  2H2O  O2
234 g 22.7 L at STP

 22700 ml of O2 gas is liberated by 68g of H2O2 solution

68 x
 x ml of O2 gas will be liberated by = g of H2O2
22700
68 x
It means that g of H2O2 will be present in 1 ml of solution.
22700
68 x 680 x
 1000 ml of solution contains H2O2 =
22700  1000 = 227
Strength (g L–1) = Molarity  Molecular weight

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680 x
so, = M  34
227
x = 11.35  M i.e., Volume strength of at STP H2O2 = 11.35  Molarity
Note : Volume strength of at NTP H2O2 = 11.2  Molarity

Oleum or fuming sulphuric acid contains SO3 gas dissolved in sulphuric acid. When water is added to
oleum, SO3 reacts with H2O to form H2SO4, thus mass of the solution increases.
SO3 + H2O  H2SO4
The total mass of H2SO4 obtained by diluting 100g of sample of oleum with desired amount of water, is
equal to the percentage labeling of oleum.
109% oleum. means we can get 109 g pure H2SO4 from 100 g oleum by adding (109-100) = 9 gm pure
water

Mass of pure H2SO4 obtained

 % labeling of oleum = x100
mass of oleum taken

(% labeling oleum - 100 )

% w/w SO3 in oleum = x80
18
% w/w H2SO 4 in oleum = 100 - % w/w SO 3 in oleum

Example 56:
Calculate the composition of 109% oleum.
Solution :
Let the mass of SO3 in the sample be ‘w’ g, then the mass of H2SO4 would be (100 – w)g.
On dilution,

SO3 + H2O  H2SO4

80g 18g

w
Moles of SO3 in oleum = = Moles of H2SO4 formed after dilution.
80

98 w
 Mass of H2SO4 formed on dilution =
80

98 w
Total mass of H2SO4 present in oleum after dilution = + (100 - w) = 109
80
 w = 40
Thus, oleum sample contains 40% SO3 and 60% H2SO4.

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1. A 500 g tooth paste sample has 0.2 g fluoride concentration. What is the concentration of F – in terms
of ppm level
(A) 250 (B) 200 (C) 400 (D) 1000
2. Density of a solution containing 14% by mass of sulphuric acid is 1.05 gm/mL. What is the molarity of
solution?
(A) 0.5 M (B) 1.0 M (C) 1.5 M (D) 2.0 M
3. A sample of clay was partially dried & then contained 50% silica & 10% water. The original clay contained
20% water. Find% of silica in original sample
(A) 44.44% (B) 31.46 (C) 52.90 (D) 29.41
4. 23.4 g of NaCl dissolved in water and weight of solution is made up to 223.4 g. What is the molality of
the NaCl solution.
(A) 4 m (B) 5 m (C) 23.4 m (D) 1.5m
5. 100 mL of 0.40 M KCl can be coverted to a solution that is 0.50 M KCl by which of the following way ,
(A) adding 0.745 g KCl (B) adding 20 mL of water
(C) adding 0.10 mole KCl (D) evaporating 10 mL water
6. Molarity of liquid HCl if density of the solution is 1.168 g/cc is
(A) 36.5 (B) 18.25 (C) 32.0 (D) 42.10
7. Solution (s) containing 40 g NaOH is:
(A) 60 gm of 80% (w/w) NaOH (B) 50 gm of 80% (w/v) NaOH (dsoln. = 1.2 g/ml]
(C) 50 gm of 20 M NaOH [dsoln. = 1 g/ml] (D) 50 gm of 5 m NaOH
8. The molarity of Cl in an aqueous solution which was 5.85% (w/V) NaCl, 22.2% (w/V) CaCl2 and 5.35%
–

(w/V) NH4Cl will be

(A) 6M (B) 5M (C) 2M (D) 4M
9. Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0.25. Hence percentage of ethyl alcohol
by weight is
(A) 54% (B) 25% (C) 75% (D) 46%
10. Assuming complete precipitation of AgCl, calculate the sum of the molar concentration of all the ions if
2 litre of 2 M Ag2SO4 is mixed with 4 litre of 1 M NaCl solution is  Ag2SO4  2NaCl  2AgCl  Na2 SO4 
(A) 4 M (B) 2 M (C) 3M (D) 2.5 M
11. 2 litre of 9.8% w/w H2SO4 (d = 1.5 g/ml) solution is mixed with 3 litre of 1 M KOH solution.The concentration
of H+ if solution is acidic or concentration of OH– if solution is basic in the final solution is
[Hint: If acid is in excess solution is acidic, if base is in excess solution is basic]

3 3 2
(A) 0 (B) (C) (D)
10 5 5

1
12. Mole fraction of glucoe in H2O is . The molality of glucose in H2O is :
11
(A) 13.88 (B) 5.55 (C) 27.78 (D) 16.8

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13. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl
alcohol in the solution ?
(A) 0.190 (B) 0.086 (C) 0.050 (D) 0.100
14. Density of 2 molal NaOH solution is 1.35 g/ml. If molar mass of NaOH is 40 find molarity?
(A) 2.25 M (B) 2.5 M (C) 1.75 M (D) 2.75M
15. Density of 2 M NaOH solution is 1.28 g/ml. If molar mass of NaOH is 40 find molality?
(A) 1.66m (B) 1.27m (C) 1.75 m (D) 2.25m
16. 500 ml of a H2O2 solution on complete decomposition produces 2 moles of H2O. Calculate the volume
strength of H2O2 solution? [Given : Volume of O2 is measured at 1 bar and 273 K]
(A) 22.4 (B) 45.4 (C) 44.8 (D) 11.35
17. A mixture is prepared by mixing 20 gm SO3 in 30 gm H2SO4. Find mole fraction of SO3.
(A) 0.2 (B) 0.45 (C) 0.6 (D) 0.8
18. Find %w/w SO3 in a sample of oleum labelled as 118%.

(A) 20% (B) 40% (C) 60% (D) 80%

(a) Dulong's and Petit's Law :

Atomic weight × specific heat (cal/g/°C)  6.4
Gives approximate atomic weight and is applicable for metals only. Take unit of specific heat as cal/gm
(b) Vapour Density of metal chloride :
Let a metal M has valency x.
The molecular formula of its chloride becomes MClx.
Molecular mass = A +35.5x (A is atomic weight of metal M)
Molecular mass = 2 × Vapour density
A + 35.5x = 2 × Vapour density
If vapour density of metal chloride is known then A can be calculated.

Example 57:
The specific heat is 1 J g–1 K–1. It forms metal oxide of type M2Ox having molucular mass 102 . Calculate
its exact atomic weight ?
Solution :
Atomic mass × specific heat (in cal)  6.4

6.4
Approximate atomic weight = (1cal = 4.18 J) = 26.75
1/ 4.18
molucular mass of M2Ox is 102

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2  26.75  16x  102 x  3.03125

but as valency can not be fractional and thus it is 3.
Now again
2  M  16  3  102 M = 27
(C) Law of isomorphism :
According to Mitscherlich, isomorphous substance are those which have the same number of atoms
similarly arranged, i.e., these have similar constitution. So, their chemical formula are similar. Isomorphous
substance form crystal which can grow in the saturated solution of each other and have the same shape.
Examples 58:
(i) ZnSO4.7H2O and MgSO4.7H2O (ii) KMnO4 and KClO4.
Application :
1. To find valency of element : The valency of two elements which form isomorphous salts are same.
If valency of one element is known, then the valency of other will be the same.
2. If elements A and B forms isomorphous compounds then
Atomic weight of element A Weight of element A that combines with a certain weight of other elements

Atomic weight of element B Weight of element B that combines with same weight of other elements.

(a) Victor Meyer's process : (for volatile substance)

Procedure : Some known weight of a volatile substance (w) is taken, converted to vapour and collected
over water. The volume of air displaced over water is given (V) and the following expressions are used.
w w
M= RT or M= RT
PV (P  P')V
If aqueous tension is not given If aqueous tension is P'
Aqueous tension : Pressure exerted due to water vapours at any given temperature.
This comes in picture when any gas is collected over water. Can you guess why?
(b) Silver salt method : (for organic acids)
Basicity of an acid : No. of replacable H+ atoms in an acid (H connected to more electronegative atom is
acidic)
Procedure : Some known amount of silver salt (w1 gm) is heated to obtain w2 gm of white shining residue
of silver. Then if the basicity of acid is n, molecular weight of acid would be

R(COOAg)n  Ag

Weight of silver salt(w 1 ) WAg (w1 )

n 
Macid  n  107 108

This is one good practical application of POAC.

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Example 59:
0.607 g of silver salt of tribasic acid on combution gave 0.37 g pure Ag. Calculate the molecular weight of
acid.
Solution:

Ag3 X  Ag
0.607 0.37
Applying POAC

0.37 0.607  3
  Macid  210.53
108 3 107  Macid
(c) Chloroplatinate salt method : (for organic bases)
Some amount of organic base is reacted with H2PtCl6 and forms salt known as chloroplatinate. If base is
denoted by B then salt formed
(i) with monoacidic base = B2H2PtCl6
(ii) with diacidic base = B2(H2PtCl6)2
(iii) with triacidic base = B2(H2PtCl6)3
The known amount (w1 gm) of chloroplatnic salt is heated and Pt residue is measured. (w2 gm). If acidity of
base is 'n'

B2(H2PtCl6)n  Pt

Weight of chloroplatnic salt(w1 ) W2

n 
2MBase  410  n 195
Example 60:
Consider a monoacidic base. It’s platinic chloride salt contains 36.93% platinum. Calculate the molecular
mass of compound.
Solution:
B  H 2PtCl6  B 2 H 2 Cl6  Pt
100 36.93

Applying POAC.

100 36.93
  MB  59 .
2MB  410 195
Example 61:
0.1015 g when vaporized in Victor Meyer’s apparatus displaced 33.6 mL of air at 0oC and 760mm pressure.
Calculate molecular mass of substance
Solution:

0.10 33.6
Number of moles = =
molecular weight 22400
Mol. wt. = 66.66 g

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1. The vapour density of a volatile chloride of a metal is 74.6. If the specific heat of the metal is 0.55, calcualte
the exact atomic weight of the metal and the formula of its chloride.

(A) 7.2 , MCl4 (B) 6.5 , MCl2 (C) 5.4 , MCl3 (D) 4.2 , MCl4

2. A sample of a metal weighing 0.22 g required 0.51 g of AgNO3 to precipitate the chloride completely. The
specific heat of the metal is 0.057. Find the molecular formula of the chloride if the metal is M.

(A) MCl2 (B) MCl3 (C) MCl4 (D) MCl

3. 7.38 g of a sample of a metal oxide is quantitatively reduced to 6.84 g of pure metal. If the specific heat of
the metal is 0.0332 cal/g, calculate the valency and the accurate atomic weight of the metal.
(A) Atomic weight (accurate) = 202.67. (B) Atomic weight (accurate) = 102.62.

(C) Atomic weight (accurate) = 182.52. (D) Atomic weight (accurate) = 302.57.

4. On dissolving 2.0 g of a metal in sulphuric acid , 4.51 grams of the metal sulphate (M2(SO4)x was formed. The
specific heat of the metal is 0.057 cal/g. What is the valency of the metal and its exact atomic weight ?

(A) 2 & 90g (B) 3 & 114.7g (C) 2 & 114.7g (D) 3 & 90g

5. 0.607 g of the silver salt of a dibasic acid on combustion deposited 0.37 g of pure silver. Calculate the
molecular weight of the acid.

(A) 110.06 (B) 310.26 (C) 210.16 (D) 510.36

6. 0.304 g of a silver salt of a dibasic acid left 0.216 g of silver on ignition. Calculate its molecular weight.
(A) 180 (B) 90 (C) 45 (D) 120

7. 0.66 g of a platinichloride of a monoacidic base left 0.150 g of platinum. Calculate its molecular weight.
(Pt = 195).

(A) 224 (B) 125 (C) 22.1 (D) 321

8. The chloroplatinate of a diacid base contains 39% platinum. What is the molecular weight of the base ?

(Pt = 195)

(A) 180 (B) 90 (C) 45 (D) 120

9. In a Victor Meyer determination 0.0926 g of a liquid gave 28.9 mL of gas collected over water and measured
at 16ºC and 753.6 mm pressure. Calculate molecular weight and vapour density of the substance. (Aq.
tension at 16ºC = 13.6 mm)

(A) 76,38 (B) 78,39 (C) 74,37 (D) 72,36

10. A Dumas bulb of capacity 821 mL weighs 22.567g filled with vapour of a substance at 227 °C and 1520 mm
Hg pressure, it weighs 25.667g. Find the molecular weight of the substance.
(A) 7.5 (B) 44 (C) 30 (D) 75

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1. Match the following prefixes with their multiples.

Prefixes Multiples

(I) MICRO (P) 106

(II) DECA (Q) 109

(III) MEGA (R) 10–6

(IV) GIGA (S) 10–15

(V) FEMTO (T) 10

2. Calculate the molecular mass of the following :

(i) H2O (ii) CO2 (iii) CH4

3. How many grams of silicon is present in 35 g-atom of silicon (Given At. wt. of Si = 28).

4. Calculate number of atoms of each of the element present in 106.5 g of NaClO3.

5. Find the total number of H, S and 'O' atoms in the following :

(a) 196 gm H2SO4 (b) 196 amu H2SO4

(c) 5 mole H2S2O8 (d) 3 molecules H2S2O6 .

6. If, from 10 moles NH3 and 5 moles of H2SO4 , all the H-atoms are removed in order to form H2 gas, then find the
number of H2 molecules formed.

7. How much weight of copper can be obtained from 100 g of copper sulphate (CuSO 4)?

8. 2 moles of He atoms at NTP occupies a volume of ...............litres

9. 6.022 x 1022 molecules of N2 at STP will occupy a volume of ........litres

10. The vapour density of pure ozone (O3) would be ............

11. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4)

12. A compound contains 25% hydrogen and 75% carbon by mass. Determine the empirical formula of the
compound.

13. Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air

(ii) 1 mole of carbon is burnt in 16 g of dioxygen

(iii) 2 moles of carbon are burnt in 16 g dioxygen.

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
14. In a reaction

A + B2  AB2

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B2 (ii) 2 mol A + 3 mol B2

(iii) 2.5 mol A + 6 mol B2

15. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following
chemical equation :

N2(g) + 3H2(g)  2NH3 (g)

(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 its mass ?

16. Li metal is one of the few substances that reacts directly with molecular nitrogen. The balanced equation for
reaction is :

6Li(s) + N2(g)  2Li3N(s)

How many grams of the product, lithium nitride, can be prepared from 3.5g of lithium metal and 8.4 g of
molecular nitrogen ? (Atomic weight Li = 7, N = 14)

17. 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 2L?

18. Calculate the mass of sodium acetate (CH 3COONa) required to make 500 mL of 0.375 molar aqueous
solution. Molar mass of sodium acetate is 82 g mol–1.

19. Calculate the concentration of nitric acid in moles per liter in a sample which has a density. 1.41gmL–1
and the mass per cent of nitric acid in it being 69%.

20. If the density of methanol is 0.793 Kg L–1, what is its volume required for making 2.5 L of its 0.25 M
solution?

21. 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)

(i) express this in percent by mass

(ii) determine the molality of chloroform in water.

22. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water
vapour would be produced ?

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1. If law of conservation of mass was to hold true, then 20.8 gm. of BaCl2 on reaction with 9.8 gm. of H2SO4 will
produce 7.3 gm. of HCl and BaSO4 equal to :–
(A) 11.65 gm. (B) 23.3 gm. (C) 25.5 gm. (D) 30.6 gm.
2. Which one of the following pairs of compounds illustrate the law of multiple proportions ?
(A) H2 O, Na2O (B) MgO, Na2O (C) Na2O, BaO (D) SnCl2, SnCl4
3. Different proportions of oxygen in the various oxides of nitrogen prove the law of -
(A) Equivalent proportion (B) Multiple proportion
(C) Constant proportion (D) Conservation of matter
4. One part of an element A combines with two parts of B (Another element).Six parts of element C combine
with four parts of element B. If A and C combine together, the ratio of their masses will be governed by
(A) Law of definite proportion (B) Law of multiple proportion
(C) Law of reciprocal proportion (D) Law of conservation of mass
5. Which of the following combination illustrate law of reciprocal proportion
(A) N2O3, N2O4, N2O5 (B) PH3, P2O5, P2S5 (C) CS2, CO2, SO2 (D) NaCl, NaBr, NaI
6. Naturally occurring boron consist of two isotopes whose atomic weights are 10.01 & 11.01. The atomic
weight of natural boron is 10.81. Calculate the relative isotopic abundance of boron(11.01).
(A) 80.2% (B) 70.6% (C) 30.9% (D) 60.5%
7. In compound A, 1.00 g nitrogen combines with 0.57 g oxygen. In compound B, 2.00 g nitrogen combines with
2.24 g oxygen. In compound C, 3.00 g nitrogen combines with 5.11 g oxygen. These results obey the
following law
(A) Law of constant proportion (B) Law of multiple proportion
(C) Law of reciprocal proportion (D) Dalton’s law of partial pressure

8. The mass of carbon present in 0.5 mole of K4[Fe(CN)6] is

(A) 1.8 g (B) 18 g (C) 3.6 g (D) 36 g
9. How many mol of electrons will have total charge equal to 4816 coulomb.
(A) 0.05 mole (B) 5 mole (C) 6.95 × 10–21 mol (D) 0.01 mole
10. Number of neutrons present in 1.7 g of ammonia is
(A) NA (B) NA/10 × 4 (C) (NA/10) × 7 (D) NA × 10 × 7
11. The number of atoms in ‘n’ mole of gas can be given by

n  Av.No. Av.No.  Atomicity

(A) n × Av. No. × atomicity (B) Atomicity (C) (D) None
n

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12. Volume occupied by one molecule of water (density = 1 g cm –3) is
(A) 5.5 × 10–23 cm3 (B) 9.0 × 10–23 cm3 (C) 6.023 × 10–23 cm3 (D) 3.0 × 10–23 cm3
13. The relation between the total number of electrons in 1.6 g of CH4 and that in 1.8 g of H2O is
(A) Double (B) Same (C) Triple (D) One fourth
14. Number of molecules in 100 g of each of O2, NH3 and CO2 at STP are
(A) in the order CO2 < O2 < NH3 (B) in the order NH3 < O2 < CO2
(C) the same (D) NH3 = CO2 < O2
15. Four one litre flasks are seperately filled with the gases hydrogen, helium oxygen and ozone at same room
temp. and pressure. The ratio of total number of atoms of these gases present in the different flasks would be
(A) 1 : 1 : 1 : 1 (B) 1 : 2 : 2 : 3 (C) 2 : 1 : 2 : 3 (D) 2 : 1 : 3 : 2
16. If V ml of the vapours of substance at NTP weight W g. Then molecular wt. of substance is

V W 1
(A) (W/V) × 22400 (B) × 22.4 (C) (W–V) × 22400 (D)
W V  22400
17. Which contains least no. molecules
(A) 1 g CO2 (B) 1 gN2 (C) 1g O2 (D) 1g H2
18. Number of oxygen atoms in 8 gms of ozone is

6.02  10 23 6.02  10 23 6.02  10 23

(A) 6.02 × 1023 (B) (C) (D)
2 3 6
19. Number of moles of water in 488 gm of BaCl2.2H2O are (Ba = 137, Cl = 35.5)
(A) 2 moles (B) 4 moles (C) 3 moles (D) 5 moles
20. Which of the following contains the least number of molecules ?
(A) 4.4 g CO2 (B) 3.4 g NH3 (C) 1.6 g CH4 (D) 3.2 g SO2
21. Which of the following contain highest number of molecules -
(A) 2.8 g of CO (B) 3.2 g of CH4 (C) 1.7 g of NH3 (D) 3.2 g of SO2
22. Which contains least no. of molecules :–
(A) 1 g. CO2 (B) 1 g N2 (C) 1 g O2 (D) 1 g H2
23. Which of the following contains largest number of atoms ?
(A) 4 g of H2 (B) 16 g of O2 (C) 28 g of N2 (D) 18 g of H2O
24. Which of the following has the highest mass ?

1
(A) 1 g atom of C (B) mole of CH4
2
(C) 10 ml of water (D) 3.011 × 1023 atoms of oxygen
25. The total number of electrons in 4.2 g of N3– ion is (NA is the Avogadro’s number)
(A) 2.1 NA (B) 4.2 NA (C) 3 NA (D) 3.2 NA
26. Four containers of 2L capacity contains dinitrogen as described below. Which one contains maximum
number of molecules under similar conditions.
(A) 2.5 g-molecules of N2 (B) 4 g-atom of nitrogen
(C) 3.01 × 1024 N atoms (D) 84 g of dinitrogen

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27. A sample of aluminium has a mass of 54.0 g. What is the mass of the same number of magnesium atoms?
(At. wt. Al = 27, Mg=24)
(A) 12 g (B) 24 g (C) 48 g (D) 96 g

28. Out of 1.0 g dioxygen, 1.0 g (atomic) oxygen and 1.0 g of ozone  O3  , the maximum number of oxygen
atoms are present in
(A) 1.0 g of atomic oxygen. (B) 1.0 g of ozone.
(C) 1.0 g of oxygen gas. (D) All contain same number of atoms.
29. A container of volume V, contains 0.28 gm of N2 gas. If same volume of an unknown gas under similar
condition of temperature and pressure weighs, 0.44 gm, the molecular mass of the gas is
(A) 22 (B) 44 (C) 66 (D) 88
30. Two flasks A & B of equal capacity of volume contain NH3 and SO2 gas respectively under similar conditions
which flask has more no. of moles –
(A) A (B) B (C) Both have same moles (D) None
31. If 32 gm of O2 contains 6.022 × 10 molecules at NTP then 32 gm of S, under the same conditions, will contain
23

(A) 6.022 × 1023 S (B) 3.011 × 1021 S (C) 1 × 1023 S (D) 12.044 × 1023S
32. 4.4 gm of an unknown gas occupies 2.24 litres of volume at NTP. The gas may be
(A) N2O (B) CO (C) CO2 (D) N2O or CO2 or Both
33. A person adds 1.71 gram of sugar (C12H22O11) in order to sweeten his tea. The number of carbon atoms added
are (mol. mass of sugar = 342)
(A) 3.6 × 1022 (B) 7.2 × 1021 (C) 0.05 (D) 6.6 × 1022
34. The number of atoms present in 0.5g-atom of nitrogen is same as the atoms in
(A) 12g of C (B) 32g of S (C) 8g of oxygen (D) 24g of Mg
35. Which has the maximum number of molecules among the following ?
(A) 8 g H2 (B) 64 g SO2 (C) 44 g CO2 (D) 48 g O3
36. Which has maximum molecules ?
(A) 7 g N2O (B) 20 g H2 (C) 16 g NO2 (D) 16 g SO2
37. 21
From 200 mg of CO2, 10 molecules are removed. How many mole of CO2 are left?
(A) 1.28 × 10–3 (B) 2.88 × 10–3 (C) 0.88 × 10–4 (D) 5.18 × 10–3
38. Consider the following data
Element Atomic Weight
X 14
Y 16
X and Y combine to form a new substance Z. If four moles of Y combine with two mole of X to give one mole
of Z, then the mass of one mole of Z is
(A) 46g (B) 23g (C) 92g (D) 138g
39. The modern atomic weight scale is based on
(A) C12 (B) O16 (C) H1 (D) C13
40. The number of water molecules in 1 litre of water is
(A) 18 (B) 18 × 1000 (C) NA (D) 55.55 NA
41. 1 mole of CH4 contains
(A) 6.02× 1023 atoms of H (B) 4 g-atom of hydrogen
(C) 1.81 × 1023 molecules of CH4 (D) 3.0 g of carbon

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42. The total number of protons in 10 g of calcium carbonate is (NA = 6.023 × 1023)
(A) 1.5057 × 1024 (B) 2.0478 × 1024 (C) 3.0115 × 1024 (D) 4.0956 × 1024
43. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?
(A) 0.02 (B) 3.125 × 10–2 (C) 1.25 × 10–2 (D) 2.5 × 10–2
44. 1 amu is equal to

1 1
(A) of C12 (B) of O16 (C) 1g of H2 (D) 1.66×10–23 kg
12 14
45. 19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered (Au =197)
(A) 100 (B) 6.02  1023 (C) 6.02  1024 (D) 6.02  1025
46. The number of molecules in 16 g of methane is
16 16
(A) 3.0  1023 (B) 6.02  1023 (C)  1023 (D)  1023
6.02 3.0
47. Number of molecules in 100 ml of each of O2,NH3 and CO2 STP are
(A) In the order CO2 < O2 < NH3 (B) In the order NH3 < O2 < CO2
(C) The same (D) NH3 = CO2 < O2
48. The number of electrons in a mole of hydrogen molecule is
(A) 6.02×1023 (B) 12.046 × 1023 (C) 3.0115×1023 (D) Indefinite
49. The numbers of moles of BaCO3 which contain 1.5 moles of oxygen atoms is

(A) 0.5 (B) 1 (C) 3 (D) 6.02  1023

50. How many molecules are present in one gram of hydrogen
(A) 6.02×1023 (B) 3.01×1023 (C) 2.5×1023 (D) 1.5×1023
51. The total number of g-molecules of SO2Cl2 in 13.5 g of sulphuryl chloride is
(A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4
52. The largest number of molecules is in
(A) 34 g of water (B) 28g of CO2 (C) 46g of CH3OH (D) 54g of N2O5
53. The number of moles of sodium oxide in 620 g of it is
(A) 1 mol (B) 10 moles (C) 18 moles (D) 100 moles
54. 2g of oxygen contains number of atoms equal to that in
(A) 0.5 g of hydrogen (B) 4g of sulphur (C) 7g of nitrogen (D) 2g of sodium
55. The number of electrons in the telluride ion. Te2–, (Z = 52) is
(A) 50 (B) 51 (C) 52 (D) 54
56. How many atoms are contained in one mole of sucrose (C12H22O11)
(A) 45 × 6.02 × 1023 atoms/mole (B) 5 × 6.62 × 1023 atoms/mole
(C) 5 × 6.02 × 1023 atoms/mole (D) None of these
57. Which one of the following statements is incorrect ?
(A) One gram atom of carbon contains Avogadro’s number of atoms.
(B) One mole of oxygen gas contains Avogadro’s number of molecules
(C) One mole of hydrogen contains Avogadro’s number of atoms.
(D) One mole of electrons contains 6.02 × 1023 electrons.

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
58. A sample of phosphorus trichloride (PCl3) contains 1.4 moles of the substance. How many atoms are there
in the sample ?
(A) 4 (B) 5.6 (C) 8.431 × 1023 (D) 3.372 × 1024
59. Which has maximum number of atoms ?
(A) 24 g of C (12) (B) 56 g of Fe (56) (C) 27 g of Al (27) (D) 108 g Ag (108)
60. Number of moles of 1 m gas at NTP are
3

(A) 44.6 (B) 40.6 (C) 42.6 (D) 48.6

61. The weight of 1 mole of a gas having density 0.1784 g/l at NTP is
(A) 0.1784 g (B) 1 g (C) 4 g (D) Can not be Calculate
62. If one mole of N2 at NTP occupies 22.4 litre the density of N2 is
(A) 1.25 g/litre (B) 0.80 g/litre (C) 2.5 g/litre (D) 1.60 g/litre
63. Vapour density of gas is 11.2 volume occupied by 2.4 gms of this at STP will be
(A) 11.2 L (B) 2.24 L (C) 22.4 L (D) 2.4 L
64. The atomic mass of an element is 27. If valency is 3, the vapour density of the volatile chloride will be
(A) 66.75 (B) 6.675 (C) 667.5 (D) 81
65. 5.6 litre of a gas at N.T.P. weighs 8 g the vapour density of gas is
(A) 32 (B) 16 (C) 8 (D) 40
66. The vapour density of a gas A is twice that of gas B. If the molecular weight of B is M, the molecular weight
of A will be

M
(A) M (B) 2M (C) 3M (D)
2
67. A nugget of gold and quartz was found to contain x g of gold and y g of quartz and has density d. If the
densities of gold and quartz are d1 and d2 respectively then the correct relation is

x y xy
(A) d + d = (B) xd1 + yd2 = (x + y) d
1 2 d

x y xy xy x x
(C) d + d = (D) + d + d =0
2 1 d d 1 2

68. A hydrocarbon contains 80% of carbon, then the hydrocarbon is

(A) CH4 (B) C2H4 (C) C2H6 (D) C2H2
69. The empirical formula of a compound of molecular mass 120 is CH2O. The molecular formula of the compound
is
(A) C2H4O2 (B) C4H8O4 (C) C3H6O3 (D) All of these
70. Insulin contains 3.4% sulphur by mass. What will be the minimum molecular weight of insulin
(A) 94.117 (B) 1884 (C) 941.17 (D) 976
71. Calculate the molecular formula of compound which contains 20% Ca and 80% Br (by wt.) if molecular
weight of compound is 200. (Atomic wt. Ca = 40, Br = 80)
(A) Ca1/2Br (B) CaBr2 (C) CaBr (D) Ca2Br

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72. The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the
compound will be
(A) C2H2 (B) C2H6 (C) C4H4 (D) C6H6
73. The empirical formula of an organic compound containing carbon and hydrogen is CH2. The mass of one litre
of this organic gas is exactly equal to that of one litre of N2 at same temperature and pressure. Therefore, the
molecular formula of the organic gas is
(A) C2H4 (B) C3H6 (C) C6H12 (D) C4H8
74. Haemoglobin contains 0.334% of iron by weight. the molecular weight of haemoglobin its approximately
67200. The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of haemoglobin is
(A) 4 (B) 6 (C) 3 (D) 2
75. A giant molecule contains 0.25% of a metal whose atomic weight is 59. Its molecule contains one atom of
that metal. Its minimum molecular weight is

100  59
(A) 5900 (B) 23600 (C) 11800 (D)
0.4

76. An oxide of sulphur contains 50% of sulphur in it. Its emperical formula is
(A) SO2 (B) SO3 (C) S2O2 (D) S2O
77. An oxide of metal M has 40% by mass of oxygen. Metal M has atomic mass of 24. The emperical formula of
the oxide is
(A) M2O (B) M2O3 (C) MO (D) M3O4
78. A compound is found to contain 80% of carbon and 20% of hydrogen, then the molecular formula of the
compound is
(A) C6H6 (B) C2H5OH (C) C2H6 (D) C2H4
79. The simplest formula of a compound containing 50% of element X(at wt. = 10) and 50% of element Y(at wt. = 20)
is
(A) XY (B) X2Y (C) XY2 (D) X3Y

80. How much oxygen is required to react with 9 gms of Al

3
2Al + O  Al2 O3
2 2
(A) 6 g O2 (B) 8 g O2 (C) 9 g O2 (D) 4 g O2
81. A mixture containing 100 gm H2 and 100 gm O2 is ignited so that water is formed according to the reaction,
2H2 + O2  2H2O; the weight of water formed will be
(A) 112.5 g (B) 50 g (C) 25 g (D) 200 g
88. 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca (OH)2. The maximum number of moles of CaSO4 formed
is Ca  OH2  H2SO4  CaSO 4  2H2O

(A) 0.2 (B) 0.5 (C) 0.4 (D) 1.5

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83. 12 litre of H2 and 11.2 litre of Cl2 are mixed and exploded. The composition by volume of mixture is
H2  Cl2  2HCl
(A) 24 litre of HCl (B) 0.8 litre Cl2 and 20.8 litre HCl.
(C) 0.8 litre H2 & 22.4 litre HCl (D) 22.4 litre HCl
84. For the reaction A + 2B  C
5 mole of A and 8 mole of B will produce -
(A) 5 mole of C (B) 4 mole of C (C) 8 mole of C (D) 13 mole of C
85. In the reaction,
2SO2  O2  2SO3
when 1 mole of SO2 and 1 mole of O2 are made to react to completion
(A) All the oxygen will be consumed (B) 1.0 mole of SO3 will be produced
(C) 0.5 mole of SO2 is remained (D) All of these

86. NH3 is produced according to the following reaction :

N2(g) + 3H2(g)  2NH3(g)
In an experiment 0.25 mol of NH3 is formed when 0.5 mol of N2 is reacted with 0.5 mol of H2. What is % yield
of the reaction ?
(A) 75% (B) 50% (C) 33% (D) 25%
87. A sample of calcium carbonate is 80% pure, 25 gm of this sample is treated with excess of
HCl. What volume of CO2 will be obtained at 1 atm & 273 K ?
(A) 2.24L (B) 3.36L (C) 4.48L (D) 6.72L
88. The mass of CO2 that shall be obtained by heating 10 kg of 80% pure limestone (CaCO3) is
(A) 4.4 kg (B) 6.6 kg (C) 3.52 kg (D) 8.8
89. How many g of copper (at. wt.=64) would be displaced from the copper sulphate solution by adding 27 g of
aluminium (at. wt. = 27)
(A) 32 (B) 64 (C) 96 (D) 160
90. 7 g of a mixture of KClO3 and KCl are strongly heated, 2.50 g of O2 is produced. The residue on analysis
contains to be only KCl. Calculate the weight fraction of KCl in the original mixture.
(A) 0.0886 (B) 0.123 (C) 0.886 (D) 0.213
91. The number of litres of air(20% oxygen by volume) required to burn 8 litres of C2H2 is approximately
(A) 40 (B) 60 (C) 80 (D) 100
92. What volume of air will be required to oxidise 210 mL of sulphur oxide to sulphur trioxide, if the air contains
21% of oxygen ?
(A) 105ml (B) 500 mL (C) 210ml (D) 1000ml
93. 15 mL of a gaseous hydrocarbon requires in 357 mL of air for complete combustion (21% of oxygen by
volume) and the gaseous products occupies 327 mL (all volumes being measured at NTP). What is the
formula of the hydrocarbon ?
(A) C3H6 (B) C3H4 (C) C2H6 (D) C3H8

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
94. One volume of a compound of carbon, hydrogen and oxygen was exploded with 2.5 volumes of oxygen. The
resultant mixture contained 2 volumes of water vapour and 2 volumes of carbon dioxide. All volumes were
measured in identical conditions. Determine the formula of the compound.
(A) C2H4O2 (B) C2H4O (C) C2H4O3 (D) C2H4O6

95. If the density of methanol is 0.793 kg L–1, what is its volume needed for make 2.5 L of its 0.25 M solution?
(A) 21.4 mL (B) 18.2 mL (C) 25.2 mL (D) 30.3 mL
96. Calculate the resulting molarity when 145mL of 6.00M H2SO4 solution is mixed with 245mL of 3.00M H2SO4.
(A) 3.22 M (B) 4.12 M (C) 5.34 M (D) 2.35 M
97. 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 ?
(A) 0.082 mol L–1 (B) 0.069 mol L–1 (C) 0.078 mol L–1 (D) 0.029 mol L–1
98. Calculate the molality of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.
(A) 2.31 mol (B) 0.11 mol (C) 3.11 mol (D) 2.11 mol
99. The total number of ions persent in 1 ml of 0.1 M barium nitrate Ba(NO3)2 solution is
(A) 6.02 × 1018 (B) 6.02 × 1019 (C) 3.0 × 6.02 × 1019 (D) 3.0 × 6.02 × 1018
100. The mass of CaCO3 produced when carbon dioxide is passed in excess through 500 ml of 0.5M Ca(OH)2
will be
Ca  OH2  CO2  CaCO3  H2 O
(A) 10 gm (B) 20 gm (C) 50 gm (D) 25 gm.
101. –1
H3PO4 (98 g mol ) is 98% by mass of solution. If the density is 1.8 g/ml, the molarity is
(A) 18 M (B) 36 M (C) 54 M (D) 0.18 M
102. The volume of water that must be added to a mixture of 250 ml of 0.6 M HCl and 750 ml of
0.2 M HCl to obtain 0.25 M solution of HCl is
(A) 750 mL (B) 100 mL (C) 200 mL (D) 300 mL
103. What volume of a 0.8 M solution contains 100 milli moles of the solute?
(A) 100 mL (B) 125 mL (C) 500 mL (D) 62.5 mL
104. 500 mL of a glucose solution contains 6.02 × 10 molecules. The concentration of the solution is
22

(A) 0.1 M (B) 1.0 M (C) 0.2 M (D) 2.0 M

105. Mole fraction of A in H2O is 0.2. The molality of A in H2O is :
(A) 13.9 (B) 15.5 (C) 14.5 (D) 16.8
106. What is the molarity of H2SO4 solution that has a density of 1.84 g/cc and contains 98% by mass of H 2SO4?
(A) 4.18 M (B) 8.14 M (C) 18.4 M (D) 18 M
107. The molarity of 10 mL of ‘22.4 V’ at NTP H2O2 is
(A) 1.79 (B) 2 (C) 60.86 (D) 6.086
108. What is the % of free SO3 in an oleum that is labelled as ‘104.5% H2SO4’?
(A) 10 (B) 20 (C) 40 (D) Noneof these
109. A mixture is prepared by mixing 20 g SO3 in 30 g H2SO4 . Determine % labelling of oleum solution.
(A) 104.5 (B) 106 (C) 109 (D) 110
110. Which of the following changes with increase in temperature ?
(A) Molality (B) Weight of fraction of solute
(C) Fraction of solute present in water (D) Mole fraction

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1. An element, X has the following isotopic composition 200X : 90% 199X : 8% 202X : 2.0%. The weighted
average atomic mass of the naturally occurring element X is closest to
(A) 201 amu (B) 202 amu (C) 199 amu (D) 200 amu
2. Use the law of multiple proportion to consider possible formula for compounds made from two elements, X
and Y, in the proportions listed below.
Compound A : 1.0g of X reacted with 2.1g of Y. Compound B : 1.0g of X reacted with 6.3g of Y.
Which of the following sets of formulas is possible ?
(A) A : X2Y3, B : X2Y9 (B) A : XY, B : XY2 (C) A : X6Y2, B : XY (D) A : XY, B : X2Y3
3. Hydrogen combines with chlorine to form HCl. It also combines with sodium to form NaH. If sodium and chlorine
also combine with each other, they may be do so in the ratio of their masses as:-
(A) 23 : 35.5 (B) 35.5 : 23 (C) 1: 1 (D) 23 : 1
4. A sample of H2O2 solution labelled as 56.75 ‘V’ at STP has density of 530 gm/L. Mark the incorrect options
representing concentration of same solution in other units. (Solution contains only H2O and H2O2)

w
(A) Molarity MH2O2  6 (B) %  17
v

1000
(C) Mole fraction of H2O2 = 0.25 (D) Molality mH2O2 
72

5. One mole of P4 molecules contains -

(A) 1 molecule (B) 4 molecules

1
(C) × 6.022 × 1023 atoms (D) 24.088 × 1023 atoms
4

6. The number of sodium atoms in 2 moles of sodium ferrocyanide Na4[Fe(CN)6], is-

(A) 2 (B) 6.023 × 1023 (C) 8 × 6.02 × 1023 (D) 4 × 6.02 × 1023
7. Which of the following contains the largest number of atoms
(A) 11g of CO2 (B) 4g of H2 (C) 5g of NH3 (D) 8g of SO2
8. The number of atoms in 0.1 mol of a tetraatomic gas is (NA = 6.02 × 1023 mol–1)
(A) 2.4 × 1022 (B) 6.026 × 1022 (C) 2.4 × 1023 (D) 3.6 × 1023
9. Which is correct for 10 g of CaCO3
(A) It contains 1g-atom of carbon (B) It contains 0.3 g-atoms of oxygen

(C) It contains 12 g of calcium (D) None of these

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10. 4.48 litres of methane at N.T.P. has
(A) 1.2 x 1022 molecules of methane (B) 0.5 mole of methane
(C) 3.2 gm of methane (D) 0.1 mole of methane
11. Two flask A & B of equal capacity of volume contain NH3 and SO2 gas respectively under similar conditions which
flask has more no. of moles
(A) A (B) b
(C) Both have same moles (D) None
12. Elements A and B form two compounds B2A 3 and B2A. 0.05 moles of B2A3 weight 9.0 gms and
0.10 mole of B2A weight 10 gms atomic weight of A and B are
(A) 20 and 30 (B) 30 and 40 (C) 40 and 30 (D) 30 and 20
13. The maximum number of molecules is present in
(A) 15 L of H2 gas at STP (B) 0.5 g of N2 gas at STP

(C) 0.5 g of H2 gas (D) 10 g of O2 gas

14. Vapour density of ammonia is 8.5, 85 gms of NH3 at NTP occupy.
(A) 22.4 litre (B) 112 litre (C) 224 litre (D) 1120 litre
15. A and B are two identical vessels. A contains 15 gm ethane at 1 atm and 298 K. The vessel B contains 75 gm
of a gas X2 at same tamperature and pressure. The vapour density of X2 is
(A) 75 (B) 150 (C) 37.5 (D) 45
16. The relative density of a gas A with respect to another gas B is 2. The vapour density of the gas B is 20, the
vapour density of the gas A is
(A) 30 (B) 40 (C) 50 (D) 60
17. The percent of N in 66% pure (NH4)2 SO4 sample is
(A) 32 (B) 28 (C) 14 (D) None of these
18. Number of Fe atoms in 100 g Haemoglobin if it contains 0.33% Fe. (Atomic mass of Fe = 56) is
(A) 0.035 × 1023 (B) 35 (C) 3.5 × 1023 (D) 7.5 ×108
19. If in a compound the masses of elements X and Y are equal. The atomic weights of X and Y are 30 and 20
respectively. Molecular formula of the compound (its molecular weightt is 120) will be
(A) X2Y2 (B) X3Y3 (C) X2Y3 (D) X3Y2
20. A compound contains 38.8% C, 16.0% H and 45.2% N. The formula of the compound will be

(A) CH3NH2 (B) CH3CN (C) C2H5CN (D) CH2(NH)2

21. The fermentation of sugar to produce ethyl alcohol occurs by the following reaction

yeast
C6H12 O6 (s)  2C2H5OH()  2CO2 (g)

How much ethyl alcohol can be made from 1.00 kg of sugar?

(A) 431g (B) 121g (C) 511g (D) 312g

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22. The equation :

3
2Al(S) + O (g)  Al2O3(S) shows that
2 2

3 7
(A) 2 mole of Al reacts with mole of O2 to produce mole of Al2O3
2 2

3 7
(B) 2gm of Al reacts with g of O2 to produce one mole of Al2O3
2 2

3
(C) 2 gm mole of Al reacts with litre of O2 to produce 1 mole of Al2O3
2

3
(D) 2 mole of Al reacts with mole of O2 to produce 1 mole of Al2O3
2

23. 12 g of alkaline earth metal gives 14.8 g  3M  N2  M3N2  of its nitride. Atomic weight of metal is

(A) 12 (B) 20 (C) 40 (D) 14.8

24. 100 ml of PH3 on decomposition produces phosphorus and hydrogen. The change in volume is

(A) 50 ml increase (B) 500 ml decrease (C) 900 ml decrease (D) None
25. Equal weight of 'X' (At. wt. = 36) and 'Y' (At. wt. = 24) are reacted to form the compound X 2Y3. Then :

 2X  3Y  X2 Y3 
(A) X is the limiting reagent
(B) Y is the limiting reagent
(C) No reactant is left over and mass of X2Y3 formed is double the mass of ‘X’ taken
(D) Can not say
26. If 1.6 gms of SO2 1.5 × 1022 molecules of H2S are mixed and allowed to remain in contact in a closed vessel
until the reaction

2H2S + SO2  3S + 2H2O,

proceeds to completion. Which of the following statement is true ?
(A) Only 'S' and 'H2O' remain in the reaction vessel. (B) 'H2S' is excess reagent
(C) 'SO2' is excess reagent (D) 'SO2' is limiting reagent
27. 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of Cl and Cl3. Calculate
the number of moles of Cl and Cl3 formed.( I = 127, Cl = 35.5)

(A) 0.1 mole, 0.1 mole (B) 0.1 mole, 0.2 mole (C) 0.5 mole, 0.5 mole (D) 0.2 mole, 0.2 mole

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28. “Superphosphate”, a water-soluble fertilizer, is a mixture of Ca(H2PO4)2 and CaSO4 on a 1 : 2 mole basis. It
is formed by the reaction : Ca3(PO4)2 + 2H2SO4   Ca(H2PO4)2 + 2CaSO4
We treat 450g of Ca3(PO4)2 with 300g of H2SO4. How many grams of superphosphate could be formed ?
(A) 515g (B) 765g (C) 635g (D) 735g
29. The crystalline salt Na2SO4.xH2O on heating lose 55.9% of its mass and becomes anhydrous. The formula
of crystalline salt is

(A) Na2SO4.5H2O (B) Na2SO4.7H2O (C) Na2SO4.2H2O (D) Na2SO4.10H2O

30. 10 mole KClO3 taken in a container undergoes parallel reaction

2KClO3(s)  2KCl(s) + 3O2(g)
4KClO3(s)  KCl(s) + 3KClO4(s)
If 7 mole O2(g) is collected then the moles of KClO4 and KCl formed inthe conatiner are
(A) 8 and 2 (B) 4 and 6 (C) 6.5 and 3.5 (D) 5 and 5
31. The mass of oxygen that would be required to produce enough CO, which completely reduces 1.6 kg
Fe2O3 (at. mass Fe = 56) is

Fe2O3  3CO  2Fe  3CO2 

(A) 240 gm (B) 480 gm (C) 720 gm (D) 960 gm
32. What weight of CaCO3 must be decomposed to produce the sufficient quantity of carbon dioxide to convert
21.2 kg of Na2CO3 completely in to NaHCO3. [Atomic mass Na = 23, Ca = 40]

CaCO3  CaO + CO2 Na2 CO3 + CO2 + H2O  2NaHCO3

(A) 100 Kg (B) 20 Kg (C) 120 Kg (D) 30 Kg

33. Calculate the weight of lime (CaO) obtained by heating 200 Kg of 95% pure lime stone (CaCO 3)
(A) 104.4 kg (B) 105.4 kg (C) 212.8 kg (D) 106.4 kg
34. A silver coin weighting 11.34 g was dissolved in nitric acid. When sodium chloride was added to the solution
all the silver (present as AgNO3) was precipitated as silver chloride. The weight of the precipitated silver
chloride was 14.35 g. Calculate the percentage of silver in the coin

(A) 4.8% (B) 95.2% (C) 90% (D) 80%

35. An iron ore that contains Fe3O4 reacts according to the reaction Fe3O4 + C 
 3Fe + 2CO2
We obtain 2.09g of Fe from the reaction of 50.0g of the ore. What is the percent Fe3O4 in the ore ?
(A) 5.78% (B) 0.78% (C) 2.12% (D) 3.13%
36. 40 mL of ammonia gas taken in an eudiometer tube was subjected to sparks till the volume did not change
any further. The volume was found to increase by 40 mL. 40 mL of oxygen was then mixed and the mixture
was further exploded. The gases remained were 30 mL. Deduce the formula of ammonia.
(A) N3H (B) NH3 (C) N2H4 (D) None

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37. 40 mL of a mixture of hydrogen, CH4 and N2 was exploded with 10 mL of oxygen. On cooling, the gases
occupied 36.5 mL. After treatment with KOH, the volume reduced by 3 mL and again on treatment with
alkaline pyrogallol, the volume further decreased by 1.5 mL. Determine the composition of the mixture.
(A) H2 : 12.50% ; CH4 :7.50% ; N2 : 80% (B) H2 : 7.50% ; CH4 :12.50% ; N2 : 80%
(C) H2 : 12.50% ; CH4 :80% ; N2 : 7.50% (D) H2 : 80% ; CH4 :7.50% ; N2 : 12.50%
38. A mixture of CH4 and C2H2 occupied a certain volume at a total pressure of 63 mm. The sample was burnt to
CO2 and H2O and the CO2 alone was collected and its pressure was found to be 69 mm in the same volume
and at the same temperature as the original mixture. What fraction of the mixture was methane ?
(A) 0.6 (B) 0.5 (C) 0.90 (D) 0.3
39. NX is produced by the following step of reactions

M + X2  M X2

3MX2 + X2  M3X8

M3 X8 + 4N2CO3  8NX + 4CO2 + M3O4

How much M (metal) is consumed to produce 206 gm of NX. (Take Atomic wt of M = 56, N=23, X = 80)

14 7
(A) 42 gm (B) 56 gm (C) gm (D) gm
3 4

40. 2PbS  3O2  2PbO  2SO2

3SO2  2HNO3  2H2O 

 3H2SO 4  2NO

According to the above sequence of reactions, how much H2SO4 will 5mol of PbS produce ?
(A) 270 g (B) 360 g (C) 400 g (D) 490 g
41. 2 mole of 50% pure Ca(HCO3)2 on heating forms 1 mole of CO2. The percent yield of CO2 is


Ca(HCO3 )2  CaO  H2O  CO 2 (unbalanced)

(A) 80% (B) 75% (C) 50% (D) 100%

42. How many mol Fe2+ ions are formed, when excess of iron is treated with 50mL of 4.0M HCl under inert
atmosphere ? Assume no change in volume - Fe  2HCl  FeCl2  H2

(A) 0.4 (B) 0.1 (C) 0.2 (D) 0.8

43. If 100 mL of 1.00 M H2SO4 solution is mixed with 200 mL of 1.00 M KOH, what is its molarity of salt
produced?
(A) 0.0333M (B) 0.0121M (C) 0.333 M (D) 0.123 M
44. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous
solution. Molar mass of sodium acetate is 82.0245 g mol–1.
(A) 15.38 g (B) 10.18 g (C) 18.32 g (D) 25.14 g

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
45. 50 mL solution of BaCl2 (20.8% w/v) and 100 mL solution of H2SO4 (9.8% w/v) are mixed then maximum

mass of BaSO4 formed is BaCl2  H2SO4  BaSO 4  2HCl

(A) 23.3 g (B) 46.6 g (C) 29.8 g (D) 11.65 g

46. 2M of 100 ml Na2 SO4 is mixed with 3M of 100 ml NaCl solution and 1M of 200 ml CaCl2 solution. Then the
ratio of the concentration of cation and anion.

1
(A) (B) 2 (C) 1.5 (D) 1
2

47. The molality of a sulphuric acid solution is 0.2. Calculate the total weight of the solution having
1000 gm of solvent.

(A) 1000 g (B) 1098.6 g (C) 980.4 g (D) 1019.6g

48. Two binary solutions have the same molarity. Which of the following statements is true ?
(A) Equal volumes of two solution contain equal number of solute molecules
(B) Equal weights of two solutions contain equal number of solute molecules
(C) two solutions, must have same molality
(D) The two solutions must have same mole fraction
49. A bottle of 1 litre capacity is labelled as 1 molar Al2(SO4)3 (aq). If the bottle is half filled and density of
solution is 1.342g/ml, molality of Al3+ (aq) in this solution will be
(A) 1 (B) 2 (C) 3 (D) 4
50. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction
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.
(A) 2.14 g (B) 0.94 g (C) 0.54 g (D) 1.12 g
51. What volume of 0.2 M Ba(OH)2 must be added to 300mL of a 0.08M HCl solution to get a solution in which
the molarity of hydroxyl (OH–) ions is 0.08 M ?
(A) 375 mL (B) 300 mL (C) 225 mL (D) 150 mL
52. Suppose you want an acidic solution to carry out a chemical reaction with 2 moles of NaOH. Which sample
of acid is the best choice for you. (At. wt. : S = 32, Cl = 35.5)

2NaOH  H2SO4  Na2SO 4  2H2O , NaOH  HCl  NaCl  H2 O

(A) 1 M H2SO4 (50 Rs per L) (B) 1 M H2SO4 (56 Rs per L)

(C) 1 M HCl (30 Rs per L) (D) 1 M HCl (27 Rs per L)

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

Directions: In the following questions, a statement of Assertion (A) is followed by a corresponding statement of
Reason (R) just below it. Of the statements, mark the correct answer :
(A) If both assertion and reason are true and reason is the correct explanation of assertion
(B) If both assertion and reason are true but reason is not the correct explanation of assertion
(C) If assertion is true but reason is false
(D) If both assertion and reason are false.
1. Assertion : No. of moles of H2 in 0.224 litre of hydrogen is 0.01 mole.
Reason : 22.4 litres of H2 at STP contain 6.023 × 1023 moles.
2. Assertion : One mole of NaCl contains 6.023 × 1023 molecules of sodium chloride.
Reason : 58.5 g of NaCl also contains 6.023 × 1023 molecules of NaCl.
3. Assertion : One mole of SO2 contains double the number of molecules present in one mole of O2.
Reason : Molecular weight of SO2 is three times to that of O2.
4. Assertion : 1 g O2 and 1 g O3 have equal number of atoms.
Reason : Mass of 1 mole atom is equal to its gram atomic mass.
5. Assertion : 1 mole oxygen and N2 have same volume at same temperature and pressure.
Reason : 1 mole gas at NTP ocupies 22.4 litre volume at STP.
6. Assertion : One mole of CO2 contains 6.023 × 1023 molecules of CO2.
Reason : 44 g of CO2 contains 6.023 × 1023 molecules of CO2.
7. Assertion : 32 g of sulphur (S8) contains 1.2048 × 1024 atoms.
Reason : Molecular mass of S8 = 32 g.
8. Assertion : At STP, 1 mL of ideal gas contains 2.69 × 1019 molecules.
Reason : One mole of a gas is the amount of gas which has a volume of 22.4 litres at STP.
9. Assertion : 16 g of methane contains 6.023 × 1023 molecules.
Reason : Molecular mass of methane = 16 g.
10. Assertion : In one mole of H2SO4, 7 × 6.023 × 1023 number of atoms are present.
Reason : 1 mole of H2SO4 contains 7 mole of atoms.
11. Assertion : 16 g oxygen contains similar number of atoms as 1 g-molecule of hydrogen.
Reason : Number of different species (atom, molecule or ion) present in same mole are different.
12. Assertion : 16 g oxygen contains similar number of atoms as 16 g-atom of oxygen.
Reason : Oxygen molecule is heteroatomic.
13. Assertion : Molecular mass = 2 × vapour density.
Reason : Vapour density is the mass ratio of 1 mole of vapour to that of hydrogen.

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
14. Assertion : The equivalent weight of reducing agent, Fe2+ is 56 (atomic weight of Fe = 56).
Reason : Fe loses 2e– to be converted into Fe2+.
15. Assertion : When a concentrated solution is diluted by adding more waver, molarity of the solution changes.
Reason : Number of moles of a solute divided by volume is equal to the normality.
16. Assertion : Strength and normality are two different methods of expressing the concentration of a solution.
Reason : Strength is equal to normality of a solution.
17. Assertion : The sum of mole fractions of all the components of a solution is unity.
Reason : Mole fraction is temperature dependent mode of concentration.
18. Assertion : For a solution, 92 g of ethanol and 144 g of water is mixed, mole fraction of ethanol and water
in the solutions are 0.2 and 0.8 respectively.
Reason : The sum of the mole fractions of all the components in a solution is unity.
19. Assertion : One molal aqueous solution of glucose contains 180 g of glucose in 100 g of water.
Reason : One molal solution is the solution containing one mole of solute in 100 g of water.
20. Assertion : In 0.5 M aqueous solution of sodium sulphate, the molarities of sodium and sulphate ions in
the solution are 1.0 M and 0.5 M respectively.
Reason : One mole of sodium sulphate gives 2 mole of sodium ions and 1 mole of sulphate ions on
dissociation.
21. Assertion : One molal aqueous solution has always lower concentration than one molar solution.
Reason : Molality of a solution depends upon the density of the solution whereas molarity is independent
of the density.
22. Assertion : Molality and normality can be calculated from each other.
Reason : Molarity is equal to the product of n and normality.
23. Assertion : Molality, mole fraction etc. are preferred over normality, molarity etc.
Reason : Temperature has no effect on molality and mole fraction.
24. Assertion : If 100 mL of 0.2 N HCl is mixed with 100 mL of 0.3 N HCl, the normality of final solution will be
0.25 N.
Reason : If two solutions of the same solute are mixed, the normalities can be added.
25. Assertion : 50 mL of 10 N HCl solution is diluted with distilled water to form one litre of the solution.
Normality of resulting solution is 0.5 N.
Reason : Normality is the number of gram equivalents of solute per litre of the solution.
26. Assertion : Normality of 0.3 M phosphorous acid (H3PO3) is 0.9 N.
Reason : It is a tribasic acid.

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1. The standard adopted for the determination of atomic weight of elements is based on [JCECE 2005]
(A) H1
(B) C
12
(C) O
16
(D) S 32

2. Law of multiple proportions is illustrated by one of the following pairs [JCECE 2005]
(A) H2S and SO2 (B) NH3 and NO2 (C) Na2S and Na2O (D) N2O and NO
3. A gas mixture contains O2 and N2 in the ratio of
1 : 4 by weight. The ratio of their number of molecules is [AFMC 2006]
(A) 1 : 8 (B) 1 : 4 (C) 3 : 16 (D) 7 : 32
4. The ability of a given subtance to assume two or more crystaline structures is called [AMU 2006]
(A) Amorphism (B) Isomorphism (C) polymorphism (D) Isomerism
5. A gas is found to have a formula [CO]x. If its vapour density is 70, the volume of x is [AMU 2006]
(A) 2.5 (B) 3.0 (C) 5.0 (D) 6.0
6. Which amount out of the following is the heaviest? [Kerala CEE 2006]
(A) 1 mole of oxygen. (B) 1 molecule of sulphur trioxide.
(C) 100 u of uranium. (D) 44 g of carbon dioxide.
7. What is the percentage of cation in ammonium dichromate ? [Kerala CEE 2006]
(A) 14.29% (B) 80% (C) 50.05% (D) 20.52%
8. The weight of one molecule of a compound C60H122 is [BHU 2007]

(A) 1.3  10 20 g (B) 5.01 10 21 g (C) 3.72  1013 g (D) 1.4  10 21 g
9. 25 g of MCI4 contains 0.5 mole chlorine then its molecular weight is [Punjab PMET 2007]
(A) 100 g mol –1
(B) 200 g mol –1
(C) 150 g mol –1
(D) 400 g mol–1
10. The crystalline salt Na 2SO 4  xH2O on heating loses 55.9% of its weight. The formula of crystalline salt is
[Kerala CEE 2007]
(A) Na2SO 4  5H2O (B) Na2SO 4  7H2O (C) Na2SO 4  2H2O (D) Na2SO 4  10H2O
11. The largest number of molecules is in [AMU 2008]
(A) 34 g of H2O (B) 28 g of CO2 (C) 46 g of CH3OH (D) 54 g of N2O5
12. Mass of 0.1 mole of methane is [KCET 2008]
(A) 1 g (B) 16 g (C) 1.6 g (D) 0.1 g
13. The maximum number of molecules are present in [BCECE 2008]
(A ) 15 L of H2 gas a STP (B) 5 L of N2 gas at STP
(C) 0.5 g of H2 gas (D) 10 g of O2 gas
14. Number of atoms of He in 100 u of He (atomic weight of He is 4 ) is [BCECE 2008]
(A) 25 (B) 100 (C) 50 (D) 100×6×10–23
15. Of two oxides of iron, the first contained 22% and the second contained 30% of oxygen by weight. The ratio
of weights of iron in the two oxides that combine with the same weight of oxygen, is [J&K CET 2008]
(A) 3 : 2 (B) 2 : 1 (C) 1 : 2 (D) 1 : 1

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
16. 1.5 g of CdCI2 was found to contain 0.9 g of Cd. Calculate the atomic weight of Cd. [EAMCET 2009]
(A) 118 (B) 112 (C) 106.5 (D) 53.25
17. One atom of an element weighs 1.8×10 –22
g. its atomic mass is [Manipal 2009]
(A) 29.9 (B) 154 (C) 108.36 (D) 18
18. The number of molecules in 18 mg of water in terms of Avogadro number NA is [J&K CET 2009]

(A) 103 NA (B) 102 NA (C) 10 1 NA (D) 10 NA

19. An organic compound made of C, H and N, contains 20% nitrogen. Its minimum molecular weight is
[WB JEE 2009]
(A) 70 (B) 140 (C) 100 (D) 65
20. In an experiment, 4 g of M2Ox oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56g
mol–1, the number of O atoms in the oxide is [AFMC 2010]
(A) 1 (B) 2 (C) 3 (D) 4
21. The number of molecules in 100 mL of 0.01 M H2SO4 is [AMU 2010]
(A) 6.02  1022 (B) 6.02  1021 (C) 6.02  1020 (D) 6.02  1018
22. The mass of one mole of electron is [CPMT 2010]

(A) 9.1 10 28 g (B) 0.55 mg (C) 9.1 10 24 g (D) 6  1012 g
23. 60 g of a compound on analysis produced 24 g carbon, 4 g hydrogen and 32 g oxygen. The empirical formula
of the compound is [BVP 2010]
(A) CH2O2 (B) CH2O (C) CH4O (D) C2H4O2
24. The total number of electrons in 18 mL of water (density = 1g ml ) –1
[KCET 2012]
(A) 6.02 × 10 25
(B) 6.02 × 10 24
(C) 6.02 × 10 × 105 23
(D) 6.02 × 1023
25. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atom?
[AFMC 2012]

(A) 0.02 (B) 3.125  10 2 (C) 1.25  10 2 (D) 2.5  10 2

26. The number of sodium atoms in 2 moles of sodium ferrocyanide is? [Manipal 2012]
(A) 12  10 23 (B) 26  1023 (C) 34  1023 (D) 48  1023
27. A metal oxide has the formula A2O3. It can be reduced by hydrogen to give free metal and water. 0.1596 g of
this metal oxide requires 6 mg of hydrogen for complete reduction. What is the atomic weight of metal
[AIIMS 2012]
(A) 52.3 (B) 57.3 (C) 55.8 (D) 59.3
28. Number of molecules in one litre of liquid water is close to [KCET 2013]

18 6.023
(A)  1023 (B) 55.5 × 6.023 × 1023 (C)  1023 (D) 18. × 6.023 × 1023
22.4 23.4
29. If 20 g of CaCO3 is treated with 100 mL of 20% (w/v) HCI solution, the amount of CO 2 produced is
[Manipal 2013]
(A) 22.4 L at NTP (B) 8.80 g (C) 4.40 g (D) 2.24 L at NTP
30. 0.56 g of a gas occupies 280 cm at NTP, then its molecular mass is
3
[OJEE 2013]
(A) 4.8 (B) 44.8 (C) 2 (D) 22.4

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
31. Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27°C in
identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be [AIPMT 2014]
(A) 8 : 16 : 1 (B) 16 : 8 : 1 (C) 16 : 1 : 2 (D) 8 : 1 : 2
32. When 22.4 litres of H2(g) is mixed with 11.2 litres of CI2(g), each at N.T.P, the mole of HCI(g) formed is equal
to [AIPMT 2014]
(A) 1 mol of HCI(g) (B) 2 mol of HCI(g) (C) 0.5 mol of HCI(g) (D) 1.5 mol of HCI(g)
33. 1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much
(At. wt. Mg = 24, O=16) [AIPMT 2014]
(A) Mg, 0.16 g (B) O2, 1.16 g (C) Mg, 0.144 g (D) O2, 0.28 g
34. The volume strength of 0.75 M H2O2 solution is [BCECE-2015]
(A) 4.8 vol. (B) 5.2 vol. (C) 8.4 vol. (D) 8.8 vol.
35. An organic compound contains C = 40%, H = 6.66% and rest is oxygen. The empirical formula of the
compound is [BCECE-2015]
(A) CH2O (B) CHO (C) C2H4O2 (D) CH3OH
36. 10g hydrogen is reacted with 64 g oxygen. The amount of water formed will be (in moles) [BCECE-2015]
(A) 3 (B) 4 (C) 1 (D) 2
37. 20.0g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium
oxide. What will be the precentage purity of magnesium carbonate in the sample ? [AIPMT-2015]
(Atomic Wt. : Mg = 24)
(A) 75 (B) 96 (C) 60 (D) 84
38. The number of water molecules is maximum in: [AIPMT-2015]
(A) 18 molecules of water (B) 1.8 gram of water (C) 18 gram of water (D) 18 moles of water
39. What is the mole fraction of the solute in a 1.00 m aqueous solution ? [AIPMT-2015]
(A) 0.177 (B) 1.770 (C) 0.0354 (D) 0.0177
40. If Avogadro number NA, is changed from 6.022 × 10 mol to 6.022 × 10 mol , this would change:
23 –1 20 –1

(A) The definition of mass in units of grams [AIPMT-2015]

(B) The mass of one mole of carbon
(C) The ratio of chemical species to each other in a balanced equation
(D) The ratio of elements to each other in a compound
41. What is the mass of the precipitate formed when 500 mL of 16.9% (w/v) solution of AgNO3 is mixed with 500
mL of 5.8% (w/v) NaCl solution ? [AIPMT-2015]
(Ag = 107.8, N = 14, O = 16, Na = 23, CL = 35.5)
(A) 28 g (B) 3.5 g (C) 7 g (D) 14 g
42. If 0.228 g of silver salt of dibasic acid gave a residue of 0.162g of silver on ignition then molecular weight of the
acid is [AIIMS-2015]
(A) 70 (B) 80 (C) 90 (D) 100
43. Suppose the elements X and Y combine to form two compound XY2 and X3Y2. When 0.1 mole of XY2 weigh
10 g and 0.05 mole of X3Y2 weighs 9g, the atomic weights of X and Y are [NEET 2016]
(A) 30, 20 (B) 40, 30 (C) 60, 40 (D) 20, 30
44. 10 mL of liquid carbon disulphide (specific gravity 2.63) is burnt in oxygen. Find the volume of the resulting
gases measured at NTP. [AIIMS 2016]
(A) 23.25 L (B) 22.26 L (C) 23.50 L (D) 20.08 L

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]
45. Which of the following is dependent of temperature? [NEET–2017]
(A) Molality (B) Molarity (C) Mole fraction (D) Weight percentage
46. In which case is the number of molecules of water maximum ? [NEET–2018]
(A) 0.00224 L of water vapours at 1 atm and 273 K (B) 0.18 g of water
(C) 18 mL of water (D) 10–3 mole of water
47. A compound possesses 8% sulphur by mass. The least molecular mass is ?
(A) 200 (B) 400 (C) 155 (D) 355
48. Calculate the molarity of one litre solution of 22.2 g of CaCl2. [JIPMER–2019]
(A) 0.4 M (B) 0.2 M (C) 0.8 M (D) 0.6 M
3
49. The density of 2 M aqueous solution of NaOH is 1.28 g/cm . The molality of the solution is
[Given that molecular mass of NaOH=40g mol–1] [NEET (Odisha)–2019]
(A) 1.20 m (B) 1.56 m (C) 1.67 m (D) 1.32 m
50. Find empirical formula of the compound, if
M = 68% (atomic mass = 34) and remaining 32% oxygen. [AIIMS–2019]
(A) MO (B) M2O (C) MO2 (D) M2O3
51. The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's
process is [NEET–2019]
(A) 20 (B) 30 (C) 40 (D) 10
175
52. The number of protons, neutrons and electrons in 71 Lu ,respectively are [NEET–2020]

(A) 104, 71 and 71 (B) 71, 71 and 104

(C) 175, 104 and 71 (D) 71, 104 and 71
53. An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option
for the empirical formula of this compound is : [Atomic wt. of C is 12,H is 1] [NEET–2021]
(A) CH (B) CH2 (C) CH3 (D) CH4
54. What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5M HCl solution according to the
following reaction ?

 CaCl2 (aq)  CO2 (g)  H2 O()

CaCO3 (s)  2HCl(aq) 

[Calculate upto second place of decimal point] [NEET–2022]

(A) 1.32g (B) 3.65g (C) 9.50g (D) 1.25g
55. Select the correct statements from the following [NEET–2023]
A. Atoms of all elements are composed of two fundamental particles :
B. The mass of the electron is 9.10939  10–31 Kg.
C. All the isotopes of a given element show same chemical properties.
D. Protons and electrons are collectively known as nucleons.
E. Dalton’s atomic theory, regarded the atom as an ultimate particle of matter.
Choose the correct answer from the options given below :
(A) A and E only (B) B, C and E only (C) A, B and C only (D) C, D and E only

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STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1. (A) 2. (D) 3. (B) 4. (D) 5. (D) 6. (B)

7. (B) 8. (A) 9. (A) 10. (D) 11. (A) 12. (D)

13. (B) 14. (D) 15. (B)

1. (B) 2. (B) 3. (C) 4. (A) 5. (C) 6. (B)

7. (B) 8. (C) 9. (C) 10. (C) 11. (D) 12. (C)
13. (A) 14. (D) 15. (C) 16. (D) 17. (A) 18. (D)
19. (B) 20. (D) 21. (A) 22. (D) 23. (A) 24. (C)

1. (C) 2. (B) 3. (B) 4. (C) 5. (D) 6. (B)

7. (B) 8. (C) 9. (C) 10. (C) 11. (B) 12. (B)
13. (D) 14. (D) 15. (D) 16. (B) 17. (C) 18. (B)
19. (C) 20. (D) 21. (B) 22. (A) 23. (B) 24. (A)

1. (C) 2. (B) 3. (D) 4. (D) 5. (A) 6. (A)

7. (A) 8. (A) 9. (D) 10. (D) 11. (B) 12. (B)

1. (A) 2. (C) 3. (A) 4. (C) 5. (B) 6. (B)

7. (C) 8. (D) 9. (B) 10. (A) 11. (A) 12. (D)
13. (A) 14. (B) 15. (A)

1. (D) 2. (A) 3. (A) 4. (C) 5. (A) 6. (B)

7. (A) 8. (D) 9. (C) 10. (B) 11. (B) 12. (C)
13. (A) 14. (B) 15. (A)

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 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1. (C) 2. (C) 3. (A) 4. (B) 5. (A) 6. (C)

7. (C) 8. (A) 9. (A) 10. (B) 11. (C) 12. (B)
13. (B) 14. (B) 15. (A) 16. (B) 17. (B) 18. (D)

1. (A) 2. (B) 3. (A) 4. (B) 5. (C) 6. (B)

7. (A) 8. (B) 9. (B) 10. (D)

1. (I - R) ; (II - T) ; (III - P) ; (IV - Q) ; (V - S)

2. (i) 18 amu (ii) 44 amu (iii) 16 amu
3. 980 g of Si
4. 6.02 × 1023 atom Na, 6.02 × 1023 atom Cl, 18.06 × 1023 atom O
5. (a) H = 4NA, S = 2N A, O = 8N A atoms (b) H = 4 atoms, S = 2 atoms, O = 8 atoms.
(c) H = 10NA, S = 10NA, O = 40 N A atoms (d) H = 6 atoms, S = 6 atoms, O = 18 atoms.
6. 20 NA
7. 39.81 g
8. 44.8
9. 2.27
10. 24
11. Na=32.4% S=22.53% and O=45.07%
12. CH4
13. (i) 44 gm (ii) 22 gm (iii) 22 gm
14. (i) B2 (ii) A (iii) A
15. (i) 2428.57g (ii) yes (iii) H2 571.43 g
16. 5.83 g of Li3 N
17. 0.029M
18. 15.375g
19. 15.44 M
20. 25.22 ml
21. (i) 1.5 × 10-3 (ii) 1.25 × 10–4 M
22. 10 VOLUMES

BIHAR SCHOOL EXAMINATION BOARD (BSEB), SINHA LIBRARY ROAD, PATNA, PIN-800017 [ PAGE # 80 ]
STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1. (B) 2. (D) 3. (B) 4. (C) 5. (C) 6. (A)

7. (B) 8. (D) 9. (A) 10. (C) 11. (A) 12. (D)
13. (B) 14. (A) 15. (C) 16. (A) 17. (A) 18. (B)
19. (B) 20. (D) 21. (B) 22. (A) 23. (A) 24. (A)
25. (C) 26. (D) 27. (C) 28. (D) 29. (B) 30. (C)
31. (A) 32. (D) 33. (A) 34. (C) 35. (A) 36. (B)
37. (B) 38. (C) 39. (A) 40. (D) 41. (B) 42. (C)
43. (B) 44. (A) 45. (D) 46. (B) 47. (C) 48. (B)
49. (A) 50. (B) 51. (A) 52. (A) 53. (B) 54. (B)
55. (D) 56. (A) 57. (C) 58. (D) 59. (A) 60. (A)
61. (C) 62. (A) 63. (D) 64. (A) 65. (B) 66. (B)
67. (A) 68. (C) 69. (B) 70. (C) 71. (B) 72. (D)
73. (A) 74. (A) 75. (B) 76. (A) 77. (C) 78. (C)
79. (B) 80. (B) 81. (A) 82. (A) 83. (C) 84. (B)
85. (B) 86. (A) 87. (C) 88. (C) 89. (C) 90. (A)
91. (B) 92. (B) 93. (D) 94. (B) 95. (C) 96. (B)
97. (D) 98. (A) 99. (C) 100. (D) 101. (A) 102. (C)
103. (B) 104. (C) 105. (A) 106. (C) 107. (B) 108. (B)
109. (C) 110. (C)

1. (D) 2. (A) 3. (A) 4. (A) 5. (D) 6. (C)

7. (B) 8. (C) 9. (B) 10. (C) 11. (C) 12. (C)
13. (A) 14. (B) 15. (A) 16. (B) 17. (C) 18. (A)
19. (C) 20. (A) 21. (C) 22. (D) 23. (C) 24. (A)
25. (C) 26. (C) 27. (D) 28. (D) 29. (D) 30. (B)
31. (B) 32. (B) 33. (D) 34. (B) 35. (A) 36. (B)
37. (A) 38. (C) 39. (A) 40. (D) 41. (C) 42. (B)
43. (C) 44. (A) 45. (D) 46. (A) 47. (D) 48. (A)
49. (C) 50. (B) 51. (D) 52. (B)

BIHAR SCHOOL EXAMINATION BOARD (BSEB), SINHA LIBRARY ROAD, PATNA, PIN-800017 [ PAGE # 81 ]
 STOICHIOMETRY-I (MOLE CONCEPT) [NEET (UG)]

1 (C) 2 (D) 3 (D) 4 (B) 5 (B) 6 (B)

7 (D) 8 (B) 9 (A) 10 (A) 11 (D) 12 (D)
13 (A) 14 (B) 15 (D) 16 (C) 17 (C) 18 (B)
19 (D) 20 (A) 21 (B) 22 (C) 23 (A) 24 (C)
25 (B) 26 (C)

1. (B) 2. (D) 3. (D) 4. (C) 5. (C) 6. (D)

7. (A) 8. (D) 9. (B) 10. (D) 11. (A) 12. (C)
13. (A) 14. (A) 15. (A) 16. (C) 17. (C) 18. (A)
19. (A) 20. (C) 21. (C) 22. (B) 23. (B) 24. (B)
25. (B) 26. (D) 27. (C) 28. (B) 29. (B) 30. (B)
31. (C) 32. (A) 33. (A) 34. (C) 35. (A) 36. (B)
37. (D) 38. (D) 39. (D) 40. (B) 41. (C) 42. (C)
43. (B) 44. (A) 45. (B) 46. (C) 47. (B) 48. (B)
49. (C) 50. (A) 51. (B) 52. (D) 53. (C) 54. (A)
55. (B)

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