CH # 1 : Fundamental Concepts of Chemistry ----- Malik Xufyan Malik Xufyan M.Sc. Applied Chemistry, B.Ed. JIAS ACADEMY Cell # 03137355727

1. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 1
i. Justify that 23 g of Na and 238 g of uranium have equal no. of
atoms.
Ans: According to mole and Avogadro’s Number concept:
23 g of Na =1 mole of Na = 6.02x1023
atoms of Na
238g of U =1 mole of U = 6.02x1023
atoms of U.
Since equal number of gram atoms (moles) of different elements contain
equal number of atoms. Hence, 1 mole (23 g) of sodium and 1 mole (238
g) of uranium contain equal number of atoms, i.e, 6.02 x1023
atoms.
ii. Why law of conservation of mass has to be obeyed during
stoichiometric calculations?
Ans: According to law of conservation of mass: the amount of each
element is conserved in a chemical reaction. Chemical equations are
written and balanced on the basis of law of conversation of mass.
Stoichiometry calculations are related with the amounts of reactants and
products in a balanced chemical equation. Hence, law of conservation of
mass has to be obeyed during stoichiometric calculations.
OR
According to the law of conservation of mass, mass cannot be created
nor destroyed but it can be converted into another shape
During Stoichiometry calculation, we use balance chemical equation in
which number of reactants are balanced with number of products. That’s
why stoichiometry calculations obey law of conservation of mass in
balanced chemical equation.
iii.No individual neon atom in the sample of element has a mass of
20.18 amu.
Ans: According to the average atomic mass: ‘’Average atomic mass
depends upon its atomic mass and its abundance in nature’’.
Since the overall atomic mass of neon in the average of the determined
atomic masses of individual isotopes present in the sample of isotopic
mixture. Hence, no individual neon atom in the sample has a mass of
20.18 amu.
Calculation:
Average atomic mass =
. . .
= 20.18 amu.

2. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 2
iv.One mg of K2CrO4 has thrice the number of ions than the
number of formula units when ionized in water.
Ans: K2Cr2O7 when ionizes in water produces two k+
ions one CrO4 ion.
Thus each formula unit of K2 Cr2O7 produces three ions in solution
.Hence one mg of K2CrO4 has thrice the number of ion than the number
of formula units ionized in water.
K2
CrO4
H2O
2K+
+ CrO4
-2
2 ions + 1 ion = 3 ions
OR
Mass of K2CrO4 = 1 mg = 10-3
g
We will find out
Total ions of K2CrO4 = ?
As we know
Molar mass of K2CrO4 = 194 g/mol
Number of moles of K2CrO4 = =
= 5.15 x 10-6
moles
Total formula units of K2CrO4 = Number of moles of K2CrO4 x NA
= 5.15 x 10-6
x 6.02 x 1023
= 3.10 x 1018
K2Cr2O7 ionizes in water
K2
CrO4
H2O
2K+
+ CrO4
-2
1 formula unit of K2CrO4 produces ions = 3 ions
3.10 x 1018
formula unit of K2CrO4 produces ions = 3 x 3.10 x 1018
= 9.30 x 1018
ions
Thus it proved that One mg of K2CrO4 has thrice the number of ions
than the number of formula units when ionized in water.
v. Mg is twice heavier than that of carbon of atoms in them.
Ans: The atomic mass of Mg is 24 which is twice as mass as compared
to the atomic mass of carbon One gram atom of different elements has
different mass. One mole of carbon is 12 g, while one mole of

3. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 3
magnesium is equal to 24 g.It means that one atom of Mg is twice in
mass than one atom of carbon.
OR
We know the molar mass of magnesium and carbon
1 mol of magnesium = 24 g = 6.02 x 1023
atoms
1 mol of carbon = 12 g = 6.02 x 1023
atoms
We can calculate the mass of magnesium and carbon
6.02 x 1023
magnesium atoms has mass = 24 g
1 -------------------------------------------- =
.
= 3.9867 x 10-23
Similarly
6.02 x 1023
carbon atoms has mass = 12 g
1 -------------------------------------------- =
.
= 1.9933 x 10-23
By comparing carbon and magnesium
Magnesium : carbon
3.9867 x 10-23 :
1.9933 x 10-23’
.
.
:
.
.
2 : 1
Thus, above this result, we can say that magnesium is twice
heavier than carbon atom.
vi.Justify many chemical reactions taking place in our surrounding
involve the limiting reactants
Ans: According to the definition of limiting reactant:
‘’A limiting reactant is one which has limited quantity and consumed
first in a chemical reaction’’.
In our surrounding many chemical reactions are taking place which
involve oxygen. In these reactions, oxygen is always in excess quantity
while other reactants are in lesser amount. Thus other reactants act as
limiting reactants.

4. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 4
Example:
1. Petrol burns in excess of oxygen present in air
2. Rusting of iron in the excess of oxygen present in air.
vii. N2 and CO have the same number of electrons, protons and
neutrons.
Ans:
For N2:
No. of electrons in N2 = 2 x 7 = 14
No. of protons in N2 = 2 x 7 = 14
No. of neutron in N2 = 2 x 7 = 14
For CO:
In CO, there are one carbon and one oxygen atoms.
No. of protons in CO = 6 C protons + 8 O proton = 14
No. of electrons in CO = 6 C electrons + 8 O electrons = 14
No. of neutron in CO = 6 C neutron + 8 O neutron = 14
For CO & N2:
Hence, N2 and CO have the same number of electrons, protons and
neutrons.
Remember that electrons, protons and neutrons of atoms remain
conserved during the formation of molecules in a chemical reaction.
viii. 180 g of glucose and 342 g of sucrose have the same number of
molecules but different number of atoms present in them.
Ans:
According to mole and Avogadro’s Number:
180 g of glucose = 1 mole of glucose = 6.02x1023
molecules of glucose
342 g of sucrose = 1mole of sucrose = 6.02x1023
molecules of sucrose
Since one mole of different compounds has the same number of
molecules.
Therefore,1 mole (180 g) of glucose and I mole (342 g) of sucrose
contain the same number (6.02x1023
) of molecules. While one molecule
of glucose (C12H22O11) contains 45 atoms and one molecules of glucose
(C6 H12O6) contains 24 atoms.
Therefore, 6.02x1023
molecules of glucose contain different atoms as
compound to 6.02x1023
molecules of sucrose. Hence, 180 g of glucose

5. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 5
and 342 g of sucrose have the same number of molecules but different
number of atoms present in them.
OR
As we know that
Molar mass of glucose = 180 g/mol
Molar mass of sucrose = 342 g/mol
According to Avogadro number, one mole contains same number of
molecules
1 mole = 6.02 x 1023
molecules
So,
180 g of glucose = 1 mol = 6.02 x 1023
molecules
342 g of sucrose = 1 mol = 6.02 x 1023
molecules
While
1 molecule of glucose (C6 H12O6) contains atoms = 24 atoms
6.02 x 1023
molecules of glucose (C6 H12O6) contains atoms = 45 x 6.02
x 1023
atoms
And
1 molecule of sucrose (C12H22O11) contains atoms = 45 atoms
6.02 x 1023
molecules of glucose (C12H22O11) contains atoms = 45 x 6.02
x 1023
atoms
Thus, it is proved that 180 g of glucose and 342 g of sucrose have the
same number of molecules but different number of atoms present in
them
ix.Define gram atom and molar volume with three examples.
Ans: Gram Atom:
The atomic mass of an element expressed in grams is called gram atom
of an element.
For example:
1 gram atom of Hydrogen = 1.008 g
1 gram atom of Carbon = 12.00 g
1 gram atom of Uranium = 238 g
Molar volume: The volume occupied by one mole of an ideal gas at
standard temperature and pressure (STP) is called molar volume. The
volume is equal to 22.414 dm3
.

6. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 6
Example
1 mole of H2 = 6.02 x 1023
molecules of H2= 2.06 g of H2 = 22.414 dm3
at S.T.P
x. Define stoichiometry and gram formula with three examples.
Ans: Stoichiometry:
Stoichiometry is the branch of chemistry which gives a quantitative
relationship between reactants and products in balanced chemical
equation.
Assumptions:
1. All the reactant are the completely changed in to the products.
2. While doing calculations law of conservation and law of definite
proportions are obeyed.
3. There is no side reaction
Gram formula:
The formula mass of an ionic compound expressed in grams is called
gram formula of the substance.
Formula:
Number of gram formula =
Example
1 gram formula of NaCl = 58.50 g
1 gram formula of Na2CO3 = 106 g
1 gram formula of AgNO3 = 170 g
Q.9 (d) 4.9 g of H2SO4 when completely ionized in water ,
have equal number of positive and negative charges but the
number of positively charged ions are twice the number of
negatively charged ions.
Ans: When one molecules of H2SO4 completely ionizes in water it
produces two H+
ion and one SO-
ion. Hydrogen ion carries a unit
positive charge whereas SO-
ion carries a double negative
charge. To keep the neutrality, the number of hydrogen is twice
than the number of sulphate ions. Similarly the ions produced by
complete ionization of 4.8g of H2SO4 in water will have equal
number of positive and negative but the numbers of positively
charged ions are twice the number of negatively charged ions.

7. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 7
H2SO4
H20
2H
+
+ SO
-
OR
Mass of H2SO4 = 4.9 g
Molar mass of H2SO4 = 98
Number of moles of H2SO4 = =
.
= 0.05 moles
H2SO4
H20
2H
+
+ SO
-
1 mol 2 mol 1 mol
0.05 moles 2 x 0.05 0.05
0.05 moles 0.1 moles 0.05 moles
Thus proved that the ions produced by complete ionization of 4.8g of
H2SO4 in water will have equal number of positive and negative
but the numbers of positively charged ions are twice the number
of negatively charged ions.
Q.9 (f) Two grams of H2, 16 g of CH4 and 44 g of CO2 occupy
separately the volumes of 22.414 dm3
, although the sizes and
masses of molecules of three gases are very different from
each other.
Ans:
2 g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP
= 22.414 dm3
16 g of CH4 =1mole of CH4 =6.02x1023
molecules of CH4 at STP
=22.414 dm3
144 g of CO2 =1mole of CO2 =6.02x1023
molecules of CO2 at STP
=22.144 dm3
Although H2, CH4 and CO2 have different masses but they
have the same number of moles and molecules. Hence the same
number of moles or the same numbers of molecules of different
gases occupy the same volume at STP. Hence 2 g of H2, 16g of
CH4 and 44 g of CO2 occupy the same volume 22.414 dm3
at
STP. The masses and the sizes of the molecules do not affect the
volumes due to 300 times greater distance between gas
molecules.

8. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 8
Q.8. Define Gram ion, Gram molecular mass and Percentage
yield with three examples.
Ans:
ii. Gram Molecular Mass
Definition
The molecular mass of a substance expressed in grams is called
Gram Molecular mass or mole of substance.
Formula
Number of Molecular substance =
Mass of molecular substance in grams
Molecular mass of molecular substance
Example
1 gram molecule of water = 18 g
1 gram molecule of H2SO4 = 98 g
1 gram molecule of sucrose = 432 g
iv. Gram ion:
Definition:
The ionic mass of an ionic specie expressed in grams is called
one gram ion or one mole of ions.
Formula
Number of gram ions =
Examples
1 gram ion of OH–1
= 17 grams
1 gram ion of SO-2
= 96 grams
1 gram ion of CO3
-2
= 60 grams
vii. Percentage yield
Definition
The yield which is obtained by dividing actual yield with theoretical
yield and multiplying by 100 is called percentage yield.
Formula
Percentage yield is efficiency of reaction which is determined by
% yield = 100

9. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 9
Q.25 (iv) One mole of H2SO4 should completely react with two
moles of NaOH. How does Avogadro, s number help to
explain it.
Ans:
H2SO4 + 2NaOH Na2SO4 + H2O
1 mole 2 moles
2 moles of H
+
ions 2 moles of OH
-
ions
2 x 6.02 x 10
23
H
+
ions 2 x 6.02 x 10
23
OH
-
ions
Hence one mole of H2SO4 consists of 2 moles of H+
ions that
contains twice the Avogadro’s number of H+
ions.
For complete neutralization it needs 2 moles of one mole of
H2SO4 should completely react with two moles of NaOH.
Q.25 (v) One mole H2O has two moles of bonds , three moles
of atoms , ten moles of electrons and twenty eight moles of
the total fundamental particles present in it.
Ans: Since one molecule of H2O has two covalent bonds between
H and O atoms.
There are three atoms, ten electrons and twenty eight total
fundamental particles present in it.
Hence, one mole of H2 O has two moles of bond, three moles of
atoms, ten mole of electrons and twenty eight moles of total
fundamental particle present in it in detail u can write that:
Bonds:
1 molecule of H20 contains bonds = 2
6.02 X 10 23
molecules contain bonds = 2 X 6.02 X 10 23
Thus 1 mole of H20 contains bonds = 2 moles
Atoms:
1 molecule of H20 contains atoms = 3
6.02 X 10 23
molecules contains atoms = 3 X 6.02 X 10 23
Thus 1 mole of H20 contains atoms = 3 moles
Electrons:
One molecule of H2O contain 2 H atoms and 1 O atom
Since

10. CH # 1 :Fundamental Concepts of Chemistry JIAS ACADEMY
Malik Xufyan
0313-7355727 10
One O atom contains electrons = 8
One H atom contains electrons = 1
two H atom contains electrons = 2
Hence 1 molecule of H2O contains = 2+8 = 10 e
6.02 X 10 23
molecules contains = 10 X 6.02 X 10 23
Thus 1 mole of H2O contains electrons =3 moles
Total Fundamental particles :
1 oxygen atom contains = 8
electrons , 8 protons , 8 necutrons
1 oxygen atom contains total fundamental particles = 8+8+8=24
1 hydrogen atom contains = 1 electron,
1 proton , 0 neutron
1 Hydrogen atom contain total fundamental particles = 1+1+0= 2
2 Hydrogens atoms contains total fundamental particles = 4
Hence
1 molecule of water contains fundamental particles = 24+4= 28
6.02 X 10 23
molecules contain fundamental particles = 28 X 6.02
X 10 23
Thus 1 mole of water contains bonds = 28 moles
Malik Xufyan
M.Sc. Applied Chemistry, B.Ed.
JIAS ACADEMY
Cell # 03137355727