2. ➢ Law of Conservation of Mass.
➢ Law of Definite Proportions.
➢ What is an Atom?
➢ What is a Molecule?
➢ The chemical formula of a molecular compound.
➢ Ionic compounds.
➢ Relative atomic mass scale.
➢ The Avogadro constant.
➢ Mole concept.
➢ Molar mass.
You will Learn
3.
4.
5.
6. i) Law of conservation of mass :- states that ‘Mass can neither be
created nor destroyed in a chemical reaction’.
This means that during a chemical reaction the sum of the masses of
the reactants and products remain unchanged.
Activity :-
Take some Sodium sulphate solution in a conical flask and some
Barium chloride solution in an ignition tube. Hang the ignition tube in
the flask by a thread and pot a cork on the flask. Find the mass of the
flask on a balance. Then tilt the flask. A chemical reaction takes place
and Sodium chloride and Barium sulphate are formed. Then find the
mass of the flask again. It will be seen that the sum of the masses of the
reactants and products remain unchanged.
BaCl2
solution
BaSO4 white
7. ii) Law of definite proportion ( Law of constant proportion )
‘ In a chemical compound the elements are always present in a definite
proportion by mass’.
Eg :- Water ( H2O) always contains two elements Hydrogen and
Oxygen combined together in the same ratio of 2:16 or 1:8 by mass.
If 9 g of water is decomposed we get 1 g of hydrogen and 8 g of
oxygen.
Ammonia (NH3) always contains two elements Nitrogen and
Hydrogen combined together in the same ratio of 14:3 by mass.
11. Lets poll!!
Q1. The percentage of Copper and Oxygen in samples of CuO
obtained by different methods were found to be same. This illustrates
the law of
a) constant proportions
b) conservation of mass
c) multiple proportions
d) reciprocal proportions
Q2.Chemical equation is balance according to the law of
a) multiple proportion
b) conservation of mass
c) reciprocal proportions
d) Constant proportion
12. Q3. Carbon dioxide in air and the Carbon dioxide dissolved in
water, both contain Carbon and Oxygen in ratio of 12:32 gram.
This shows:
a. Law of conservation of mass
b. Law of constant proportion
c. Law of combination
d. Law of conservation of energy
Q4. The ratio of H:O by mass in Hydrogen peroxide (H202) is
a) 16:1
b) 1:8
c) 1:32
d) 1:16
13. Q5. Assertion: In any chemical reaction ,total number of molecules are
conserved.
Reason: Law states that mass can neither be created nor destroyed in
a chemical reaction.
a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
b) Both assertion (A) and reason (R) are true but reason (R) is not the correct
explanation of assertion (A).
c) Assertion (A) is true but reason (R) is false.
•
d) Assertion (A) is false but reason (R) is true.
e) Both assertion and Reason are false
•
14. Poll Answers
Q1. The percentage of Copper and Oxygen in samples of CuO
obtained by different methods were found to be same. This illustrates
the law of
a) constant proportions
b) conservation of mass
c) multiple proportions
d) reciprocal proportions
Q2.Chemical equation is balance according to the law of
a) multiple proportion
b) conservation of mass
c) reciprocal proportions
d) Constant proportion
15. Q3. Carbon dioxide in air and the Carbon dioxide dissolved in
water, both contain Carbon and Oxygen in ratio of 12:32 gram.
This shows:
a. Law of conservation of mass
b. Law of constant proportion
c. Law of combination
d. Law of conservation of energy.
Q4. The ratio of H:O by mass in Hydrogen peroxide (H202) is
a) 16:1
b) 1:8
c) 1:32
d) 1:16
16. Q5. Assertion: In any chemical reaction ,total number of molecules are
conserved.
Reason: Law states that mass can neither be created nor destroyed in
a chemical reaction.
a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
b) Both assertion (A) and reason (R) are true but reason (R) is not the correct
explanation of assertion (A).
c) Assertion (A) is true but reason (R) is false.
•
d) Assertion (A) is false but reason (R) is true.
e) Both assertion and Reason are false
•
17. ANSWER THE FOLLOWING QUESTIONS
1. In an experiment ,5.3 grams of Sodium carbonate
reacted with 6 grams of Ethanoic acid to form 8.2 grams
Sodium ethanoate,2.2grams of Carbon dioxide and 0.9
grams of water. Show that this data verifies the law of
conservation of mass.
2. .In an experiment, when 10g of Calcium carbonate is
decomposed completely then 5.6 g of Calcium oxide is
formed .Calculate the mass of Carbon dioxide formed.
Which law of chemical combination will you use in
solving the problem.
18. 3. State the law of conservation of mass. In a chemical reaction 4.2g of
Na
2
CO
3
reacted with 3.0g of Hydrochloric acid. The products obtained
were 4.1 g of Sodium chloride, 0.9g of water and 2.2g of Carbon dioxide.
Show that these observations are in agreement with the law of
conservation of mass.
4. Carbon and Oxygen reacts in the ratio of 3:8 by mass to form Carbon
dioxide. Calculate the amount of Oxygen required to burn 6g of Carbon.
State the law of chemical combination governs your answer.
5. In an experiment 1.288g of Copper oxide on reduction gave 1.03 g of
Copper. In another experiment 3.672g of Copper oxide gave 2.938 g of
Copper. Show that these figures verify the law of constant proportion.
19. ANSWERS
1.Na2CO3 + CH3COOH → CH3COONa + CO2+ H2O
5.3 g + 6 g → 8.2 g + 2.2g + 0.9 g
11.3 g = 11.3g
2.We use law of conservation of mass for solving the
problem
Mass of products = mass of reactants
Mass of CaO + mass of CO2 =mass of CaCO3
mass of CO2 = x
5.6 g + x =10g
x= 10 - 5.6
x=4.4g
20. 3. a) Mass can neither be created nor destroyed in a chemical reaction.
Sodium hydrogen carbonate + Hydrochloric Acid -> Sodium chloride + water + CO
2
4.2g 3g 4.1g 0.9g 2.2g
7.2g → 7.2g
These observations are in agreement with the law of conservation of mass in a chemical
reaction. Total Mass of reactants is equal to total mass of products.
4. a) 3 : 8 :: 6 : X
3X = 6 * 8
x = 16g
21. 5. EXPERIMENT 1
Mass of Copper oxide =1.288 g
Mass of Copper =1.03 g
Mass of Oxygen = CuO – Cu
= 1.288-1.03
= 0.258g
In the first sample of Copper oxide compound
Mass of Cu : mass of Oxygen
1.03 : 0.258
4:1
Experiment 2
Mass of Copper oxide =3.672
Mass of Copper =2.938
Mass of Oxygen = CuO – Cu
=3.672-2.938
= 0.734 g
In the second sample of CuO compound
Mass of Copper : Mass of Oxygen
2.938: 0.734
4 : 1
22.
23. Dalton’s atomic theory :-
i) Matter is made up of tiny particles called atoms.
ii) Atoms are indivisible and cannot be created or destroyed
in a chemical reaction
iii) Atoms of a given element are similar in mass and
properties.
iv) Atoms of different elements have different masses and
properties.
v) Atoms combine in small whole number ratios to form
compounds.
vi) In a given compound the relative number and kind of
atoms are constant.
24.
25. Drawbacks of Dalton’s atomic theory
• It does not account for subatomic particles
Dalton’s atomic theory stated that atoms were indivisible.
However ,the discovery of subatomic particles (such as protons, electrons and
neutrons ) disproved this postulate.
• It does not account for isotopes
As per Dalton’s atomic theory, all atoms of an element have identical masses
and densities. However, different isotopes of elements have different atomic masses
• It does not account for isobars:
This theory states that the masses of the atoms of two different elements must
differ. However, it is possible for two different elements to share the same mass
number. Such atoms are called isobars.
• Elements need not combine in simple whole number ratios to form compounds.
Certain complex organic compounds do not feature simple ratios of constituent
atoms. Example :Sugar/Sucrose (C12H22O11)
26. Atom :-
An atom is the smallest particle of an element that may or
may not exist independently and retains all its chemical
properties.
Atoms are very small in size and smaller than anything we
can imagine or compare with.
Atomic radius is measured in nano metres (nm)
1 nanometer = 10 -9 m or 1 meter = 109 nm
Eg :- The atomic radius of an atom of hydrogen is 10-10 m.
The radius of a molecule of water is 10-9 m.
27.
28.
29. Symbols of atoms of different elements :-
Symbols of some common elements :-
Element Symbol Element Symbol Element Symbol
Aluminium Al Copper Cu Nitrogen N
Argon Ar Fluorine F Oxygen O
Barium Ba Gold Au Potassium K
Boron B Hydrogen H Silicon Si
Bromine Br Iodine I Silver Ag
Calcium Ca Iron Fe Sodium Na
Carbon C Lead Pb Sulphur S
Chlorine Cl Magnesium Mg Uranium U
Cobalt Co Neon Ne Zinc Zn
32. Molecule :-
A molecule is the smallest particle of an element or
compound which exists independently and shows all the
properties of that substance.
A molecule is a group of two or more elements that are held
together by attractive forces.
Atoms of the same element or different elements can join
together to form molecules.
33. Lets poll!!
Q1. Which of the following statements is not true about the atom ?
a) Atoms are not be exist independently
b) Atoms are the basic units from which molecules and ions are formed
c) Atoms are always neutral in nature
d) Atoms aggregate in large numbers to form the matter that we can
see, feel or touch
Q2. Which among the following is not a postulate of Dalton’s atomic theory ?
a) Atoms can not be created or destroyed
b) Atoms of different elements have different sizes ,masses and
chemical properties.
c) Atoms of same elements can combine in only one ratio to
produce more than any one compound
d) Atoms are very tiny particles which cannot be further divided
34. Q3.The symbol of Cadmium is
a) Ca b) Cu
c) Cm d) Cd
Q4. The chemical symbol for nitrogen gas is
a) Ni b) N2
c) N+ d) N
35. POLL ANSWERS
Q1. Which of the following statements is not true about the atom ?
a) Atoms are not able to exist independently
b) Atoms are the basic units from which molecules and ions are formed
c) Atoms are always neutral in nature
d) Atoms aggregate in large numbers to form the matter that we can
see, feel or touch
Q2. Which among the following is not a postulate of Dalton’s atomic theory ?
a) Atoms can not be created or destroyed
b) Atoms of different elements have different sizes ,masses and
chemical properties.
c) Atoms of same elements can combine in only one ratio to
produce more than any one compound
d) Atoms are very tiny particles which cannot be further divided
36. Q3.The symbol of Cadmium is
a) Ca b) Cu
c) Cm d) Cd
Q4. The chemical symbol for nitrogen gas is
a) Ni b) N2
c) N+ d) N
37.
38. Atomic mass :-
Mass of atom is called atomic mass.Since ,atoms are very small
consequently actual mass of an atom is very small. For example the actual
mass of one atom of Hydrogen is equal to 1.673 x10-24 g. This equal to
0.000000000000000000000001673.To deal with such small number is
difficult. Thus for convenience relative atomic mass is used.
Carbon-12 is considered as unit to calculate atomic mass.Carbon-12 is an
isotope of Carbon. The relative mass of all atoms are found with respect to
C-12.
One atomic mass=1/12th of the mass of one atom of C-12
This means atomic mass unit =1/12th of Carbon 12
Thus atomic mass is the relative atomic mass of an atom with respect to
1/12th of the mass of Carbon-12 atom. ‘amu’ is the abbreviation of Atomic
mass unit, but now it is denoted just by ‘u’.
The atomic mass of Hydrogen atom =1u
This means one hydrogen atom is 1 times heavier than 1/12th of the Carbon-
12 atom.
39. Atomic masses of some elements :-
Element Atomic mass (u) Element Atomic mass (u)
Hydrogen 1 Magnesium 24
Carbon 12 Aluminium 27
Nitrogen 14 Sulphur 32
Oxygen 16 Chlorine 35.5
Sodium 23 Calcium 40
40. Molecule- Molecule represents a group of two or more
atoms (same or different) chemically bonded to each other
and held tightly by strong attractive forces.
Ex:
Hydrogen molecule (H2) contains two atoms of Hydrogen
in one molecule.
One molecule of Carbon dioxide (CO2) contains one atom of
Carbon and two atoms of Oxygen.
Atomicity of an element :-
The number of atoms present in a molecule of an element.
41. Non metal Argon Ar 1 – Monoatomic
Non metal Helium He 1 – Monoatomic
Non metal Oxygen O2 2 – Diatomic
Non metal Hydrogen H2 2 – Diatomic
Non metal Nitrogen N2 2 – Diatomic
Non metal Chlorine Cl2 2 – Diatomic
Npn metal Phosphorus P4 4 – Phosphorus
Non metal Sulphur S8 Poly atomic
Metal Sodium Na 1 – Monoatomic
Metal Iron Fe 1 – Monoatomic
Metal Aluminium Al 1 – Monoatomic
Metal Copper Cu 1 – Monoatomic
Atomicity of some elements :-
42. ii) Molecule of compounds :-
Molecule of a compound contains atoms of two or more different types of
elements.
Molecules of some compounds :-
Compound Combining elements Number of atoms of each
elements
Water – H2O Hydrogen, Oxygen 2 - Hydrogen, 1 - Oxygen
Ammonia – NH3 Nitrogen, Hydrogen 1 - Nitrogen, 3 - Hydrogen
Carbon dioxide
CO2
Carbon, Oxygen I - Carbon, 2 - Oxygen
Hydrochloric acid
HCl
Hydrogen, Chlorine 1 - Hydrogen, 1 - Chlorine
Nitric acid
HNO3
Hydrogen, Nitrogen, Oxygen 1 - Hydrogen, 1 - Nitrogen,
3 - Oxygen
Sulphuric acid
H2SO4
Hydrogen, Sulphur, Oxygen 2 - Hydrogen, 1 - Sulphur,
4 - Oxygen
45. Lets poll!!
Q1. The atomicity of K2Cr2O7 is
a) 9 b) 11
c) 10 d) 12
Q2. The total number of atoms represented by the compound CuSO4.5H2O is:
a) 27 b) 21
c) 5 d) 8
Q3. 1 u or 1 amu means
(a) 1/12 th mass of C-12 atoms
(b) Mass of C-12 atom
(c) Mass of O-16 atom
(d) Mass of Hydrogen molecule
Q4. The relative atomic mass of Na2S2O3.5H2O is
a) 250 amu
b) 250g
c) 248 amu
d) 248 g
49. cy Symbol
1 Sodium Na+ Hydrogen H + Ammonium
NH 4
+
1 Potassium K+ Hydride H - Hydroxide
OH -
1 Silver Ag+ Chloride Cl - Nitrate
NO3
-
1 Copper (I) Cu+ Bromide Br - Hydrogen carbonate
HCO3
-
2 Magnesium Mg 2+ Iodide I -
2 Calcium Ca 2+ Oxide O 2- Carbonate
CO3
2-
2 Zinc Zn 2+ Nitride N 3- Sulphite
SO3
2-
2 Iron (II) Fe 2+ Sulphate
SO4
2-
2 Copper (II) Cu 2+
3 Aluminium Al 3+ Phosphate
PO4
3-
3 Iron (III) Fe 3+
Some common ions and their valencies :-
50. .
Monoatomic ions/radicals
A monoatomic ion is an ion consisting of exactly one
atom.
ex: Cl
-, Na+
Polyatomic ions/radicals
They are a group of atoms carrying a charge and behaving as a
single unit.
ex; Carbonate (CO3)2-, Hydroxide (OH
- ),Ammonium (NH4
+)
51. Writing chemical formulae :-
i) Write the symbols / formula of the elements or ions so that the
symbol of the metal or positive ion is on the left and symbol / formula
of the non metal or negative ion is on the right.
ii) Write the valencies of the elements or ions below the elements or
ions.
iii) Cross over the valencies of the combining ions.
iv) Polyatomic ions should be enclosed in bracket before writing the
formula.
Examples :-
i) Formula of Hydrogen chloride ii) Formula of Hydrogen sulphide
Symbol H Cl Symbol H S
Valency 1 1 Valency 1 2
Formula HCl Formula H2S
iii) Formula of Magnesium chloride iv) Formula of Carbon tetrachloride
Symbol Mg Cl Symbol C Cl
Valency 2 1 Valency 4 1
Formula MgCl2 Formula CCl4
52. v) Formula of Calcium oxide vi) Formula of Aluminium oxide
Symbol Ca O Symbol Al O
Valency 2 2 Valency 3 2
Formula Ca2O2 = CaO Formula Al2O3
vii) Formula of Sodium nitrate viii) Formula of Calcium hydroxide
Symbol/ Na NO3 Symbol/ Ca OH
Formula Formula
Valency 1 1 Valency 2 1
Formula NaNO3 Ca(OH)2
ix) Formula of Sodium carbonate x) Formula of Ammonium sulphate
Symbol/ Na CO3 Symbol/ NH4 SO4
Formula Formula
Valency 1 2 Valency 1 2
Formula Na2CO3 (NH4)2SO4
53. Molecular mass
The molecular mass of a substance is the sum of the atomic masses of all
the atoms in a molecule of the substance.
Molecular mass is expressed in atomic mass unit (amu) or Unified mass
(u).
Gram Molecular Mass
The molecular mass expressed in grams which is numerically equal to the
molecular mass in u.
Eg:- Molecular mass of water – H2O
Atomic mass of H = I u
Atomic mass of O = 16 u
Molecular mass of H2O = 1x2+10 = 2+16 = 18 u
Molecular mass of Nitric acid – HNO3
Atomic mass of H = 1 u
Atomic mass of N = 14 u
Atomic mass of O = 16 u
Molecular mass of HNO3 = 1+14+16x3 = 1+14+48 = 63 u
54. Formula Unit Mass
The formula unit mass of a substance is a sum of the atomic masses of all
the atoms present in the formula of a respective compound. Formula unit
mass is calculated in the same manner as we calculate the molecular mass.
The only difference is that we use the word formula unit for those substances
whose constituent particles are ions. In such cases ,the aggregate of ions is
known as a formula unit. For example, one formula unit of Sodium chloride is
constituted by one Na+ ion and one Cl- ion. In the ionic compounds, the
molecular mass is known as gram formula unit mass.
Q. Calculate the formula unit mass of CaCl2.
Atomic mass of Ca + (2 x atomic mass of Cl)
=40 +2 x 35.5=40+71=111u
55. Q1.An element X is divalent and another element Y is tetravalent. The
compound formed by these two elements will be:
(a) XY
(b) XY2
(c) X2Y
(d) XY4
Q2.The molecular formula of potassium nitrate is ________.
a) KNO3
b) KNO
c) KNO2
d) KON
Lets poll!!
56. Q3.Formula of a Chloride of a metal M is MCl3, the formula of the Phosphate
of metal M will be:
a.MPO4
b.M2PO4
c.M3PO4
d.M2(PO4)3
(a) (NH4)3PO4
(b) (NH4)3PO3
(c) (NH4)3P
(d) (NH4)PO4
Q5.The relative atomic mass of Na2SO3.5H2O is
a) 250 amu
b) 250g
c) 248 amu
d) 248 g
Q4. The formula of Ammonium phosphate is
57. Q1.An element X is divalent and another element Y is tetravalent. The
compound formed by these two elements will be:
(a) XY
(b) XY2
(c) X2Y
(d) XY4
Q2.The molecular formula of Potassium nitrate is ________.
a) KNO3
b) KNO
c) KNO2
d) KON
Q3.Formula of a Chloride of a metal M is MCl3, the formula of the Phosphate of
metal M will be:
a.MPO4
b.M2PO4
c.M3PO4
d.M2(PO4)3
POLL ANSWERS
58. (a) (NH4)3PO4
(b) (NH4)3PO3
(c) (NH4)3P
(d) (NH4)PO4
Q4. The formula of Ammonium phosphate is
Q5.The relative atomic mass of Na2SO3.5H2O is
a) 250 amu
b) 250g
c) 248 amu
d) 248 g
63. MOLE CONCEPT
• We use the term ‘mole’ to indicate the
number of atoms or molecules.
• A mole means 6.023 x 1023 particles.
• This number is also called the Avagadro
number. (N0 )
64.
65.
66.
67. Mole concept :-
A mole of a substance is that amount of the substance which contains the
same number of particles (atoms, molecules or ions) that are present in 12g of
Carbon – 12.
The number of particles (atoms) present in 12g of Carbon – 12 is
6.022 x 1023 . This number is called Avagadro Number or Avagadro Constant.
A mole represents two things :-
i) It represents a definite number of particles (atoms, molecules or ions) equal to
6.022 x 1023 .
ii) It represents a definite mass of a substance equal to the gram atomic mass of
an element or the gram molecular mass of a compound.
Gram atomic mass of an element :- is its atomic mass expressed in grams.
Eg :- Gram atomic mass of Oxygen =16 g
Gram molecular mass of a compound :- is its molecular mass expressed in
grams.
Eg :- Gram molecular mass of water = H2O = 1 x 2 + 16 = 2 + 16 = 18g.
Relationship between number of moles (n), mass (m), molar mass (M), Number
of atoms or molecules (N), and Avagadro number (NO).
m N m
n = ---- , n = ---- , m = n x N , M = ---
M NO n
68. Calculation of the Number of Moles
from mass
i) 52 g of He (Finding mole from mass)
No of moles=n
Given mass = m
Avogadro number of particles = N0
Atomic mass of He =4 u
Molar mass of He =4g
Thus, the number of moles=given mass/molar mass
n= m/M =52/4 =13 mol
69. Calculation of Mass
i) 0.5 mole of N2 gas (mass from mole of molecule)
mass = molar mass x number of moles
m = M x n
= 28 x 0.5 = 14 g
ii) 0.5 mole of N atoms (mass from mole of atom)
mass = molar mass x number of moles
14 x 0.5 =7 g
iii) 3.011x1023 number of N atoms (mass from number)
The number of moles, n=given number of particles /Avogadro number
n = N/N0
mass = molar mass x number of moles
14 x 3.011 x1023 x 6.022 x10 23 =7 g
70.
71. Calculation of the number of
Particles
i) 46 g of Na atoms (number of atoms from mass )
The number of Atoms
=given mass /molar mass * Avagadro number
N =m/M * N0
N= 46/23 * 6.022 * 10 23
12.044 * 1023
72.
73. Calculation of the number of
particles from mass
ii) 8g O2 molecules ( number of molecules from mass)
The number of molecules
=given mass /molar mass * Avagadro number
N = m/M * N0
N = 8/32 *6.022 * 10 23
N = 1.5 1 * 10 23
76. Q1. Which of the following represents the correct relation between Avogadro's number
(No), number of particles (N) and moles (n)?
(a) n = N / No
(b) n = No / N
(c) n = N No
(d) all are correct
Q2.The atomic mass of sodium is 23. The number of moles in 46g of sodium is
________.
(a) 4
(b) 2
(c) 0
(d) ½
Lets poll!!
77. Q3. Which of the following correctly represents 360g of water?
i)2 moles of water
ii) 20 moles of water
iii) 6.022 × 1023 molecules of water
iv) 1.2044 × 1025 molecules of water
a i) b) i) and iv)
c) ii) and iii) d) ii) and iv)
Q4. Which of the following has maximum number of atom?
a) 18 g of H2O b) 18g of O2
c) 18 g of CO2 d)18g of CH4
Q5. Calculate the weight of 2.5 moles of CaCO3
a) 200g b) 230 g
c) 240 g d) 250 g
78. Q1. Which of the following represents the correct relation between Avogadro's number
(No), number of particles (N) and moles (n)?
(a) n = N / No
(b) n = No / N
(c) n = N No
(d) all are correct
Answer. (a) n = N / No
Q2. The atomic mass of sodium is 23. The number of moles in 46g of sodium is
________.
(a) 4
(b) 2
(c) 0
(d) ½
Answer. (b) 2
Poll Answers
79. Q3. Which of the following correctly represents 360g of water?
i)2 moles of water
ii) 20 moles of water
iii) 6.022 × 1023 molecules of water
iv) 1.2044 × 1025 molecules of water
a i) b) i) and iv)
c) ii) and iii) d) ii) and iv)
Q4. Which of the following has maximum number of atom?
a) 18 g of H2O
b) 18g of O2
c) 18 g of CO2
d)18g of CH4
Q5. Calculate the weight of 2.5 moles of CaCO3
a) 200g b) 230 g
c) 240 g d) 250 g
80. MOLE CONCEPT
CALCULATIONS
Q1. Which has more number of atoms 100g of Magnesium
or 100 g of Carbon.
Q2. Calculate the mass of nitrogen(N
2
) which contains the same
number of molecules as are present in 4.4 grams of Carbon
dioxide( CO
2
).
Q3. What weight of Calcium contains the same number of atoms
as are present in 3.2g of Sulphur?
(atomic mass of Ca=40u , S=32u)
81. ANSWERS
Q1. no of mole of Mg = 100g/24g= 4.16 mol
No of mole of Carbon = 100g/12g= 8.33 mol
So 100g of Carbon has more number of atoms.
Q2.no. of molecules present in 44grams of Carbon dioxide=6.022x1023
So, molecules present in 4.4 grams Carbon dioxide
=6.022x1023x4.4/44 =6.022x1022
Now, mass of 6.022x1023 molecules of N2 =28g
So, mass of 6.022x1022 molecules of N2=28X6.022x1022 / 6.022x1023 =2.8g.
.
82. Q3. Gram atomic mass of Sulphur = 32 gm
3.2 gm of Sulphur contains = 6.022 X 1023 x 3.2 / 32 = 6.022 x 1022
Gram atomic mass of Ca = 40 g
6.022 x 1023 atoms of Ca weigh = 40g
6.022 x 1022 atoms of Ca weigh = 40 x 6.022 x 1022/6.022 x 1023= 4 g
.
83.
84.
85.
86.
87.
88.
89. Video LInks
• https://www.youtube.com/watch?v=mcnga-bbNXk - law of conservation of
masshttps://www.youtube.com/watch?v=08-96_wkUi8 - Law of definite
proportions
• https://www.youtube.com/watch?v=D9tzdUlfDzg - Dalton’s atomic theory
and demerits
• https://www.youtube.com/watch?v=B-qsoJzWSF0 – formula writing of a
compound
• https://www.youtube.com/watch?v=4q2elWPfB6A –mole concept
