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7 inch
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2.5 inch
MOLE CONCEPT
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7 inch
2.5 inch
2.5 inch
2.5 inch
MOLE CONCEPT
I N D E X
Topic Page No.
PHYSICAL CHEMISTRY
MOLE CONCEPT
01. Mole 01
02. Methods to calculate mole 03
03 Laws of chemical combination 06
04. Significanceofchemical equation 09
05. Limitingreagent 10
06. Problems related with mixture 11
07. PercentageYield 11
08. Percentage purity 12
09. Problems related with sequential reaction 13
10. Problems related parallel reaction 14
11. Principle of atom conservation 15
12. Average atomic Mass 16
13. Average molecular mass 16
14. Empiricalandmolecular formula 17
15. Experimental methodsfor determination 19
MOLE CONCEPT
Page 3


7 inch
2.5 inch
2.5 inch
2.5 inch
MOLE CONCEPT
I N D E X
Topic Page No.
PHYSICAL CHEMISTRY
MOLE CONCEPT
01. Mole 01
02. Methods to calculate mole 03
03 Laws of chemical combination 06
04. Significanceofchemical equation 09
05. Limitingreagent 10
06. Problems related with mixture 11
07. PercentageYield 11
08. Percentage purity 12
09. Problems related with sequential reaction 13
10. Problems related parallel reaction 14
11. Principle of atom conservation 15
12. Average atomic Mass 16
13. Average molecular mass 16
14. Empiricalandmolecular formula 17
15. Experimental methodsfor determination 19
MOLE CONCEPT
Topic Page No.
16. Eudiometry 23
17. V olume expansion and contraction in the edudiometer tube 24
18. General reactions for combustion of organic compounds 24
19. Analysis ofgaseous mixture 25
20. V olume -V olume analysis 26
21. Determination of molecular formula of gaseous hydrocarbon 26
22. Determination of molecular formula of gases 27
23. Concentration Terms 28
24. Methods of expressing concentration of solution 28
25. Molarityof ionic compounds 35
26. Mixingordilutionofsolution 36
27. Mixing of acid and base solutions 36
28. Problems involving precipitation 36
29. Some typical concentration terms 37
30. Relationship between different concentration terms 38
31. Solved examples 40
32. Exercise - 1 53
Exercise - 2 59
Exercise - 3 65
Exercise - 4 72
33. Answer Key 74
34. Hints/Solution 76
Page 4


7 inch
2.5 inch
2.5 inch
2.5 inch
MOLE CONCEPT
I N D E X
Topic Page No.
PHYSICAL CHEMISTRY
MOLE CONCEPT
01. Mole 01
02. Methods to calculate mole 03
03 Laws of chemical combination 06
04. Significanceofchemical equation 09
05. Limitingreagent 10
06. Problems related with mixture 11
07. PercentageYield 11
08. Percentage purity 12
09. Problems related with sequential reaction 13
10. Problems related parallel reaction 14
11. Principle of atom conservation 15
12. Average atomic Mass 16
13. Average molecular mass 16
14. Empiricalandmolecular formula 17
15. Experimental methodsfor determination 19
MOLE CONCEPT
Topic Page No.
16. Eudiometry 23
17. V olume expansion and contraction in the edudiometer tube 24
18. General reactions for combustion of organic compounds 24
19. Analysis ofgaseous mixture 25
20. V olume -V olume analysis 26
21. Determination of molecular formula of gaseous hydrocarbon 26
22. Determination of molecular formula of gases 27
23. Concentration Terms 28
24. Methods of expressing concentration of solution 28
25. Molarityof ionic compounds 35
26. Mixingordilutionofsolution 36
27. Mixing of acid and base solutions 36
28. Problems involving precipitation 36
29. Some typical concentration terms 37
30. Relationship between different concentration terms 38
31. Solved examples 40
32. Exercise - 1 53
Exercise - 2 59
Exercise - 3 65
Exercise - 4 72
33. Answer Key 74
34. Hints/Solution 76
BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P .I.A., Kota-05
ACC- CH-MOLE CONCEPT 1
MOLE CONCEPT
1. MOLE
A mole is the amount of substance that contains as many species [Atoms, molecules, ions or other
particles] as there are atoms in exactly 12 gm of C-12.
species 1
23
10 6.022 mole ? ?
2.1 Atomic mass
Atomic mass of an element can be defined as the number which indicates how manytimes the mass of
one atom of the element is heavier in comparison to
12
1
th part of the mass of one atom of Carbon-12.
Atomic mass =
] 12 - carbon of atom an of Mass [
12
1
] element the of atom an of Mass [
?
=
amu 1
amu in atom an of Mass
2.2Atomic mass unit (amu) or Unified mass (u)
The quantity [
12
1
× mass of an atom of C–12] is known as atomic mass unit.
The actual mass of one atom of C-12 = 1.9924 × 10
–26
kg
? 1 amu = kg
12
10 9924 . 1
26 ?
?
= 1.66 × 10
–27
kg = 1.66 × 10
–24
gm =
A
N
1
gm
2.3 Gram atomic mass
The gram atomic mass can be defined as the mass of 1 mole atoms of an element.
e.g., Mass of one oxygen atom = 16 amu =
A
N
16
gm.
Mass of N
A
oxygen atom = A
A
N .
N
16
= 16 gram
Page 5


7 inch
2.5 inch
2.5 inch
2.5 inch
MOLE CONCEPT
I N D E X
Topic Page No.
PHYSICAL CHEMISTRY
MOLE CONCEPT
01. Mole 01
02. Methods to calculate mole 03
03 Laws of chemical combination 06
04. Significanceofchemical equation 09
05. Limitingreagent 10
06. Problems related with mixture 11
07. PercentageYield 11
08. Percentage purity 12
09. Problems related with sequential reaction 13
10. Problems related parallel reaction 14
11. Principle of atom conservation 15
12. Average atomic Mass 16
13. Average molecular mass 16
14. Empiricalandmolecular formula 17
15. Experimental methodsfor determination 19
MOLE CONCEPT
Topic Page No.
16. Eudiometry 23
17. V olume expansion and contraction in the edudiometer tube 24
18. General reactions for combustion of organic compounds 24
19. Analysis ofgaseous mixture 25
20. V olume -V olume analysis 26
21. Determination of molecular formula of gaseous hydrocarbon 26
22. Determination of molecular formula of gases 27
23. Concentration Terms 28
24. Methods of expressing concentration of solution 28
25. Molarityof ionic compounds 35
26. Mixingordilutionofsolution 36
27. Mixing of acid and base solutions 36
28. Problems involving precipitation 36
29. Some typical concentration terms 37
30. Relationship between different concentration terms 38
31. Solved examples 40
32. Exercise - 1 53
Exercise - 2 59
Exercise - 3 65
Exercise - 4 72
33. Answer Key 74
34. Hints/Solution 76
BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P .I.A., Kota-05
ACC- CH-MOLE CONCEPT 1
MOLE CONCEPT
1. MOLE
A mole is the amount of substance that contains as many species [Atoms, molecules, ions or other
particles] as there are atoms in exactly 12 gm of C-12.
species 1
23
10 6.022 mole ? ?
2.1 Atomic mass
Atomic mass of an element can be defined as the number which indicates how manytimes the mass of
one atom of the element is heavier in comparison to
12
1
th part of the mass of one atom of Carbon-12.
Atomic mass =
] 12 - carbon of atom an of Mass [
12
1
] element the of atom an of Mass [
?
=
amu 1
amu in atom an of Mass
2.2Atomic mass unit (amu) or Unified mass (u)
The quantity [
12
1
× mass of an atom of C–12] is known as atomic mass unit.
The actual mass of one atom of C-12 = 1.9924 × 10
–26
kg
? 1 amu = kg
12
10 9924 . 1
26 ?
?
= 1.66 × 10
–27
kg = 1.66 × 10
–24
gm =
A
N
1
gm
2.3 Gram atomic mass
The gram atomic mass can be defined as the mass of 1 mole atoms of an element.
e.g., Mass of one oxygen atom = 16 amu =
A
N
16
gm.
Mass of N
A
oxygen atom = A
A
N .
N
16
= 16 gram
2 ACC- CH-MOLE CONCEPT
BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P .I.A., Kota-05
Illustration
(a) What is the mass of one atom of Cl? (b) What is the atomic mass of Cl?
(c) What is the gram atomic mass of Cl?
Sol. (a) Mass of one atom of Cl = 35.5 amu.
(b) Atomic mass of Cl =
1amu
amu in atom an of Mass
=
1amu
35.5amu
= 35.5
(c) Gram atomic mass of Cl = [Mass of 1 Cl atom × N
A
]
= 35.5 amu × N
A
=
A
N
35.5
× N
A
gram = 35.5 gram
Exercise
(a) What is the mass of one atom of S?
(b) What is the atomic mass of S ?
(c) What is the gram atomic mass of S?
Ans. (a) 32 amu, (b) 32, (c) 32 gram
3.1 Molecular mass
Molecular mass is the number which indicates how manytimes one molecule of a substance is heavier in
comparison to
th
12
1
of the mass of one atom of C-12.
Molecular mass =
] 12 - C of atom an of Mass [
12
1
amu) (in substance the of molecule one of Mass
?
=
amu 1
amu) in ( substance the of molecule one of Mass
3.2 Gram Molecular mass
Gram molecular mass can be defined as the mass of 1 mole of molecules.
e.g., Mass of one molecule of O
2
= 32 amu =
gram
N
32
A
.
Mass of N
A
molecules of O
2
=
gm N
N
32
A
A
?
= 32 gm
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FAQs on Mole Concept, Class 11, Chemistry - NEET

1. What is the mole concept in chemistry?
Ans. The mole concept is a fundamental concept in chemistry that relates the mass of a substance to the number of particles it contains. It is based on the Avogadro's law, which states that one mole of any substance contains 6.022 x 10^23 particles, also known as Avogadro's number.
2. How is the mole concept used in chemical calculations?
Ans. The mole concept is used in chemical calculations to determine the amount of a substance in a given sample. It allows us to convert between the mass, number of particles, and volume of a substance using the molar mass, Avogadro's number, and molar volume, respectively. These conversions are essential in stoichiometry, which involves balancing chemical equations and determining the quantities of reactants and products.
3. How do you calculate the molar mass of a compound?
Ans. The molar mass of a compound is calculated by summing up the atomic masses of all the atoms in the compound. The atomic masses can be found on the periodic table. For example, to calculate the molar mass of water (H2O), we would multiply the atomic mass of hydrogen (1.008 g/mol) by 2 and add it to the atomic mass of oxygen (16.00 g/mol) to get a molar mass of 18.02 g/mol.
4. What is the relationship between moles and grams?
Ans. The relationship between moles and grams is determined by the molar mass of a substance. The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). To convert moles to grams, you can multiply the number of moles by the molar mass. Conversely, to convert grams to moles, you divide the mass in grams by the molar mass.
5. How does the mole concept apply to gas calculations?
Ans. The mole concept is crucial in gas calculations as it allows us to relate the volume, pressure, and temperature of a gas to the number of moles. The ideal gas law, PV = nRT, combines these variables, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. By manipulating this equation, we can solve for any of the variables when the others are known.
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