JEE Exam  >  JEE Notes  >  Additional Study Material for JEE  >  Mole Concept and Basic Concepts

Mole Concept and Basic Concepts | Additional Study Material for JEE PDF Download

MOLE CONCEPT 
Definition of mole: One mole is a collection of those many entities as there are number of atoms exactly in 12 gm of C - 12 isotope.

Mole Concept and Basic Concepts | Additional Study Material for JEE

Fig. C-12 isotope

Or 1 mole = collection of 6.02 × 1023 species
6.02 × 1023 = NA = Avogadro's No.
1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion and 1 mole of molecule termed as 1 gm-molecule.

Methods of Calculations of mole: 
(a) If no. of some species is given, then no. of moles
Mole Concept and Basic Concepts | Additional Study Material for JEE
(b) If weight of a given species is given, then no of moles

Mole Concept and Basic Concepts | Additional Study Material for JEE
or
 Mole Concept and Basic Concepts | Additional Study Material for JEE

(c) If volume of a gas is given along with its temperature (T) and pressure (P)

use
Mole Concept and Basic Concepts | Additional Study Material for JEE

where R = 0.0821 liter-atm/mol-K (when P is in atmosphere and V is in litre.)
1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litre.
1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litre.
Atom:  Atom is smallest particle which can not be divided into its constituents.

Mole Concept and Basic Concepts | Additional Study Material for JEE

Fig. Structure of atom


Atomic weight: It is the weight of an atom relative to one twelfth of weight of 1 atom of C-12 isotope.

Relationship between gram and amu: 
⇒ 1 amu = 1/12 wt of one C - 12 atom
⇒ for C, 1 mole C = 12 gm = 6.023 × 1023 atoms
⇒ wt. of 6.023 × 1023 atoms = 12 gm

Mole Concept and Basic Concepts | Additional Study Material for JEE
(NA → Avogadro's number = 6.23 × 1023)
⇒ 1 amu = 1/12 wt of one C - 12 atom

Mole Concept and Basic Concepts | Additional Study Material for JEE

⇒ 1 amu = 1/NA g
So, 1 amu = 1.66 x 10-24
= 1.66 x 10-27 kg .

Elemental Analysis:

For 'n' mole of a compound (C3H7O2)

Moles of C = 3n

Moles of H = 7n

Moles of O = 2n

Example. Find the weight of water present in 1.61 g of Na2SO4. 10H2O
Solution. Moles of Na2SO4.10H2O =  0.005 moles
Moles of water = 10 × moles of Na2SO4. 10H2O = 10 × 0.05 = 0.05
wt. of water = 0.5 × 18 = 0.9 gm 

Average atomic weight:
= % of isotope x molar mass of isotope.
The % obtained by above expression (used in above expression) is by number (i.e. its a mole%).

Molecular weight:
It is the sum of the atomic weight of all the constituent atoms.
(a) Average molecular weight= where ni = no. of moles of any compound and
mi = molecular mass of any compound.
Shortcut for % determination if average atomic weight is given for X having isotopes XA & XB.
Try working out of such a shortcut for XA, XB, X

EMPIRICAL FORMULA and MOLECULAR FORMULA 
Empirical formula: Formula depicting constituent atom in their simplest ratio.
Molecular formula: Formula depicting an actual number of atoms in one molecule of the compound.
Relation between the two: Molecular formula= Empirical formula × n
Check out the importance of each step involved in calculations of empirical formula.

Mole Concept and Basic Concepts | Additional Study Material for JEE

Fig. Mole Concept


Example. A molecule of a compound has 13 carbon atoms, 12 hydrogen atom, 3 oxygen atoms and 3.02 × 10-23 gm of other element. Find the molecular wt. of compound. 
Solution. Wt. of the 1 molecule of a compound = 13 × 12 × 3 × 3.02 × 10-23
= 234.18 / NA = 234 amu.

Density: 
(a) Absolute density
(b) Relative density

Mole Concept and Basic Concepts | Additional Study Material for JEE

Vapor density: It is defined only for gas.
It is a density of gas with respect to H2 gas at same temp & pressure.

Mole Concept and Basic Concepts | Additional Study Material for JEE

Mole Concept and Basic Concepts | Additional Study Material for JEE

Density of Cl2 gas with respect to O2 gas

Mole Concept and Basic Concepts | Additional Study Material for JEE

The document Mole Concept and Basic Concepts | Additional Study Material for JEE is a part of the JEE Course Additional Study Material for JEE.
All you need of JEE at this link: JEE
22 videos|162 docs|17 tests

Top Courses for JEE

FAQs on Mole Concept and Basic Concepts - Additional Study Material for JEE

1. What is the mole concept in JEE?
Ans. The mole is a fundamental unit in chemistry that is used to measure the amount of a substance. The mole concept in JEE involves the calculation of the number of moles of a chemical substance. One mole of a substance is defined as the amount of that substance that contains the same number of entities as there are atoms in 12 grams of pure carbon-12. The number of entities in a mole is known as Avogadro's number, which is equal to 6.022 x 10^23.
2. What are the basic concepts of the mole in JEE?
Ans. The basic concepts of the mole in JEE include the molar mass, the molarity, and the molality. Molar mass is the mass of one mole of a substance and is expressed in grams per mole. Molarity is the number of moles of solute per liter of solution, while molality is the number of moles of solute per kilogram of solvent.
3. How do you calculate the number of moles in JEE?
Ans. The number of moles of a substance can be calculated by dividing the given mass of the substance by its molar mass. The formula for calculating the number of moles is: Number of moles = Given mass of substance / Molar mass For example, if the given mass of a substance is 20 grams and its molar mass is 40 g/mol, then the number of moles would be: Number of moles = 20 g / 40 g/mol = 0.5 mol
4. What is the relationship between the mole and the gas laws in JEE?
Ans. The mole concept is closely related to the gas laws in JEE. According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain the same number of molecules. This means that the number of moles of gas is directly proportional to its volume. The ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas, is also based on the mole concept.
5. How is the mole concept used in stoichiometry problems in JEE?
Ans. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. The mole concept is used extensively in stoichiometry problems in JEE. By knowing the balanced chemical equation for a reaction and the number of moles of one reactant or product, it is possible to calculate the number of moles of the other reactants or products involved in the reaction. This allows for the determination of the limiting reactant, the theoretical yield, and the percent yield of a reaction.
22 videos|162 docs|17 tests
Download as PDF
Explore Courses for JEE exam

Top Courses for JEE

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Previous Year Questions with Solutions

,

Objective type Questions

,

ppt

,

pdf

,

Viva Questions

,

MCQs

,

Mole Concept and Basic Concepts | Additional Study Material for JEE

,

Semester Notes

,

Extra Questions

,

Mole Concept and Basic Concepts | Additional Study Material for JEE

,

Mole Concept and Basic Concepts | Additional Study Material for JEE

,

study material

,

practice quizzes

,

Exam

,

shortcuts and tricks

,

video lectures

,

Free

,

Sample Paper

,

Important questions

,

past year papers

,

Summary

,

mock tests for examination

;