1st Year Chemistry Notes Free Download Easy Notes Education Helper junaidansaari.blogspot.com

1st Year Chemistry Notes

11th class chemistry Notes Free Download pdf 

Here You Can View 11th class chemistry Notes If you Like it You Can Download It In Chapter 2 Download bottom is available in chapter 2

CHAPTER – 1

Atom

It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.

Molecule

When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.

Formula Weight

It is the sum of the weights of the atoms present in the formula of a substance.

Molecular Weight

It is the sum of the atomic masses of all the atoms present in a molecule.

Chemistry

It is a branch of science which deals with the properties, composition and the structure of matter.

Empirical Formula

Definition
It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.

Examples
The empirical formula of benzene is “CH”. It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.
The empirical formula of glucose is “CH2O”. This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.

Determination of Empirical Formula

To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.

Illustrated Example of Empirical Formula

Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.

Step I – Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.

Step II – Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm

Step III – Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%

Step IV – Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5

Step V – Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO

Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.

Molecular Formula

Definition

The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.

OR

It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.

Examples
The molecular formula of benzene is “C6H6″. It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.
The molecular formula of glucose is “C6H12O6″. The formula represents that
1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.

Determination of Molecular Formula

The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula)n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
n = Molecular Mass / Empirical Formula Mass

Example
The empirical formula of a compound is CH2O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH2O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH2O)6
= C6H12O6

Molecular Mass

Definition

The sum of masses of the atoms present in a molecule is called as molecular mass.

OR

It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.

Example
The molecular mass of CO2 may be calculated as
Molecular mass of CO2 = Mass of Carbon + 2 (Mass of Oxygen)
= 12 + 2 x 16
= 44 a.m.u
Molecular mass of H2O = (Mass of Hydrogen) x 2 + Mass of Oxygen
= 1 x 2 + 16
= 18 a.m.u
Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
= 1 + 35.5
= 36.5 a.m.u

Gram Molecular Mass

Definition
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.

Examples
1. The molecular mass of H2O is 18. If we take 18 gm H2O then it is called 1 gm molecular mass of H2O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.

Mole

Definition

It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.

OR

The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.

Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.
From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO2 is 44. If we take 44 gm of CO2 it is called one mole of CO2 or the molecular mass of H2O is 18. If we take 18 gm of H2O it is called one mole of H2O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contain 6.02 x 10(23) atoms
While
One gram molecule contain 6.02 x 10(23) molecules.

Avagadro’s Number

An Italian scientist, Avagadro’s calculated that the number of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10(23). This number is known as Avogadro’s number and represented as N(A).

Example
1 gm mole of Na contain 6.02 x 10(23) atoms of Na.
1 gm mole of Sulphur = 6.02 x 10(23) atoms of Sulphur.
1 gm mole of H2SO4 = 6.02 x 10(23) molecules H2SO4
1 gm mole of H2O = 6.02 x 10(23) molecules of H2O
On the basis of Avogadro’s Number “mole” is also defined as
Mass of 6.02 x 10(23) molecules, atoms or ions in gram is called mole.

Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance

Example 1
Calculate the number of atoms in 9.2 gm of Na.

Solution
Atomic mass of Na = 23 a.m.u
If we take 23 gm of Na, it is equal to 1 mole.
23 gm of Na contain 6.02 x 10(23) atoms
1 gm of Na contain 6.02 x 10(23) / 23 atoms
9.2 gm of Na contain 9.2 x 6.02 x 10(23) /23
= 2.408 x 10(23) atoms of Na

Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance

Example 2
Calculate the mass in grams of 3.01 x 10(23) molecules of glucose.

Solution
Molecular mass of glucose = 180 a.m.u
So when we take 180 gm of glucose it is equal to one mole So,
6.02 x 10(23) molecules of glucose = 180 gm
1 molecule of glucose = 180 / 6.02 x 10(23) gm
3.01 x 10(23) molecules of glucose = 3.01 x 10(23) x 180 / 6.02 x 10(23)
= 90 gm

Stoichiometry

(Calculation Based On Chemical Equations)

Definition
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called stoichiometry.

In this study we always use a balanced chemical equation because a balanced chemical equation tells us the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass – Mass Relationship
2. Mass – Volume Relationship
3. Volume – Volume Relationship

1. Mass – Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.

Example
Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.

Solution
Step I – Write a Balanced Equation
CaCO3 —-> CaO + CO2

Step II – Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO3 —-> CaO + CO2

Method I – MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO3 gives 1 mole of CO2
0.5 mole of CaCO3 will give 0.5 mole of CO2
Mass of CO2 = Moles x Molecular Mass
= 0.5 x 44
= 22 gm

Method II – FACTOR METHOD
From equation we may write as
100 gm of CaCO3 gives 44 gm of CO2
1 gm of CaCO3 will give 44/100 gm of CO2
50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2
= 22 gm of CO2

2. Mass – Volume Relationship

The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm3 volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.

Example
Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.

Solution
Step I – Write a Balanced Equation
CH4 + 2 O2 —-> CO2 + 2 H2O

Step II – Write Down The Molecular Masses And Moles Of Reactant & Product
CH4 + 2 O2 —-> CO2 + 2 H2O

Method I – MOLE METHOD
Convert the given mass of CH4 in moles
Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4
From Equation
1 mole of CH4 gives 1 moles of CO2
1.25 mole of CH4 will give 1.25 mole of CO2
No. of moles of CO2 obtained = 1.25
But 1 mole of CO2 at S.T.P occupies 22.4 dm3
1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25
= 28 dm3

Method II – FACTOR METHOD
Molecular mass of CH4 = 16
Molecular mass of CO2 = 44
According to the equation
16 gm of CH4 gives 44 gm of CO2
1 gm of CH4 will give 44/16 gm of CO2
20 gm of CH4 will give 20 x 44/16 gm of CO2
= 55 gm of CO2
44 gm of CO2 at S.T.P occupy a volume 22.4 dm3
1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3
55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44
= 28 dm3

3. Volume – Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac’s law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH4 + 2 O2 —-> CO2 + 2 H2O
In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to give one volume of CO2 and two volumes of H2O

Examples
What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4 (ethylene)?

Solution
Step I – Write a Balanced Equation
C2H4 + 3 O2 —-> 2 CO2 + 2 H2O

Step II – Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 —-> 2 CO2 + 2 H2O

According to Equation
1 dm3 of C2H4 requires 3 dm3 of O2
500 dm3 of C2H4 requires 3 x 500 dm3 of O2
= 1500 dm3 of O2

Limiting Reactant

In stoichiometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stoichiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.

Problem
We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N2 + 3 H2 —-> 2 NH3

Solution
In this problem moles of N2 and H2 are as follows
Moles of N2 = Mass of N2 / Mol. Mass of N2
= 50 / 28
= 1.79
Moles of H2 = Mass of H2 / Mol. Mass of H2
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N2 + 3 H2 —-> 2 NH3

1 mole of N2 gives 2 moles of NH3
1.79 mole of N2 gives 2 x 1.79 moles of NH3
= 3.58 moles of NH3

Mass of NH3 = Moles of NH3 x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH3

Click here for download full book notes of 1st Year Chemistry Notes

Three States Of Matter

CHAPTER – 2

Three States Of Matter

Matter

It is defined as any thing which has mass and occupies space is called matter.

Matter is composed of small and tiny particles called Atoms or molecules. It exist in three different states which are gaseous, liquid & solid.

Properties of Gas

1. It has no definite shape.
2. It has no definite volume, so it can be compressed or expanded.
3. A gas may diffuse with the other gas.
4. The molecules of a gas are in continuous motion.

Properties of Liquids

1. A liquid has no definite shape.
2. It has a fixed volume.
3. The diffusion of a liquid into the other liquid is possible if both of the liquids are polar or non-polar.
4. It can be compressed to a negligible.

Properties of Solids

1. A solid has a definite shape.
2. It has a fixed volume.
3. The rate of diffusion of solid with each other is very slow.
4. It cannot be compressed easily.

Kinetic Theory of Gases

It was an idea of some scientist like Maxwell & Bolzmann that the properties of gases are due to their molecular motion. This motion of the molecules is related with the kinetic energy, so the postulates give by the scientist about the behaviour of gases are collectively known as kinetic molecular theory of gases.
The postulates of kinetic molecular theory are as follows.
1. All gases consists of very large number of tiny particles called molecules.
2. These molecules are widely separated from each other and are so small that they are invisible.
3. The size of the molecules is very small as compared to the distance between them.
4. There is no attractive or repulsive force between molecules so they can move freely.
5. The molecules are very hard and perfectly elastic so when they collide no loss of energy takes place.
6. The gas molecules are in continuous motion they move in a straight path until they collide. The distance between two continuous collision is called Mean Free Path.
7. During their motion these molecules are collided with one another and with the walls of the container.
8. The collision of the molecules are perfectly elastic. When molecules collide they rebound with perfect elasticity and without loss or gain of energy.
9. The pressure of the gas is the result of collision of molecules on the walls of the container.
10. The average kinetic energy of gas molecules depends upon the absolute temperature. At any given temperature the molecules of all gases have the same average kinetic energy (1/2 mv2).

Kinetic Theory of Liquids

This theory is bases on the following assumptions.
1. The particles of a liquid are very close to each other due to which a liquid has fixed volume.
2. The particles in a liquid are free to move so they have no definite shape.
3. During the motion these molecules collides with each other and with the walls of the container.
4. These molecules possess kinetic which is directly proportional to its absolute temperature.

Kinetic Theory of Solids

The assumptions of kinetic theory for solids are as follows.
1. The particles in a solid are very closely packed due to strong attractive forces between the molecules.
2. These molecules are present at a fixed position and are unable to move.
3. They have definite shape because the particles are arranged in a fixed pattern.
4. They possess only vibrational energy.

Mean Free Path

The distance which a molecule of a gas travels before its collision with the other molecule is called free path. This distance between the collision of the molecules changes constantly so the average distance which a molecule travels before its collision is called mean free path.

Boyle’s Law

A relationship of volume with external pressure was given by Boyle’s in the form of law. This law is known as Boyle’s Law which states,
For a given mass of a gas the volume of the gas is inversely proportional to its pressure provided the temperature is kept constant.
Mathematically it may be written as
V ∞ 1 / P
Or V = K / P
Or PV = K
On the bases of the relation, Boyle’s law can also be stated as
The product of the pressure and volume of a given mass of a gas is always constant at constant temperature.

Explanation
Consider for a given mass a gas having volume V1 at pressure P1, so according to Boyle’s Law we may write as
P1V1 = K1 (constant)
If the pressure of the above system is changed from P1 to P2 then the volume of the gas will also change from V1 to V2. For this new condition of the gas we can write as,
P2V2 = K2 (constant)
But for the same mass of the gas.
K1 = K2
P1V1 = P2V2
This equation is known as Boyle’s Equation.

Charle’s Law

We know that everything expand on heating and contract cooling. This change in volume is small in liquids and solids but gases exhibit enormous changes due to the presence of large intermolecular spaces.
Change of volume of a gas with the change of temperature at constant pressure was studied by Charles and was given in the form of a law. which states,

Statement
For a given mass of a gas the volume of the gas is directly proportional to its absolute temperature provided the pressure is kept constant.
Mathematically this law may be written as
V ∞ T
V = K T
OR
V / T = K
This relation shows that the ratio of volume of a given mass of a gas to its absolute temperature is always constant provided the pressure is kept constant. On this bases Charles Law may also be defined as,
If the pressure remains constant for each 1ºC change of temperature the volume of the gas changes to 1/273 of its original volume.
On the bases of this statement
V1 / T = K & V2 / T2 = K
V1 / T1 = V2 / T2
This equation is known as Charle’s equation.
The volume temperature relationship can be represented graphically. When volume of a given mass of gas is plotted against temperature, a straight line is obtained.
Graph Coming Soon

Absolute Scale Of Temperature

There are different scales for the measurement of temperature such as Celsius ºC and Fahrenheit ºC. Similarly another scale known as absolute scale or Kelvin scale is determined on the basis of Charle’s law.
On the basis of Charle’s law we known that the volume of the gas changes to 1/273 times of its original volume for each 1 ºC change of temperature. It suggests that the volume of a gas would theoretically be zero at -273ºC. But this temperature has never been achieved for any gas because all the gases condense to liquid at a temperature above this point. So the minimum possible temperature for a gaseous system is to be -273ºC. This temperature is referred as absolute zero or zero degree of the absolute scale or Kelvin scale.
To form an absolute scale thermometer if the equally spaced divisions of centigrade thermometer are extended below zero and when the point -273ºC is maked then this point is called as absolute zero and the scale is called as absolute scale. It shows that for the conversion of centigrade scale into Kelvin scale 273 is added to the degrees on the centigrade scale.
K = 273 + ºC

Avogadro’s Law

In 1811, a scientist Avogadro’s established a relationship between the volume and number of molecules of the gas, which is known as Avogadro’s law.

Statement
Equal volume of all gases contains equal number of molecules under the same condition of temperature & pressure.
Mathematically it may be represented as
V ∞ n
OR
V = K n
On the basis of the above statement we can say that
1 dm3 of O2 gas will contain the same number of molecules as 1 dm3 of H2 or N2 or any other gas at same temperature and pressure.
It was also observed that 22.4 dm3 of any gas at S.T.P contain 1 mole of that gas, so 22.4 dm3 volume at S.T.P is called as molar volume or the volume of 1 mole of the gas and the mass present in 22.4 dm3 of any gas will be equal to its molar mass or molecular mass. It can also be explained on the basis of following figures.

Determination of Unknown Molecular Mass of a Gas With the Help of Avogadro’s Law
Suppose we have two gases (i) Oxygen (ii) CO
The volume of these two gases are equal which are 1 dm3.
The mass of 1 dm3 of oxygen is 1.43 gm
The mass of 1 dm3 of Co is 1.25 gm
According to Avogadro’s law we know that 1 dm3 of CO at S.T.P contain the same number of molecules as 1 dm3 of O2 under similar condition. Hence a molecule of CO has 1.25 / 1.43 times as much as a molecule of O2 and we know that the molecular mass of oxygen is 32 so the molecular mass of CO would be
1.25 / 1.43 x 32 = 28 g / mole

General Gas Equation (Ideal Gas Equation)
To give a relation between the volume, pressure and number of moles of n gas, Boyle’s law, Charle’s law and Avogadro’s law are used.
According to Boyle’s law | V ∞ 1 / P
According to Charle’s law | V ∞ T
According to Avogadro’s law | V ∞ n
By combining these laws we get
V ∞ 1 / P x T x n
OR
V = R x 1 / P x T x P
OR
P V = n R T
This equation is known as general gas equation n is also known as equation of state because when we specify the four variables = pressure, temperature, volume and number of moles we define the state for a gas.
In this equation “R” is a constant known as gas constant.

Value of R

1. When Pressure is Expressed in Atmosphere and Volume in Litres or dm3
According to general gas equation
P V = n R T
OR
R = PV / nT
For 1 mole of a gas at S.T.P we know that
V = 22.4 dm3 or litres
T = 273 K (standard temperature)
P = 1 atm (standard pressure)
So,
R = PV / nT
= 1 atm x 22.4 dm3 / 1 mole x 273 K
= 0.0821 dm3 K-1 mol0-1

2. When Pressure is Expressed in Newtons Per Square Metre and Volume in Cubic Metres
For 1 mole of a gas at S.T.P
V = 0.0224 m3 ………. ( 1 dm3 = 10-3 m3)
n = 1 mole
T = 273 K
P = 101200 Nm-2
So,
R = PV / nT
= 101300 Nm-2 x 0.0224 m3 / 1 mole x 273 K
= 8.3143 Nm K-1 mole-1
= 8.3143 J K-1 mol-1

Derivation of Gas Equation
According to general gas equation
P V = n R T
For 1 mole of a gas n = 1
P V = R T
OR
P V / T = R
Consider for a known mass of a gas the volume of the gas is V1 at a temperature T1 and pressure P1. Therefore for this gas we can write as
P1 V1 / T1 = R
If this gas is heated to a temperature T2 due to which the pressure is changed to P2 and volume is changed to V2. For this condition we may write as
P2 V2 / T2 = R
P1 V1 / T1 = P2 V2 / T2 = R
P1 V1 / T1 = P2 V2 / T2
This equation is known as gas equation.

Graham’s Law of Diffusion

We know that gas molecules are constantly moving in haphazard direction, therefore when two gases are placed separated by a porous membrane, they diffuse through the membrane and intermix with each other. The phenomenon of mixing of molecules of different gases is called diffusion.
In 1881, Graham established a relationship between the rates of diffusion of gases and their densities which is known as Graham’s law of diffusion.

Statement
The rate of diffusion of any gas is inversely proportional to the square root of its density.
Mathematically it can be represented as
r ∞ 1 / √d
r = K / √d
Graham also studied the comparative rates of diffusion of two gases. On this basis the law os defined as
The comparative rates of diffusion of two gases under same condition of temperature and pressure are inversely proportional to the square root of their densities.
If the rate of diffusion of gas A is r1 and its density is d1 then according to Graham’s law
r1 ∞ 1 / √d1
OR
r1 = K / √d1
Similarly the rate of diffusion of gas B is r2 and its density is d2 then
r2 ∞ 1 / √d2
OR
r2 = K / √d2
Comparing the two rates
r1 / r2 = (K / √d1) / (K / √d2)
r1 / r2 = √d2 / d1 ………………. (A)
But density d = mass / volume
Therefore,
For d1 we may write as
d1 = m1 / v1
And for d2
d2 = m2 / v2
Substituting these values of d1 & d2 in equation (A)
r1 / r2 = √(m2 / v2) / (m1 / v1)
But v1 = v2 because both gases are diffusing in the same volume.
Therefore,
r1 / r2 = √m2 / m1
Hence Graham’s law can also be stated as,
The comparative rates of diffusion of two gases are inversely proportional to the square root of their masses under the same condition of temperature and pressure.
It means that a lighter gas will diffuse faster than the heavier gas. For example compare the rate of diffusion of hydrogen and oxygen.
Rate of diffusion of H2 / Rate of diffusion of O2 = √Mass of O2 / Mass of H2 = √32/ 2 = √16 = 4
It shows that H2 gas which is lighter gas than O2 will diffuse four times faster than O2.

Dalton’s Law of Partial Pressures

Partial Pressure
In a gaseous mixture the individual pressure oxerted by a gas is known as partial pressure.
When two or more gases which do not react chemically are mixed in the same container each gas will exert the same pressure as it would exert if it alone occupy the same volume.
John Dalton in 1801 formulated a law which is known as Dalton’s Law of partial pressure and stated as.

Statement
The total pressure of a gaseous system is equal to the sum of the partial pressures of all the gases present in the system.
Suppose in a system three gases A, B & C are present. The partial pressure of these gases are
PA = Partial pressure of gas A
PB = Partial pressure of gas B
PC = Partial pressure of gas C
Then Dalton’s law may be mathematically written as
PT = PA + PB + PC
Where PT is the total pressure of the system.
To calculate the individual pressures of gases in the above example suppose the number of moles of A, B & C in the container are nA, nB and nC. So the total number of moles in the container will be
n = nA + nB + nC
Apply the general gas equation
P V = n R T
PT = n R T / V
Since R, T and V are same for gases A, B and C, therefore the partial pressure of these gases are as follows.
Partial pressure of gas A | PA = n(A)RT / V ……… (2)
Partial pressure of gas B | PB = n(B)RT / V ……… (3)
Partial pressure of gas C | PC = n(C)RT / V ……… (4)
Now divide equation (2) by (1)
PA / PT = (nA RT/V) / (nRT/V)
OR
PA / PT = nA / nT
Therefore,
P(gas) = P1 x n(gas) / n(total)

Application of Dalton’s Law
In an inert mixture of gases the individual gas exerts its own pressure due to collision of its molecules with the walls of the container but the total pressure produced on the container wall will be the sum of pressure of all the individual gases of the mixture.
On this basis the number of moles formed during a chemical reaction can be measured. For this purpose a gas produced in a chemical reaction is collected over water. The gas also contains some of water vapours. So the pressure exerted by the gas would be the pressure of pure gas and the pressure of water vapours.
Therefore the pressure of the system may be represented as
P(moist) = P(dry) + P(water vapour)
So,
P(dry) = P(moist) – P(water vapour)
In this way we can obtain the pressure of the gas and by using general gas equation we can calculate the number of moles of the prepared gas.

Ideal Gas

A gas which obeys all the gas laws at all temperatures and pressures is known as ideal gas.
It means that the product of pressure and volume must be constant at all pressures.
Similarly the rate of V/T will remain constant for an ideal gas.
But there is no gas which is perfectly ideal because of the presence of the force of attraction or repulsion between the molecules.

Gas Laws on the Basis of Kinetic Theory

Boyle’s Law
According to Boyle’s law the volume of a given mass of a gas is inversely proportional to its pressure at constant temperature.
It means that when the volume of the gas is decreased the pressure of the gas will increase.
According to kinetic molecular theory of gases the pressure exerted by a gas is due to the collisions of the molecules with the walls of the container. If the volume of a gas is reduced at constant temperature, the average velocity of the gas molecules remains constant so they collide more frequently wit the walls which causes higher pressure.

Charle’s Law
According to Charles law the volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure.
According to kinetic molecular theory the average kinetic energy of gas molecules is directly proportional to its absolute temperature so if the temperature of the gas is increased the average kinetic energy of the gas molecules is also increased due to which the sample of the gas expanded to keep the pressure constant. It is accordance with the law.

Graham’s Law
According to Graham’s Law
r1 / r2 = √m2 / m1
The rate of diffusion of a gas is directly proportional to the velocity of the molecules so,
v1 / v2 = √m2 / m1

Liquefaction
According to kinetic theory, the kinetic energy of the molecules is low for lower temperature. These slower moving molecules become subject to inter molecular attraction. At a sufficiently low temperature these attractive forces are capable of holding the molecules with one another so the gas is changed into liquid and the process is called liquefaction.

Liquid State
It is one of the state of matter. In this state, the kinetic energy of the molecule is very high due to which the molecules of the liquid are able to move but due to compact nature liquids are not compressible. On this basis we can say that the volume of a liquid is always constant but its shape can be changed.

Behaviour of Liquids
The main properties of liquids are as follows.

Diffusibility
The diffusion of one liquid into another liquid is possible but its rate is slow as compared with the rate of diffusion of gases. Example of diffusion of liquids is mixing of alcohol in water.

Explanation of Diffusion in Terms of Kinetic Energy
As the molecular of a liquid are in cluster form they are very close to each other but these molecules are movable so they can mix with the other molecules. Since the intermolecular distance are smaller due to which the rate of diffusion of liquids is slow.

Compressibility
The space between liquid molecules are very small due to strong Van der Waals forces. When the pressure is applied, they can be compressed but to a very little extent.

Expansion
When a liquid is heated, the kinetic energy of its molecules also increases so the attraction between the molecules becomes weaker due to which they go further apart and hence the liquid expands.

Contraction
When a liquid is cooled its kinetic energy is lowered and the attraction among the molecules becomes stronger so they comes close to each other and hence the liquid contract.

Viscosity

Definition
The internal resistance in the flow of a liquid is called viscosity.

Liquids have the ability to flow, but different liquids have different rates of flow. Some liquids like honey mobil oil etc. flow slowly and are called viscous liquids while ether, gasoline etc. which flow quickly are called less viscous.

Explanation
The viscosity of liquid can be understood by considering a liquid in a tube, a liquid in a tube is considered as made up of a series of molecular layer. The layer of the liquid in contact with the walls of the tube remains stationary and the layer in the center of the tube has highest velocity as shown.
Each layer exerts a drag on the next layer and causes resistance to flow.

Factors on Which Viscosity Depends

1. Size of Molecules
The viscosity of a liquid depends upon the size of its molecules. If the size of the molecules is bigger the viscosity of the liquid is high.

2. Shape of Molecules
Shape of the molecules affects the viscosity. If the shapes of the molecules are spherical they can move easily but if the shapes of the molecules are irregular such as linear or trigonal then the molecules will move slowly and its viscosity will be high.

3. Intermolecular Attraction
If the force of attraction between the molecules of a liquid is greater the viscosity of the liquid is also greater.

4. Temperature
Viscosity of a liquid decreases with the increase of temperature.

Units of Viscosity

Viscosity of a liquid is measured in poise, centipoise or millipoise & S.I unit.
1 poise = 1 N.s.m(-2)
1 centipoise = 10(-2) N.s.m(-2)

Surface Tension

Definition
The force acting per unit length on the surface of a liquid at right angle direction is called surface tension.

Explanation
Consider a liquid is present in a beaker. The molecules inside the liquid are surrounded by the other molecules of the liquid. So the force of attraction on a molecule is balanced from all direction. But the force of attraction acting on the molecules of the surface from the lower layer molecules is not balanced.
The molecules lying on the surface are attracted by the molecules present below the surface Due to this downward pull the surface of the liquid behave as a membrane which tends to contract to a smaller area and causes a tension on the surface of the liquid known as surface tension.
Diagram Coming Soon

Factors on Which Surface Tension Depends

1. Molecular Structure of the Liquid
If the force of attraction between the molecules is greater, the surface tension of the liquid is also greater. Those liquids in which hydrogen bond formation take place will have more surface tension.

2. Temperature
Surface tension of a liquid is inversely proportional to the temperature.

Units
1. Dynes / cm
2. Ergs / cm2

Capillary Action
The fall or rise of a liquid in a capillary tube is called capillary action.
When a capillary tube is dipped in a liquid which wets the wall of the tube, the liquid will rise in the capillary tube, to decrease the surface area due to surface tension. The liquid will rise in the capillary tube until the upward force due to surface tension is just balanced by the downward gravitational pull. This is called capillary action.

Vapour Pressre

Definition
The pressure exerted by the vapours of a liquid in its equilibrium state with the pure liquid at a given temperature is called vapour pressure.

Explanation
Consider a liquid is present in a bottle as shown.
Diagram Coming Soon
In the beginning the atmosphere above the surface of liquid is unsaturated but due to continuous evaporation the molecule of the liquid are trapped in the bottle and the air present above the surface of the liquid is becomes saturated and after it the molecules present in the vapour state may hit the liquid again and rejoin it by condensing into liquid. Thus in this closed vessel two process are going on simultaneously which are evaporation and condensation of vapours. When the rates of these two processes becomes equal at this point the pressure exerted by vapours is called vapour pressure.

Units of Vapour Pressure

The units for vapour pressure are
1. Millimeter of Hg
2. Atmosphere
3. Torr
4. Newton / m(2)

Factors for Vapour Pressure

1. Nature of Liquid
Vapour pressure of a liquid depends upon the nature of the liquid. Low boiling liquid exert more vapour pressure at a given temperature.

2. Temperature
Vapour pressure of a liquid also depends upon temperature. The vapour pressure of the liquid increases with the increase of temperature due to the increase of average of kinetic energy.

3. Intermolecular Forces
Those liquids in which the intermolecular forces are weak shows high vapour pressure.

Explanation of Evaporation on the Basis of Kinetic Theory
According to this theory the molecules of a liquid collide with each other during their motion. Due to these collisions some of the molecules acquire greater energy than Van der Walls forces which binds the molecules of the liquid together so these molecules of higher energy escapes from the surface into the air in the form of vapours.

Evaporation is a Cooling Process
In liquids, due to collision between molecules some molecules acquire higher energy and escapes from the surface of the liquid in the form of vapours. The kinetic energy of the remaining molecules decreases due to which the temperature of the liquid also decreases and hence we can say that evaporation is a cooling process.

Boiling Point

Definition
The temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure is called boiling point.

When a liquid is heated the rate of evaporation of the molecules also increases with the increase in temperature. When the pressure of the vapours becomes equal to the atmospheric pressure the liquid starts boiling and this temperature is known as boiling point.
If the external pressure on a liquid is changed the boiling point of the liquid also change. The increase in external pressure on a liquid increases the boiling point while the decreases of external pressure decrease the boiling point.

Solid State

It is a state of matter which posses both definite shape and definite volume. In solids the particles are very close to each and tightly packed with a greater force of attraction.

Properties of Solids

1. Diffusibility
Diffusion also occurs in solids but its rate is very slow. If a polished piece of zinc is clamped with a piece of copper for a long time. After few years we will see that some particles of zinc are penetrated into copper and some particles of copper are penetrated into zinc. It shows that the diffusion in solids is possible but it occurs with a slow rate.

2. Compressibility
In solids the molecules are close to each other so it is not easy to compress a solid. In other words we can say that the effect of pressure on solids is negligible.

3. Sublimation
It is a property of some solids that on heating these solids are directly converted into vapours without liquification. This property of solids is known as sublimation.

4. Melting
When solids are heated, they are changed into liquids and the property is called melting of the solids.

5. Deformity
Solids may be deformed by high pressure. When a high pressure is applied on solids due to which some particles are dislocated the force of attraction is so strong that the rearranged atoms are held equally well with their new neighbours and hence the solid is deformed.

Classification of Solids

Solids are classified into two main classes.
1. Crystalline
2. Amorphous

1. Crystalline Solids
In a solid if the atoms are attached with each other with a definite arrangement and it also possesses a definite geometrical shape. This type of solid is called crystalline solid.
e.g. NaCl, NiSO4 are crystalline solids.

2. Amorphous Solids
In these solids there is no definite arrangement of the particles so they do not have a definite shape. The particles of such solids have a random three dimensional arrangement. Examples of amorphous solids are glass, rubber, plastic etc.
The properties of crystalline and amorphous solids are quite different from each other. These differences in properties are given below.

Difference of Geometry

1. Crystalline Solids
In crystalline solids particles are arranged in a definite order due to which it possesses a definite structure.

2. Amorphous Solids
In amorphous solids particles are present without any definite arrangement so they do not have definite shape.

Difference of Melting Point

1. Crystalline Solids
Crystalline solids have sharp melting point due to uniform arrangement.

2. Amorphous Solids
Amorphous solids melts over a wide range of temperature.

Cleavage and Cleavage Plane

1. Crystalline Solids
When a big crystal is broken down into smaller pieces the shape of the smaller crystals is identical with the bigger crystal. This property of crystalline solids is called cleavage and the plane from where a big crystal is broken is called cleavage plane.

2. Amorphous Solids
Amorphous solids do not break up into smaller pieces with an identical shape.

Anisotropy & Isotropy

1. Crystalline Solids
It is a property of crystalline solid that they show different physical properties in different direction. For example graphite can conduct electric current only through the plane which is parallel to its layers. This property is called anisotropy.

In amorphous solids the physical properties are same in all directions. This property of solids is called isotropy.

Symmetry in Structure
1. Crystalline solids are symmetric in their structure when they are rotated about an axis, their appearance remains same so they are symmetric in structure.
2. Amorphous Solids
Amorphous solids are not symmetric.

Types of Crystals

There are four types of crystals.
1. Atomic crystals
2. Ionic crystals
3. Covalent crystals
4. Molecular crystal

1. Atomic Crystals
Metals are composed of atoms. These atoms are combined with each other by metallic bond and the valency electrons in metals can move freely throughout the crystal lattice. This type of solid is called atomic crystal.
The properties of atomic crystals are
1. High melting point.
2. Electrical and thermal conductivity.
3. These are converted into sheets so these are malleable.
4. These are used as wire so these are ductile.

2. Ionic Crystals
Those solids which consists of negativity and positively charged ions held together by strong electrostatic force of attraction are called ionic crystals. Ionic crystalline solids possesses the following properties.
1. The melting and boiling point of ionic crystals is high.
2. They conduct electricity in molten state.
3. Ionic crystals are very hard.
4. Indefinite growth of crystals is also a property of ionic crystals.

3. Covalent Crystals
In covalent solids, the atoms or molecules are attached with each other by sharing of electrons. Such type of solids are called covalent solids e.g. diamond is a covalent solid in which carbon atoms are attached with each other by covalent bond. The other examples of covalent crystals are sulphur, graphite etc.
Covalent crystals possesses the following properties.
1. High melting point.
2. High refractive index.
3. Low density.

4. Molecular Crystals
Those solid in which molecules are held together due to intermolecular forces to form a crystal lattice are called molecular crystals e.g. iodine and solid CO2 are molecular crystals. The general properties of molecular crystals are as follows.
1. Low melting and boiling point.
2. Non – conductor of heat and electricity.

Isomorphism

When two different substance have same crystalline structure, they are said to be isomorphous and the phenomenon is called isomorphism.e.g. ZnSO4 and NiSO4 are two different substances but both are orthorhombic similarly the structure of CaCO3 and NaNO3 is frigonal.

Polymorphism

If a substance exist in more than one crystalline form it is called polymorphous and the phenomenon is known as polymorphism. E.g. sulphur exist in rhombic and monoclinic form similarly CaCO3 exist in trigonal and orthorhombic form.

Unit Cell

The basic structural unit of a crystalline solid which when repeated in three dimensions generates the crystal structure is called a unit cell.

A unit cell of any crystalline solid has a definite geometric shape and distinguish from other crystals on the basis of length of the edges and angle between the edges.

Crystal Lattice

In crystalline solids atoms, ions or molecules are arranged in a definite order and form a three dimensional array of particles which is known as crystal lattice.