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Basic concepts of chemistry

 

UNIT 1 SOME BASIC CONCEPTS OF CHEMISTRY

 Chemistry may be defined as the branch of science which deals with the study

of various forms of matter, especially about their composition, methods of preparation, properties and their reactions with other substances.

SOME BRANCHES OF CHEMISTRY

Ø  Inorganic Chemistry Organic Chemistry

Ø  Physical Chemistry Analytical Chemistry

Ø  Polymer Chemistry Biochemistry

Ø  Medicinal Chemistry Industrial Chemistry

Ø  Hydrochemistry Electrochemistry

Ø  Green Chemistry etc.

IMPORTANCE OF CHEMISTRY :

1.      In medical and drugs synthesis- cisplatin  and texol is used in cancer therapy. AZIDOTHYMIDINE for AIDS

2.      AGRICULTURE :  New effective fertilizer , insecticide and pesticides have been developed. Like DDT

3.      INDUSTRIAL :.  It help in manufacturing of goods like acid , alkalies, dyes , polymer etc

4.      In construction :  cements are mixture of various compound : CaO, CaCO3 , silica etc .

 

MATTER

 Matter may be defined as anything which occupies space and has mass and can be perceived by our senses.

 Eg: Air, Water, Stone etc.

CLASSIFICATION OF MATTER:

STATES OF MATTER

v  Solid state

v  Liquid state

v  Gaseous state

v  Plasma state

v  Bose-Einstein condensate

1. PHYSICAL CLASSIFICATION

On the basis of rigidity, volume and shape, matter can exist in three physical states.

a) Solids

v  Solids are rigid substances which have definite shape and definite volume.

v  Least intermolecular space

v  Strong intermolecular attraction force.

v  Least compressible state

v  High density, melting point and boiling point .

b) Liquids

v  Liquids are not rigid, but have definite volume.

v  They do not have a definite shape.

v  Their particles can move around.

v  They take the shape of the container in which they are placed.

c) Gases

v  Gases have neither definite volume nor definite shape.

v  They occupy the shape and volume of the container.

v  Movement of particles is easy and fast than liquid and solid.

v  Highly compressible because of large intermolecular space.

v  Highest diffusion capacity

Solid ,  liquids and gases are inter convertible to one another By temparature and pressure change.

  Thermal energy directly proportional to  temparature

 

CHEMICAL CLASSIFICATION

PURE SUBSTANCE :   All constituents particles are  same in chemical nature . Constituents particles of pure substance has fixed composition.

Eg Cu ,Au Zn , Glucose etc

a)      Elements : Elements  is a pure substance and consists  only one type of atoms.

Ø  These particles  may be atom or molecules.

Ø  All atoms of an element are identical.

Ø  Atoms of different elements are different

Ø   It cannot be broken down into two or more simpler substances by physical and  Chemical means.

 Eg: Copper, Silver, Gold, Iron etc.

b)      Compounds : A compound is a pure substance made up of two or more elements combined Chemically in fixed proportion by mass.

Ø  Their constituents can not be separated into simpler substance by physical methods.

Ø  Property of compound are totally different from it’s constituents atoms.

Eg: H₂O, NH₃, CH₄ etc.

Hydrogen : highly combustible  and oxygen is supporter for combustion but their product -Water is exitnguisher of fire.

c)       Mixtures :A mixture is made up of two or more pure substance mixed in any ratio.

Constituents particles never loss it’s identity.

1.       HOMOGENEOUS MIXTURE :

Ø  mixture have single phase.

Ø  HOMOGENEOUS mixture is also called Solution.

Ø  It’s components completely mixed with each other

Ø  Constituent particles are distributed uniformly throughout the mixture.

Ø  Particles can not be seen by Naked eyes and microscope.

Eg  : gold+Cu, gasoline ( mixture of hydrocarbons), iodine in benzene, sugar in water, air etc .

        Brass is a mixture of Cu and Zn,

         Bronze is a mixture of Cu and Sn.

2.      HETEROGENEOUS MIXTURE

Ø  Composition is not uniform throughout the mixture

Ø  Different components may visible

Ø  Components can be separated by simple physical mathods .

                                    Eg.  Gun powder is a mixture of charcoal, sulphur and nitre.  Muddy water, Chalk powder in water , smoke, soil etc.

Physical Properties :

 Properties which can be measured or observed without changing the identity or The composition of the suPropertie

 Eg: Colour, odour, melting point, boiling point, density etc.

Chemical Properties

 Properties in which a chemical change occur.

Eg.  Acidity, basicity , combustibility  etc.

MASS AND WEIGHT

Mass of a substance is the amount of matter present in it.

The mass of a substance is constant.  The SI unit of mass is kg.

Weight is the force exerted by gravity on an object.

Weight may vary from one place to another due to change in gravity.

DENSITY

Density of a substance is its amount of mass per unit volume.

                                           Density. =

The SI unit of density is kg/m³

TEMPERATURE :

Temperature is the degree of hotness of a body.

Three common scales are used to measure temperature.

Ø  They are degree Celsius (º C), degree Fahrenheit (º F) and Kelvin (K).

Ø  The Celsius scale is represented between zero degree to hundred degree.

Ø  Celsius and Fahrenheit scale is represented between 32º To 212ºF.

Ø  The Fahrenheit scale is related to Celsius scale as follows

                         º F =

Ø  The Kelvin scale is related to Celsius scale as follows

                             K = º C + 273.15

ACCURACY :

 It is the agreement of a particular value to the true value of the result.

PRECISION

Precision is the closeness of various measurements for the same quantity.

Eg: Value of A is precise but not accurate, values of B is neither accurate nor precise but values of C IS precise as well as accurate.

SIGNIFICANT FIGURES

Significant figures are meaningful digits which are known with certainty.

RULES FOR DETERMINING THE SIGNIFICANT FIGURES

Ø  All non zero digits are significant.

Eg: 235 has 3 significant figures.

Ø  Zero’s to the left of the first non zero digit are not significant.

Eg: 0.002 has only one significant figure.

Ø  Zero’s between non zero digits are significant.

Eg: 3.02 has 3 significant figures.

Ø  Zero’s to the right of the decimal point are significant.

Eg:2.00 has 3 significant figures.

2.00                         3

0.00001                  1

1                            Infinity

1.2003.                  5

6.022 × 10^{23         .}     4

LAWS OF CHEMICAL COMBINATION

 

1.      LAW OF CONSERVATION OF MASS

 

 This law was put forth by Antoine Lavoisier. (Father of modern chemistry)

“The law states that matter can neither be created nor be destroyed.”

 

               C    +    O_2. --à   _CO2

                       _{12gm.         2×16=32gm.     12+32= 44gm}    

: 12 g carbon combines with 32 g oxygen to form 44 g CO2.

Question 1:     4.9 g KCl0₃ when pyrolysed gives the residue of 2.96 g. Calculate the mass of oxygen released, using law of conservation of Mass.

Here the total mass of the reactants is equal to the total mass of the products.

Ans:

 

2 .LAW OF DEFINITE PROPORTION (Law of Definite composition )

 

This law was put forth by Joseph Proust.

 

“The law states that, the same compound always contain the same elements Combined in the same fixed proportion by mass. “

Example: natural and synthetic sample of CuO is analysed and found that composition by mass is identical in both cases.

Eg : NaCl may be obtained from sea water         Or   It is also prepared by chemical reactions between NaOH and HCl.

 These samples on analysis are found to contain Na and Cl in the ratio 23:35.5 By mass.

Questions 2: If the law of constant composition is true, what weights of calcium, carbon and oxygen are present in 1.5 g calcium carbonate? Given that the sample of calcium carbonate contains 40% Ca; 12% Carbon; 48% Oxygen.

 

Ans:   mass of atom. =

 

Mass of  Ca   =  

Mass of  C.  =.

Mass of oxygen =. 

1.      LAW OF MULTIPLE PROPORTION

This law was put forth by Dalton.

 

“ The law states that when two elements combines to form more than one  Compound, the different masses of one of the elements which combines with a Fixed mass of the other element are in the ratio of simple whole numbers. “

              Eg: Carbon combines with oxygen to form two different oxides CO and CO₂ Under different conditions.  

Mass of atoms à	C	O
CO	12	16
CO2	12	2× 16= 32

 In CO_2. , 12 g of carbon combines with 32 g of oxygen and in carbon monoxide  12 g carbon combines with 16g of oxygen. In these two cases, the mass of oxygen combining with the fixed mass (12gm) of Carbon are in the ratio 16:32 or 1:2.

Example:

	H (gm)	O (gm)
H20	2	1× 16
H2O2	2	2 × 16 = 32

 

Ratio of oxygen combining with 2 gm of hydrogen bear a simple Ratio – 1:2 or 16:32

 

 

2.      GAY LUSSAC’S LAW OF GASEOUS VOLUMES ( law of definite proportion by volume)

 

 This law was put forth by Gay Lussac.

“ The law states that, when gases combine to form gaseous products, a simple Ratio exists between the volumes of the reactants and the products at constant Temperature and pressure. “

H₂      +        Cl₂             à            2HCl

1 volume   1 volume.               2 volume

Ratio    =    1:1:2

 

 

 

 

3.      AVOGADRO’S hypothesis

 

 This law was put forth by Avogadro.            

 The law states that “ equal volume of all gases under the same conditions of Temperature and pressure contains the Equal number of molecules. “

 Application of Avogadro hypothesis-

1.      Atomicity :

It is number of atoms present in one molecule of an molecule or elementry gas. 

2.      Molecular mass and vapour density

 

Vapour density =  

 

DALTONS ATOMIC THEORY

 In 1808, John Dalton proposed an atomic theory.

 POSTULATES :

1.       Matter is made up of small indivisible particles called atoms.

2.       Atoms of the same element are identical in mass and other properties.

3.       Atoms of different elements are different in properties.

4.       Only reorganization of atoms occurs in chemical reactions.

5.       Atoms can neither be created nor be destroyed.

6.       Since atoms are indivisible, they combine in small whole numbers to form compound atoms called molecules.

Limitations : He couldn’t explain that  why atoms combine.

 

ATOMS  :

 Atom is the smallest particle of substance that may or may no be exist in free state but Participate in chemical combination.

MOLECULES :    molecules are the smallest particle of a substance that can exist in free state but don’t participate in chemical combination  in molecular form.

Ø  A molecule has all the properties of that substance.

TYPES OF MOLECULES

 Based on the type of atoms, there are two types of molecules.

Homonuclear molecule  ( homoatomic)

A molecule containing only one type of atom is called homonuclear molecule.

Eg. H₂, O₂, N₂, O₃ etc.

Heteronuclear molecule. (Heteroatomic) :

Heteronuclear molecules contain different types of atoms.

E.g. CO₂, H₂O, C₆H₁₂O₆, NH₃ etc.

Based on the no. of atoms there are three types of molecules.

Ø  Monoatomic, diatomic and polyatomic molecules.

Ø  Monoatomic molecules contain only one atom.

Eg. All metals, noble gases like He, Ne, Ar etc.

Ø  Diatomic molecules contain 2 atoms.

E.g. H2, O2, N2, halogens (F2, Cl2, Br2 and I2)

Ø  Polyatomic molecules contain more than two atoms. Eg. Ozone (O₃), Phosphorus (P₄), Sulphur (S₈) etc.

ATOMIC MASS

Atomic mass of an element is defined as a number which expresses how many Times the mass of one atom of the element is greater than 1/12^Th the mass of an atom of Carbon-12 atom.

 Mass of one atom of the element

= Relative Atomic mass =.  amu

One amu = 1.66× 10 ^-24.   gm

 Atomic mass =

AVERAGE ATOMIC MASS

The average atomic mass of an element is the sum of the masses of its

Isotopes, each multiplied by its natural abundance  We can calculate an average atomic mass of an element by considering the Atomic mass of the isotopes and their relative abundance.

·        Av.atomic mass =  atomic mass of isotop  ×  fraction of that ratio + ……

·        Av. Atomic mass =  

Eg: Chlorine has two isotopes ³⁵Cl and ³⁷Cl in the ratio 3:1.

 So the average atomic mass Cl = (3 x 35 + 1 x 37) / 4 = 35.5 amu

ATOMIC MASS UNIT  :  One atomic mass unit is defined as mass exactly equal to 1/12^th The mass of one Carbon-12 atom. Amu is also known as unified mass. (u)

1 amu = 1.66056 x 10^―24 g

Example:

 

Example:  find average atomic mass of C using given  data :

 Ans :

 

MOLECULAR MASS : Molecular mass is the sum of atomic masses of the elements present in a

Molecule.

It is obtained by multiplying the atomic mass of each element by the number of Its atoms and adding them together.

Molecular mass = no of atoms of elements + it’s atomic mass + ……

Eg.  Na2SO4 = 2×23+32×1+16×4

                       =46 + 32 + 64 = 142 amu

GRAM ATOMIC MASS OR GRAM ATOM  

Atomic mass expressed in grams is called gram atomic mass or simply gram Atom.

Mass in grams =  No. of gram atoms × Gram atomic mass

GRAM MOLECULAR MASS OR GRAM MOLE

 The molecular mass expressed in grams is called gram molecular mass or Simply gram mole.

Mass in gram.=.   No. of gram moles ×. Gram molacular mass

FORMULA MASS

: It is the sum of atomic masses of all atom present in one formula unit of an ionic compounds

 

Eg: The formula mass of NaCl = Atomic mass of sodium + Atomic mass of Cl  = 23 + 35.5 = 58.5

 

MOLE CONCEPT -:

·        Mole is the unit Of amount of substance .

·        One mol is  mass in gram which contains  6.022× 10^23.    Atoms ,ions or molecules Particles .

·        Mole is represented by the symbol ‘mol’.

HOW TO FIND THE NUMBER OF MOLES?

 

       Avogadro number = 6.022× 10²³

 Molar Mass

Example 37. How many molecules and atoms of oxygen are present in 5.6 litres of oxygen (0₂) at NTP?

Solution: We know that 22.4 litres of oxygen at NTP contain 6.02 x 10 23 molecules of oxygen.

So, 5.6 litres of oxygen at NTP contain

                                                                       

x 6.02 x 10 ²³molecules

 

= 1.505 x 10 23 molecules

1 molecule of oxygen contains

= 2 atoms of oxygen

So, 1.505 x 10 23 molecules of oxygen contain

23

= 2 x 1.505 x 10 atoms

23

= 3.01 x 10 atoms

Example . How many electrons are present in 1.6 g of methane ?

Solution: Gram-molecular mass of methane (CHO -12 + 4 =16 g

 

Number of moles in 1.6 g of methane =

 

Number of molecules of methane in 0.1 mole

0. 1 x 6.02 x 10 ^23.     Oxygen atoms

6.02 x 10 ^{22.  Atoms}

One molecule of methane has = 6 + 4 = 10 electrons

So, 6.02 x 10 22 molecules of methane have

 

 

 

 

 

Example 37. How many molecules and atoms of oxygen are present in 5.6 litres of oxygen (0₂) at NTP?

Solution: We know that 22.4 litres of oxygen at NTP contain 6.02 x 10 23 molecules of oxygen.

So, 5.6 litres of oxygen at NTP contain

                                                                       

x 6.02 x 10 ²³molecules

 

= 1.505 x 10 23 molecules

1 molecule of oxygen contains

= 2 atoms of oxygen

So, 1.505 x 10 23 molecules of oxygen contain

23

= 2 x 1.505 x 10 atoms

23

= 3.01 x 10 atoms

Example . How many electrons are present in 1.6 g of methane ?

Solution: Gram-molecular mass of methane (CHO -12 + 4 =16 g

 

Number of moles in 1.6 g of methane =

 

Number of molecules of methane in 0.1 mole

0. 1 x 6.02 x 10 ^23.     Oxygen atoms

6.02 x 10 ^{22.  Atoms}

One molecule of methane has = 6 + 4 = 10 electrons

So, 6.02 x 10 22 molecules of methane have

 

= 10 6× .02 10²² electrons

= 6.02 x 10²³ electrons

Example . Calculate the number of moles in 25 g of calcium carbonate and number of oxygen atoms.

Solution:  Do itself

 

Stoichiometry =. Stoicheion + metron. ( Element + measure).

It deals with the measurements of masses and volume of reactants and products involve in a chemical reaction.

·       To calculate , the chemical equations must be balanced and states of reactants and products must be written.

·  

 

Limiting reagent

The reactants that gets consumed first and  limit the amount of product formed is called limiting reagent.