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SOME BASIC CONCEPT OF
CHEMISTRY
SUBSTANCE (MATTER)
Anything having some mass and volume is called as substance
...
: NA = 6
...




 Atomic weight of an element =



massof oneatomof theelement
1
massof one C12  atom
12

mass of itsonemolecule
 Sum of atomic masses of all atoms
Molecular weight = 1
12
 massof one C  atom
12
Example : Molecular weight of H2SO4 = (1 × 2) + (32 × 1) + (16 × 4) = 98

 Moles of atom =
Example :

Noof atoms
weight of elemental samplein gm

 No
...

Gram atomic weight
For sample of oxygen having weights 1
...
6

 0
...
10 × 6
...
022 × 1022

16

 Moles of molecule = No
...
of gram  molecules
Avogadro ' s no
...

[1]

[2]

Some Basic Concept of Chemistry

Example :

For a sample of 3
...
011024
Moles of H3PO4 =

6
...


volumeof gasinlit
...
T
...

Moles of molecules =
Example :



Calculate moles of molecule of NH3 having 2
...
T
...

2
...
10
22
...
T
...
It is 2
...

Molecular weight
Equivalent Weight : =
n  factor
Solution :



22
...


Moles of NH3 =

 Calculation of n-factor :
(i)

For any acid it is the number H+ replaced or e-pair gained by its one molecule, i
...
its basicity
...




n-factor for H3BO3 = 1
(ii) For any base, it is the no of OH– lost or H+ gained or electron pair donated by its one molecule, i
...
its acidity
...

Example :
For Al2(SO4)3 n-factor = 6
(iv) For any radical, it is the charge over it
...

Example :

K2Cr2 O7 
Cr3
n-factor = 6

[3]

Some Basic Concept of Chemistry

(vi) For any reducing agent it is the no of electrons gained by it’s one molecule
...
wt
...


Laws of equivalence : Gram equivalent of every reactant used is the same and it is equal to the gram equivalent
of every product, formed
...

molesof solute
w 1000
(i) Molarity (M) :
Moles of solute in one lit solution = vol
...
What is the molarity
...
1 M

Solution :

Molarity =

w
M

(ii) Normality (N) :

0



1000
4 1000
 
1
volume(ml) 40 100

No
...

g
...
of solute
w 1000
= vol
...
eq
...
= N × V (mL) = 1/20 × 50 = 2
...
If density of the solution is 1
...
34
(3) 0
...
4
Weight of solute = 10 g
Volume of solution = 250 mL
Density of solution = 1
...
2 = 300 g
 Weight of solvent = weight of solution – weight of solute = 300 – 10 = 290 g
w 1000  10  1000  0
...
5 290

[4]

Some Basic Concept of Chemistry

(iv) % Composition :
(a) w/w = wt of solute in gm per 100 gm of solution =

wt
...
of solutionin gm

Example : What is the weight percentage of NaCl solution in which 20g NaCl is dissolved in 60 g of
water
...
of solutein ml

(b) v/v = vol
...
of solution in ml

100

Example : A solution is prepared by mixing of 10 ml ethanol with 120 ml of methanol
...

(1) 10%
(2) 7
...
7%

wt
...
of solution in ml 100
Example :What is weight / volume percentage of a solution in which 7
...

(1) 7
...
5%
(3) 50%
(4) none
Solution : 7
...
7
...

7
...
5%
100

NOTE :
(i)

If for any solution percentage (w/w) is ‘x’, density is ‘d’ g/ml and molar mass of the soute is ‘M0’ then
Molarity 

10x
...
Specific gravity (density) of solution is 1
...

Find its Molarity
...
71
 80 13
...

1000d
Molarity = M
Solution : M 

10  d  percent
GMM

M

0

[5]

Some Basic Concept of Chemistry

Molarity of pure water =

1000 1

 55
...
Strength (g/lit) =

wt
...
of solution in lit
Molarity × Mol Mass of solute = g/lit
Normality × Eq
...


Note :
(i) Due to dilution of any solution moles of solute, present in solution remain the same but concentration is
decreased
M1
V1

M2
V2

Before dilution
M1V1 = M2 V2 = moles of solute
...
1 M NaOH solution which in diluted till its concentration become
0
...
Calculate volume of H2O added
...
1
 1000 ml
0
...
of water added = 1000 – 100 = 900 ml
(ii) If from any solution any small amount is taken out, concentration of the small amount remain exactly the
same
...


Empirical formula :

It represent atoms present in any molecule in the simplest ratio
...
The Law of Conservation of Mass (Lavoiser 1744) : This law states “matter can neither be created nor
destroyed or in a chemical reaction, the mass of the reactants is equal to the mass of the products”
...


[6]

Some Basic Concept of Chemistry

Example :
2Ca

*








O2  2CaO

80 g
32 g
112 g
Total mass reactant = Total mass product = 112 g
For a complete irreversible reaction total mass of reactants before reaction in equal to the total mass of Products
after reaction
...
The Law of Constant Composition or Definite Proportion (Proust in 1799) : This law states that “ All
pure sample of the same chemical compound contain the same elements combined in the same proportion by
mass irrespective of the method of preparation”
...

Similarly in water ratio of weight of hydrogen to oxygen is 1 : 8
...
The Law of Multiple Proportion (Dalton)
This law states that :when two elements A and B combine together to from more than one compound, then
several, masses of A which separately combine with a fixed mass of B, are in a simple ratio
...
e
...
The Law of Reciprocal Proportions (Richer in 1792 - 94)
This law states that “when two elements combines separately with third element and form different types of
molecules their combining ratio is directly reciprocated if they combine directly
...
In CO2 12 g of C reacts with 32g of O,
whereas in CH4 12 g of reacts with 4g of H
...
e
...
The same is found to be true in H2O molecules
...

5
...
Or in other words volume of
reacting gases and product gases have a simple numerical ratio to one another
...

1 unit vol
2 unit vol
...
The Avogadro Law
This law states that “equal voume of all gaseous under similar conditions of temperature and pressure contain
equal number of molecules”
...

1 vol
...
ratio 2 : 1 : 2
It provides a relationship between vapour density and molecular mass of substances
...




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