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Chapter

1

Chemical Arithmetic
Chemistry is basically an experimental science
...
The results of measurement can we reported in two steps,
(a) Arithmetic number, (b) Unit of measurement
...
The closeness of the
set of values obtained from identical measurement called precision and a
related term, refers to the closeness of a single measurement to its true
value called accuracy
...
Number of significant figures in a physical
quantity depends upon the least count of the instrument used for its
measurement
...
All non zero digits are significant
...
All zeros on the right of the non zero digit are not
significant
...
Again
x  378000 has three significant figures
...
All zeros on the right of the last non zero digit become
significant, when they come from a measurement
...
It has four significant figures
...
050 km or 3
...
In all these expressions, number of
significant figures continues to be four
...
By changing the position of the decimal point, the
number of significant digits in the results does not change
...
The reverse is also true
...
If the digit to be dropped is less than 5, then the preceding
digit is left unchanged
...
Again x  189
has only three significant figures
...
All zeros occurring between two non zero digits are
significant
...
82 is rounded off to 7
...
94 is
rounded off to 3
...


Example : x  1007 has four significant figures
...
0809 has five significant figures
...
In a number less than one, all zeros to the right of decimal
point and to the left of a non zero digit are not significant
...
87 is rounded off to 6
...
78 is rounded
off to 12
...


Example : x  0
...
Again,
x  1
...
This is on account of rule 2
...
All zeros on the right of the last non zero digit in the
decimal part are significant
...
351 is rounded off to 16
...
758 is
rounded off to 6
...

Rule 4
...

Example : x = 3
...
2 on rounding off, again
x  12
...
6 on rounding off
...
00800 has three significant figures 8, 0, 0
...
00 has three significant figures
...
If the digit to be dropped is more than 5, then the preceding
digit is raised by one
...
If the digit to be dropped is 5 followed by digits other than
zero, then the preceding digit is raised by one
...
If digit to be dropped is 5 or 5 followed by zeros, then the
preceding digit is raised by one, if it is odd
...
750 is rounded off to 3
...
150 is
rounded off to 16
...

(3) Significant figure in calculation
(i) Addition and subtraction : In addition and subtraction the
following points should be remembered
(a) Every quantity should be changed into same unit
...

(c) The result obtained after addition or subtraction, the number of
figure should be equal to that of least, after decimal point
...

(b) The product or division of two significant figures, will contain
the significant figures equal to that of least
...
G
...
System : Length (centimetre), Mass (gram), Time (second)
(2) M
...
S
...
P
...
System
: Length (foot), Mass (pound),
Time
(second)
(4) S
...
System : The 11th general conference of weights and measures
(October 1960) adopted International system of units, popularly known as the SI
units
...
The standard prefixes which helps to reduce the basic units are now
widely used
...
The
units for such quantities can be obtained by defining the derived quantity in
terms of the base quantities using the base units
...
So the unit is m / s or ms 1
...


The chosen standard of measurement of a quantity which has
essentially the same nature as that of the quantity is called the unit of the
quantity
...
1 Seven basic S
...
units
Length

Mass

Time

Temperature

Electric Current

Luminous Intensity

Amount of
substance

metre (m)

Kilogram (kg)

Second (s)

Kelvin (K)

Ampere (A)

Candela (Cd)

Mole (mol)

Table 1
...
3 Standard prefixes use to reduce the basic units
Multiple

Prefix

Symbol

Submultiple

10

yotta

Y

10

zetta

Z

10

exa

E

10

peta

P

10

tera

T

24

21

18

15

12

Prefix

Symbol

10

deci

d

10

centi

c

10

milli

m

10

micro



10

nano

n

–1

–2

–3

–6

–9

10

giga

G

10

pico

p

10

mega

M

10

femto

f

10

kilo

k

10

atto

a

10

hecto

h

10

zeto

z

10

deca

da

10

yocto

y

9

6

3

2

1

–12

–15

–18

–21

–24

Table 1
...
s
...
= 3
...
4 L at STP

1 dyne = 10 N

1 mole a substance = N molecules

1 atm = 101325 Pa

1 g atom = N atoms

1 m = 39
...
184 J

1 inch = 2
...
602  10 J

1 litre = 1000 mL

1 eV/atom =96
...
79 L

1 amu = 931
...
59237 g

1 kilo watt hour = 3600 kJ

1 litre atm = 101
...
1536  10 s

1Å = 10 m

1 joule = 10 erg

1 debye (D) = 1  10 esu cm

1nm = 10 m

–10

–5

–19

1 newton =1 kg m s

–2

1 J = 1 Nm =1 kg m s
2

–2

7

–1

0

5

0

o

–2

–3

7

Laws of chemical combination
Various chemical reactions take place according to the certain laws,
known as the Laws of chemical combination
...
According to this law, Matter is neither created nor

destroyed in the course of chemical reaction though it may change from one
form to other
...

(2) Law of constant or definite proportion : It was proposed by
Proust
...

(3) Law of multiple proportion : It was proposed by Dalton and
verified by Berzelius
...

(4) Law of equivalent proportion or law of reciprocal proportion : It
was proposed by Ritcher
...

(5) Gay-Lussac’s law : It was proposed by Gay–Lussac and is
applicable only for gases
...
The Gay-Lussac’s law, was based
on experimental observation
...

Postulates of Dalton's hypothesis is as followes,
(i) Each element is composed of extremely small particles
called atoms which can take part in chemical combination
...
e
...


o

–18

–3

–10

–9

(iii) Atoms of different elements possess different properties
(including different masses)
...
e
...

(v) Atoms of elements take part to form molecules i
...
, compounds
are formed when atoms of more than one element combine
...

(2) Modern atomic hypothesis : The main modifications made in
Dalton’s hypothesis as a result of new discoveries about atoms are,
(i) Atom is no longer considered to be indivisible
...


e
...
, isotopes of oxygen O16 , O17 , and O18
...
e
...
,
isobars Ca 40 and Ar 40
...
In many nuclear reactions, a
certain mass of the nucleus is converted into energy in the form of , 
and  rays
...


e
...
, in sucrose (C12 H 22 O11 ) , the elements carbon, hydrogen and oxygen
are present in the ratio of 12 : 22 : 11 and the ratio is not a simple whole
number ratio
...

When applied to law of combining volumes, this hypothesis predicts that
atoms are divisible and hence it is contrary to Dalton's hypothesis
...
” Avogadro hypothesis has been found to explain as follows,
(i) Provides a method to determine the atomic weight of gaseous
elements
...
D
...

Molecular mass  2  vapour density

(iii) It helps in the determination of mass of fixed volume of a
particular gas
...
D
...
0000897 gm
...
D
...
0000897 gm
...
D
...
e
...
D
...
D
...
0000897

Valency of the element =

gm
...
T
...

(v) It helps in determination of molecular formulae of gases and is
very useful in gas analysis
...


Atomic, Molecular and Equivalent masses
(1) Atomic mass : It is the average relative mass of atom of element
as compared with an atom of carbon –12 isotope taken as 12
...
Since the atomic mass is a ratio, it has no units and is
=
m n
expressed in amu, 1 amu = 1
...


Mass of an element
GAM

(ii) Mass of an element in gm
...
of gm
...
02

 GAM

 10 23

 10 23

Atomic mass = Equivalent mass of metal

Mass
=
GAM

 6
...
02  10
Atomic mass
23

(iv) Number of atoms in 1gm of element =

 Valency

(iii) Specific heat method : It is suitable only for gases
...
Their ratio is known as  whose value is constant (1
...
40 for diatomic and 1
...

Atomic mass of a gaseous element 

Molecular mass
Atomicity

(iv) Isomorphism method : It is based on law of isomorphism which
states that compounds having identical crystal structure have similar
constitution and chemical formulae
...
7 H 2 O, MgSO4
...
7 H 2 O

(valency of Zn, Mg, Fe = 2)
...
Molecular mass is a ratio and hence has
no units
...
m
...

Molecular mass 

Mass of one molecule of the substance
1 / 12  Mass of one atom of C - 12

Actual mass of one molecule = Mol
...
66 10 24 gm
...
of GAM  6
...
5

g
...

12

(i) Number of gram atoms =

Molecular mass of chloride
Equivalent mass of chloride

has

 Molar mass of a gas or its 1 mole occupies 22
...


(iv) It also helps in the determination of molar volume at N
...
P
...
) 
6
...

Gram molecular mass (GMM) and Gram molar volume : Molecular
mass of an element or compound when expressed in gm
...

Number of gm molecules =

Mass of substances
GMM

Mass of substances in gm = No
...
molecules  GMM
Volume occupied by one mole of any gas at STP is called Gram
molar volume
...
4 litres
...
4 litres

law
Atomic mass  Specific heat = 6
...
)
Atomic mass (approx
...
4
Specific heat (in cals
...

Atomic mass = Equivalent mass  Valency

Expression for mass and density
Mass of 11
...
D
...

Density of a gas at NTP =

Mol
...

22400 ml

Important generalisations

Number of atoms in a substance

= Number of GMM

 6
...
02  10  Number of electrons
23

(v) EM of an oxidising agent

Methods of determination of molecular mass
Following methods are used to determine molecular mass,
(i) Diffusion method (For gases) : The ratio of rates of diffusion of
two gases is inversely proportional to the square root of their molecular
masses
...
N
...

Methods of determination of equivalent mass

M2
M1

(ii) Vapour density method (For gases only) : Mass of a fixed volume
of the vapour is compared with the mass of the same volume of hydrogen
under same conditions
...


(i) Hydrogen displacement method : The mass of metal which
displaces 11200 ml of hydrogen at NTP from an acid, alkali or alcohol is the
equivalent mass of the metal
...

22400 ml of vapours of a substance = Molecular mass of that
substance
(iv) Colligative property method (For non-volatile solids)
Discussed in colligative properties of solutions
...
008 g
 1
...
(ml) of H 2 displaced at STP

This method is useful for metals which can displace hydrogen from
acids or can combine with hydrogen (Mg, Zn, Na, Ca etc
...


Mass of metal
8
Mass of oxygen

(a) Equivalent mass of metal =
(b) Equivalent mass of metal
=

A1 , A2 , A3
...
etc
...
etc
...

(3) Equivalent mass : The number of parts by mass of a substance
that combines with or displaces 1
...
0
parts of oxygen or 35
...

On the other hand quantity of a substance in grams numerically equal to its

equivalent mass is called its gram equivalent mass (GEM) or gram
equivalent
...
5 gm
...

(a) Equivalent mass of metal 

(i) EM of an element 
(ii) EM of an acid 

Atomic mass
Valency

Molecular mass
Basicity

(Basicity of acid is the number of replaceable hydrogen atoms in one
molecule of the acid)
...


Mass of metal
 35
...
of O 2 at S
...
P
...
of Cl 2 (in ml
...


Equivalent mass of acid (or base) 

W
VN

Where ,

W = Mass of acid or base in gm
...
of base or acid in litre required for neutralisation
N is Normality of base or acid
(v) Metal displacement method : It is based on the fact that one gm
...

Mass of metal added
Eq
...
mass of metal displaced

Element

W1
E
 1
W2
E2

% Relative no
...
wt
...

Gram equivalent mass = Electrochemical equivalent  96500
The ratio of masses of two metals deposited by the same
quantity of electricity will be in the ratio of their equivalent masses
...
of atoms : Divide the percentage of each element
present in compound by its at
...
This gives the relative no
...


W1
E
 1
W2
E2

It the simplest ratio obtained are not complete integers, multiply
them by a common factor to get integer values of simplest ratio
...
of atoms derived in simplest ratio
...


Mass of compound AB Eq
...
mass of B

Mass of compound AD Eq
...
mass of D

Molecular formula : Molecular formula  n  empirical formula
where ' n' is the whole no
...
of atoms and
divide each value of relative no
...


Mass of salt taken (W1 )
Eq
...
obtained (W2 ) Eq
...
(E 2 )

(viii) Conversion method : When one compound of a metal is
converted to another compound of the same metal, then

Mass of compound I (W1 ) E  Eq
...
mass of radical II
(E = Eq
...

That means quantitative calculations of chemical composition and reaction
are referred to as stoichiometry
...


2  V
...
of Chloride
 35
...


2  V
...
of Chloride
2  V
...


Eq
...
5

Equivalent Mass of acid =

molecular weight of compound
empirical formula weight of compound

contains

a

fixed

number

(6
...
Thus it is
23

correct to refer to a mole of helium, a mole of electrons, or a mole of Na  ,
meaning respectively Avogadro’s number of atoms, electrons or ions
...
If the molecular mass of a compound
is M and B is the mass of an element in the molecule, then

Mass of element
X
Percentage of element 
 100 
 100
Molecular mass
M
(2) Determination of empirical formula : The empirical formula of a
molecule is determined using the % of elements present in it
...


There is no borderline, which can distinguish the set of laws
applicable to gravimetric and volumetric analysis
...
e
...

But in actual practise, the problems on gravimetric involves simpler
reactions, thus mole concept is convenient to apply while volumetric
reactions being complex and unknown (unknown simple means that it is
not known to you, as it’s not possible for you to remember all possible
reactions), equivalent concept is easier to apply as it does not require the
knowledge of balanced equation
...

Problems Involving Mass-Mass Relationship
Proceed for solving such problems according to the following
instructions,
(i) Write down the balanced equation to represent the chemical
change
...
Also write the relative weights of the reactants and products
(calculated from the respective molecular formula), below the respective
formula
...

Problems Involving Mass-Volume Relationship
For solving problems involving mass-volume relationship, proceed
according to the following instructions,
(i) Write down the relevant balanced chemical equations (s)
...

(iii) Gases are usually expressed in terms of volumes
...
T
...
), convert it into N
...
P
...


(iii) Iodiometric titrations : This is a simple titration involving free
iodine
...
Let the volume of sodium
thiosulphate is V ml
...
wt
...


Equivalents of I 2  Equivalent of Na 2 S 2 O 3

 Equivalents of I 2  N  V  10 3

Calculate the unknown factor by unitary method
...

2



(i) Write down the relevant balanced chemical equation
...
4 litres at N
...
P
...


(iv) Iodometric titrations : This is an indirect method of estimation
of iodine
...
The
oxidising agent oxidises I  to I 2
...


Oxidising Agent

Take the help of Avogadro’s hypothesis “Equal volume of different

2 Na 2 S 2 O3
( A)  KI  I 2 
 2 NaI  Na 2 S 4 O 6

gases under the similar conditions of temperature and pressure contain the
same number of molecules”
...
For the analysis a standard solution is required
...
)

Let the normality of Na 2 S 2 O 3 solution is N and the volume of
thiosulphate consumed to V ml
...
Different types of titrations
are possible which are summerised as follows,

Moles of I 2 liberated from KI 

N  V  10 3
2

 N  V  10 3

 254  g
...

Examples:

N  V  10 3
2

(v) Precipitation titrations : To determine the anions like


CN , AsO33  , PO43  , X  etc, by precipitating with AgNO3 provides

K 2 Cr2 O7  4 H 2 SO 4  K 2 SO 4  Cr2 (SO 4 )3  4 H 2 O  3[O]
[2 FeSO 4  H 2 SO 4  O  Fe2 (SO 4 )3  H 2 O]  3
6 FeSO 4  K 2 Cr2 O7  7 H 2 SO 4  3 Fe(SO 4 )3  K 2 SO 4  Cr2 (SO 4 )3 7 H 2 O
2 KMnO4  3 H 2 SO 4  K 2 SO 4  2 MnSO 4  3 H 2 O  5[O]
[2 FeSO 4  H 2 SO 4  O  Fe2 (SO 4 )3  H 2 O]  5
10 FeSO 4  2 KMnO4  8 H 2 SO 4  5 Fe2 (SO 4 )3  K 2 SO 4  2 MnSO 4  8 H 2 O

Similarly with H 2 C 2 O4
2 KMnO4  3 H 2 SO 4  5 H 2 C2 O4 
K 2 SO 4  2 MnSO 4  8 H 2 O  10CO 2 etc
...

Example: NaOH  HCl  NaCl  H 2 O
and NaOH  CH 3 COOH  CH 3 COONa  H 2 O etc
...


NaCl  AgNO3  AgCl   NaNO 3
KSCN  AgNO3  AgSCN   KNO 3
End point and equivalence point : The point at which
titration is stopped is known as end point, while the point at which
the acid and base ( or oxidising and reducing agents ) have been
added in equivalent quantities is known as equivalence point
...

Normal solution : A solution containing one gram equivalent
weight of the solute dissolved per litre is called a normal solution ;
e
...
when 40 g of NaOH are present in one litre of NaOH solution,
the solution is known as normal ( N ) solution of NaOH
...
For
example, a solution of NaOH containing 20 g (1/2 of g eq
...
) of
NaOH dissolved per litre is a sub-normal solution
...
5 N solution
...
wt
...
wt
...
Some of these excess substances will therefore be left
over when the reaction is complete; the reaction stops immediately as soon
as one of the reactant is totally consumed
...
The
other reactant present in excess are called as excess reagents
...

2

(4) Molarity

Moles of solute
Volume in litre

(5) Number of moles 

Wt
...
wt
...
4

2 H 2 (g)  O2 (g)  2 H 2O (v)
Moles before reaction

10

7

0

Moles after reaction

0

2

10

The reaction stops only after consumption of 5 moles of O as no
further amount of H is left to react with unreacted O
...

2

2

(6) Number of millimoles 

Wt
...
wt
...


(7) Number of equivalents



Wt
...
of moles  Normality  Volume in litre
Eq
...


(8) Number of milliequivalents (meq
...
in gm  1000
 normality  Volume in ml
...
wt
...
of millimoles
 x  Molarity 

where x 

Strength in gm litre1
Eq
...


Mol
...

, x = valency or change in oxi
...

Eq
...


(10) Normality formula, N1 V1  N 2 V2

 Law of conservation of mass does not hold good for nuclear
reactions
...
g
...

0
...
g
...

2

2

 The term atom was introduced by Ostwald and the term molecule
was introduced by Avogadro
...

 The number of atoms present in a molecule of a gaseous element is
called Atomicity
...
of solvent
 100
(11) % by weight 
Wt
...
of solvent
 100
(12) % by volume 
Vol
...
Instead,

have no units
...
of solvent
 100
(13) % by strength 
Vol
...

4

2

4

 Minimum molecular mass of a macromolecular substance can be


(15) Formality 

(because density of H O = 1 g/cc)
2

(14) Specific gravity

2

Wt
...
of 1 ml
...
of solution

Wt
...
of solute Vin l

(16) Mol
...
= V
...

Minimum molecular mass is obtained when it is supposed that one
molecule of the macromolecule contains only one atom or molecule
of the minor component

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