CLASS-
11th
SOME BASIC
CONCEPTS OF
CHEMISTRY
Designed By -
Bharat Panchal Sir
Do% Bharat Panchal -
Chemistry Guruji 2.0

•
Classification of matter.
Properties of matter
•
Units for measurement •
scientific Notation
•
significant figures
•
Precision & Accuracy
•
laws of chemical combination
•
law of conservation of mass •
law of constant comp.
• law of multiple proposition •
Gay-Lussac law
•
Mole concept
•
Avogadro law .
•
Percentage composition
•
Empirical and Molecular formula
•
Expressing concentration of solution
•
stoichiometry
matter
.
-
BOB Bharat Panchal -
Chemistry Guruji 2.0
Anything which has mass and occupies space is#
alter.
Chemistry
It is the branch of science which deals with the study of
substances ,
their properties,
structure and their
transformation.
at
organic chemistry b) Physical chemistry c) Inorganic chemistry
it Pure substance It is a substance in
which all the constituent
particles are
of same nature
ii ) Element It consist only one teebe of
-
atoms
e.g
Na ,
ca,
Ag etc
iin (0M¥
It consist of two or more
atoms of different element combined in a definite ratio by mass
e.g Hao, Mtg,
CO
,
etc.
ith
Mixture :
It consist
of two or more substance mixed in
any ratio

v1 Homogeneousmixture The composition is uniform throughout C single
phaseleg sugar in water .
"it HÉmI¥ :
such type of mixtures in which composition
is not uniform throughout e.g sand in water
•
Physical Classification Of Matter Interconversion of states
of Matter
Gao Bharat Panchal -
Chemistry Guruji 2.0
Physical Properties
which can be measured without changing the chemical
composition of the substance is known as physical property e.g mass, volume,
density
chemical Properties property which require a chemical change to
occur is known as chemical property e.
g acidity
or basicity , combustibility.
Difference between compound $ Mixture
Mads MH
ed volume
qeÑgjgs
mass ✗
acceleration
pisH¥me
lxb

Scientific Notation :
"
Proper Representation of number in
exponential form
"
N ✗ 10h
← any
=
integer
C)a non zero
number to the
left of decimal
e.
g
126.40 1.2640×102
0-00043 4.3×10-4
Significant figures :
"
Au certain digits in a measurement
plus one
doubtful digit called significant figure
"
e.g
12.4 cm where 12 is certain $44s doubtful
Rules 1-1 All non -
zero
digits are
significant.
e.g 394 → 3 Sif
2.) Zeros between the non -
zero
digits are significant.
e.g 5005 →
Asf
Bo% Bharat Panchal -
Chemistry Guruji 2.0
Iii) Preceding zero or zeros to the left of non zero are no
ns.re.g
0.0045 → 2 5. F
in zeros in exponential form are not
significant figure.
e.g 4×102 → 1 Sif
2. 5×103 → 2 Sir
4) Zero at the end or
sight of a number are
significant
while decimal is present.
e.g 0.200 → 3 5. F
Yi ) Terminal zeros are not Sif if there is no decimal point
e.g 126000 → 3 Sif
Xiii Exact numbers have infinite
s.FI
books →
infinite Sif.

Prefixes used in Numerical # Precision and Accuracy
Precision
Prefix Symbol Multiple
deci d to
-1
It refers to the closeness of
anti c to
-
a
various measurements too the
Milli m lo
-3 same quantity
micro he 10-6
Accuracy It is the
agreement of particular value to
nano h 10-9
pico p so
-
in the true value of result "
when the values
of
deca da lo
different measurements are close to the
hecto h 102
true or accurate value.
kilo K 103
106
me
"
m
µ
. Go☒ Bharat Panchal -
Chemistry Guruji 2.0
Angstrom A.
e.
g
True value is
2.00g and three students
I II
student 'A
'
1.95 1.93 Precise but not accurate
student
'
B
'
1.93 2.06 Neither Precise nor accelerate
student
'
C'
/2.01/1.99 both Precise and accurate
Laws Of Chemical Combination :
LAW OF CONSERVATION Of mass :
v0 Antonie Lavoisier
"
matter can neither be created
nor destroyed
"
In other words ,
total mass of the reactant equal to the total
mass of product
"
limitation It is not applicable to the nuclear reactions because
during a reaction mass is not conserved .
Q.
when 4.2g of Matteo}
is added to a solution of acetic acid (CH
,
COOH)
weighing log it is observed that
2.2g of CO2 is released into the
atmosphere . The residue left behind is
found to weigh 12g . Show
that these observations are in
aggrement with law of conservation
of Mab.
Matteo,
+
CHZCOOH -7
CHZCOONA -11%0+10
,
e.g 4.
2g 10g kg 2.2g

Total mass of reactant =
14.2g
Total mass
of product =
14.2g
These observations are in agreement with law of conservation of mass .
Law of Definite Proportion :
v0 BY Joseph Proust
"
A given compound always contains exactly
the same proportion of elements
by mass and it does not depend on
source.
CO2 can be obtained from different sources but the ratio of
C $0 remains same
c :O
2 : 32
3 : 8
Limitations
;) The law is not always true as element may combine
in the same Matto but compound formed may be different .
e.
g
C : H :O = 12 :3 :S by mass
may form two compound ↳ Hs OH $ CH
]
OCH }
both having same molecular formula ↳ Hoo,
ii) It is not applicable when isotopes of an element are involved
in the formation of compound.
eg In CO2 ,
c :O C : 0
carbon has two 12:32 14 : 32
isotopes
3.) Law Of Multiple Proportion "
When two element can combine
to form more than one compound ,
the different masses of one element that combine other element
bear a simple ratio to one another
oxide Mass of G Mass of 0
CO 12 16
(02 12 32
ratio of masses of oxygen -16 :32-
1 :&
N -
Phosphorous and chlorine form two compounds the first contains 22.54
by mass of phosphorous and the second 14.88%
of phosphorous .
Show
that these data are consistent with law of multiple proportions
compound -
I mass of D=
22.54g
Mass of A =
77.46g
22.54g of P combine with =
77.46g a
19 - =
72-2%-4 =
3.4384

Compound-
II
mass of P =
14.88g
Man of Cl =
85.12g
14.88g of P combines with =
85.12
gu
1g of p combines with =
8Yj.gg#--5.72gofCl
ratio of different masses of a
3. 43 =
5,7¥
1:16
3 :S
The given data illustrates law of multiple
propositions
•
Gay Lussac 's Law of combining volumes
Acc. to this law
"
at
given temperature and pressure the volumes of all
gaseous
reactants and products bear a simple whole number ratio to
each other.
e.
g Ha + Uz →
2116cg ) ratio of
(g) (g)
11101.
I v01.
21101 . v01 .
of gases
1 : I :L
•
Avogadro's law "
* states that equal volume of gases contain
equal .
Number
'
of molecules
at the same temperature and
pressure
ra Dalton 's Atomic Theory :
John Dalton proposed atomic theory
of matter .
The main points of Dalton Atomic
theory are
e) Matter is made up of extremely small ,
indivisible particles called
atoms .
•
1 Atom of same element are identical in all respects i.e
they posses
same size, shape, mass etc .
e) Atoms of different element are different in all respect
• ' Atom is the smallest particle that take part in reaction
•
7 Atoms can neither be created nor destroyed i. e atoms are indestructible
Mole Concept
One Mole is equal to Avogadro Number (Na = 6-022×1023 particles)
"
A mole is defined as the amount of substance that contains
as
many particles as these are carbon atoms in
exactly
12g of c
'
?

g. Define 1 am -
u C atomic mass unit)
AI Hz the mass of one atom of C- 12 isotopes
*
I a. m.cl = 1.66 ✗ to -24g
"
According to IUPAC it is now written as
'
u
'
Lama is also called 1 Dalton
*
Gram Atomic Mass : Atomic mass of an element in grams
= = =
called gram atomic mass
Man of 1 atom in g
= Gram atomic mass
NA .
# Molecular Mass : It can be defined as the mass
of one molecule
of the compound relative to the c-atom taken
as
exactly.
# Gram Molecular Mass : Molecular mass
of a compound in
grams .
called gram molecular mass
"
6ham molecular mass of any compound is the mass of
G. 022 ✗ 1023 molecules
of the compound.
"
Man of 1 Mole
E-
is
n=W_m =
^¥ =
¥-4b
9. Find total no -
of moles in
18g water
Sol. n =
? w= 18 M= 18
9) h =
% =
187g = I Mot.
Percentage composition
•
To determine %
age of each element in compound
"
Man %
of an element =
Man of that element ✗ too
Molar man of compound
G. Determine the percentage composition of ferric sulphate redcoats
CAT.
Man of Fe = 56 , 5=32 , 0=16 )
8¥ Moi .
Mass of Fey (Soa) }
= 400 he
i.
of Fe =
III.•
✗ too =
28%
%
of 5 =
96-4. •
✗ too = 24%
%
90 =
1%-5×1- •
=
48%

Molecular and Empirical formula
* E¥
Its the formula which express
* M¥
formula which represents the actual
the smallest whole no .
ratio of no .
of each atom in
any molecule
elements
→ Glucose → ↳ His 06 ( Mrf ) $ CHAO CE -
f)
→ Benzene → ↳ Hf ( Mrf) $ CH C E. f)
9. Determine the empirical formula and Molecular formula of naphthalene
has following % carbon =
93.71% ,
hydrogen =
6.29% $ Mol. Man is 128
whole no .
MY
elements T
.at man hcnocfomoled simplest
ratio ratio
c 93.71 12 93^-412=7.8 76.8-29 =
1.24 1.24×4--5
H G. 29 I
6-2911 =
629 6
?÷g :| 1×4--4
a) empirical formula =
G- Hq
b) Mol .
Formula = n C Emp. formula )
n=m÷?ñ→
Do% Bharat Panchal -
Chemistry Guruji 2.0
•
Expressing concentration of solution :
i , mass percentage It is defined as the mass
of one component
present in
100g of solution
Massy.
of component = Mass of component ✗ too
Man of solution
9 Determine the mass percentage of the
22g of CHU
] $
122g CCI 4 .
ii) volume percentage as It is defined as the volume of one
component present in in parts of solution
V01 . %
of component : V01 .
of component
Volume of Solan
✗ 100
Iii) PPMC Parts per million ) It is defined as the parts of
solute present in per million
parts of solution
ppm =
number of parts of component ✗ 106
number of parts of solution

•
Malefactions: -
It is the ratio of number of moles
of
one component to the total number of
moles of all component .
XA =NA_
hath☐
1
"B
=n?n%-n,
RA= Mole fraction of solvent
§
ha -
- no -
of moles of solvent
KB : mole fraction of solute hrs = no
of moles of solute
mole fraction of solution = 1 RATKB =L
•
)
Molarity (M) It is the number of moles
of solute
present in per litre of solution called
molarity. It is denoted by 'M !
Molarity =
no -
of moles
of Solute
v01 .
of solution in like
Mef¥
case lil
M=wmB_☐✗,÷y or
M=%☐×Y÷↳
case lie ) when two solutions are mixed but with
same no .
of moles
MY =
Mdk
case ciiil when two solutions with different cone-
are mixed , then the
molarity of resulting
solution can be calculated as .
Milli -1 Milk =
My (4+11)
Q How
many moles and how many gram of sodium
chloride C NaCl ) are present in 250 MI
of a 0.50M
Nacl solution ?
AE n= ? WB =
? 4--250 MI ,
M -
-0.50M
n=
WI
,
{
M
:#☐
✗ '
÷÷,
h =
7-3158.5=011-5 0.50 =
wgg-s.tl;÷
Wrs =
-73g

9 Calculate molarity of a solution of ethanol in water
in which mole fraction of ethanol is 0-040 .
Sdi KB = 0.040
nA ( no -
of moles wa
of water )=_ma :
'
÷
✗B =
MBNA
+ hrs =
55.55
0.040 = hB_ Mol
55.55 + hrs
9 NB = 2.31 MOI
molarity of solution =
¥4m
- I = 2.31 Moll-1
Hit A solution is 25% water, 25% ethanol $50 % acetic
acid by mass .
Calculate the mole fraction of each
component.
Hid what is the molarity of the resulting solution
obtained by mixing 2.5L of 0.5M Urea solution
and 500 mL
of 2M uoea solution . ?
•
Molality ( m) It is the number of moles
of solute
dissolved in
1kg of solvent.
Its unit is moi kg
"
or molal or m
molality ( m) =
wmB_☐✗%¥g)
Q And the molality of a 15% Solution of Hasan
C density of Hasoq solution = 1.020 g c. m
-
3)
Sofa mass
of solute ( WB) =
15g
Molar mass of 112504 ( Mg) =
98g Mot
'
molality (m ) =
?
Mass
of solvent( WA) =
85g
m=wm÷×"w%-g, ¥g✗'%É-
= 1.8 Mold.

Q Out of molarity and
molality which one is unaffected
with increase in temperature ?
Aus Molarity decreases with the increase in temp.
because it involves volume . On the other hand .
Molality remains unaffected , because it involves
man of solvent only.
# STOICHIOMETRY #
"
stoichiometry means to measure an element
"
when a balanced chemical reaction is written .
It
gives quantitative relationship b/w various reactants
and products in terms of mass
,
moles ,
molecules
and volumes e.g
Caco
} -12 HCl 7 Call,
-1 CO2 + Hao
1m01 2m01 1 Mol 1m01 Im-1
100g 73g 111g 44g 188 Coefficient
* coefficient of a balanced chemical eqn are called stoichiometric -
Q If 20g of Caco
,
is treated with
20g HU .
How
many grams
of CO2 will be produced ?
Sots Caco
,
+ 2116 > call,
+ CO2 -1110
loog 73g 44g
100g Caco,
produce =
-44g CO2
1g Caco,
produce =
9% g coz
20g Caco
,
produce =
4,4-0×20=8-8g coz
9 Calculate the mass
of graphite that must be
burnt to produce 13.2g of CO2.
80¥ C +02 →
cog
12g 44g
-44g CO2 produced on combustion of =
12g goop
1g =
12144 "
"
13^2
=
1-4×13.2 =3.bg

#
Limiting Reagent # some reactants are totally
consumed called limiting
seeagent . Reaction stops when it is consumed .
* Excess Reagent# The reactant which is not consumed
completely in the reaction called
excess reagent.
Q Assume that you have 139m01
of Nz and 3. 44m01.
of the
e) How many moles of NH
, can you make
it) How
many grams of which reactant will be left over
?
Aye Ms -13112 > 2MHz
1m01 3m01 2m01
it 1m01 of N,
reacts with = 3 Mol do Ha
" 39 Mol
of Ng reacts with = 3×139--4.17 mad of Ha
But we have 3.44 moi of Ha .
So this limiting
it)•) 3m01 Ha produce =
2 Mol of NH
,
Reagent .
1m01 of H2 produce =
f- Mol
of MHz
3.44 MOI of Ha Produce '
=
2-3×3.44 =
2.29 Mol of Mtf
•
) ] Mot tf react with =L Mol of N
,
I Mot Hz react with =
tgrnol of Nff
3. 44 MOI - =
tg ✗3.44 = 1.15 Molq Nz
no -
of Moles
of Nc left
= 1. zq -1.15=0.24 Mot
dona
h =
Im g
w = NXM
= 0.24×28
=
6.72g of Nc left
Thank
yoke