some basics concept of chemistry.pdf

Some basic concepts of chemistry class 11 jee mains and neet

1. CLASS
-
11thSOMEBASIC
CONCEPTSOF
CHEMISTRY
DesignedBy
-
BharatPanchalSir
Do%BharatPanchal
-
ChemistryGuruji
2.0

2. •
Classification
ofmatter
.
Properties
ofmatter
•
Unitsformeasurement
•
scientificNotation
•
significantfigures
•
Precision&Accuracy
•
lawsof
chemicalcombination
•
lawofconservationofmass
•
law
of
constantcomp
.
•
lawofmultipleproposition
•
Gay-Lussaclaw
•
Moleconcept
•
Avogadro
law
.
•
Percentagecomposition
•
Empirical
andMolecularformula
•
Expressing
concentrationofsolution
•
stoichiometry
matter
.
-
BOBBharatPanchal
-
ChemistryGuruji
2.0
Anythingwhichhas
massandoccupiesspaceis#
alter
.Chemistry
It
is
thebranchofsciencewhich
dealswith
the
study
of
substances
,
theirproperties
,
structureandtheir
transformation
.
atorganicchemistry
b)Physicalchemistry
c)
Inorganicchemistry
itPuresubstance
<z
It
is
a
substance
in
whichalltheconstituent
particles
areofsamenature
ii)
Element
<z
Itconsist
only
one
teebeof
-
atomse.gNa
,
ca
,
Ag
etc
iin(
0M¥
<z
It
consistoftwo
ormore
atomsofdifferent
element
combinedin
a
definiteratioby
mass
e.gHao
,
Mtg
,
CO
,
etc
.
ithMixture
:
Itconsistoftwo
ormore
substancemixedinanyratio

3. v1
Homogeneousmixture
The
compositionisuniformthroughout
C
single
phaselegsugarin
water
.
"
it
HÉmI¥
:
suchtypeofmixtures
in
whichcomposition
is
not
uniformthroughoute.gsandinwater
•
PhysicalClassificationOfMatterInterconversionofstates
ofMatter
GaoBharatPanchal
-
ChemistryGuruji
2.0
PhysicalProperties
which
canbemeasuredwithout
changing
thechemical
compositionofthesubstanceisknown
as
physicalproperty
e.g
mass
,
volume
,
density
chemicalPropertiesproperty
which
require
a
chemicalchange
to
occur
is
known
as
chemicalproperty
e.
gacidity
or
basicity
,
combustibility
.
Differencebetweencompound$Mixture
Mads
MH
ed
volumeqeÑgjgs
mass✗
accelerationpisH¥me
lxb

4. Scientific
Notation
:
"
ProperRepresentationofnumber
in
exponentialform
"
N✗
10h
←
any
=
integer
C)
anonzero
numbertothe
leftofdecimal
e.
g126.40
<z
1.2640×102
0-00043
<z
4.3×10-4
Significantfigures
:
"
Aucertain
digitsin
a
measurement
plus
one
doubtfuldigit
calledsignificantfigure
"e.g
12.4
cm where12
iscertain
$44sdoubtful
Rules
1-1
All
non
-
zero
digits
are
significant
.
e.g394
→
3Sif
2.)
Zerosbetweenthe
non
-
zero
digits
are
significant
.e.g5005
→
Asf
Bo%BharatPanchal
-
ChemistryGuruji
2.0
Iii)Preceding
zeroorzeros
tothe
left
of
nonzeroareno
ns.re.g
0.0045
→
25.F
in
zeros
inexponentialform
are
not
significantfigure
.
e.g4×102
→1Sif
2.5×103
→ 2Sir
4)Zeroatthe
end
or
sightof
a
number
are
significant
while
decimal
ispresent
.
e.g0.200
→
35.F
Yi)
Terminal
zerosare
notSif
if
there
is
no
decimal
point
e.g126000
→3Sif
Xiii
Exactnumbershave
infinite
s.FI
books
→
infinite
Sif
.

5. PrefixesusedinNumerical
#
Precision
and
Accuracy
Precision
Prefix
SymbolMultiple
decid
to
-1
It
refers
totheclosenessof
anti
c to
-a
various
measurementstoothe
Milli
m lo-3
same
quantity
micro
he
10-6
Accuracy
It
is
the
agreementofparticularvalue
to
nanoh
10-9
pico
p so
-in
thetruevalueofresult
"
whenthevaluesof
decada
lo
different
measurements
areclosetothe
hectoh
102
true
or
accuratevalue
.
kilo
K
103
106
me"
m
µ
.
Go☒BharatPanchal
-
ChemistryGuruji
2.0
AngstromA.
e.
g
Truevalue
is2.00g
andthreestudents
III
student
'A
'
1.951.93
Precise
butnotaccurate
student
'
B
'
1.93
2.06
NeitherPrecise
nor
accelerate
student
'
C'
/
2.01/1.99
bothPreciseandaccurateLaws
OfChemicalCombination
:
LAWOFCONSERVATIONOfmass
:
v0AntonieLavoisier
"
mattercan
neither
becreated
nor
destroyed
"Inotherwords
,
total
mass
ofthereactantequal
tothetotal
mass
ofproduct
"
limitation
It
is
not
applicabletothenuclearreactionsbecauseduring
a
reactionmass
is
notconserved
.Q
.
when4.2g
ofMatteo
}
isadded
toa
solutionofaceticacid
(
CH
,
COOH
)
weighinglogit
is
observedthat
2.2g
ofCO2
is
releasedintothe
atmosphere
.
Theresidueleft
behindisfoundtoweigh12g
.
Show
thattheseobservations
are
in
aggrement
withlawofconservation
ofMab
.
Matteo
,
+
CHZCOOH
-7
CHZCOONA-11%0+10
,
e.g
4.2g 10g kg
2.2g

6. Total
mass
ofreactant
=
14.2g
Total
mass
ofproduct
=
14.2g
Theseobservations
are
inagreementwithlaw
of
conservationof
mass
.
LawofDefiniteProportion
:
v0BYJosephProust
"
A
givencompoundalwayscontainsexactly
thesameproportionof
elements
by
mass
anditdoesnot
depend
on
source
.
CO2can
beobtainedfromdifferent
sourcesbuttheratioof
C$0remains
same
c:O
2
:
32
3
:
8Limitations
;)
Thelawis
notalways
true
as
elementmay
combine
in
the
sameMattobutcompoundformed
maybe
different
.e.gC:H:O
=
12:3:S
by
mass
mayformtwo
compound
"¥
Hs
OH$
CH
]
OCH
}
both
having
same
molecularformula
"¥
Hoo
,
ii)It
is
not
applicable
whenisotopesof
an
element
are
involved
inthe
formationofcompound
.
eg
InCO2
,
c:O
C
:
0
carbonhastwo 12:32 14
:
32
isotopes
3.)
LawOfMultipleProportion
"
Whentwoelement
can
combine
to
form
more
than
one
compound
,
thedifferent
massesof
oneelementthatcombine
otherelement
bear
a
simple
ratioto
one
another
oxideMassof
GMassof0
CO 12 16
(02
12
32
ratioof
massesof
oxygen-16:32-
1
:&
N
-
Phosphorousand
chlorine
formtwocompounds
thefirstcontains22.54
by
mass
ofphosphorousandthesecond
14.88%ofphosphorous
.
Show
thatthesedataare
consistentwithlawofmultiple
proportions
compound
-
I mass
ofD=
22.54g
MassofA
=
77.46g
22.54g
ofPcombinewith
=
77.46g
a
19
- =
72-2%-4
=
3.4384

7. Compound
-
II
massofP
=
14.88g
ManofCl
=
85.12g
14.88gofP
combines
with
=
85.12
gu
1g
ofpcombineswith
=
8Yj.gg#--5.72gofCl
ratioofdifferent
masses
ofa
3.43
<z
=
5,7¥
<z
1:16
<z
3:S
Thegiven
dataillustrateslawofmultiple
propositions
•
Gay
Lussac
's
Lawofcombiningvolumes
Acc
.
to
thislaw
"at
giventemperature
andpressurethevolumesofall
gaseous
reactantsandproducts
beara
simple
wholenumberratio
to
eachother
.
e.
gHa
+
Uz
→
2116cg
)ratio
of
(g)(g)
11101
.
Iv01
.
21101
.
v01
.
ofgases
<z
1
:I
:L
•
Avogadro
's
law
<z
"
*statesthatequal
volumeofgasescontainequal
.
Number
'
of
molecules
at
thesame
temperatureand
pressure
ra
Dalton
's
AtomicTheory
:
JohnDaltonproposedatomictheory
ofmatter
.
Themainpoints
of
Dalton
Atomictheory
are
e)Matterismadeupofextremelysmall
,
indivisible
particlescalled
atoms
.
•
1
Atomofsameelement
are
identicalin
allrespects
i.
e
they
posses
samesize
,
shape
,
mass
etc
.e)
Atomsof
differentelement
are
differentin
allrespect
•'
Atom
is
thesmallestparticlethattakepart
in
reaction
•
7
Atoms
can
neitherbecreated
nor
destroyed
i.
e
atoms
are
indestructible
MoleConcept
OneMole
isequal
to
AvogadroNumber
(
Na
=
6-022×1023particles)
"
Amole
isdefined
as
theamountofsubstancethat
contains
as
manyparticles
as
these
arecarbonatomsinexactly
12gof
c
'
?

8. g.Define
1am
-
u
Catomic
mass
unit)
AI
Hzthe
mass
of
one
atomof
C-12
isotopes
*
Ia.m.cl
=
1.66✗to
-24g
"
According
toIUPACitis
nowwritten
as
'
u
'Lamaisalsocalled1
Dalton
*
GramAtomicMass
:
Atomic
mass
of
an
element
ingrams
= =
=
calledgramatomic
mass
Manof1atom
in
g
=
Gram
atomic
mass
NA
.
#MolecularMass
:
It
canbe
defined
as
the
mass
of
one
molecule
ofthecompoundrelativetothec-
atomtaken
as
exactly
.#
GramMolecularMass
:
Molecular
massof
a
compoundin
grams
.
calledgrammolecular
mass
"
6hammolecular
mass
ofany
compoundis
themass
of
G.022
✗
1023
moleculesof
the
compound
.
"Manof1Mole
E- isn=W_m
=
^¥
=
¥-4b
9.Findtotal
no
-
ofmolesin18gwater
Sol
.
n
=
?w=
18M=18
9)h
=
%
=
187g
=
IMot
.
Percentagecomposition
•
Todetermine
%
ageofeachelementincompound
"
Man%
of
an
element
=
Manofthatelement
✗too
Molarmanofcompound
G.Determine
thepercentagecompositionofferricsulphateredcoats
CAT
.
Man
ofFe
=
56
,
5=32
,
0=16)
8¥
Moi
.
MassofFey(Soa)
}
=
400
he
i.
of
Fe
=
III.
•
✗too
=
28%
%of
5
=
96-4
.•
✗
too
=
24%
%
90
=
1%-5×1
-•
=
48%

9. Molecular
and
Empiricalformula
*
E¥
Itstheformula
whichexpress
*
M¥
formulawhichrepresents
theactual
thesmallestwhole
no
.
ratio
of
no
.
ofeachatominanymolecule
elements
→
Glucose
→
"¥
His06(
Mrf
)$CHAOCE
-
f)
→
Benzene
→
"¥
Hf
(Mrf)$
CHCE.
f)
9.Determinethe
empiricalformula
andMolecularformulaofnaphthalene
hasfollowing
%carbon
=
93.71%
,
hydrogen
=
6.29%
$
Mol
.
Manis128
whole
no
.
MYelementsT.at
man
hcnocfomoledsimplest
ratioratio
c
93.7112
93^-412=7.8
76.8-29
=
1.241.24×4--5
H G.29I
6-2911
=
629
6
?÷g
:|
1×4--4
a)empiricalformula
=
G-Hq
b)Mol
.
Formula
=
nCEmp
.
formula)
n=m÷?ñ→
Do%BharatPanchal
-
ChemistryGuruji
2.0
•
Expressingconcentrationofsolution
:
i,
masspercentage
<z
It
isdefined
as
themass
of
one
component
presentin
100gofsolution
Massy
.
ofcomponent
=
Mass
of
component
✗too
Man
of
solution
9
Determine
the
mass
percentageofthe
22g
of
CHU
]
$
122g
CCI4
.ii)volumepercentage
as
It
isdefined
as
thevolumeof
onecomponentpresent
in
in
partsofsolution
V01
.
%of
component
:
V01
.
ofcomponent
VolumeofSolan
✗100
Iii)
PPMCPartsper
million)
<z
It
is
defined
as
thepartsof
solutepresentin
permillion
partsofsolution
ppm
=
number
of
partsof
component
✗
106
numberofpartsofsolution

10. •
Malefactions
:
-
It
is
theratioofnumberofmoles
of
one
componenttothe
totalnumber
of
molesofallcomponent
.
XA=NA_
hath
☐
1
"B
=n?n%-n,
RA=
Mole
fraction
ofsolvent§ha
-
-
no
-
ofmolesofsolvent
KB
:
mole
fractionofsolutehrs
=
noofmolesofsolute
mole
fraction
of
solution
=
1
RATKB
=L
•
)
Molarity(M)
It
is
thenumberof
molesof
solute
presentinperlitreofsolutioncalled
molarity
.
It
isdenotedby'M!
Molarity
=
no
-
of
molesofSolute
v01
.
ofsolutioninlike
Mef¥
case
lil
M=wmB_☐✗,÷y
or
M=%☐×Y÷"¥
caselie)whentwosolutions
are
mixedbutwith
sameno
.
ofmoles
MY
=
Mdk
case
ciiil
whentwosolutionswithdifferent
cone
-
are
mixed
,
then
the
molarityof
resulting
solution
canbe
calculated
as
.
Milli
-1
Milk
=
My
(4+11)
Q
Howmany
molesand
howmanygramofsodium
chloride
CNaCl)
are
presentin
250MIof
a
0.50M
Naclsolution
?
AE
n=
?
WB
=
?
4--250MI
,
M
-
-0.50M
n=
WI
,
{
M
:#
☐
✗
'
÷÷
,
h
=
7-3158.5=011-5
0.50
=
wgg-s.tl;÷
Wrs
=
-73g

11. 9Calculatemolarityof
a
solutionofethanolin
water
in
whichmolefraction
ofethanolis
0-040
.
Sdi
KB
=
0.040
nA(
no
-
ofmoles
wa
ofwater
)=_ma
:
'
÷
✗
B
=
MBNA
+
hrs
=
55.55
0.040
=
hB_
Mol
55.55+
hrs
9
NB
=
2.31MOI
molarity
ofsolution
=
¥4m
-
I
=
2.31Moll
-1
Hit
Asolutionis
25%water
,
25%ethanol$50
%
acetic
acid
by
mass
.
Calculatethemole
fractionofeach
component
.
Hid
whatis
the
molarity
ofthe
resulting
solution
obtained
by
mixing
2.5Lof0.5M
Urea
solution
and
500mLof2Muoea
solution
.
?
•
Molality
(
m
)
It
is
thenumberofmolesofsolute
dissolvedin1kgofsolvent
.
Itsunit
is
moikg
"or
molal
or
mmolality
(
m
)
=
wmB_☐✗%¥g
)
QAndthemolalityof
a
15%
Solution
ofHasan
C
densityofHasoqsolution
=
1.020g
c.m
-
3)
Sofa
mass
of
solute(WB)
=
15g
Molarmass
of112504(Mg)
=
98g
Mot
'molality
(
m
)
=
?
Mass
ofsolvent(
WA)
=
85g
m=wm÷×"w%-g
,
<z
¥g✗'%É-
=
1.8
Mold
.

12. QOutofmolarity
and
molality
which
one
isunaffected
withincreaseintemperature
?
AusMolaritydecreaseswiththe
increaseintemp
.
becauseitinvolvesvolume
.
Ontheotherhand
.
Molalityremains
unaffected
,
because
itinvolves
manofsolventonly
.
#STOICHIOMETRY
#
"
stoichiometry
means
to
measureanelement
"
when
a
balanced
chemical
reaction
iswritten
.
It
givesquantitativerelationship
b/wvariousreactants
andproducts
in
termsof
mass
,
moles
,
molecules
andvolumese.g
Caco
}
-12HCl
7
Call
,
-1
CO2
+
Hao
1m01 2m01
1
Mol 1m01
Im-1
100g 73g 111g
44g
188
Coefficient
*
coefficientof
a
balancedchemicaleqn
are
calledstoichiometric
-Q
If20gofCaco
,
istreatedwith20g
HU
.
Howmanygrams
ofCO2willbeproduced
?
SotsCaco
,
+
2116
>
call
,
+CO2
-1110
loog73g 44g
100gCaco
,
produce
=
-44g
CO2
1g
Caco
,
produce
=
9%
g
coz
20gCaco
,
produce
=
4,4-0×20=8
-8gcoz
9Calculatethe
mass
ofgraphite
thatmustbe
burnttoproduce13.2g
ofCO2
.
80¥
C
+02
→
cog
12g 44g
-44gCO2produced
on
combustionof
=
12g
goop
1g
=
12144
"
"13^2
=
1-4×13.2
=3
.bg

13. #
LimitingReagent
#somereactants
are
totally
consumedcalled
limiting
see
agent
.
Reactionstops
whenit
isconsumed
.
*ExcessReagent
#Thereactantwhich
is
notconsumed
completelyin
thereactioncalled
excess
reagent
.
Q
Assumethat
youhave139m01ofNz
and3.44m01
.
ofthe
e)HowmanymolesofNH
,
canyoumake
it)Howmanygramsofwhichreactantwillbeleft
over
?
Aye
Ms-13112
>
2MHz
1m013m01 2m01
it
1m01ofN
,
reactswith
=
3Mol
do
Ha
"
39Molof
Ng
reactswith
=
3×139--4.17madofHa
But
we
have3.44moi
of
Ha
.
So
thislimiting
it)•)3m01Haproduce
=
2MolofNH
,
Reagent
.1m01ofH2produce
=
f-
Molof
MHz
3.44MOIofHaProduce
'
=
2-3×3.44
=
2.29Molof
Mtf
•
)]Mot
tfreact
with=LMolofN
,
I
MotHzreactwith
=
tgrnol
of
Nff
3.44MOI
- =
tg
✗
3.44
=
1.15MolqNz
no
-
ofMolesofNc
left
=
1.
zq-1.15=0.24Motdona
h
=
Im
g
w
=
NXM
=
0.24×28
=
6.72g
ofNcleft
Thankyoke