Some Basic Concepts of Chemistry ( रसायन विज्ञान के मूलभूत सिद्धांत ) @irfanullah_mehar.pdf

Chemistr
 
Notes
o
 
IrfanullahMehar
 
[email protected]
 
Oirfanullah_mehar
 
WorldofWisdom

hied
 
'Read
noo,est
latcA
–youkutureis
 
wadekingdou.
 
SomeBostcCanceplsohChanisy
 
RGIDHR (Atomicaesn
 
goatsig
(Equivalesd
Mas)
 
moleCneeot
 
Q.relateoltoGloichi'omet
 
29))0)2025
 
LimitingReogeut
 
Prity
 
Reactan)

1
 
12
 
Q.based
 
Ayciocaben
 
Jois
(Radfcblo)
 
Atomic
masc)
 
-fasa]
 
(RelahveVeue)
 
)2
 
Yao
 
25.5

He
 
Atomic(ss
=
 
dre
 
Be g
 
Scume
 
12
 
F
19
 
maes
 
1
 
Mass
 
atem>)
 
xc
 
Ng 22
 
A)
2
 
Si 2%
 
P2)
 
S 32
 
C35.5
 
Ar40
 
Ca
40
 
One
 
Cr
52
 
mn54
 
63,
5
 
Zh65
 
B 80

ho.)
a
 
AA=
 
(AAM) SU))
 
Aloncoplasc
 
NHAT|
 
G-Tat
,
xAlumíc
Mag
 
AAm
 
98
.
 
Totul

.3|
 
Ane
 
3%X122X14
 
CO37
 
total319NA
 
7.
ntsl
 
75.
 
25%.
 
totulY,
 
|17620
 
31
 
75X35t25
X33
 
35X3137X)
 
3t)
 
105137
 
42
35.5

46.1
 
Totad
 
10D
 
l610=16XH1800-18
 
95
 
ol6=
g5
%h

(msteeulasmase)(MaleculasWweipht)
 
NOTE
ad
gas
(sfgyhT)
 
anfearg
 
molecula
 
=
2Ht
lstOxgn
 
=
2X)t1x32+4XJ
6
 
msS
 
Ti
 
Cb0ild)
 
QEd)
 
Pas
 
AternfoMasc
 
-2+32
u
 
3
mole(6.023x6)

K
sOy.Al2(SOu)2.2uH,0
 
furaa
(Pottuch
aln)
 
Fesoy:NMy) 30y6n,0
 
1moLeco,
 
MoasMes
 
(6rammoleoulas
 
maes)
 
Douksle
 
Selt
 
enal))

foz
a
lLemevd)
 
(3g0
 
355
 
1
 
(Equvalnt
mase)
 
Ht
-
1
 
n-feetey
 
grammoleculamasS
 
35.5
 
meslumaes
 
B
 
A)t3

ncidfo
 
(1)HCI
 
n-fccteY
 
mfaatr)
 
Acid fop
 
E
eqwvaleud
 
T
 
36.5
 
mess
 
Reox
Rxy
 
1moleoule
 
Reaetmt
 
nfuetor
 
36.5

(1)
 
(m)
 
H,G04
 
H
 
coDh,
 
1)
 
Oxalfeaoid
 
COOH
 
(tumato)
 
(6n)(6n
 
7-factor
 
63
 
1
 
623)

Hclna
 
ofve
chasge
 
(i)
 
D
 
facter
 
H
 
nfactur
 
-
 
Gmm
 
nt
 
)(morobasfeacfd
 
1
 
blc

(1)Nab
 
Calo1),_
 
mglom),
 
()
 
h-etos
 
()FA
=
Lí,
Ng,k,Rb,
Cc,Fr1
 
Sot
Cb]
E
 
()IA
B,Mg
1
4,S,
BaRa
 
|iv)Fe,fe3
 
-Puctr
 
)

tv)F,c,B,
 
, -
 
-2.
 
(vi)Soy
 
NOg
 
(i))No2,NO,,
N N
 
(ix)c,
o
 
Cxalate
 
(i)
Na
CI
 
Co
 
52.5
 
1
 
(aKa)
 
Eb.5)

(3)GCz
 
(i)A1cla
 
=2X1t2X2=8
 
1
 
1Fet2
42
Nh,
 
Gnm
 
(v)Fesoy.(NMy)Soy
6h,
o
 
IX2+2XI
 
--65:5)
 
2Su
 
1))
 
>X2-y
 
2
 
9X
28
 
E
 
3

9elo]2025
 
KMnOy
 
Redlox5]
qouis (e)
 
Reclox
 
n
fatus
 
moleouleReetut
ŠRT
 
6tOaiclaticnho.)

Acic'cMeum
 
ho.e
trunaes
 
nfuatosG)
 
Adle
medlium
 
Cyt3
 
(9)KyCs
 
C
 
cbJ
 
Neuctral
oxBasfc
 
Medium
 
3
 
Trouneye,
 
1:factos
 
Bassc
 
Ctrong
eoufc
 
Medun
 
[X=6)
 
Bosfehecig

+3
 
2c13 2ct
9ele)
 
E C,O4
 
ele1
 
but
 
)totclele.
 
iele
 
1
ele
 
m-
faeter)
 
2.
 
Ghmm
 
2

2X*0
 
fuctés
 
FeCaOy
 
fet3
 
te
 
t3
 
t3X1|=0
 
6=
-3)
 
-2
 
3
ee
=
 
8ele
 
ngoctor
 
LutK,
 
thranáfet

NONO,
 
L2X-2=-1
 
2A3
 
2
O
 
Sele
 
6
 
feetor-
29
 
+5
 
ele
 
E
 
nfacter-)
 
2
 
(2x2-y ele)
 
(3Y0-2yele)
 
Gmn
 
20

5)Mole
 
mole
 
OR
 
Cut2
 
ct1 at2
x1
=
cle)
 
so
 
6
 
6.028X103101/77O
 
(ix
 
6.022X)023
 
Csncapt
 
loele
=
n(act
 
Actual
i
G.09914D76X1023aj
 
G+028X10°3
 
10
 
Coleetion
 
STPT
NTP
Uy
 
SAttD2.y
Lete

NOTE
 
GTP=
 
SiancleredTempvefwe Preseut
 
NTP= NoomalSemprutwe
4Dseagwe
 
P
 
T
 
1
molet
 
atn
 
293
K
 
caleulat
 
(sTP/NpR)

Rarnbaan
Fomula
 
NOT
E
O
 
mole=
 
Gmm
 
No
 
NA
 
29.4
 
N
3TGNIST
UT
 
=6.023X10s
 
M)
EU)J mle
,
No

Q.6)
 
L'.
 
densíty(ear)
 
gcoee
 
t2nya
(ntaaicity)
 
Ob)
 
99,4
 
Ceiz
 
O
molecule
atom tem moleoulefio
 
nel
 
caleulat
 
nole
x
Ghmm
1
X
180
=
18O
m

mole
=
9
 
Le-342
 
Gm
 
SuCTOSe/ane
 
Gm
 
No
 
3.42
 
342
 
Ne
 
LONA
 
QucTOce
 
O.01mole
 
-
D.01XNA=
 
0.01Nq
 
OR0.IX6023X103

Gm
 
No
 
NA
 
SucrOLe
 
220y
 
no. H
 
OR
 
NA
 
3.2
 
342
 
20.4
 
No
 
=0.0)X
220
y
 
=
0.224Ltr
 
c=12/)Catom
XSucroCe
meslecule
&
 
-12
X
G023
X101
 
i.
 
20X.61
NAO.22
Na
 
22
X
6.023
X10
 
SUCYE
 
atobecule
 
HH
atemsXno.f

.
 
1
X
O.01No
 
H
etums
 
O.22Nq
 
O.13Ng
 
Sutroce
melecde
 
.11
Ng

Q.8
512
 
(im)gatom
 
1i)lesle
=
 
tn)No-Satoms
 
mol
No
 
(ifi)9,Molecule
 
–
()me
 
(iv)S,móeeule
 
S
 
5)2
 
256
 
2X2Np
 
=2mele
 
Satom
 
2
 
())Nos
 
moleol
-
fos)6
N
 
6
 
23
 
+2
N)n=2X6.023X)o

NOTE
m
-nole
culeholeculeamele
 
fm
-afem
 
Gran
ples
 
moleculeoxygen
>
 
2
gm
-mosleeule
 
Ogen
 
Atonn
 
VOs
 
Dgm
meslecule
v,Og
 
(3T210)
 
cLtums
 
-5
X 2
NA=10
NA
 
No
 
atoms

Q.10
 
(0)
16
 
No
 
19,
 
Gmm
 
Gmm
 
16
 
XNA
 
2
 
16
 
NA
 
I6
 
16NA
 
l6
 
32
 
V
 
NA
22.4
 
2X% NA
 
5N
 
atom
 
(O
312)

4)5.6
Lkr
 
Gmm
 
He
 
Heliun
 
Hatm
-
2X2
N=
(4
NA
 
Helium
 
No
 
Og
 
Nn
 
atom
 
=2
NAH,379
 
(mensetanfe)(e
 
No
XNA 9
 
Gmm
 
X
Na XNn
-(Na)

Gmm
 
5.6LtrOg
 
Mole
 
No
 
32
 
X
Na
 
Gmm
 
No
 
Nn
 
VL
 
XNA
 
22,4
 
0-ctem3X.
 
22.4
 
S.b
 
22.4
 
Nn
 
XNA
 
X
 
)6
 
Tevs-A

Reet2o23
 
Solh
 
Q13)
 
4gmn
He 6
m
No 0.46gm
Ca
 
No
NA
 
40
 
V
 
-N
O.D1Na
 
NA
 
Boneene
 
8.9mebenzene ele.
pxoton
 
Benzene
 
Noy6
 
m
d
XV 2X3.9
 
X3.9
7m
 
m
 
(Ans-9)
 
22.4
 
X

Q.4
 
mole=
 
No
 
hmm
 
X
Na
 
XNA
 
mole
 
No
 
No
 
ch
5
p=42
X
b.Nno2
Na
 
p=2X 0.1
N
4.2Na
 
N36
X
0.1Na
 
N
 
4,2
g1»Nitride(Ne)ele
 
Gmm
 
Gmm
 
O.1NnCoHe
 
No
 
NA
 
9.6NA
 
4o2.
 
X
Na
 
e-42
 
N=36

Mole
 
hmm
 
O.1
 
18
 
N=)8x 01
 
1
 
NA
 
Nn
 
No
 
NA
 
Na
 
-l0.055
Na

NDAHyeis)a(SmplectintegoWalue)yo
 
l5,
2
3,8.8,
2
+5
 
X
 
Que,besed
om
 
Gxaumples
 
X
 
becase
 
eteichiomety
 
X
 
LHS RHS)
 
t-
 
hatmee

(Tleal)
 
6
 
1
mole.
N2,
 
L
 
N
t3H,
 
Q2.4Ltr
 
3mde.
 
3meleh, Rcnettoea mole
 
éapab
 
Fe)
 
Prgmotor
(ardA5,aels)
 
9mole
 
3y
 
2Na
 
2
X
22.yo

Que
 
Teal
 
Acceveling
 
Q1
 
stufehiomehg
 
to
 
Na
 
Solve
 
J.J.Gr
 
Nz
 
3mdle
 
methec
 
met
hoo
 
2Nn,
 
9
m
N

50
m Nà çAT 34
 
yNn,t50,
 
mole
 
2%
 
4X14
 
Na
 
GX50
=(GO.7
 
NO
 
yNGt
6H50
 
yX30
 
AmNO

Ans-
 
+|2
 
(CacO2)
 
Cacog
+
2hc1
 
(A)0.22tn
 
100gm
 
Cac03
 
CaC,.Co,th0
 
224
me
 
994
 
20
 
CO,b]31)4d

100
m
 
Q.20 ContaatPrseec
(TEUS
ae)
 
280.
 
22U00
 
100
 
80pm
60
m
 
I
mol
 
2S03
 
4om

29.u
 
n
3tsona(ReageutT
 
100
 
(Censuoe)
 
224
 
22+4
 
2
 
-0
 
Reget)
 
Reactant)Bfeay
 
SNËLimitingRcageud
(3)
 
L.R.

Que
 
L.R.
 
)
 
9
ER.(
excessRengend/emfegafs)
 
Que.
 
1)
L.R.
 
Tleal
 
fna
)
9
 
Tleal
 
Bshe
 
helpfea
 
LR

A
 
Lleal
 
(i)LA=9
 
N3
 
Imle
 
Value
=
 
3mole
 
2melE
 
Nz
 
H
 
2%
 
6
 
-
x10-y6
-66

T50gm
N,
N,8ife)cs(3nfelo
 
as)
 
feal
 
LR
 
02
 
jmoe
 
Y.8
 
1.6g
m
 
4O2.
 
Ime
 
32
 
-R
 
6
 
2mol
 
80

32
 
1.6
 
excessValue
 
32
 
1
 
LR
 
29
 
32
 
30
 
X1,6
 
22
 
X

given
 
4mole
 
160
 
I
 
3y
 
R
 
+
50,
 
O2
 
5mole
 
ER=
NMG
LR=O,)
 
ER=
Nha
 
NHa
 
NO
 
yNDt6H,0
 
>,UNO+6H,O
 
ume
 
120
 
|60
 
)66
 
X6
2-69

Q.24/
 
Q,D5
 
24
 
)60
 
T5
%
 
120
 
l20
 
16
0
 
120
 
L60X6 5gn
 
ActuslValue
 
Theonhcodhalue
 
X)00
 
18
 
X18=75)

Ckemist
 
ActualCaCo3
 
1
 
Ca
COz
 
Joco
 
|50
 
CaC03
 
22.y
 
C02
(V)
 
22y
 
22:901
 
1
 
7-8
 
1moie
 
1.3,
 
900XI5
 
=
)50gm

9)
Que
 
Casbem)
 
l)
 
)7d)
 
Last
 
Base)m
 
+2O2
 
2
 
yasecanbon
 
2H,0
 
2.c0,+3
H,0
 
2.00,+31,0
 
erangles
 
Belance529))

Q.96afaLPG
 
Dden)
 
(m)
 
Relese
 
favaat-feu
-
 
50
 
5mde
 
5X2.44it.
 
(K2),%.)
 
?
 
gmole
 
5)
 
3X22,4o.

(M)
 
(ue)
 
O,(V)
 
1
 
5
X
22:4
 
5X22y
 
ly
4001m 5X22.4
 
-JHG
=
5
X11,200
 
3X22
y
 
-)(6926
 
S6,000t

2,2
 
ti
y
Cnut20,
 
lmole
 
,3
 
+505
 
22.406).
 
Accerding.+0Coue
 
3,)
 
Co
2
,
0
 
20.46).
 
1.5,2.5
 
).5+75
 
Cny(o)

Ng
 
20.4
 
5e.
 
Avogaaboule
(&reTaIE,5Tfa)
 
3mole
 
3X9264
 
OR
 
2NHg
 
Qmole
 
X12.4

2,
 
2,2
 
|.5,35
 
CHyt20,
 
92.y
 
Can
 
>15
 
1mee
 
S65=1|
 
t0,
 
3+17.5
=
2o.5
 
elukle
 
5mole
 
5X22:4
 
Q.29)ejo(Ssooctane)
 
.

2.
Ca
Hat250,
 
211|4
m
 
caleuleLtebHJà)T
 
SX||4
 
8Co,tB0
 
d
-)
 
16Co2
 
6etrol
iorloras
A:
d
 
16mole
 
16X.22.
y
ht
 
Co,(V)
 
|6X22
 
16X2
2-y
 
27)14
 
=
|140
m

Q.20
 
H
 
Poreendage
 
1,05]
Gmm
 
2x100
 
X
100
 
(Gmm)
 
11,11%
 
Vapto.
 
Black
White
 
Cermbírahon.
 
o9)1o)20)5
 
/sfa9I()
 
5:08
PR)
 
o.-
 
(Tue)
 
X100lps
 
16
 
X)00

40
 
9n,o
H o
 
(Emphencal
lamula)(M6enulasfoaub)
 
-(i
 
Atomic
m
ass
 
(m)arg,
HIST

Jalue
 
dec
 
m)n-
value
 
Atome
asc
 
X
 
Y
 
N
 
J.5X
=3
 
lo
33
X
3
=

Dob0etal
 
deq
 
)mo@)mo mo,
 
50
 
50
 
6c66
 
oxfde meta
T
Atumic
 
53.33
 
2/At.
masc
 
56
 
J,6
 
=
6625
 
3125;
 
41.masS
 
u=3.33
 
g3.33,-333
 
Valup
 
min-Value
 
6.25
 
8,125
 
8ol25
 
3,)25
 
3.33
 
3.33
 
3.33
 
3.33
 
=2
 
(m,o)

C
 
30
 
Q=
40
H3,23,N= 46e67
%
 
40,
 
13.33
 
X
 
mo-I
 
Oriele
 
6
 
333
 
1333/=1333
 
46-67
 
-3.33
 
tlaulue
 
Be33
 
mo-I
 
3.33
 
)3.33
 
333.
 
3,33
 
333
 
minvaluse
 
=)
 
-)

fomula
 
200
 
2X442
 
kenmuula
,o, mo-TT
 
lusc+336207e
 
Atmaes
 
At,maes
 
metcla)nkK
 
X100
 
6>C=
336
 
Dl336G)

35
 
Ane
 
724
 
276
 
1
 
56
 
99.6
 
O.005m->
 
56
 
(Sheet
 
Ssutfon
 
,29
o
 
1629
 
l725
 
)0.075 )O.005 o.
05
 
1000O
 
|629
 
=
J33
 
aro.36.
Ba(oH),O.D65
X
O•2
 
1ocop
 
x
3mOy
 
=O.00
30)

Q.961gm
Ca
C03
 
given
 
100
 
mole
 
No
 
Gmm
 
22-y
 
224
 
l
 
No
 
0.224o5
 
127
 
loi
e
 
(Bme-c)

Ghmm
 
1
 
(A)G.02X\02atoms
 
(n2)
 
NCabon
 
6.02X)DIO
 
atm
 
23
 
Nn
b0237103
 
6.2
X10atums
O
 
2h
 
-(4N,J
 
35,5
 
afm
 
NO,Mlebul
 
C
 
224
 
|2
 
127
 
XNn(D
N
 
atemn
 
Nn
 
X
 
(ane-9)
 
2
NA
 
X
N
 
noleouly
 
=1me

(Valenceeleotns)
 
Moie
 
N3
 
STP
 
201
 
No
 
NA
 
Gmm
N,2240
 
Gmm
 
2.10msle
 
-(03
N
 
0.2
NA
X

20u
m
Hydgn
 
A0.65
m
 
Znt
H,
S0y
 
Imole
 
Ime
 
0,224
 
o.6o
93)
 
Jmele
 
65
 
224
 
fo.65gm)
 
100

5-6
 
1
mle
Nodis)
 
móle
 
ymole
 
5-y
 
NO 6
H,Ô
 
I
mele
n
 
6rmele

10
0
 
6720
 
2
 
224
 
56
 
Fe
 
6.334
 
X6720
 
4

Ane
 
Qou5
 
Y.
 
13,33
 
(A)181
 
mole
 
Qeu69.85
m,
Bo
 
2.24
 
13,33
 
)
 
46.67
 
21
 
=3.33
 
-
13,33
 
=3.33
 
Velue/m'h-ele
 
3.23
 
3.33
 
333
 
(12)
 
32
 
Ane-c

9.5
 
9.5
 
Gmm
o
BaC03
 
BaCo237
112tug19
 
224
 
197
 
Q.
Pensxiolaseanhyus
 
)I.568
X
10
 
O.5%ST
 
+
CO2
 
(112
 
8)j5G8
X
103
 
(-)
 
enmea
se
 
(9.R-8.y2)

O.5
 
79.4
 
(9)
 
2
 
7gmNg
 
20
 
l6
m
Oz
 
Moe
 
Gmm
 
XNg
 
2xNn
NA
 
Cnagme
 
O.5
 
No
 
V
 
29.4
 
l6
 
6
 
5
680'|.568
X16
 
Ane-A
 
l6gmNO,
 
32XNnN

Q.49
 
Haber
 
30
 
50
y.f
SUSAAHrfzua
 
(A)
20
 
Prscs
 
50.
 
NN
t3H,
 
Consutne
 
nsume
 
3X22:y
 
30l
 
r5
6,
 
CmewUme

Q.50
 
e
 
15
L
nagas
sTP
 
2
 
30
 
NaO25Na
 
O33Ng
 
22.y
 
),5)
 
2
 
3
 
0.66N,
 
Cascbem

2Al
 
Q521L
 
Al202
 
Balance
ondrf)
-
 
+3C
 
3112
?m
 
2T0Y|000
 
A)
 
3X12
 
3X12
 
3
 
32
x
270Al000
 
X
 
Ane-)

5L
 
Imol
 
16
 
Nn
 
5X):4
 
3X10
 
Gmm No22-y
 
cr2)
 
-23
 
e)5.5X1023Cm
 
dm
 
3l633

Q.59)
b.5
 
RT
 
pba)
Chledd
faba
 
)o.ol|
 
Pbo
 
Pbot2Hcl
 
Imole
 
h
 
.2
 
9mele
 
923
 
223tid
 
3
 
HO
 
PLO)
Checlctrà
 
223
 
3
 
Imlie
 
278

Q55
 
ghuen
 
32
 
Gu
 
O.029
 
10gn
 
2
H,tO2
 
2mole
 
(LoR.
 
36
 
36
 
26
 
32
 
2
 
PboPbC),
 
OO2g
 
3.6
 
12
 
6 5
 
223
 
O-029
 
B
 
fa

Q.56|
 
2].6
 
21:
6
 
44.8ii
 
2
302
 
2X22:y
 
2X22:y
 
21•6
 
3X22.4
 
3X22.412)6
 
2)·6
 
29it
 
Sg)
 
(0)67.2
 
286HA.
 
21.6gm
 
6
 
-3X22.4
 
faaa)

Q.57
 
2A)
 
2.5X102
 
1
 
0-aton
 
inthe
 
C2he4eme
 
Gmm
 
92.4
 
0.02
 
2
 
LPOM),
 
Q.sbSTP
4R
O,NH, co,H
l00
e
 
O.25
 
1mole
 
29.4XNA
 
700
 
O
 
2-125
X10
 
3.125
 
<
 
D7y00
XNA
 
)"
 
2
 
3612571D)

Q.53)
far
 
Qi60)
 
No
 
34
 
XN
 
Nn
 
Nh
 
V
 
22.9
 
-02
Nn(,
 
O.05nole.A,A3) 9ym
 
10 20
 
lo6
 
100

Mole
 
No
 
0.05
 
28+3A
=
120
 
Q.62)fol)
 
10
 
Cmm
 
+l00BA)
 
-180
 
=10
 
)S t.5U5
1o
 
2
A=%0 A=uo
 
X
G023
X103

Q63)
 
2
 
Gmn
 
-No
 
WX22.y
 
X
 
6)
 
&m
 
T
ST
P0R2.24
 
3.2
 
16
 
224
 
3o2xNnbb5
Nn
 
X
 
10

Ateribyauet
 
Q.65
 
A
 
Gmm
 
No
 
NA
 
Hsoy
 
22
 
22.4t
 
)
0X103
 
Xbo023X)0
 
3.01
X
1020
ed
 
=6.623X100molecul
 
tt,soy3. 0)X)63
 
x
)20
 
1020/6.023-3.0)
 
No

Qo66
 
No
 
N
 
3,01
XA20
 
2
 
Rme-
NA
 
lo3
 
(A)6.
o
2
X10
 
H2
 
Gmm
 
Gymn,
 
2
 
Na
 
XNA
 
e)
6o02.A103
 
32
 
22.4
 
NA
 
2.
 
Ans-A
 
6o027)O
 
atum
 
NA

Qo6O.1
 
Q.69
 
Na
Nn/10xy
 
No
=
 
Ha
 
XN
O.|
NA
 
lo
 
Ans-c

l6
 
32
 
2%
 
2
 
Nn
N
O,
nolecule)
 
6.02
XIo
 
Goy
atom
 
naom
e
eule
 
atem(3NA
 
)6.02
X
105
30X6.02X
16
 
O3o
(1bcowal)

Totuline
 
(4)
Jmee
 
Q.72
4o25
 
-3X106022X1
 
3
Ba(NOs),'X
34(NO2),
 
=3.0
X
6oo2X105
 
(4)
)X1o23
 
Y6e62
3X)0'3
 
l6
 
Dus-cl
 
66X1022

2C
 
1oe
 
P=
6
XNg
 
10.N
 
Nn
 
6
 
12
 
12
Xbe023X103
 
-23
 
I.08x1025
 
bott3X)623

b20
)023
 
No
 
Gm
m
 
N
 
Gmm WX22.y
 
(00
 
)X22HOo
 
294
 
oofm6o023
Xi0 )
 
55
XJ023
 
(10
m
usete
 
22

224ml
 
12md
 
I
+6 +1CH
 
1moe
 
386ml
 
17:
2
 
mee
X24

50mole
oOa
 
I25moleo
O2
 
12•5
 
114
 
moe
 
9i00
mole.okO,
 
125
moe
 
m=dXV
=
0
X126

'Q.79)
Cy2+
63.51
 
Cyt H,S
 
moe
 
63.5
 
63.5
 
)63.5
 
Q.B0/
 
Imoe
 
)0.6
 
34y
 
34
 
63.5
 
34
 
(3.5
 
X63.5
 
CO,
 
S3)-75
 
Co14 2c0
 
Cus
 
Ame-c

CO
 
27an
 
2
 
2

Q.s2
Cacos
 
1
 
2.0l6
.
1.008off.
 
10
X
9O:
 
COtCO,
 
22.y
 
22:4
 
co
 
2
 
22-4
x
2.0l6c7
 
hoAne-A)
 
Bq
(PU4),

Q.B4)
 
3en
 
Q.35
 
Imee2mel
 
mo
 
50
 
Mt16
 
m
l6
 
4
mle
s
 
m,O
 
06D
 
6
 
2
 
-2,5
 
2-5.
 
2.5
1
 
2,5
 
1
 
),5

1.
 
2.
 
3.
 
5.
 
6.
 
7
 
8,
 
ThenumberofmolesofOH-in0.3litreof0.005M
 
Ba(OH),is
-
 
0.005
M
Ba(OH), 0.3ccrÀOH- i
 
(1)0.075
 
(3)0.045
 
(1)1litre
 
(3)0.224litre
 
ThevalumeofCO,atSTPobtainedbyheating
1gm
of
 
CaCO,willbe
 
(1)4
ghydrogen
 
(3)48gmagnisum
 
OnemoleofC0,contains
 
(1)6.02x102atomsof
C
 
Chemistry-Somebasicconceptsofchemistry
 
Fromthefollowingthenumberofatomsisgreaterin
 
(2)6.02
x10atomsofO
 
(1)2.1N,
 
(3)1.6N,
 
(2)0.005
 
(4)0.003
 
(3)18.1
x10moleculesofCO,
 
(1)2.10mole
 
(4)3grammoleculesofCo,
 
(3)0.186mole
 
(2)22.4litres
 
Thetotalnumberofvalenceelectronsin4.2gm
ofN*
 
ionis-
 
(4)11,2litre
 
4.2gmN-AIYThet#hI
gaciH
 
(Valenceelectrons)
 
(1)0.65gm
 
(3)65gm
 
(2)71
gchlorine
 
Inthereaction
 
(4)127
g
iodine
 
AtSTPconditionsthenumberofmolesofoxygenina
 
litreofairwhichhas21%oxygenbyvolumeis
 
(2)4.2N,
 
(4)2.4N,
 
Howmuchamountofzincisrequiredtoreactwith
 
diluteH,sO,forobtaining224mlhydrogenatSTP
 
STPqT224mlIgguITV
 
(2)0.0093
 
(4)0.21mole
 
(2)6.5gm
 
(4)0.065gm
 
forg
 
[AIPMT-96]
 
9,
 
10
 
11
 
12
 
13
 
4NH,(g)
+
50,(e)>4NOg)
+
6H,0()when
1moleof
 
ammoniaand
1moleofO,aremadetoreactto
 
completion.
 
aGI
4NH,(e)
+
50,(8)
4NO(B)+
6H,0\e)
 
(1)1moleofH,0isproduced/1
 
(2)1moleofNowillbeproduced/1H|eNO
T
 
(3)Allthe
oxygenwillbeconsumed/quf
ifcfrT
 
(4)Alltheammoniawillbeconsumed/yut
r-nT
 
Haemoglobin(Molecularweight
=
67200)has0.3349%
 
ironbyweight.Onemoleculeofhaemoglobin
 
containshowmuchatomsofiron-
 
ÉTH AT(30T4TR=67200)GET0.334%HRT
 
(1)1
 
(3)3
 
dH,0gTTI
 
PercentageofC,Hand
Naregivenasfollows
 
C=40%,H=13.33%,N
=
46.67%Theempirical
 
formulawillbe-
 
(1)CH,N
 
C=40%,H=13.33%,N=46.67
%
HeI4TGt
a
 
(3)CH,N
 
(2)2
 
(4)4
 
(1)1:1
 
(3)
4
:1
 
Moleratioof
H,
andO,gasis8:1.Whatwillbethe
 
ratio
ofwt.:
 
(2)C,H,N
 
(4)CH,N
 
(2)2:1
 
(4)1:2
 
(At.wtofBa
=
137)
 
A
compoundcontainC,HandO.IfC=40%andH=
 
6.67%thenempiricalformulaofcompoundwillbe-
 
yf
C=40%
T
H=6.67%
fooI
 
(1)CH,0 (2)CH,0 (3)CH,0, (4)CHO
 
VolumeofCO,atNTPobtainedbythecomplete
 
decompositionof9.85gm.BaCO,is-
 
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14.
 
15.
 
16.
 
17.
 
13.
 
9.85gm.Baco,
 
BT3ATUTT
 
(1)2.24lit
 
(4)0.56lit
 
PercentageofSeinperoxidaseanhydrousenzymeis
 
0.5%byweight(at.wt.
=
78.4)thenmin.mol.wt.of
 
peroxidaseanhydrousenzymesis:
 
(1)1.568
x104
 
(3)15.68
 
0.5%HIAT(4.HR
=
78.4)yRGÍSUI
 
(1)7gmN,
 
(3)16gmNO,
 
Chemistry-Somebasicconceptsofchemistry
 
(Ba
T
4.YR
=
137)
 
utfatt
HSTP4RYITCO,
 
(2)1.12lit(3)0.84lit
 
Whichhasmaximummolecules
:
 
(2)1.568
x103
 
InHaberprocess30litresofdihydrogenand
 
(4)2.136
x10+
 
30litresofdinitrogenweretakenforreactionwhich
 
yieldedonly50%oftheexpectedproduct.Whatwill
 
bethecompositionofgaseousmixtureunderthe
 
aforesaidconditionintheend:
 
(1)20litresammonia,20litresnitrogen,20litres
 
hydrogen
 
(2)0-5g
ofH,gas
 
(2)2gmH,
 
(2)10litresammonia,25litresnitrogen,15litres
 
hydrogen
 
(1)5LofN,gasatSTP
 
(3)10g
of0,
gas
 
(4)16gm
0,
 
(3)20litresammonia,10litresnitrogen,30litres
 
hydrogen
 
:
 
(4)20litresammonia,25litresnitrogen,15litres
 
hydrogen
 
Themaximumnumberofmoleculesispresentin
 
(4)15
L
ofH,gasatSTP
 
Themassofcarbonanodeconsumed(givingonly
 
carbondioxide)intheproductionof270Kgof
 
aluminiummetalfrombauxitebytheHallprocessis
:
 
19
 
20
 
21
 
22
 
23
 
(1)180Kg
 
(2)270Kg
 
(3)240Kg
 
(4)90Kg
 
Whatvolumeofoxygengas(0,)measuredat0°Cand
 
1atm,isneededtoburncompletely1L,ofpropane
 
gas(c,H,)measuredunderthe
sameconditions:
 
ILTTH
(C,H,)(9HOIYTT0c
TeT1atm
 
(1)5
L
 
(3)7
L
 
Volumeoccupiedbyonemoleculeofwater
 
(density=lgcm")is:
 
(1)3.0
x10
2cm
 
(2)5.5x102
cm
 
(39.0
x
102cm3
 
(4)6.023
x
102
cm
 
Howmanymolesoflead
(I1)chloridewillbeformed
 
from
a
reactionbetween6.5g
ofPbOand3.2g
ofHCI
 
?(Atomicwt.ofPb=207)
 
6.5
gPbOqd3.2gHCI
T
ufafsb4IERTGS()
 
iRIGfociHliq?(Pb
T
TAR=207)
 
(1)0.011
 
(3)0.044
 
(2)10
L
 
(4)6L
 
10gofhydrogenand64
g
ofoxygenwerefilledina
 
steelvesselandexploded.Amountofwater
 
producedinthisreactionwillbe
:
 
(1)1mol
 
(3)3mol
 
10g813$GTTT
64
g
3TciT
#
Ia
 
(2)0.029
 
(4)0.333
 
(1)44.8lit.
 
(3)89.6lit.
 
(2)2mol
 
Whatvolumeofhydrogengasat273Kand1atm
 
pressurewillbeconsumedinobtainig21.6
gmof
 
elementalboron(atomicmass=
10.8)fromthe
 
reductionofborontrichloridebyhydrogen-
 
(4)4mol
 
(2)22.4lit
 
(4)67.2Iit.
 
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24.
 
25.
 
26
 
27
 
Howmanymolesofmagnesiumphosphate,
 
Mg,(PO,),willcontain0.25moleofoxygenatoms?
 
(1)2.5
x
102
 
(3)3.125
x
102
 
Chemistry-Somebasicconceptsofchemistry
 
(2)0.02[AIEEE2006]
 
(4)1.25
x102
 
Onepartofan
element
A
combineswithtwopartsof
 
anotherelementB.Sixpartsoftheelement
C
 
combineswithfourpartsoftheelement
B.IfAand
C
 
combinetogathertheratiooftheirweightswillbe
 
governedby
 
HYToGITd
C
6IT
TCg
B 4HT41
 
(1)Lawofdefiniteproportion/ferrg4TGf4
 
GIRI
 
(2)Lawofmultipleproportion/yfrIGSTGfr44
 
(3)Lawofreciprocalproportion/sH
T
 
(4)Lawofconservationofmass/HTHRAU
 
Twoelements
X(at-mass16)and
Y(at-mass14)
 
combinetoformcompoundsA,BandC.Theratioof
 
differentmassesof
Y
whichcombineswith
afixed
 
massof
X
inA,
B
andCis
1:3:5.
If32partsbymass
 
of
Xcombineswith84partsbymassof
YinB,thenin
 
C
16partsbymassof
X
willcombinewith:
 
Td
X
(4RHI
R
16)
RY(4RATHIR14)
 
HyTsIoRfrca
A,
B TNc
ItŽIY¢
 
SH HN
A,
B N C
HYTaIt
 
51fBÀ
X
I
32HITTR
 
(1)14partsbymassofY/4R
 
(2)42partsbymassofY/HIR
 
(3)70partsbymassofY/YR
 
(4)84partsbymassof
Y
 
(1)Co,,CH,H,0
 
1:3:
 
Y¢84HrT
 
14HTT
Y
HT
 
42HTTY¢HRT
 
70TT
Y
HIT
 
HNH8441T
Y
HT
 
Lawofreciprocalproportioncanbevertifiedby
 
followingcompounds
 
28
 
29
 
30
 
31
 
32.
 
33
 
(2)N,0,N,0,N,0,
 
(3)NaCl,Na,co,,NaOH
 
(4)H,0,
HCI,NaCl
 
Numberofmoleculesin100mlofeachof0,,NH,and
 
CO,atSTPare
 
STPYTO,,NH,
3ÌNCo,yo 100fMi.
 
(1)intheorderCO,
<0,<NH,
 
(2)intheorderNH,<0,<CO,
 
(3)the
same
 
(4)NH,
=co,<0,
 
Whichofthefollowingcombinationillustratelawof
 
reciprocalproportion
:
 
(1)N,O,N,0,N,0,
 
(2)PH,P,0,,P,S,
 
(3)CS,,C0,,SO,
 
(4)NaCl,NaBr,Nal
 
Whichofthefollowingcontainstheleastnumberof
 
molecules
?
 
(1)4.4gmCo,
 
(3)1.6gmCH,
 
ElementsAand
BformtwocompoundsB,A,andB,A.
 
0.05molesofB_A,weight9.0gmsand
 
0.10moleofB,Aweight10gmsatomicweightofA
 
andBare
 
dca
AqB
T
E
B,A,TT
B,A
Gt
0.05
 
HR10
YTH
Š|AT
B
YRHTyHRBT
 
(1)20and30
 
(3)40and30
 
(2)3.4gmNH,
 
(4)3.2gm
S0,
 
4.25UIHNH,
 
(1)1.505
x103
 
(3)6.02x
1023
 
Thenumberofmoleculein4.25gmsofNH,s
 
3U3t
HAIBf
 
Whichofthefollowing
 
molecules-
 
(1)2.8gmsofCo
 
(3)1.7gmsofNH,
 
(2)30and40
 
(4)30and20
 
(2)3.01x
1023
 
(4)Noneofthese
 
containhighestnumberof
 
(2)3.2gms
ofCH,
 
(4)3.2gmsofSO,
 
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34.
 
35.
 
36.
 
37
 
38.
 
39.
 
40.
 
41.
 
4.4gm
of
anunknown
gasoccupies2.24litresof
 
volumeatSTP.Thegasmaybe:
 
4.4gm3A5IIT
 
(1)N,0
 
(3)CO,
 
(1)1gm.CO,
 
(3)1gm.O2
 
Whichcontainsleast
no.
ofmolecules:
 
(1)0.1X103
 
(3)1.66
X103
 
(1)1g
Co,
 
If3.01x1020moleculesareremovedfrom98mg.
of
 
H,SO,,thenthenumberofmolesofH,SO,leftare:
 
i
98mg.
H,so,
H3.01
x1020
3uBIfy
 
(3)1gO,
 
(1)6.02x
1023
 
(2)
 
(3)
 
Whichcontainsleast
no.molecules:
 
(4)
 
6.02x1023
 
2
 
6.02x1023
 
Numberofoxygenatomsin8gmsofozoneis-
 
Chemistry-Somebasicconceptsofchemistry
 
3
 
6.02x1023
 
(1)N,
 
6
 
H
STP4R2.24iIrU ÈN
 
(2)CO
 
(3)(N/10)
x
7
 
(4)1&3both
 
(1)4gm.ofH,
 
(2)1gm.N,
 
(4)1gm.H2
 
(3)28gm.ofN,
 
(2)0.5x103
 
(4)9.95
X10-2
 
Numberofneutronspresentin1.7gmsofammonia
 
is-
 
(2)1gN,
 
(4)1gH,
 
Whichofthefollowingcontainslargestnumberof
 
atoms?
 
(2)N,/10
x4
 
(4)N,
x
10
x7
 
(2)16gm.of0,
 
(4)18gm.
ofH,0
 
Whichofthefollowingcontainsgreatestnumbeof
 
OXygenatoms
 
42
 
43
 
44
 
45
 
46
 
(1)1g
ofO
 
(2)1gofO,
 
(3)1g
of0,
 
(4)allhavethe
samenumberofatoms
 
Thetotalnumberofionspersentin1mlof0.1M
 
bariumnitratesolutionis-
 
0.1
M
ayH
TIRREfaeyg
1fA.
it.
 
(1)6.02
x1018
 
(3)3.0
x
6.02
x1019
 
(1)12gofC
 
Thenumberofatomspresentin0.5gatomof
 
nitrogenissameas
theatomsin-
 
(3)8gofoxygen
 
1moleculeofoxygen
 
1atomofNitrogen
 
Rearrangethefollowing(|toM)intheorderof
 
increasingmassesandchoosethecorrectanswer.
 
(Atomicmasses
:
N=14,0
=
16,Cu
=
63)
 
Cu
=
63)
 
(2)6.02x
1019
 
II
1x100(gmmolecularweight
ofoxygen)
 
(4)3.0
x
6.02
x1018
 
IV1x1010(gmatomicweightofcopper)
 
(1)Il<I<Il
<IV
 
3NHST
JRgRI
(4RHIYYRN=14,0
=
16,
 
(3)Il
<IIl<|<
IV
 
(2)32gof
S
 
(4)24gofMg
 
IV1x1o10o4RTAHUHIVE
 
(1)1x1023
 
(3)2x1023
 
(1)1.8
 
No.ofatomsin4.25gm.ofNH,isapprox:
 
(2)IV
<
IIl<Il<|
 
4.25
UTHNH,
À
YRHIysI IAT4T
 
(4)Ill<
IV<|<
 
(2)1.5
>X
1023
 
(4)
6x
1023
 
Totalnumberofprotons,electronsandneutronsin
 
12gm
ofC
is:
 
(2)12.044
x1023
 
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47.
 
48.
 
49.
 
50
 
51
 
52.
 
53.
 
(3)1.084
x
105
 
(1)18gm
 
Theactualweightof
a
moleculeofwateris-
 
(2)2.99x102
gm
 
(4)Noneofthese
 
(3)both(1)&(2)arecorrect
 
If224mlof
a
triatomic
gashas
amassof1gat273
K
 
and
1atmpressure,thenthe
massofoneatomis
 
afayofayrarfrqgtÀ
 
atm|4RJHT
1gm
 
(1)8.30x10-gm
 
(2)2.08
x
10
gm
 
(3)5.53x10-"gm
 
(4)6.24
x
10-"gm
 
Chemistry-Somebasicconceptsofchemistry
 
(1)1gatomof
C
 
(2)½moleofCH,
 
Whichofthefollowinghasthehighestmass?
 
(3)10mlofwater
 
0.01
 
(4)3.011
x
10atomsofoxygen
 
(1)224ml
 
(4)10.84
×1023
 
(3)336ml
 
0.01moleofiodoform(CHI,)reactswithAgto
 
produce
agaswhosevolumeatNTPis
 
(1)50moleof0,
 
224ml35T273
KBTN1
 
(3)125moleof0,
 
(1)63.5g
 
SGTÍ
(CHI,),
Ag
 
(3)34
g
 
Tty64HUTT
 
Theequation
:
 
Assumingthatpetrolisoctane(C,H,)andhasdensity
 
0.8g/ml,1.425litreofpetrol
oncomplete
 
combustionwillconsume.
 
0.8g/ml,
1.425iieRtgloquiGETfay
 
fhI
pRGtH
 
(2)112ml
 
(4)Noneofthese
 
TheminimumquantityingramsofH,Sneededto
 
precipitate63.5
gofCuwillbenearly:
 
(2)100moleof0,
 
(4)200moleof0,
 
(2)31.75
g
 
(4)20g
 
54
 
55
 
56.
 
3
 
2AI(S)
+
,9,(8)
Al,O,(S)showsthat
:
 
3
 
2AI(S)
+0,(8)
Al,0,(5)qaftt
fh:
 
2
 
(1)2moleofAlreactswithmoleof0,toproduce
 
mole
ofAl,O,/AIFT2
l
0,
 
(2)2gmof
Alreactswith
g
of0,toproduceone
 
moleofAl,o,/2gmAl,gm
0,fb4T GI
Ž
 
2
 
(1)0.6litreCo
 
3
 
(3)2gmmoleofAlreactswithlitreof0,to
 
(2)0.8IitreC0,
 
3
 
2
 
(4)None
 
produce1moleofAl,0,/2
gm
HITA,leR0,
 
moleofAl,O/2iT
AI,
 
(1)27gm
 
3
 
(3)54gm
 
(4)2moleofAlreactswithmoleof0,toproduce1
 
2
 
(3)0.6litreCO,and0.8litreCO
 
(1)2.016litres
 
3
 
3
 
OnelitreofCO,ispassedoverhotcoke.Thevolume
 
becomes1.4litre.Thecompositionofproductsis
 
2
 
2
 
2
 
3
 
If1/2molesofoxygencombinewithAluminimumto
 
formAl,0,thenweightofAluminiummetalusedin
 
thereactionis(Al=27)-
 
2
 
(2)18gm
 
(4)40.5gm
 
CaCO,is90%pure.VolumeofCO,collectedatSTP
 
when10gmsofCaC0,isdecomposedis-
 
Caco,,
90%
g|
GG10
YTHCaco,t34efca
 
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57.
 
58
 
59.
 
(2)1.008litres
 
(3)10.08litres
 
(4)20.16litres
 
If0.5molesofBaCl,ismixedwith0.2molesof
 
Na,PO,themaximumnumberofmolesofBa(PO),
 
thatcanbeformedis-
 
afa
0.5
TBaCl,
T
0.24eTNa,PO,HA
 
(1)0.7
 
(3)0.3
 
ForthereactionA+2B
 
Chemistry-Somebasicconceptsofchemistry
 
moleof
Bwillproduce
 
(1)5moleofC
 
(3)8moleof
C
 
3AMSDTA+2Bc,*fTy,
A 5
HGTT
 
2H,S+
SO,
 
(2)0.5
 
(4)0.1
 
2H,S+
S0,
 
C,5 mole
of
A
and8
 
(2)4moleof
C
 
If1.5gmsofSO,1.5
x
102moleculesofHSaremixed
 
andallowedtoremainincontactaclosedvessel
 
untilthereaction
 
(4)13moleof
C
 
3S+
2H,0,
 
proceedstocompletion.Whichofthefollowing
 
statementistrue
?
 
ufG1.6UTHSO,TTHS 1.5
x1022
3yet
 
35+
2H,0,
 
60
 
61
 
62.
 
(1)Only'S'and'H,0'remaininthereactionvessel./
 
(2)'H,S'willremain
inexcess/'H,S'3AHITÀgf
 
(3)'So,'willremaininexcess/'So,'feoGIT
f
 
(4)None/
ÉT
 
If0.5molofBaCl,ismixedwith0.1moleofNa,PO,,
 
themaximumnumberofmoleofBa,(PO,),that
can
 
beformed
is:
 
uft
0.5
ii
BaCl, 0.1
li
Na,PO,
H
 
(1)0.7
 
(3)0.30
 
(1)X,Y,
 
(3)X,Y,
 
(2)0.05
 
Acompoundof
Xand
Y
hasegualmassofthem.
If
 
theiratomicweightsare30and20respectively.
 
Molecularformulaofthatcompound(itsmol.wt.is
 
120)couldbe
 
(4)0.10
 
(1)MO,
 
(3)M,0
 
(2)X,Y,
 
(4)X,Y,
 
Twooxidesof
a
metalcontains50%and40%ofthe
 
metalrespectively.TheformulaofthefirstOxideis
 
MO.Thentheformulaofthesecondoxide
is
 
(2)M,O,
 
(4)M,0,
 
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Question1
 
Answer
 
Question11
 
Answer
 
Question21
 
Answer
 
Question31
 
Answer
 
4
 
Answer
 
Question41
 
Answer
 
3
 
Question51
 
Question61
 
UTKARSH
 
Answer3
 
Chemistry-Somebasicconceptsofchemistry
 
3
 
12
 
22
 
4
 
32
 
1
 
42
 
3
 
52
 
3
 
62
 
2
 
13
 
2
 
23
 
1
 
33
 
13
 
3
 
53
 
4
 
63
 
4
 
1
 
14
 
1
 
24
 
3
 
34
 
44
 
1
 
54
 
3
 
64
 
4
 
15
 
25
 
35
 
1
 
45
 
4
 
55
 
2
 
65
 
2
 
16
 
26
 
36
 
2
 
46
 
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Some Basic Concepts of Chemistry ( रसायन विज्ञान के मूलभूत सिद्धांत ) @irfanullah_mehar.pdf

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Some Basic Concepts of Chemistry ( रसायन विज्ञान के मूलभूत सिद्धांत ) @irfanullah_mehar.pdf

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