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I N D E X
Topic Page No.
PHYSICAL CHEMISTRY
MOLE CONCEPT
01. Chemical Classification of Matter 02
02. Laws of Chemical Combinations 02
03. Methods to calculate moles 06
04. Basic Reaction Stoichiometry 08
05. Concentration terms 14
06. Methods of expressing concentration of solution 15
07. Solved Example 21
08. Problems Related with Mixing& Dilution 27
09. Summary 29
09. Exercise - 1 30
Exercise - 2 41
Exercise - 3 44
Exercise - 4 45
10. Answer Key 46
11. Hints/Solution 48
MEDICAL ENTRANCE
MOLE CONCEPT
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ACC_CHEMISTRY_Mole Concept 1
Syllabus :
• General Introduction: Importance and scope of chemistry.• Laws of chemical combination, Dalton’s atomic theory: concept of elements, atoms and molecules.• Atomic and molecular masses. Mole concept and molar mass; percentage composition and empirical
and molecular formula; concentration terms, stoichiometry and calculations based on stoichiometry.
Contents :
• Chemical Classification of Matter
• Laws of Chemical Combinations
• Mole concept
• Basic Reaction Stoichiometry
• Concentration terms
• Methods of expressing concentration of solution
• Solved Example
• Exercise
MOLE CONCEPT
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ACC_CHEMISTRY_Mole Concept 2
MOLE CONCEPT
CHEMICALCLASSIFICATION OFMATTER
LAWS OFCHEMICALCOMBINATIONS Law of Conservation of Mass (Lavoisier)
Mass cannot be created or destroyed. In any physical or chemical process, the total mass of the systemremain conserved. This law is not applicable to the nuclear processes, where mass and energy are inter-
conversable by the Einstein’s equation, E =m.c2, (c is the speed of light)
Ex. 1.7 gram of silver nitrate dissolved in 100 gram of water is taken. 0.585 gram of sodium chloridedissolved in 100 gram of water is added to it and chemical reaction occurs. 1.435 gm ofAgCl and 0.85
gm NaNO3are formed. Show that these results illustrate the law of conservation of mass.
Sol. Total mass before chemical change = mass ofAgNO3+ Mass of NaCl + Mass of water
= 1.70 + 0.585 + 200 = 202.285 gram
Total mass after the chemical reaction = mass ofAgCl + Mass of NaNO3+ Mass of water
= 1.435 + 0.85 + 200 = 202.285 gram
Thus in the given reaction
Total mass of reactants = Total mass of the products.
Law of Constant Composition or Definite Proportion (Proust)
The composition of a compound always remains fixed and it is independent to the source from which the
compound is obtained. This law cannot be applied to the compound obtained by using different isotopesof the elements, selectively.
Special Highlights
BRAIN TEASERS: GENERAL MISTAKE:
TEACHER’S ADVICE:
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ACC_CHEMISTRY_Mole Concept 3
Ex. Thefollowingareresultsof analysisof twosamplesof thesameor twodifferentcompoundsofphosphorus
andchlorine.Fromtheseresults,decidewhether the twosamplesare fromthesameordifferentcompounds.
Also state the law, which will be obeyed by the given samples.
Compound AmountP Amount Cl
CompoundA 1.156 g 3.971 g
Compound B 1.542 g 5.297 g
Sol. The mass ratio of phosphorus and chlorine in compoundA,
mP
: mCl
= 1.156 : 3.971 = 0.2911 : 1.000
the mass ratio of phosphorus and chlorine in compound B,
mP
: mCl
= 1.542 : 5.297 = 0.2911 : 1.000
As the mass ratio is same, both the compounds are same and the samples obey the law of definite
proportion.
Law of Multiple Proportion (Dalton)
If two elements combine to form more than one compound, then for the fixed mass of one element, the
mass of other element combined will be in simple ratio.
e.g. CO and CO2
12:16 12:32
ratio = 16 : 32
1 : 2
H2S and H2S2
2:32 2:64
ratio = 32 : 64
1 : 2
Law of Reciprocal Proportion (Richter)
If the two elements combine separately with a third element, the mass ratio of the first two elements
combined with a fixed mass of the third element will be equal to or in simple ratio to the mass ratio of first
two elements in a compounds formed by their direct combination.
e.g. C combines with O to form CO2 and with H to form CH4. In CO2 12g of C reacts with 32g of O,
whereas in CH4 12g of C reacts with 4g of H. Therefore when O combines with H, they should combine
in the ratio of 32 : 4 (i.e. 8:1) or in simple multiple of it. The same is found to be true in H2O molecules.
The ratio of weight of H and O in H2O is 1:8.
Gay Lussac’s Law of Volume Combination
This law is applicable to only those chemical reactions in which at least two reaction components aregaseous. According to this law the volumes of all gaseous reactants reacted and the volumes of all
gaseous products formed, measured at the same pressure and temperature, bear a simple ratio.
e.g. H2 (g) + Cl2 (g) 2HCl (g)
1 unit vol. 1 unit vol. 2 unit vol.
ratio = 1 : 1 : 2
N2 + 3H2 2NH3
1 unit vol. 3 unit vol. 2 unit vol.
ratio = 1 : 3 : 2
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ACC_CHEMISTRY_Mole Concept 4
"WONDER THESE LAWS ARE USEFUL?"
"These are no longer useful in chemical calculations now but gives an idea of earlier
methods of analysing and relating compounds by mass."
Can you think of the cases where the above laws will not be applicable.
MOLE CONCEPT
MoleA mole is the amount of substance that contains as many species [Atoms, molecules ions or otherparticles] as there are atoms in exactly 12 gm of C-12.
species106.0221 mole 23
Ex 2 mole of Cl– ion = 2 × 6.022 × 1023 ions of Cl–
3 moles of H2O= 3 × 6.022 × 1023 Molecules of H
2O
1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion
and 1 mole of molecule termed as 1 gm-molecule.
Atomic mass –Atomic mass of an element can be defined as the number which indicates how many
times the mass of one atom of the element is heavier in comparison to12
1th part of the mass of one atom
of Carbon-12.
Atomic mass =]12-carbonofatomanofMass[
12
1
]elementtheofatomanofMass[
=amu1
amuinatomanofMass
Atomic mass unit (amu) – The quantity [12
1× mass of an atom of C–12] is known as atomic mass
unit. The actual mass of one atom of C-12 = 1.9924 × 10–26 kg
So, 1amu = kg12
109924.1 26= 1.66 × 10–27 k.g. = 1.66 × 10–24 gm =
AN
1gm.
Be clear in the difference between 1 amu and 1 gm.
Gram atomic mass – The gram atomic mass can be defined as the mass of 1 mole atoms of an element.It is also called molar mass of an element.
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ACC_CHEMISTRY_Mole Concept 5
e.g. Mass of one oxygen atom = 16 amu =AN
16gm.
Mass of NA
oxygen atom = A
A
N.N
16= 16 gram
Ex (a) What is the mass of one atom of Cl?
(b) What is the atomic mass of Cl?
(c) What is the gram atomic mass of Cl?
Sol. (a) Mass of one atom of Cl = 17 amu.
(b) Atomic mass of Cl =1amu
amuinatomanofMass=
amu1
amu17= 17.
(c) Gram atomic mass of Cl = [Mass of 1 Cl atom × NA]
= 17 amu × NA
=AN
17× N
Agram = 17 gram
Ex (a) What is the mass of one atom of He ?
(b) What is the atomic mass of He ?
(c) What is the gram atomic mass of He ?
Sol. (a) Mass of one atom of He = 4 amu.
(b) Atomic mass of He =1amu
amuinatomanofMass=
amu1
amu4= 4.
(c) Gram atomic mass of He = [Mass of 1 He atom × NA]
= 4 amu × NA
=AN
4× N
Agram = 4 gram
Molecular mass : Molecular mass is the number which indicates how many times the mass of one
molecule of a substance is heavier in comparison to th12
1part of the mass of one atom of C-12.
Molecular mass =]12-CofatomanofMass[
12
1
substancetheofmoleculeoneofMass
=amu1
substancetheofmoleculeoneofMass
Gram Molecular mass – Gram molecular mass can be defined as the mass of 1 mole of molecules.
e.g. Mass of one molecule of O2= 32 amu = gram
N
32
A
.
Mass of NA
molecules of O2= gmN
N
32A
A
= 32 gm
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ACC_CHEMISTRY_Mole Concept 6
Ex (a) What is the mass of one molecule of HNO3?
(b) What is the molecular mass of HNO3?
(c) What is the gram molecular mass of HNO3?
Sol. (a) Mass of one molecule of HNO3= (1 + 14 + 3 × 16) amu = 63 amu.
(b) Molecular mass of HNO3= 63
amu1
amu63
(c) Gram molecular mass of HNO3= Mass of 1-molecule of HNO
3× N
A
= 63 amu × NA
=AN
63gm × N
A= 63 gram
Methods to calculate moles :
Volum
eW
eight
Mole
No. ofparticles
(i) From number of particles
AN
molecules]es [atoms/of ParitclGiven no.eNo. of mol
Ex. A piece of Cu contains 6.022 × 1024 atoms. How many mole of Cu does it contain?
Sol. No. of mole =A
24
N
10022.6 = 23
24
10022.6
10022.6
= 10 mole
(ii) From given Mass –(a) For atom :
No. of mole =massatomicGram
gm)substance(theofmassGiven
(b) For molecules :
No. of mole =massmolecularGram
gm)substance(theofmassGiven
Ex. How many molecules of water are present in 9 gram of water?
Sol.2310023.6
18
9 = 3.011 × 1023
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ACC_CHEMISTRY_Mole Concept 7
(ii) From the given volume of a gasOne mole of Gas occupies 22.4 L at NTP and 22.7 L at STP.
22.7
e)K (in litr273&bar1gas atvolume ofn at S.T.P..
OR
22.4
(in litre)m and 273Kgas at 1atvolume ofn at N.T.P..
If volume is given under any other condition of temperature and pressure, then use the ideal gas equationto find the no. of mole.
RT
PV)n(moleof.No (P in atm, V in L, R = 0.0821 atmLK–1mol–1, T in K)
1 mole of any gas at STP occupies 22.7 litre.
Units of Pressure1 atm = 76 cm Hg = 760 torr = 1.01325 bar = 1.01325 × 105 pa. (N/m2)
= 760 mm of Hg
Units of temp :
273CK
Units of R :R = 0.0821 litre-atm/mole.K
= 8.314 J/mole.K= 1.987 cal/mole.K
Units of volume :
ml10m10litre1cm10dm1 33333 3
Ex. A sample of He gas occupies 5.6 litre volume at 1 atm and 273 K. How many mole of He are present inthe sample?
Sol. No. of mole = 25.04
1
4.22
6.5
Gases do not have volume. What is meant by "Volume of gas"?
Do not use this expression (PV = nRT) for solids/liquids.
How would I calculate moles if volume of a solid is given?
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ACC_CHEMISTRY_Mole Concept 8
Significance of Chemical Equations &Avogadro’s Law :
Achemical equationdescribes thechemicalprocessbothqualitativelyandquantitatively.Thestoichiometric
coefficients in the chemical equation give the quantitative information of the chemical process. These
coefficients represent the relative number of molecules or moles of the reactants and products, e.g.,
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
2 molecules 2 molecules 3 molecules
or 2 N molecules 2 N molecules 3 N molecules
or 2 moles 2 moles 3 moles
Avogadro’s principle states that under the same conditions of temperature and pressure, equal volumes
of gases contain the same number of molecules. Thus, for homogeneous gaseous reactions, the
stoichiometric coefficients of the chemical equation also signify the relative volumes of each reactant and
product under the same conditions of temperature and pressure, e.g.,
H2(g) + I2(g) 2 HI(g)
1 molecule 1 molecule 2 molecule
or 1 mole 1 mole 2 mole
or 1 volume 1 volume 2 volume
(T & P constant)
or 1 pressure 1 pressure 2 pressure(T & V constant)
BASIC REACTION STOICHIOMETRY :
Suppose a Reaction :
2NH3(g) N2(g) + 3H2(g)
2 mole 1 mole 3 mole
34 gm 28 gm 6 gm
On the basis of above reaction we can say :
2 moles of NH3 will react to form one mole of N2 and 3 mole of H2
34 gm of NH3 on decomposition will give 28 gm of N2 and 6 gm of H2
Like if we have to answer following question :
Ex How many moles of H2 will be produced on decomposition of 0.8 moles of NH3?
Sol. As with 2 moles of NH3 , 3 moles of H2 are produced
with 0.8 moles of NH3 ,2
3× 0.8 = 1.2 moles of H2 will be produced.
Ex. During decomposition of NH3 , 14 gm of N2 is produced, what weight of H2 gas will be produced?
Sol.. With 28 gm of N2 , 6 gm of H2 are produced
with 14 gm of N2, 1428
6 = 3 gm of H2 will be produced
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ACC_CHEMISTRY_Mole Concept 9
Limiting Reagent :
The reactant which gives least amount of product on being completely consumed is known as limitingreagent.
OR
Limiting reagent is the reactant that is completely consumed when a reaction goes to completion.
It comes into the picture when reaction involves two or more reactants. For solvingsuch reactions, first step is to calculate Limiting Reagent.
Amount of product formed or amount of other reactant left is determined with respectto limiting reagent only.
Calculation of Limiting Reagent:
(i) By calculating the required amount by the equation and comparing it with given amount.
[Useful when only two reactant are there]
(ii) By calculating amount of any one product obtained taking each reactant one by one irrespective of other
reactants. The one giving least product is limiting reagent.
(iii) If initially weight or volume of reactant is given then convert them in moles.
(iv) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting
reagent. [Useful when number of reactants are more than two.]Ex. If 20gm of CaCO3 is treated with 20gm of HCl, how many grams of CO2 can be generated according
to followingreaction?CaCO3(g) + 2HCl(aq) CaCl2(aq) + H2O() + CO2(g)
Sol. Mole of CaCO3 = 2.0100
20
Mole of HCl = 274.073
20
tcoefficientricStoichiome
Molefor CaCO3 = 2.0
1
2.0
tcoefficientricStoichiome
Molefor HCl = 137.0
2
274.0
So HCl is limiting reagent
According to reaction : 73 gm of HCl gives 44gm of CO2
20 gm HCl will give 2073
44 =12.054 gm CO2
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ACC_CHEMISTRY_Mole Concept 1 0
Ex. 2 moles ofA, 3 moles of B and 4 moles of C are taken in vessel where they reacts according to reaction.
A + 2B + 3C 5D + 6E
Find the maximum number of moles of ‘E’. Which can produced and moles of reactant left?
Sol. A + 2 B + 3C 5D + 6E
moles 2 3 4
now divide with respective stoichiometric coefficient.
Ratio : 2/1 3/2 4/3
least ratio
C is limitingreagent
Now amount of product formed or amount of reactant left will be determined corresponding to C as C
is limitingreagent
From reaction stoichiometry
3 moles of C give, 6 moles of E
With 4 moles of C , 43
6 = 8 moles of E will be formed.
Also, 3 moles of C react with 1 mole ofA
4 moles of C will react with 43
1 =
3
4moles ofAA
Moles of A left = 2 –3
4=
3
2moles
Problems related with mixture
Ex. 4 grams of a mixture of CaCO3 and Sand (SiO2) is treated with an excess of HCl and 0.88 gm of CO2is produced. What is the percentage of CaCO3 in the original mixture?
Sol. CaCO3 + 2HCl CaCl2 + H2O + CO2SiO2 + HClNo reactionCaCO3 = x gm100 gm CaCO3 gives 44 gm CO2x gm CaCO3 gives0.88 gm CO2
88.0
44
x
100 x = 2 gram
% CaCO3 = 1004
2 = 50%
Percentage yield :In general when a reaction is carried out in the laboratory we do not obtain the theoretical amount ofproduct. The amount of product that is actually obtained is called the actual yield. Knowing the actualyield and theoretical yield, the % yield can be calculated by the following formula–
Percent yield = 100yieldlTheoritica
yieldActual
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ACC_CHEMISTRY_Mole Concept 1 1
Ex. For the reaction
CaO + 2HCl CaCl2 + H2O
1.23 gram of CaO is reacted with excess of hydrochloric acid and 1.85 gm CaCl2 is formed. What is the
% yield of the reaction?
Sol. CaO + 2HCl CaCl2 + H2O
56 gm CaO will produce 111 gm CaCl2
1.23 gram of CaO will produce 23.156
111 = 2.43 gm CaCl2
Thus Theoretical yield = 2.3 gm
Actual yield = 1.85 gm
% yield = 10043.2
85.1 = 76%
Percent purity :
Depending upon the mass of the product, the equivalent amount of reactant present can be determinedwith the help of given chemical equation. Knowing the actual amount of the reactant taken and theamount of reactant calculate with the help of a chemical equation the purity can be determined.
% purity = 100akenreactant tofamountActual
equationchemicalthefromcalculatedreactantofAmount
The actual amount of any limiting reagent consumed in such incomplete
reactions is given by [% yield × given moles of limiting reagent][For reversible reactions]
For irreversible reaction with % yield less than 100, the reactants is
converted to product (desired) and waste.
Ex. Calculate the amount of (CaO) in kg that can be produced by heating 200 kg lime stone that is 90% pureCaCO3.
Sol. Pure CaCO3 = kg180100
90200
CaCO3 CaO + CO2
100 k.g. 56 k.g.
180 x
x
56
180
100
n x = 100.8 kg
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ACC_CHEMISTRY_Mole Concept 1 2
Principle of atom conservation
The principle of conservation of mass, expressed in the concepts of atomic theorymeans the conservationof atoms. And if atoms are conserved, moles of atoms shall also be conserved. This is known as theprinciple of atom conservation. This principle is in fact the basis of the mole concept.
Ex. All carbon atoms present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How manygram of CO2 were obtained?
Sol. KH3(C2O4)2.2H2O CO2
Apply P.O.A.C. on carbon atom
4 × mole of KH3(C2O4)2.2H2O = 1 × mole of CO2
254
2544 = 1 × mole of CO2
mole of CO2 = 4
wt. of CO2 = 4 × 44 = 176 gram
Average atomic mass
Average atomic mass = moletotal
masstotal
Let a sample contains n1 mole of atoms with atomic wt M1 and n2 mole of atoms with atomic wt M2.then
21
2211av nn
MnMnM
Ex. Find the average atomic mass of a mixture containing 25% by mole Cl37 and 75% by mole Cl35 ?
Sol. n1 = 25 M1 = 37
n2 = 75 M2 = 35
5.357525
35753725Mav
Average molecular mass :(Weight of one mole of mixture)
Average molecular mass = moletotal
masstotal
Letasamplecontainsn1 moleofmoleculeswithmolecularwtM1 andn2 moleofmoleculeswithmolecular
wt. M2 , then :-
21
2211av nn
MnMnM
Ex. Air is a mixture of O2 and N2 in which O2 is present 20% by mole and N2 is present 80% by mole. Findout the average M. wt. of air ?
Sol. n1 = 20 M1 = 32
n2 = 80 M2 = 28
)nn(
MnMnM
21
2211av
gram8.28)8020(
28803220
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ACC_CHEMISTRY_Mole Concept 1 3
Chemical Formula : It is of two types -
(i) Molecular formula : Chemical formula that indicate the actual number and type of atoms in a
molecule are called molecular formula eg. - Molecular formula of benzene is C6H
6
NaCl i not a molecular formula but actually is a formula unit.
(ii) Empirical formulae : The chemical formulae that give only the relative number of atoms of each
type in a molecule are called empirical formulae eg. - empirical formula of benzene is CH.
Check out the importance of each step involved in calculations of empirical formula.
Steps for writing the empirical formula
The percentage of the elements in the compound is determined by suitable methods and from the data
collected, the empirical formula is determined by the following steps:
(i) Divide the percentage of each element by its atomic mass. This will give the relative number
of moles of various elements present in the compound.
(ii) Divide the quotients obtained in the above step by the smallest of them so as to get a simple
ratio of moles of various elements.
(iii) Multiply the figures, so obtained by a suitable integer if necessary in order to obtain a whole
number ratio.
(iv) Finally write down the symbols of the various elements side byside and put the above number as
the subscripts to the lower right hand corner of each symbol. This will represent the empirical
formula of the compound.
If percentage is not given than from whatever given information try to calculate number
of moles of each element and then take their ratio to determine empirical formula.
Steps for writing the molecular formula
(i) Calculate the empirical formula as described above.
(ii) Find out the empirical formula mass by adding the atomic masses of all the atoms present in the
empirical formula of the compound.
(iii) Divide themolecularmass (determinedexperimentallybysomesuitablemethod)by theempirical
formula mass and find out the value of n.
Molecular formula = n × Emperical formula
where n = weightformulaEmperical
weightMolecular
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ACC_CHEMISTRY_Mole Concept 1 4
Vapour density :Some times in numericals molecular mass of volatile substance is not given, instead vapour density isgiven. Vapour density can be defined as
PandTsameatHofDensity
PandTgivenaatgasofDensity.D.V
2
For gases : PV = nRT = RT.M
w
RTV
wPM
= dRTT d for a gas =
.RT
M.P gas
RT
M.PRT
M.PD.V
2H
gas =
2H
gas
M
M
2
MV.D
gas
Ex. A compound of nitrogen and oxygen was found to contain 7 : 16 by mass N and O respectively.Calculate molecular formula of the compound if V.D. is 46 ?
Sol. Let mass of N = 7 K gramMass of O = 16 K gram
Element wt. Atomic wt.Atomic
wtSimple ratio
N 7 K 14 K5.014
K7 1
K5.0
K5.0
O 16 K 16 K16
K16 2
K5.0
K
N : O = 1 : 2Empirical formula= NO2(Empirical formula)n = Molecularweight
46
D.V2
.wtformulaempirical
wt.Mn
=
46
462= 2
Molecular formula = (NO2)2 = N2O4
CONCENTRATION TERMS
A solution is a homogeneous mixture of two or more substances, the composition of which may vary
within limits. “Asolution is aspecial kindof mixture inwhichsubstances are intermixed so intimately that
they can not be observed as separate components”. The substance which is to be dissolved is called
solute while the medium in which the solute is dissolved to get a homogeneous mixture is called the
solvent.Asolution is termed as binary and ternary if it consists of two and three components respectively.
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ACC_CHEMISTRY_Mole Concept 1 5
METHODS OF EXPRESSING CONCENTRATION OF SOLUTION :
Concentration of solution is the amount of solute dissolved in a known amount of the solvent or solution.
The concentration of solution can be expressed in various ways as discussed below.
What is solute and solvent in a solution?
If the mixture is not homogeneous, then none of the concentration
term is applicable.
General concentration terms :
(a) Density =Volume
Mass, Unit : g/cm3
(b) Relative density = substanceereferencDensity of
bstancef any suDensity o
(c) Specific gravity =C4water atDensity of
bstancef any suDensity o
(d) Vapour density = essurere and prtemperatusamegas atf HDensity o
end pressurerature afied tempat specif vapourDensity o
2
Which of these are temperature dependent?
Classify each of them as w/w, w/v, v/v ratio.
Percentage : It refers to the amount of the solute per 100 parts of the solution. It can also be called as
parts per hundred (pph). It can be expressed by any of following four methods.
(i) Weight to weight percentage (%w/w) =
100gutionWt. of sol
guteWt. of sol
e.g., 10% Na2CO
3solution w/w means 10g of Na
2CO
3is dissolved in 100g of the solution. (It means
10g Na2CO
3is dissolved in 90 of solvent)
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ACC_CHEMISTRY_Mole Concept 1 6
(ii) Weight to volume percent (% w/v) =
100)(cmsolutionVolume of
guteWt. of sol3
e.g., 10% Na2CO
3(w/v) means 10g Na
2CO
3is dissolved in 100 cm3 of solution.
(iii) Volume to volume percent (% v/v) = 100cmsolutionVolume of
cmsoluteVolume of3
3
e.g., 10% ethanol (v/v) means 10cm3 of ethanol dissolved in 100cc of solution.
(iv) Volume to weight percent (%v/w) = 100solutionofWt.
soluteof.Vol
e.g., 10% ethanol (v/w) means 10cm3 of ethanol dissolved in 100g of solution.
Classify each of the given ratio as w/w, w/v, v/v and comment on their
temperature dependence.
Ex.: Concentrated nitric acid used as laboratory reagent is usually 69% by mass of nitric acid. Calculate thevolume of the solution which contains 23 g nitric acid. The density of concentrated acid is 1.41 g cm–3.
Sol. Given HNO3 is 69% by mass;density of HNO3 = 1.41 g cm–3.
Thus (i) 69 g HNO3 is present in conc. HNO3 = 100 g
23 g HNO3 is present in conc. HNO3 =69
100× 23 = 33.33 g
(ii) Volume of solution required =Density
Mass=
41.1
33.33= 23.64 mL
Parts per million (ppm) and parts per billion (ppb) : When a solute is present in very small quantity,
it is convenient to express the concentration in parts per million and parts per billion. It is the number of
parts of solute per million (106) or per billion (109) parts of solution. It is independent of the temperature.
ppm = 610solutionofmassTotal
componentsoluteofMass
ppm = 610solventofmassTotal
componentsoluteofMass (For dilute solution)
ppb =910
solutionofmassTotal
componentsoluteofMass
ppb =910
solventofmassTotal
componentsoluteofMass (For dilute solution)
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ACC_CHEMISTRY_Mole Concept 1 7
Ex. Calculate the parts per million of SO2 gas in 250 ml water (density 1 g cm–3) containing5 × 10–4 g of SO2 gas.
Sol. Mass of SO2 gas = 5 × 10–4 g;Mass of H2O = Volume × Density = 250 cm3 × 1 g cm–3 = 250 g
Parts per million of SO2 gas =6
4
10g250
105
= 2
Strength :The strength of a solution is defined as the amount of solute in grams present in one litre (or
dm3) of the solution. It is expressed in g/litre or (g/dm3)
litresinsolutionofVolume
gramsinsoluteofMassStrength
Mole fraction ( ) :
Mole fraction may be defined as the ratio of number of moles of one component to the total number of
moles of all the components (solvent and solute) present in the solution. It is denoted by the letter . It
may be noted that the mole fraction is independent of the temperature. Mole fraction is dimensionless. If
a solution contains the components A and B and suppose that WA
gram of A and WB
gram of B are
present in it.
Number of moles ofAis given by,A
AA
M
Wn and the number of moles of B is given by,
B
BB
M
Wn
where MA
and MB
are molecular mass ofAand B respectively.
Total number of moles ofAand B = nA
+ nB
Mole fraction ofA,BA
AA
nn
n
Mole fraction of B,BA
BB
nn
n
The sum of mole fractions of all the components in the solution is always one.
1nn
n
nn
n
BA
B
BA
ABA
BA = 1
Thus, if we know the mole fraction of one component of a binary solution, the mole fraction of the other
can be calculated.
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ACC_CHEMISTRY_Mole Concept 1 8
Ex. Find out the weights of acid and water required to prepare CH3COOH solution of 0.3 mole fraction.
Sol. 3.0COOHCH 3
7.03.01OH2
Wt. of CH3COOH = COOHCH3
× mol. wt.(CH3COOH) = 0.3 × 60 = 18
Wt. of water = OH2 × mol. wt. (H
2O) = 0.7 × 18 = 12.6 g
Mass fraction : Mass fraction of a component in a solution is the mass of the component divided by the
total mass of the solution. For a solution containing wA
gm ofAand wB
gm of B
Mass fraction ofBA
A
ww
wA
;
Mass fraction ofBA
B
ww
wB
Molarity (M) : Molarity of a solution is the number of moles of the solute per litre of solution (or
number of millimoles per ml. of solution). Unit of molarity is mol/litre or mol/dm3. For example, a molar
(1M) solution of sugar means a solution containing 1 mole of sugar per litre of the solution. Solutions in
terms of molarity is generally expressed as,
1M = One molar solution, 2M = Molarity is two,2
Mor 0.5M = Semimolar solution,
Mathematically, molarity can be calculated byfollowing formulas :
(i)litresinsolutionofVol.
solute(n)ofmolesof.NoM
(ii)mlinsolutionofVol.
1000
soluteofwt.Mol.
gm)solute(inof.WtM
(iii)mlinsolutionofVol
soluteofmillimolesof.NoM
(iv) If molarity and volume of the solution are changed from M1, V
1to M
2,V
2. Then,
M1V
1= M
2V
2
(v) In balanced chemical equation, if n1moles of reactant-1 react with n
2moles of reactant-2 . Then,
n1A + n
2B Product
2
22
1
11
n
VM
n
VM
(vi) If two solutions of the same solute are mixed then molarity (M) of resulting solution
)VV(
VMVMM
21
2211
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ACC_CHEMISTRY_Mole Concept 1 9
Molality (m) :
It is the number of moles of the solute per 1000g of the solvent. Unit of molality is mol/kg. For example,
a 0.2 molal (0.2m) solution of glucose means a solution obtained by dissolving 0.2 mole of glucose in
1000 gm of water. Molality (m) does not depend on temperature since it involves measurement of
weightof liquids.
Mathematically molality can be calculated byfollowingformulas,
(i)t in kgthe solvenWeight of
he solutemoles of tNumber ofm
(ii) 1000t in gmthe solvenWeight of
he solutemoles of tNumber ofm
(iii)gmsolvent inWeight of
1000
tef the soluMol. wt. o
uteWt. of solm
(iv)vent in gmWt. of sol
solutelimoles ofNo. of milm
Note : Molarity is temperature dependent but molality is temperature independent.
Ex. Calculate the molarity and molality of a solution of H2SO
4(sp. gr. = 1.98) containing 27% H
2SO
4by
mass.
Sol. Vol of 100 g of 27% H2SO
4=
d
.wt=
198.1
100ml
)litre(solutionof.vol
.wt.mol/.wtM
42SOH
=10098
1000198.127
= 3.3 mol L–1
kgsolventof.wt
.wt.mol/.wtm
42SOH
=98)27100(
100027
= 3.77 mol L–1
Formality (F) :Formalityof solution may be defined as the number of gram formula masses of the ionic
solute dissolved per litre of the solution. It is represented by F. Commonly, the term formality is used to
express the concentration of the ionic solids which do not exist as molecules but exist as network of ions.
A solution containingone gram formula mass of solute per litre of the solution has formality equal to one
and is called Formal solution. It may be mentioned here that the formality of a solution changes with
change in temperature.
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ACC_CHEMISTRY_Mole Concept 2 0
litresinsoluionofVolume
soluteofmassesformulagramofmberNu)F(Formality
)(solutionVolume ofsolutela mass ofgram formu
(g)nic soluteMass of io
l
Ex. What will be the formality of KNO3 solution having strength equal to 2.02 g per litre ?Sol. Strength of KNO3 = 2.02 gL–1
g formula weight of KNO3 = 101 gWe know that :
Formality of KNO3 =3
-1
KNOof.wtformula.g
ginstrength l=
101
02.2= 0.02 F
Note : Although formality is different from molarity term but we generally use molarity termat place of formality as well
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ACC_CHEMISTRY_Mole Concept 2 1
SOLVED EXAMPLE
Ex.1 How many carbon atoms are present in 0.35 mol of C6H12O6
Sol. 1 mol of C6H12O6 has = 6 NA atoms of C
0.35 mol of C6H12O6 has
= 6 × 0.35 NA atoms of C
= 2.1 NA atoms
= 2.1 × 6.023 × 1023 = 1.26 × 1024 carbon atoms
Ex.2 8 litre of H2 and 6 litre of Cl2 are allowed to react to maximum possible extent. Find out the finalvolume of reaction mixture. Suppose P and T remains constant throughout the course of reaction-
Sol. H2 + Cl2 2 HCl
Volume before reaction 8 lit 6 lit 0
Volume after reaction 2 0 12
Volume after reaction = Volume of H2 left + Volume of HCl formed
= 2 + 12 = 14 lit
Ex.3 How many years it would take to spend Avogadro's number of rupees at the rate of 1 millionrupees in one second -
Sol. 106 rupees are spent in 1sec.
6.023 × 1023 rupees are spent in
= 6
23
10
10023.61 sec
=36524606010
10023.616
23
years = 19.098 × 109 year
Ex.4 Naturally occurring chlorine is 75.53% Cl35 which has an atomic mass of 34.969 amu and 24.47%Cl37 which has a mass of 36.966 amu. Calculate the average atomic mass of chlorine-
Sol. Average atomic mass
=100
massatomicitsisotopeIIof%massatomicitsisotopeIof%
=75 53 34 969 24 47 36 96
100
. . . .x x
= 35.5 amu.
Ex.5 The density of O2 at NTP is 1.429g / litre. Calculate the standard molar volume of gas-
Sol. 1.429 gm of O2 gas occupies volume = 1 litre.
32 gm of O2 gas occupies =32
1429.= 22.4 litre/mol.
Ex.6 How many molecules are in 5.23 gm of glucose (C6H12O6)
Sol. 180 gm glucose has = N molecules
5.23 gm glucose has =180
10023.623.5 23= 1.75 × 1022 molecules
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ACC_CHEMISTRY_Mole Concept 2 2
Ex.7 How many g of S are required to produce 10 moles and 10g of H2SO4 respectively?
Sol. 1 mole of H2SO4 has = 32g S
10 mole of H2SO4 has = 32 × 10 = 320 g S
Also, 98g of H2SO4 has = 32 g S
10 g of H2SO4 has = (32 × 10)/98
= 3.265 g S
Ex.8 P and Q are two elements which form P2Q3, PQ2 molecules. If 0.15 mole of P2Q3 and PQ2
weighs 15.9 g and 9.3g, respectively, what are atomic weighs of P and Q?
Sol. Let at. wt. of P and Q be a and b respectively,
Mol. wt. of P2Q3 = 2a + 3b
and Mol. wt. of PQ2 = a + 2b
(2a + 3b) × 0.15 = 15.9
and (a + 2b) × 0.15 = 9.3
Ex.9 The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27ºC. Calculate the moleof NO2 in 100 mole mixture.
Sol. Mol. wt. of mixture of NO2 and N2O4 = 38.3 × 2 = 76.6
Let a mole of NO2 be present in 100 mole mixture
= wt. of NO2 + wt. of N2O4 = wt. of mixture,
a × 46 + (100 – a) × 92 = 100 × 76.6
a = 33.48 mole
Ex.10 How many moles of potassium chlorate to be heated to produce 5.6 litre oxygen at STP?
Sol. 2KClO3 2KCl + 3O2
Mole ratio for reaction 2 mole 2 mole 3 mole
3 × 22.4 litre O2 is formed by 2 mole KClO3
5.6 litre O2 is formed by4.223
6.52
= 1/6 mole KClO3
Ex.11 Zinc and hydrochloric acid react according to the reaction :
Zn(s) + 2HCl (aq.) ZnCl2(aq.) + H2(g)
If 0.30 mole of Zn are added to hydrochloric acid containing 0.52 mole HCl, how many moles
of H2 are produced?
Sol. Zn(s) + 2HCl(aq.) ZnCl2(aq.) + H2(g)
Initial moles 0.30 0.52 0 0
HCl is limiting reagent
Final moles )4.02
52.30.0( 0 0 0.26
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ACC_CHEMISTRY_Mole Concept 2 3
Ex.12 How many molecules are present in 1m of water vapours at STP -
Sol. 22.4 litre water vapour at STP has
= 6.023 × 1023 molecules
1 × 10–3 litre water vapours at STP has
=4.22
10023.6 23× 10–3 = 2.69 × 10+19
Ex.13 A mixture of FeO and Fe3O4 when heated in air to constant weight gains 5% in its weight. Findout composition of mixture.
Sol. Let weight of FeO and Fe3O4 be a and b g, respectively.
2FeO +2
1O2 Fe2O3
2Fe3O4 +2
1O2 3Fe2O3
144 g FeO gives 160 g Fe2O3
a gm of FeO gives 160 × a /144 g Fe2O3
Similarly, weight of Fe2O3 formed by b gm of Fe3O4 =464
b3160
Now if a + b = 100 ; then 105464
b3160
144
a160
Solving these two equations : a = 21.06 and b = 78.94
Percentage of FeO = 21.06% and percentage of Fe3O4 = 78.94%
Ex.14 A plant virus is found to consist to uniform cylindrical particles of 150Å in diameter and 5000Ålong. The specific volume of the virus is 0.75 cm3/g. What is the molecular weight of virus.
Sol. Volume of virus = r2l =316816 cm10994.010500010
2
150
2
150
7
22
Weight of one virus = g10178.1g75.0
10884.0 1616
Mol. wt of virus = 1.178 × 10–16 × 6.023 × 1023 = 7.095 × 107
Ex.15 8g of sulphur are burnt to from SO2, which is oxidised by Cl2 water. The solution is treated withBaCl2 solution. The amount of BaSO4 precipitated is :
Sol.4
BaCl–24
Cl2
O BaSOSOSOS 222
One mole of S will give 1 mole of BaSO4
Mole of BaSO4 formed = moles of sulphur
=32
8=
4
1
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ACC_CHEMISTRY_Mole Concept 2 4
Ex.16 An unknown compound contains 8% sulphur by mass. Calculate
(a) Least molecular weight of the compound and
(b) Molecular weight if one molecule contains 4 atoms of “S”
Sol. 100 gm molecule 8 gm sulphur
forminimum molecular weight at least 32 gm of sulphur
M.M compound =8
100× 32 = 400
If one molecule contains 4 atomsM.M. = 4 × 400 = 1600 Ans.]
Ex.17 If molecular weight of glucose-1-phosphate is 260 and its density is 1.5g/mL. What is the averagevolume occupied by each molecule of this compound ?
Sol. Volume occupied by one molecule =density
moleculeoneofmass=
5.1
1066.1260 –24= 2.88 × 10–22 mL
Ex.18 Amixture of 1.65 × 1021 molecules of X and 1.85 × 1021 molecules ofYweighs 0.638 g. If the molecular
weight of X is 42, what is the molecular weight ofY
Sol. mmix
= mx+ m
y
=yAxA
wt.mol.N
moleculesof.nowt.mol.
N
moleculesof.no
or, 0.638 = y23
21
23
21
M10023.6
101.8542
10023.6
1065.1
or, My
= 170.27
Ex.19 Calculate the molarityof the following solutions :(a) 4g of caustic soda is dissolved in 200 mL of the solution.(b) 5.3 g of anhydrous sodium carbonate is dissolved in 100 mL of solution.(c) 0.365 g of pure HCl gas is dissolved in 50 mL of solution.
Sol. (a) M = 5.0510
1
200
1000
40
4
(b) M = 5.0100
1000
106
3.5
(c) M =50
1000
5.36
365.0 = 2.0
5
10
55.36
365
Ex.20 Density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Then molarity of solutionwillbe
Sol. M = 1000100
09.1
98
13 = 1.445
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ACC_CHEMISTRY_Mole Concept 2 5
Ex.21 Thedensityofasolutioncontaining40%bymassofHCl is1.2g/mL.Calculate themolarityof thesolution.
Sol. M = 1000100
2.1
5.36
40 = 15.13
365
12040
Ex.22 15 gof methyl alcohol is present in 100 mLof solution. If densityof solution is 0.90 g mL–1. Calculate themass percentage of methyl alcohol in solution
Sol. g15W OHCH3
lnsoW = 100 × 0.9 = 90
Wsolvent = 90–15 = 75
% of solute =90
15× 100 =
3
50
6
100 = 16.67
Ex.23 A 6.90 M solution of KOH in water contains 30% by mass of KOH. What is density of solution ingm/ml.
Sol. 30 g of solute in 100 g of solution6.9 mole of KOH 1000 mL of solution6.9 × 56 g 1000 mL of solution
30 g =569.6
1000
× 6.9 × 56
= 23 × 56 × 10–3 = 1.288 g ml–1
Ex.24 The average concentration of Na+ ion in human body is 3 to 4 gm per litre. The molarity of Na+ ion isabout.
Sol. M1 =21
3=
7
1= 0.12
M2 = 19.021
4
Avg. =
2
MM 21= 0.15
Ex.25 What is the concentration of chloride ion, in molarity, in a solution containing 10.56 gm BaCl2.8H2O perlitre of solution ? (Ba = 137)
Sol. BaCl2 Ba+2 + 2Cl¯1M 1 M 2M
¯ClM = 100014471137
100056.102
=
352
12.21= 0.06
Ex.26 The mole fraction of solute in aqueous urea solution is 0.2. Calculate the mass percent of solute ?Sol. 0.2 mole solute = 0.2 × 60 = 12 g
0.8 mole solvent = 0.8 × 18 = 14.4
Mass percentage =4.1412
12
× 100
= 45.45 %
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ACC_CHEMISTRY_Mole Concept 2 6
Ex.27 The concentration of Ca(HCO3)2 in a sample of hard water is 500 ppm. The density of water sample is1.0 gm/ml. Calculate the molarity of solution ?
Sol. M = 6
3
10
101
162
500 = 3 × 10–3
Ex.28 0.115 gm of sodium metal was dissolved in 500 ml of the solution in distilled water. Calculate themolarity of the solution ?
Sol. M =500
1000
23
115.0
=23
10230 3= 0.01
Ex.29 How much BaCl2 would be needed to make 250 ml of a solution having the same concentration of Cl–
as one containing 3.78 gm NaCl per 100 ml ? (Ba = 137)Sol. M of NaCl = 0.646
M of BaCl2 =2
646.0
Wt of BaCl2 = 16.8 g
Ex.30 Calculate molality (m) of each ion present in the aqueous solution of 2M NH4Cl assuming 100%dissociation according to reaction.
NH4Cl (aq) 4NH (aq) + Cl– (aq)
Given : Density of solution = 3.107 gm / ml.Sol. 2M = 2
M = 5.532107.31000
10002
=1070.3107
2000
=3000
2000=
3
2= 0.667
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ACC_CHEMISTRY_Mole Concept 2 7
PROBLEMS RELATED WITH MIXING& DILUTION
Ex.31 Find out the volume of 98% w/w H2SO4 (density = 1.8 gm/ ml), must be diluted to prepare 12.5 litresof 2.5 M sulphuric acid solution.
Sol. M1V1 = M2V2
100010098
8.198
V = 2.5 × 12.5
1 × 18 × V = 2.5 × 12.5V = 1.736
Ex.32 500 ml of 2 M NaCl solution was mixed with 200 ml of 2 M NaCl solution. Calculate the final volumeand molarityof NaCl in final solution if final solution has density 1.5 gm/ml.
Sol. M1V1 + M2V2 = MV500 × 2 + 2 × 200 = M (700)1000 + 400 = M (700)
700
1400= M
M= 2
Ex.33 Calculate the molarity and molality of a solution of H2SO
4(sp. gr. = 1.98) containing 27% H
2SO
4by
mass.
Sol. Vol of 100 g of 27% H2SO
4=
d
.wt=
198.1
100ml
)litre(solutionof.vol
.wt.mol/.wtM
42SOH
=10098
1000198.127
= 3.3 mol L–1
kgsolventof.wt
.wt.mol/.wtm
42SOH
=98)27100(
100027
= 3.77 mol L–1
Ex.34 Derive the relation between molality (m) and mole fraction of solute, 2
Sol. Molality, m means, m mole of solute in 1000 g of solvent which is equal to 1000/M1 molwhere M1 = molar mass of the solvent.
Mole fraction, 2 = solventofMolessoluteofMoles
soluteofmoles
=1000mM
mM
M
1000m
m
1
1
1
Or11
1
2 mM
10001
mM
1000mM1
12
2
12 mM
10001
mM
10001
1
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ACC_CHEMISTRY_Mole Concept 2 8
Or12
1
mM
1000
1 = mole fraction of solvent [
1 + 2 =1
1 = 1 – 2]
Hence m =11
2
M
1000
Ex.35 The molality and molarity of a solution of H2SO
4are 94.13 and 11.12 respectively. Calculate the density
of the solution.
Sol. d = M
1000
.wt.mol
m
1= 11.12
1000
98
13.94
1= 1.2079 g/ml
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ACC_CHEMISTRY_Mole Concept 2 9
SUMMARY
The study of chemistry is very important as its domain encompasses every sphere of life. Chem-ists study the properties and structure of substances and the changes undergone by them.All substancescontain matter which can exist in three states –solid, liquid or gas. The constituent particles are held indifferent ways in these states of matter and they exhibit their characteristic properties. Matter can also beclassified into elements, compounds or mixtures.An element contains particles of only one type whichmay be atoms or molecules. The compounds are formed where atoms of two or more elementscombine in a fixed ratio to each other. Mixtures occur widely and many of the substances present aroundus are mixtures.
When the properties of a substance are studied, measurement is inherent. The quantification ofproperties requires a system of measurement and units in which the quantities are to be expressed. Manysystems of measurement exist out of which the English and the Metric Systems are widely used. Thescientific community, however, has agreed to have a uniform and common system throughout the worldwhich is abbreviated as SI units (International System of Units).
Since measurements involve recordingof data which are always associated with a certain amountof uncertainty, the proper handling of data obtained by measuring the quantities is very important. Themeasurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, aconvenient system of expressing the numbers in scientific notation is used. The uncertainty is takencare of by specifying the number of significant figures in which the observations are reported. Thedimensional analysis helps to express the measured quantities in different systems of units. Hence, it ispossible to interconvert the results from one system of units to another.
The combination of different atoms is governed by basic laws of chemical combination – thesebeing the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Propor-tions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law.All these laws led to the Dalton’satomic theory which states that atoms are building blocks of matter. The atomic mass of an element isexpressed relative to 12C isotope of carbon which has an exact value of 12u. Usually, the atomic massused for an element is the average atomic mass obtained by taking into account the natural abundanceof different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of theatomic masses of different atoms present in a molecule. The molecular formula can be calculated bydetermining the mass per cent of different elements present in a compound and its molecular mass.
The number of atoms, molecules or any other particles present in a given system are expressedin the terms of Avogadro constant (6.022 1023). This is known as 1 mol of the respective particles orentities.
Chemical reactions represent the chemical changes undergone by different elements and compounds.Abalanced chemical equation provides a lot of information. The coefficients indicate the molar ratios andthe respective number of particles taking part in a particular reaction. The quantitative study of thereactants required or the products formed is called stoichiometry. Using stoichiometric calculations,the amounts of one or more reactant(s) required to produce a particular amount of product can bedetermined and vice-versa. The amount of substance present in a given volume of a solution is expressedin number of ways, e.g., mass per cent, mole fraction, molarity and molality.
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ACC_CHEMISTRY_Mole Concept 3 0
EXERCISE-1
[SINGLE CORRECT CHOICE TYPE]
Q.1 Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2% Hydrogenrespectively. The data illustrates –
(1) Law of conservation of mass (2) Law of constant proportions
(3) Law of reciprocal proportions (4) Law of multiple proportions
Q.2 If water samples are taken from sea, rivers, clouds, lake or snow, they will be found to contain H2 andO2 in the fixed ratio of 1 : 8. This indicates the law of –
(1) Multiple proportion (2) Definite proportion
(3) Reciprocal proportion (4) None of these
Q.3 The law of conservation of mass holds good for all of the following except:
(1)All chemical reactions (2) Nuclear reactions
(3) Endothermic reactions (4) Exothermic reactions.
Q.4 If law of conservation of mass was to hold true, then 20.8 gm of BaCl2 on reaction with 9.8 gm ofH2SO4 will produce 7.3 gm of HCl and BaSO4 equal to -
(1) 11.65 gm (2) 23.3 gm (3) 25.5 gm (4) 30.6 gm
Q.5 Percentage of copper and oxygen in sample of CuO obtained by different methods were found to besame. This proves the law of -
(1) Constant proportion (2) Multiple proportion
(3) Reciprocal proportion (4) None of these
Q.6 The haemoglobin of most mammals contains approximately0.33%of iron bymass. The molecularmassof haemoglobin is 67200. The number of iron atoms in each molecule of haemoglobin is-
(1) 3 (2) 4 (3) 2 (4) 6
Q.7 The atomic mass of calcium is 40. How many atoms are present in 2g calcium ?
(1) 3.0115 × 1022 (2) 3.1115 × 1022 (3) 4.0115 × 1022 (4) 3.1125 × 1022
Q.8 The point given as full stop at the end of a sentence by a graphite pencil weighs 2 × 10–12 g. How manycarbon atoms are present in such a point ?
(1) 1.003 × 1010 (2) 1.004 × 1011 (3) 1.002 × 1012 (4) 1.001 × 10–11
Q.9 Assume the human body contains 80% water. Calculate the number of molecules of water that are in thebody of a person who has a mass of 65 kg.
(1) 17.39 × 1026 (2) 17.39 × 1027 (3) 17.39 × 1226 (4) 18.39 × 1026
Q.10 Morphine contains 67.3% carbon, 4.6 % nitrogen (by mass) and remaining are the other constituents.Calculate the relative number of carbon and nitrogen atoms in morphine.
(1)1
17(2)
17
1(3)
1
16(4)
1
15
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ACC_CHEMISTRY_Mole Concept 3 1
Q.11 Total number of atoms of all elements present in 1 mole of ammonium dichromate [(NH4)2Cr2O7] is
(1) 14 (2) 19 (3) 6 × 1023 (4) 114 × 1023
Q.12 11.2 litre of a gas at NTP weighs 14 g. The gases would be :
(1) N2 (2) CO (3) N2O (4) B2H6
Q.13 The number of molecules of water in 333g ofAl2(SO4)3.18H2O is
(1) 18.0 × 6.02 × 1023 (2) 9.0 × 6.02 × 1023
(3) 18.0 (4) 36.0
Q.14 The atomic weights of two elementsAand B are 40 and 80 respectively. If x g ofAcontains y atoms,
how many atoms are present in 2x g of B
(1)2
y(2)
4
y(3) y (4) 2y
Q.15 One mole of CO2 contains :
(1) 6.023 × 1023 g-atom of C (2) 3.02 × 1023 atom of oxygen
(3) 18.1 × 103 molecule of CO2 (4) 6.023 × 1023 atom of carbon
Q.16 Which has the maximum number of atoms :
(1) 24 g C (12) (2) 56 g Fe (56) (3) 27 gAl (27) (4) 108 g Ag (108)
Q.17 Mass of 1 atom of Hydrogen is -
(1) 1.66×10–24g (2) 10–22 g (3) 10–23 g (4) 10–25 g
Q.18 Which of the following contains the largest number of atoms -
(1) 11g of CO2 (2) 4g of H2 (3) 5g of NH3 (4) 8g of SO2
Q.19 The total number of electrons present in 18 mL water (density 1 g/mL) is -
(1) 6.023 × 1023 (2) 6.023 × 1024 (3) 6.023 × 1025 (4) 6.023 × 1021
Q.20 One mole of P4 molecules contains -
(1) 1 molecule (2) 4 molecules
(3) 1/4 × 6.022 × 1023 atoms (4) 24.088 × 1023 atoms
Q.21 The total number of protons , electrons and neutrons in 12gm of 6C12 is -
(1) 1.084 × 1025 (2) 6.022 × 1023 (3) 6.022 × 1022 (4) 18
Q.22 The number of sodium atoms in 2 moles of sodium ferrocyanide Na4[Fe(CN)6], is-
(1) 2 (2) 6.023 × 1023
(3) 8 × 6.02 × 1023 (4) 4 × 6.02 × 1023
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Q.23 Number of Ca+2 and Cl– ion in 111 g of anhydrous CaCl2 are -
(1) NA, 2NA (2) 2NA, NA (3) NA, NA (4) None
Q.24 16 g of SOx occupies 5.6 litre at NTP.Assuming ideal gas nature, The value of x is :
(1) 1 (2) 2 (3) 3 (4) None of these
Q.25 How many molesof magnesium phosphateMg3(PO4)2will contain 0.25 mole of oxygen atoms:
(1) 0.02 (2) 3.125 ×10–2 (3) 1.25 ×10–2 (4) 2.5 × 10–2
Q.26 How many moles of electron weigh one kilogram :
(1) 6.023 × 1023 (2)108.9
1× 1023 (3)
108.9
10023.6 54(4)
023.6108.9
1
×108
Q.27 2 moles of H2 at NTP occupy a volume of
(1) 11.2 litre (2) 44.8 litre (3) 2 litre (4) 22.4 litre
Q.28 4.48 litres of methane at N.T.P. correspond to-
(1) 1.2 x 1022 molecules of methane (2) 0.5 mole of methane
(3) 3.2 gm of methane (4) 0.1 mole of methane
Q.29 Mol. wt. = vapour density × 2, is valid for -
(1) metals (2) non metals (3) solids (4) gases
Q.30 5.6 litre of a gas at N.T.P. weighs equal to 8 gm, the vapour density of gas is -
(1) 32 (2) 16 (3) 8 (4) 40
Q.31 Equal masses of O2 , H2 and CH4 are taken in a container. The respective mole ratio of these gases incontainer is -
(1) 1 : 16 : 2 (2) 16 : 1 : 2 (3) 1 : 2 : 16 (4) 16 : 2 : 1
Q.32 The vapour density of gasAis four times that of B. If molecular mass of B is M, then molecular mass ofA is -
(1) M (2) 4M (3)M
4(4) 2M
Q.33 The density of air is 0.001293 gm/ml at S.T.P. It’s vapour density is -
(1) 147 (2) 14.7 (3) 1.47 (4) 0.147
Q.34 Nitric acid is manufactured by the Ostwald process, in which nitrogen dioxide reacts with water.
3 NO2 (g) + H2O (l) 2 HNO3 (aq) + NO (g)
How many grams of nitrogen dioxide are required in this reaction to produce 25.2 gm HNO3 ?
(1) 27.6 gm (2) 26.7 gm (3) 72.6 gm (4) None
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Q.35 Flourine reacts with uranium to produce uranium hexafluoride, UF6, as represented by this equation
U(s) + 3F2(g)UF6(g)
How many fluorine molecules are required to produce 2.0 mg of uranium hexafluoride, UF6, from anexcess of uranium ? The molar mass of UF6 is 352 gm/mol.
(1) 2.0 ×1019 (2) 1.0 ×1019 (3) 1.0 ×1219 (4) 1.2 ×1019
Q.36 X gm ofAg was dissolved in HNO3 and the solution was treated with excess of NaCl. When 2.87 gm
ofAgCl was precipitated the value of x is -
(1) 1.08 gm (2) 2.16 gm (3) 2.70 gm (4) 1.62 gm
Q.37 What mass of calcium chloride in grams would be enough to produce 14.35 gm ofAgCl.
(Atomic mass : Ca = 40, Ag = 108) -
(1) 5.55 gm (2) 8.295 gm (3) 16.59 gm (4) 11.19 gm
Q.38 What is the maximum amount of nitrogen dioxide that can be produced by mixing 4.2 gm of NO(g) and3.2 gm of O2(g) ?
(1) 4.60 gm (2) 2.30 gm (3) 3.22 gm (4) 6.44 gm
Q.39 The percentage by volume of C3H8 in a gaseous mixture of C3H8, CH4 and CO is 20. When 100 mLof the mixture is burnt in excess of O2 , the volume of CO2 produced is :
(1) 90 mL (2) 160 mL (3) 140 mL (4) none of these
Q.40 A gaseous mixture contains CO2(g) and N2O(g) in a 2 : 5 ratio by mass. The ratio of the number of
molecules of CO2(g) and N2O(g) is :
(1) 5 : 2 (2) 2 : 5 (3) 1 : 2 (4) 5 : 4
Q.41 The percentage of nitrogen in urea is about –
(1) 38.4 (2) 46.7 (3) 59.1 (4) 61.3
Q.42 The mass of carbon present in 0.5 mole of K4[Fe(CN)6] is –
(1) 1.8 gm (2) 18 gm (3) 3.6 gm (4) 36 gm
Q.43 1.2 gm of Mg (At. mass 24) will produce MgO equal to –
(1) 0.05 mol (2) 40 gm (3) 40 mg (4) 4 gm
Q.44 Insulin contains 3.4% sulphur by mass. What will be the minimum molecularweight of insulin-
(1) 94.117 (2) 1884 (3) 941 (4) 976
Q.45 The percent of N in 66% pure (NH4)2SO4 sample is -
(1) 32 (2) 28 (3) 14 (4) None of these
Q.46 For the reaction, 2Fe(NO3)
3+ 3Na
2CO
3 Fe
2(CO
3)
3+ 6NaNO
3
Initially if 2.5 mole of Fe(NO3)
2and 3.6 mole of Na
2CO
3is taken. If 6.3 mole of NaNO
3is obtained
then % yield of given reaction is :
(1) 50% (2) 84% (3) 87.5% (4) 100%
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Q.47 The conversion of oxygen to ozone occurs to the extent of 15% only. The mass of ozone that can be
prepared from 67.2 L of oxygen at S.T.P. will be -
(1) 14.4 gm (2) 96 gm (3) 640 gm (4) 64 gm
Q.48 An element (1) (at wt = 75) and another element (2) (at. wt. = 25) combine to form a compound. Thecompound contains 75% (1) by weight. The formula of the compound will be -
(1)A2B (2)A3B (3)AB3 (4)AB
Q.49 The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of thecompound will be -
(1) C2H2 (2) C3H3 (3) C4H4 (4) C6H6
Q.50 Carbon reacts with chlorine to form CCl4. 36 gm of carbon was mixed with 142 g of Cl2. Calculatemass of CCl4 produced and the remaining mass of reactant.
(1) wc = 24 gm ;4CClw = 154 gm (2) wc = 21 gm ;
4CClw = 145 gm
(3) wc = 14 gm ;4CClw = 164 gm (4) None
Q.51 Sulphuric acid is produced when sulphur dioxide reacts with oxygen and water in the presence of acatalyst : 2SO2(g) + O2 (g) + 2 H2O(l)2 H2SO4 . If 5.6 mol of SO2 reacts with 4.8 mol of O2 anda large excess of water, what is the maximum number of moles of H2SO4 that can be obtained?
(1) 6.5 (2) 5.6 (3) 4.6 (4) 3.6
Q.52 A mixture containing 100 gm H2 and 100 gm O2 is ignited so that water is formed according to thereaction, 2H2 + O2 2H2O; How much water will be formed -
(1) 112.5 gm (2) 50 gm (3) 25 gm (4) 200 gm
Q.53 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca(OH)2. The maximum number of moles of CaSO4formed is -
(1) 0.2 (2) 0.5 (3) 0.4 (4) 1.5
Q.54 For the reaction : A + 2BC
5 mole ofA and 8 mole of B will produce -
(1) 5 mole of C (2) 4 mole of C (3) 8 mole of C (4) 13 mole of C
Q.55 In the reaction 4A+ 2B + 3CA4B2C3 , what will be the number of moles of product formed, startingfrom one mole of A, 0.6 mole of B and 0.72 mole of C ?
(1) 0.25 (2) 0.3 (3) 0.24 (4) 2.32
Q.56 The amount of zinc required to produce 224mL of H2at NTP on treatment with dilute H
2SO
4will be –
(1) 0.65g (2) 6.5g (3) 65g (4) 0.065g
Q.57 When 0.5 mol of BaCl2 is added to 0.2 mol of Na3PO4, the number of moles of Ba3(PO4)2 formed is:
(1) 0.10 (2) 0.20 (3) 0.40 (4) 0.15
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Q.58 Density of water is 1 g/m L. The concentration of water in mol/litre is
(1) 1000 (2)18 (3) 0.018 (4) 55.5
Q.59 How many grams of NaOH will be needed to prepare 250 mL of 0.1 M solution
(1) 1g (2) 10g (3) 4g (4) 6g
Q.60 The molarity of a glucose solution containing 36 g of glucose per 400 mL of the solution is
(1) 1.0 (2) 0.5 (3) 2.0 (4) 0.05
Q.61 The molality of 15% (wt./vol.) solution of H2SO4 of density 1.1 g/cm3 is approximately
(1) 1.2 (2) 1.4 (3) 1.8 (4) 1.6
Q.62 1000 gram aqueous solution of CaCO3 contains 10 gram of carbonate. Concentration of solution is
(1) 10ppm (2) 100ppm (3) 1000ppm (4) 10,000ppm
Q.63 200 ml of a solution contains 5.85 g dissolved sodium chloride. The concentration of the solution will be
(Na = 23; Cl = 35.5)
(1) 1 molar (2) 2 molar (3) 0.5 molar (4) 0.25 molar
Q.64 How many grams of NaOH will be required to neutralize 12.2 grams of benzoic acid
(1) 40 gms (2) 4 gms (3) 16 gms (4) 12.2 gms
Q.65 If one mole of a substance is present in 1 kg of solvent. Then
(1) It shows molar concentration (2) It shows molal concentration
(3) It shows normality (4) It shows strength gm / gm
Q.66 A mixture has 18g water and 414g ethanol. The mole fraction of water in mixture is (assume ideal
behaviourof the mixture)
(1) 0.1 (2) 0.4 (3) 0.7 (4) 0.9
Q.67 What is the molarity of H2SO4 solution, that has a density 1.84 gm/cc at 35°C and contains solute 98%
byweight
(1) 4.18 M (2) 8.14 M (3) 18.4 M (4) 18 M
Q.68 The maximum amount of BaSO4 precipitated on mixing 20 mL of 0.5 M BaCI2 with 20 mL of 1 M
H2SO4 is
(1) 0.25 mole (2) 0.5 mole (3) 1 mole (4) 0.01 mole
Q.69 The number of moles of a solute in its solution is 20 and total number of moles are 80. The mole fraction
of solute is :
(1) 2.5 (2) 0.25 (3) 1 (4) 0.75
Q.70 2.5 L of 1 M NaOH solution mixed with another 3Lof 0.5 M NaOH solution. Then find out molarity ofresultant solution.
(1) 0.80 M (2) 1.0 M (3) 0.73 (4) 0.50 M
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Q.71 A sample of 98% acid has a density of 1.9 g cm–3. The amount of pure acid present in 2 L of the acid is:
(1) 3.724 kg (2) 372.4 g (3) 3.724 g (4) 100 g
Q.72 What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of0.10 M Ba(OH)2?
(1) 0.12 M (2) 0.10 M (3) 0.40 M (4) 0.0050 M
Q.73 25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. Ifsodium carbonate dissociates completely, molar concentration of sodium ion, Na+ and carbonate ions,CO3
2– are respectively (Molar mass of Na2CO3 = 106 g mol–1)
(1) 0.477 M and 0.477 M (2) 0.955 M and 1.910 M
(3) 1.910 M and 0.955 M (4) 1.90 M and 1.910 M
Q.74 How much NaNO3 must be weighed out to make 50 ml of an aqueous solution containing 70 mg Na+
per ml ?
(1) 12.934 g (2) 1.29 g (3) 0.129 g (4) 1.59 g
Q.75 How many ions are present in 1 L of 0.6 M K4[Fe(CN)6]? (Avogadro's Number = 6 × 1023)
(1) 1.8 × 1024 (2) 1.8 × 1023 (3) 3.6 × 1023 (4) 3.6 × 1024
Q.76 360g of water is present in a one Lmixture of ethanol and water. Molarity of water in the mixture is –
(1) 20.0 (2) 36.0 (3) 18.0 (4) None of these
Q.77 Mole fraction of solvent in aqueous solution of NaOH having molality of 3 is :
(1) 0.3 (2) 0.05 (3) 0.7 (4) 0.95
Q.78 What volume of 0.1M H2SO
4is needed to completely neutralize 40mL of 0.2M NaOH solution -
(1) 10mL (2) 40mL (3) 20mL (4) 80 mL
Q.79 171 g of cane sugar (C12H22O11) is dissolved in 1 litre of water. The molarity of the solution is
(1) 2.0 M (2) 1.0 M (3) 0.5 M (4) 0.25 M
Q.80 Equal volumes of 0.1 M AgNO3 and 0.2 M NaCl are mixed. The concentration of NO3– ions in the
mixturewillbe
(1) 0.1 M (2) 0.05 M (3) 0.2 M (4) 0.15 M
Q.81 20 ml of HCl solution requires 19.85 ml of 0.01 M NaOH solution for complete neutralization. The
molarity of HCl solution is
(1) 0.0099 (2) 0.099 (3) 0.99 (4) 9.9
Q.82 At a particular place, 1000 mL of air weighs 1.248 g. The specific gravity of air in terms of hydrogen if
1g H2 occupies 11200 mLvolume will be :
(1) 13.98 (2) 10 (3) 16 (4) 2.47
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Q.83 Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g/mL. Volume
of acid required to make 1L of 0.1 M H2SO4 solution is :
(1) 16.65 mL (2) 22.20 mL (3) 5.55 mL (4) 11.10 mL
Q.84 The charge on 1 gram ion ofAl3+ is :
(1)27
1NAe coulomb (2)
3
1NAe coulomb
(3)9
1NAe coulomb (4) 3 × NAe coulomb
Q.85 The density in grams per litre of a mixture containing an equal number of moles of methane and ethane at
NTP is :
(1) 1.03 (2) 1.10 (3) 0.94 (4) 1.20
Q.86 Equal weights of ethane and hydrogen are mixed in an empty vessel at 25°C. The fraction of the total
pressure exerted by hydrogen is :
(1)2
1(2)
1
1(3)
16
1(4)
16
15
Q.87 6 gm of carbon combines with 32 gm of sulphur to form CS2. 12 gm of C also combine with 32 gm ofoxygen to form carbondioxide. 10 gm of sulphur combines with 10 gm of oxygen to form sulphurdioxide. Which law is illustrated by them -
(1) Law of multiple proportions (2) Law of constant composition
(3) Law of Reciprocal proportions (4) Gay Lussac's law.
Q.88 The volume of 1.0 g of Hydrogen in liters at N.T.P. is -
(1) 2.24 (2) 22.4 (3) 1.12 (4) 11.2
Q.89 The specific gravity of the stainless steel spherical balls used in ball-bearings are 10.2. How many ironatoms are present in each ball of diameter 1 cm if the balls contain 84% iron, by mass ? The atomic massof iron is 56.
(1) 4.12 × 1021 (2) 4.82 × 1022 (3) 3.82 × 1022 (4) None
Q.90 The molar mass of N2O as well as CO2 is 44g mol–1.At 25°C and 1 atm, N2O contains n molecules of
gas. The number of molecules in 2.0 L of CO2 under the same conditions is :
(1)4
n(2)
8
n(3) 4n (4) n
Q.91 Two flasks P and Q of the same capacity contain helium and hydrogen respectively at 27°C and 1
atmospheric pressure. Flask P contains
(1) the same number of atoms as Q does (2) half the number of atoms as Q does
(3) twice the number of atoms as Q does (4) gas of the same weight as Q does
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Q.92 Samples of 1.0 g ofAl are treated separately with an excess of sulphuric acid and an excess of sodium
hydroxide. The ratio of the number of moles of the hydrogen gas evolved is :
(1) 1 : 1 (2) 3 : 2 (3) 2 : 1 (4) 9 : 4
Q.93 A mixture of gas ''X'' (mol. wt. 16) and gas "Y" (mol. wt. 28) in the mole ratio a : b has a mean molecularweight 20. What would be mean molecular weight if the gases are mixed in the ratiob : a under identical conditions (gases are non reacting).
(1) 24 (2) 20 (3) 26 (4) 40
Q.94 Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and56 lit. O2 at 1 atm & 273 K according to given reaction, is
12C(s) + 11H2(g) + 11/2O2 (g) C12H22O11(s)
(1) 138.5 (2) 155.5 (3) 172.5 (4) 199.5
Q.95 An elementAis tetravalent and anotherelement B is divalent.The formula of the compound formed fromthese elements will be-
(1)A2B (2)AB (3)AB2 (4)A2B3
Q.96 Two elements X (At. mass 16) andY(At. mass 14) combine to form compoundsA, B and C.The ratio ofdifferent masses ofYwhich combine with a fixed mass of X inA, Band C is 1 :3 : 5. If 32parts bymass ofX combineswith 84 partsbymass ofYin B, then in C, 16partsbymass ofX will combine with-
(1) 14 parts by mass of Y (2) 42 parts by mass of Y
(3) 70 parts by mass of Y (4) 84 parts by mass of Y
Q.97 If one mole of ethanol (C2H5OH) completely burns to form carbon dioxide and water, the weight ofcarbon dioxide formed is about -
(1) 22 gm (2) 45 gm (3) 66 gm (4) 88 gm
Q.98 The moles of O2 required for reacting with 6.8 gm of ammonia.
(..... NH3 +..... O2 ..... NO + ..... H2O) is
(1) 5 (2) 2.5 (3) 1 (4) 0.5
Q.99 If isotopic distribution of C–12 and C–14 is 98% and 2% respectively, then the number of C–14 atomsin 12 gm of carbon is -
(1) 1.035 × 1022 (2) 3.01 × 1022 (3) 5.88 × 1023 (4) 6.02 × 1023
Q.100 If 3.01 × 1020 molecules are removed from 98 mg of H2SO4, then the number of moles of H2SO4 leftare –
(1) 0.1 × 10–3 (2) 0.5 × 10–3 (3) 1.66 × 10–3 (4) 9.95 × 10–2
Q.101 The density of mercury is 13.6 g/mL. Calculate the diameter of an atom of mercury assuming that eachatom of mercury is occupying a cube of edge length equal to the diameter of the mercury atom(Hg = 200)
(1) 9.2 × 10–8 cm (2) 2.9 × 10–6 cm (3) 2.7 × 10–8 cm (4) 2.9 × 10–8 cm
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Q.102 If the density of water is 1 gm/cm3, then the volume occupied by one molecule of water is approximately-
(1) 18 cm3 (2) 22400 cm3 (3) 6.02×10–23 cm3 (4) 3.0×10–23 cm3
Q.103 How many gram glucose burn completely to produce sufficient CO2 needed for complete conversion of12 g NaOH into Na2CO3 :
(1) 5.5 g (2) 4.5 g (3) 6.5 g (4) 3.5 g
Q.104 1.35 gm of pureAmetal was quantitatively converted into 1.88 gm of pureAO. What is atomic weightof A –
(1) 40.75 (2) 50 (3) 60 (4) 70
Q.105 Mass of H2O in 1000 kg CuSO4.5H2O is - (Cu = 63.5)
(1) 360.7 kg (2) 36.05 kg (3) 3605 kg (4) 3.605 kg
Q.106 0.8 mole of a mixture of CO and CO2 requires exactly 40 gram of NaOH in solution for completeconversion of all the CO2 into Na2CO3. How manymoles more of NaOH would it require for conversioninto Na2CO3, if the mixture (0.8 mole) is completely oxidised to CO2
(1) 0.2 (2) 0.6 (3) 1 (4) 1.5
Q.107 4.4 gm of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of moleculespresent in the container will be -
(1) 6.022 × 1023 (2) 1.2046 × 1023 (3) 2 moles (4) 6.023 × 1024
Q.108 Find the volume of CO2 obtained at N.T.P. on heating 200 gm of 50% pure CaCO3 –
(1) 11.2 litre (2) 22.4 litre (3) 44.8 litre (4) None of these
Q.109 Asample ofAlF3 contains 3.0 × 1024 F– ions. The number of formula units in this sample are –
(1) 9.0 × 1024 (2) 3.0 × 1024 (3) 0.75 × 1024 (4) 1.0 × 1024
Q.110 The mass of 70% pure H2SO4 required for neutralisation of 1 mol of NaOH -
(1) 49 gm (2) 98 gm (3) 70 gm (4) 34.3 gm
Q.111 Assuming that petrol is iso-octane (C8H18) and has density 0.8 gm/ml. 1.425 litre of petrol on complete
combustion will consume oxygen -
(1) 50 L (2) 125 L (3) 125 mol (4) 50 mol
Q.112 What volume of air at STP containing 21% of oxygen by volume is required to completely burn sulphur(S
8) present in 200 g of sample, which contains 20% inert material which does not burn. Sulphur burns
according to the reaction :
1/8S8(s) + O
2(g) SO
2(g)
(1) 23.52 litre (2) 320 litre (3) 112 litre (4) 533.33 litre
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Q.113 The percentagebymole of NO2 ina mixture ofNO2(g) and NO(g)havingaverage molecularmass 34 is :(1) 25% (2) 20% (3) 40% (4) 75%
Q.114 26 c.c. of CO2 are passed over red hot coke. The volume of CO evolved is -
(1) 15 c.c (2) 10 c.c. (3) 32 c.c. (4) 52 c.c.
Q.115 An iodized salt contains 0.5 % of NaI.Aperson consumes 3 gm of salt everyday. The number of iodideions going into his body everyday is(1) 10–4 (2) 6.02 ×10–4 (3) 6.02 × 1019 (4) 6.02 × 1023
Q.116 The mass of CO2 produced from 620 gm mixture of C2H4O2 & O2, prepared to produce maximumenergy is (Combustion reaction is exothermic)(1) 413.33 gm (2) 593.04 gm (3) 440 gm (4) 320 gm
Q.117 The mass of P4O10 produced if 440 gm of P4S3 is mixed with 384 gm of O2 isP4S3 + O2 P4O10 + SO2
(1) 568 gm (2) 426 gm (3) 284 gm (4) 396 gm
Q.118 Two oxides of Metal contain 27.6% and 30% oxygen respectively. If the formula of first oxide is
M3O4 then formula of second oxide is -
(1) MO (2) M2O (3) M2O3 (4) MO2
Q.119 Assuming complete precipitation ofAgCl, calculate the sum of the molar concentration of all the ions if 2lit of 2MAg2SO4 is mixed with 4 lit of 1 M NaCl solution is :(1) 4M (2) 2M (3) 3 M (4) 2.5 M
Q.120 What volumes should you mix of 0.2 M NaCl and 0.1 M CaCl2 solution so that in resulting solution theconcentration of positive ion is 40% lesser than concentration of negative ion.Assuming total volume ofsolution 1000 ml.(1) 400 ml NaCl , 600 ml CaCl2 (2) 600 ml NaCl, 400 ml CaCl2(3) 800 ml NaCl, 200 ml CaCl2 (4) None of these
Q.121 Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After thecompletion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate wascompletely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At.mass of calcium = 40)(1) 200 cm3 (2) 500 cm3 (3) 400 cm3 (4) 300 cm3
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ACC_CHEMISTRY_Mole Concept 4 1
EXERCISE-2
[PREVIOUS YEAR'S QUESTIONS]
Q.1 Pressure in a mixture of 4g of O2and 2g of H
2confined in a bulb of 1L at 0OC is (A.I.I.M.S. 1999)
(1) 15.210 atm (2) 25.215 atm (3) 31.205 atm (4) 45.215 atm
Q.2 12g of alkaline earth metal gives 14.8g of its nitrite.Atomic weight of metal is - (A.I.I.M.S. 2000)
(1) 12 (2) 20 (3) 40 (4) 14.8
Q.3 Volume of CO2obtained by the complete decomposition of 9.85 g BaCO
3is- (CPMT-2000)
(1) 2.24L (2) 1.12L (3) 0.84L (4) 0.56L
Q.4 2.5L NaOH of 1M solution is mixed with 3L NaOH of 0.5 M solution. What is the molarity of the
resultingsolution- (CBSE 2002)
(1) 0.80 M (2) 1.0 M (3) 0.73 M (4) 0.50 M
Q.5 Avogadro's number is the number of molecules present in : (AFMC 2003)
(1) 1L of molecule (2) 1g of molecule
(3) gram molecular mass (4) 1g-atom of molecules
Q.6 In Haber process, 30L of dihydrogen and 30L of dinitrogen were taken for reaction which yielded only
50% of the expected product. What will be the composition of gaseous mixture under the aforesaid
condition in the end? (CBSE AIPMT 2003)
(1) 20 L ammonia, 10 L nitrogen, 30 L hydrogen
(2) 20 L ammonia, 25 L nitrogen, 15 L hydrogen
(3) 20 L ammonia, 20 L nitrogen, 20 L hydrogen
(4) 20 L ammonia, 25 L nitrogen, 15 L hydrogen
Q.7 The number of sodium atoms in 2 moles of sodium ferrocyanides is : (BHU 2004)
(1) 12 × 1023 (2) 26 × 1023 (3) 34 × 1023 (4) 48 × 1023
Q.8 The amount of H2S required to precipitate 1.69 g BaS from BaCl
2solution is :
(Haryana PMT 2005)
(1) 3.4 g (2) 0.24 g (3) 0.34 g (4) 0.17 g
Q.9 The oxygen obtained from 72 kg of water is : (RPMT 2005)
(1) 72 kg (2) 45 kg (3) 50 kg (4) 64 kg
Q.10 Total number of protons in 10g of calcium carbonate is (NA
= 6.023 × 1023) : (RPMT 2005)
(1) 3.01 × 1024 (2) 4.06 × 1024 (3) 2.01 × 1024 (4) 3.24 × 1024
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ACC_CHEMISTRY_Mole Concept 4 2
Q.11 Total number of atoms represented by the compound CuSO4· 5H
2O is : (BHU 2005)
(1) 27 (2) 21 (3) 5 (4) 8
Q.12 A gas is found to have a formula [CO]x. If its vapour density is 70, the value of x is : (AMU 2006)
(1) 2.5 (2) 3.0 (3) 5.0 (4) 6.0
Q.13 A gas mixture contains O2and N
2in the ratio of 1 : 4 by weight. The ratio of their number of molecules
is (AFMC 2006)
(1) 1 : 8 (2) 1 : 4 (3) 3 : 16 (4) 7 : 32
Q.14 2.76 g of silver carbonate on being strongly heated yield a residue weighing: (BHU 2006)
(1) 2.16 g (2) 2.48 g (3) 2.64 g (4) 2.32 g
Q.15 The weight of one molecule of a compound C60
H122
is : (BHU 2007)
(1) 1.3 × 10–20g (2) 5.01 × 10–21 g (3) 3.72 × 1013 g (4) 1.4 × 10–21 g
Q.16 The largest number of molecules is in : (AMU 2008)
(1) 34 g of H2O (2) 28 g of CO
2(3) 46 g of CH
3OH (4) 54 g of N
2O
5
Q.17 How many moles of lead (II) chloride will be formed from a reaction between 6.5g of PbO and
3.2 g of HCl? (CBSE AIPMT 2008)
(1) 0.044 (2) 0.333 (3) 0.011 (4) 0.029
Q.18 What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1L of
propane gas (C3H
8) measured under the same conditions? (CBSE AIPMT 2008)
(1) 7L (2) 6L (3) 5L (4) 10L
Q.19 What volume of CO2will be liberated at NTP, if 12g of carbon is burnt in excess of oxygen?
(AFMC 2009)
(1) 11.2 L (2) 22.4 L (3) 2.24 L (4) 1.12 L
Q.20 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water
produced in this reaction will be : (CBSE AIPMT 2009)
(1) 3 mol (2) 4 mol (3) 1 mol (4) 2 mol
Q.21 60 g of a compound on analysis produced 24 gm carbon, 4 gm hydrogen and 32 gm oxygen. The
empirical formula of the compound is : (BVP 2010)
(1) CH2O
2(2) CH
2O (3) CH
4O (4) C
2H
4O
2
Q.22 Volume of a gas at NTP is 1.12 × 10–7 cm3 . The number of molecules in it is : (Manipal 2010)
(1) 3.01 × 1012 (2) 3.01 × 1024 (3) 3.01 × 1023 (4) 3.01 × 1026
Q.23 The mass of one mole of electron is : (CPMT 2010)
(1) 9 × 10–28 g (2) 0.55 mg (3) 9.1 × 10–24 g (4) 6 × 10–12 g
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ACC_CHEMISTRY_Mole Concept 4 3
Q.24 In an experiment, 4 g of M2O
xoxide was reduced to 2.8 g of the metal. If the atomic mass of the metal
is 56 g mol–1 , the number of O atoms in the oxide is : (AFMC 2010)
(1) 1 (2) 2 (3) 3 (4) 4
Q.25 Number of atoms present in 4.25 g of NH3is : (AFMC 2010)
(1) 6.023 × 1023 (2) 4 × 6.023 × 1023 (3) 1.7 × 1024 (4) 4.25 × 6.023 × 1023
Q.26 Which has the maximum numberof molecules amongthe following? [CBSE (Final) 2011]
(1) 8 g H2
(2) 64 g SO2
(3) 44 g CO2
(4) 48 g O3
Q.27 Mole fraction of the solute in a 1.00 molal aqueous solution is [CBSE (Pre) 2011]
(1) 1.7700 (2) 0.1770 (3) 0.0177 (4) 0.0344
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ACC_CHEMISTRY_Mole Concept 4 4
EXERCISE-3
[ASSERTION & REASON TYPE QUESTIONS]
Each of the questions given below consist of Assertion and Reason. Use the following Key to choose
the appropriate answer.
(A) If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
(B) If both Assertion and Reason are correct, and Reason is not the correct explanation of Assertion.
(C) If Assertion is correct and Reason is incorrect.
(D) IfAssertion is incorrect but Reason is correct.
(E) If bothAssertion and Reason are incorrect.
Q.1 Assertion :Approximate mass of 1 atom of O16 in gms is (16 / NA)
Reason : 1 atom of O16 weighs 16 a.m.u & 1 a.m.u = (1 / NA) g.
(1)A (2) B (3) C (4) D (5) E
Q.2 Assertion : During a chemical reaction total moles remains constant.
Reason : During a chemical reaction total mass remains constant.
(1)A (2) B (3) C (4) D (5) E
Q.3 Assertion : For the reaction producing Fe & CO2 by the reaction of Fe2O3 and C the ratio
of stoichiometric coefficients of Fe2O3 : Fe is 1 : 2.
Reason : During a chemical a reaction atoms can neither be created nor be destroyed.
(1)A (2) B (3) C (4) D (5) E
Q.4 Assertion : 2A + 3B C
4/3 moles of 'C' are always produced when 3 moles of 'A' & 4 moles of 'B' are added.
Reason : 'B' is the liming reactant for the given data.
(1)A (2) B (3) C (4) D (5) E
Q.5 Assertion : Molality of pure ethanol is lesser than pure water.Reason :As density of ethanol is lesser than density of water.
[Given : dethanol = 0.789 gm/ml; dwater = 1 gm/ml](1)A (2) B (3) C (4) D (5) E
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ACC_CHEMISTRY_Mole Concept 4 5
EXERCISE-4
[SUBJECTIVE BOARD QUESTIONS]
Very Short Answer Question :Q.1 What is the S.I. unit of density?
Q.2 What is one a.m.u. or one ‘u’?
Q.3 Which isotope of carbon is used for getting relative atomic masses?
Q.4 Write down the empirical formula of acetic acid.
Short Answer Question :Q.5 What will be the mass of one atom of C-12 in grams?
Q.6 Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca3(PO4)2.
Q.7 If two elements can combine to form more than one compound, the masses of one element that combinewith a fixed mass of the other element, are in whole number ratio.(a) Is this statement true ?(b) If yes, according to which law ?(c) Give one example related to this law
Q.8 Calculate the average atomic mass of hydrogen using the following data :Isotope % Natural abundance Molar mass1H 99.985 12H 0.015 2
Long Answer Question :Q.9 A vessel contains 1.6g of dioxygen at STP (273.15K, 1atm pressure). The gas is now transferred to
another vessel at constant temperature, where pressure becomes half of the original pressure.Calculate : (i) volume of the new vessel (ii) number of molecules of dioxygen.
Q.10 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction?CaCO3 (s) + 2HCl (aq) CaCl2(aq) + CO2(g) + H2O (l)
What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 100 g of CaCO3?Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.
Q.11 A box contains some identical red coloured balls, labelled as A, each weighing 2 gm. Another boxcontains identical blue coloured balls, labelled as B, each weighing5 gm. Consider the combinationsAB,AB2,A2B andA2B3 and show that law of multiple proportions is applicable.
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ACC_CHEMISTRY_Mole Concept 4 6
EXERCISE-1
Q.1 4 Q.2 2 Q.3 2 Q.4 2
Q.5 1 Q.6 2 Q.7 1 Q.8 2
Q.9 2 Q.10 1 Q.11 4 Q.12 1
Q.13 2 Q.14 3 Q.15 4 Q.16 1
Q.17 1 Q.18 2 Q.19 2 Q.20 4
Q.21 1 Q.22 3 Q.23 1 Q.24 2
Q.25 2 Q.26 4 Q.27 2 Q.28 3
Q.29 4 Q.30 2 Q.31 1 Q.32 2
Q.33 2 Q.34 1 Q.35 2 Q.36 2
Q.37 1 Q.38 4 Q.39 3 Q.40 2
Q.41 2 Q.42 4 Q.43 1 Q.44 3
Q.45 3 Q.46 3 Q.47 1 Q.48 4
Q.49 4 Q.50 1 Q.51 2 Q.52 1
Q.53 1 Q.54 2 Q.55 3 Q.56 1
Q.57 1 Q.58 4 Q.59 1 Q.60 2
Q.61 4 Q.62 4 Q.63 3 Q.64 2
Q.65 2 Q.66 1 Q.67 3 Q.68 4
Q.69 2 Q.70 3 Q.71 1 Q.72 2
Q.73 3 Q.74 1 Q.75 1 Q.76 1
Q.77 4 Q.78 2 Q.79 3 Q.80 2
Q.81 1 Q.82 1 Q.83 3 Q.84 4
Q.85 1 Q.86 4 Q.87 3 Q.88 4
Q.89 2 Q.90 4 Q.91 2 Q.92 1
Q.93 1 Q.94 2 Q.95 3 Q.96 3
Q.97 4 Q.98 4 Q.99 1 Q.100 2
Q.101 4 Q.102 4 Q.103 2 Q.104 1
Q.105 1 Q.106 2 Q.107 2 Q.108 2
Q.109 4 Q.110 3 Q.111 3 Q.112 4
Q.113 1 Q.114 4 Q.115 3 Q.116 3
Q.117 2 Q.118 3 Q.119 2 Q.120 4
Q.121 2
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ACC_CHEMISTRY_Mole Concept 4 7
EXERCISE-2
Q.1 2 Q.2 3 Q.3 2 Q.4 3
Q.5 3 Q.6 4 Q.7 4 Q.8 3
Q.9 4 Q.10 1 Q.11 2 Q.12 3
Q.13 4 Q.14 1 Q.15 4 Q.16 1
Q.17 4 Q.18 3 Q.19 2 Q.20 2
Q.21 2 Q.22 1 Q.23 2 Q.24 3
Q.25 1 Q.26 1 Q.27 3
EXERCISE-3
Q.1 1 Q.2 4 Q.3 1 Q.4 4
Q.5 2
EXERCISE-4
Q.1 Kg m–3
Q.2 1 a.m.u.
Q.3 C–12
Q.4 CH2O
Q.5 1.992 × 10–23 gm
Q.6 Ca = 38.71%, P = 20.0%, O = 41.29%
Q.7 (a)Yes, the given statement is true(b) True, Law of multiple proportions.(c) Carbon and oxygen combine to form CO and CO2
2C + O2 2CO2 × 12g 32g carbon monoxideC + O2 CO212g 32g carbon dioxide
Q.8 1.00015u
Q.9 (i) 2.24L (ii) 3.011 × 1022
Q.10 HCl will be limiting reagent, 0.095 mol
Q.11 Combination AB AB2 A2B A2B3Mass ofA(g) 2 2 4 4Mass of B(g) 5 10 5 15
Masses of B which combine with fixed mass ofA(say 4 g) will be 10 g, 20 g, 5 g and 15 g i.e., theyare in the ratio 2 : 4 : 1 : 3 which is a simple whole number ratio. Hence, the law of multiple proportionsis applicable.
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ACC_CHEMISTRY_Mole Concept 4 8
EXERCISE-1
1. Same ammount of oxygen combine with different % of H.
Law of multiple proportion.
2. Ratio of H and O is fixed which is law of definite proportions.
3. For nuclear reation law of conservation of mass does not hold good.
4. BaCl2 + H2SO4 BaSO4 + 2HCl
0.1 mole 0.1 mole 0 0
0 0 0.1 0.1
Mass of reactant = Mass of product
20.8 + 9.8 = 7.3 + x
x = 23.3 gm
5. % is fixed so it proves law of constant proportion.
6. % of Fe = MassMolar
protonsof.NoAtomFeof.wt × 100
=67200
56x× 100 =
3
1
x =3
1×
56
672= 4
7. Moles of Ca =M
W=
AN
N
N =40
2× 6.022 × 1023 = 3.0115 × 1022
8.AN
N=
M
W
N =12
102 12× 1023 × 6.022 = 1.004 × 1011
9. Mass of H2O = 65 × 0.8 = 52 Kg
N =18
1052 3× 6.022 × 1023 = 17.39 × 1026
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ACC_CHEMISTRY_Mole Concept 4 9
10. atomNof.No
atomCof.No=
14/6.4
12/3.67=
1
17
11. Total number of atoms = 19 × 6 × 1023 atoms
12. ngas =4.22
2.11=
2
1mole
2
1=
M
W= M = 28 gm
13. n =666
333= 0.5
number of H2O = 0.5 × 18 × 6.02 × 1023 = 9 × 1023
14.40
x× NA = y
80
x2× NA = y
15. 1 mole (CO2) 6.023 × 1023 atom of C
16. C has max moles so will have maximum number of atoms.
17. 1 amu = 1.67 × 10–24 gm.
18. H2 has maximum moles so it will have maximum atoms.
19. P = 1 =V
M= M = 18 gm
OH2n =
18
18= 1
Total number of e– = 10 × NA
20. 1 mole of P4 contain 4 mole of P or 4 NA atoms of P.
21. nC =12
12= 1
Total number of e + p + n = 18 × 6.022 × 1023
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ACC_CHEMISTRY_Mole Concept 5 0
22. Total number of Na atom = 2 × NA
23.2CaCln =
111
111= 1
Number of Ca2+ = NA
Number of Cl– = 2NA
24.xSOn =
4.22
6.5=
4
1=
x1632
16
=4
1=
x2
1
= n = 2
25. O moles are present in 1 mole Mg3 (PO4)3
4
1mole will be present in
8
1×
4
1=
32
1= 3.125 × 10–2
26. 9.1 × 10–31 Kg weight mean 1 e–
1 Kg weight mean = 31101.9
1
e–
Moles of e– =022.61.9
1010 2331
=023.61.9
108
27.2HV at NTP = 2 × 22.4 = 44.8 liter..
28.4CHn =
4.22
48.4=
16
W
W = 3.2 gm.
30. n =4.22
6.5=
VD2
8
VD = 16
31. O2 : H2 : CH4
32
32:
2
32:
16
32
1 : 16 : 2
32. VDA = 4 VDB2 × VDA = 4 × 2 (VD)B
MA = 4 × MB
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ACC_CHEMISTRY_Mole Concept 5 1
33. PM = dRT
M =1
2730021.0293.1 = 28.9 gm
VD =2
M= 14.49.
34.3HNOn =
63
2.25= 0.4
2NOn =2
3× 0.4
2NOW =2
3× 0.4 × 46 = 27.6 gm
35. Using POSe
2 ×2Fn = 6 ×
6uFn
2Fn =352
1023 3
Molecules of F2
=352
1023 3× N
A= 1 × 1019
36. Ag + HNO3 AgNO3 + NaCl Ag Cl
108
x
108
x=
5.143
87.2
x =5.143
87.2× 108 = 2.16 gm.
37. POAC on Cl
CaCl2 AgCl
2 ×2CaCln =
5.143
35.14
2 ×111
w= 0.1
w = 5.55 gm.
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ACC_CHEMISTRY_Mole Concept 5 2
38. NO +2
1O2 NO2
4.2 3.2
Mole1
14.0
5.0
1.0
0.14 0.2
NO is LR So
2NOn = 0.14
2NOw = 0.14 × 46 = 6.44 gm.
39. C H3 8
O23CO2
20 60 ml
CH4
O2CO2
x x
COO2 CO2
60 – x 80 – x
2COv = 60 + x – 180 – x = 140 ml.
40.ON
CO
2
2
w
w=
5
2
ON
CO
2
2
n
n=
544
442
=
5
2
41. urea NH – C – NH22
O
% of N =60
142× 100 = 46.66
42. Mass of C = 0.5 × 6 × 12 = 36 gm.
43. Mg +2
1O2 MgO
24
2.1= 1 × nMgO
nMgO = 0.5
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ACC_CHEMISTRY_Mole Concept 5 3
44. Attleast one sulphur will be there in 1 molecule
3.4 = 100M
32
M = 941
45. % of N = %14100142132
66
46. 2Fe(NO3)
3+ 3Na
2CO
3 Fe
2(CO
3)
3+ 6NaNO
3
2
5.2
3
6.3
1.25 1.2
Na2CO
3is LR
% yield =6.32
3.6
× 100 = 87.5%
47. 2O2
3 O
3
4.22
2.67= 3mole 2 mole
mole3.015.02n3O
gm4.14483.0w3O
49. EMn
EM
massMolecular
massEmpirical
n
1
78
13
n = 6
MF = (CH)6 = C6H6
50. C(s) + 2Cl2 CCl4
1
mole3
2
mole2
4CCln = 1 mole = 12 0+ 1 + 35.5 × 4 = 154 gm.
51. 2SO2 (g) + O2 (g) + 2H2O () 2H2SO4
2
6.5
2
8.4
SO2 is LR
Mole of H2SO4 = 5.6
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ACC_CHEMISTRY_Mole Concept 5 4
52. 2H2 (g) + O2 (g) 2H2O
50 mole 3.125
O2 is LR
OH2n = 2 × 3.125
OH2w = 18 × 2 × 3.125 = 112.5 gm
53. H2SO4 + Ca(OH)2 CaSO4 + 2H2O
0.5 0.2
Ca(OH)2 is LR
4CaSOn = 0.2
54. A + 2B C
1
5
2
8
B is LR
Moles of C = 4
55. 4A + 2B + 3C AuB2C3
4
1
2
6.0
3
72.0
0.25 0.3 0.24
C is LR
324 CBAn =3
72.0= 0.24
56. Zn + H2SO
4 ZnSO
4+ H
2
0.1 mole22400
224= 0.01 mole
wZn
= 0.1 0+ 65 = 0.65
57. 3BaCl2 + 2Na3PO4 24Cl POB
3
3
5.0= 1.66
2
2.0= 0.1
243 POBan = 0.1 mole
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ACC_CHEMISTRY_Mole Concept 5 5
58. 1 gm H2O is present in 1 ml of water
M =18
1×
1
1000= 55.5 M
59. M =v
n NaOH× 100
1 =25040
w
× 100
w = 1 gm
60. M =180
36×
400
1000= 0.5 M
61. 15 gm is present in 100 ml solutions
d =V
M= Msolution = 1.1 × 100 = 110 gm
Mass of solvent = 110 – 15 = 95 gm
m =9598
15
× 1000 = 1.6 m
62. Conrains =1000
10× 106 = 104 ppm
63. Molarity =5.58
85.5×
200
1000= 0.54
64. C6H5CO2H + NaOH C6H5CO2Na + H2O
0.1 mole 0.1 mole = 4 gm
65. Molar concentration.
66. OH2x =
46
414
18
18
18/18
=
91
1
= 0.1
67. M =M
dx10=
98
9884.110 = 18.4 M
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ACC_CHEMISTRY_Mole Concept 5 6
68. BaCl2 + H2SO4 BaSO410 mm 20 mm
LR 10 mm
69. Msolute =Total
solute
n
n=
80
20= 0.25
70. M1V1 + M2V2 = M3V3
M3 =5.5
5.0315.2 = 0.73
71. Mass of solution = 1.9 × 2000 = 3800 gm
Mass of acid = 0.98 × 3800 = 3724 gm
72. 2HCl + Ba(OH)2 BaCl2 + 2H2O
1 3
LR
mm OH– = 2 × 2.5
molerity of OH– =50
5.22= 0.1 M
73.32CONaM =
106
3.25×
250
1000= 0.955 M
NaM = 1.9 M
23CO
M = 0.955 M
74. NaM =
3NaNOH
23
101070 33
=85
10w 3
w = 12.934 gm
75. Number of ion = 5 × 0.6 × 6 × 1023 = 1.8 × 1024
76. OH2M =
18.0
360×
1000
1000= 20 M
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ACC_CHEMISTRY_Mole Concept 5 7
77. 3 mole of NaOH in 1000 gm of water
XNaOH
=
18
10003
3
= 0.5
OH2X = 0.95
78. H2SO
4+ 2NaOH Na
2SO
4+ 2H
2O
X 8
4 = M ×42SOHV
42SOHV =1.0
4= 40 ml
79. M =342
171×
1000
1000= 0.5 M
80. AgNO3 + NaCl AgCl + NaNO30.1 × v 0.2 × v
3NO
M =v2
v1.0 = 0.05 M
81. HCl + NaOH
20 × v = 19.85 × 0.01
v = 0.0099 ml
82. Specific gravity (in terms of H) = Hydrogenofdensity
Gasofdensity
=
11200/1
1000/248.1= 13.98
83. H2SO4 required = 1 × 0.1 = 0.1 moles = 9.8 gr.
from information 98 gr. H2SO4 is present in = 100 gr. solution
=8.1
100mL .solution
9.8 gm H2SO4 is present in =98
8.9
8.1
100 mL
= 5.55 mL.
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ACC_CHEMISTRY_Mole Concept 5 8
84. 1 gm ion ofAl3+ = 1 molAl3+ = NA ions
charge = NA × (3e) coulomb
85. Mav =2
130161 = 23 g/mol
d =4.22
23= 1.03 g/L
86. Mole fraction of H2 =
30
w
2
w
2/w
=
16
15
from dalton's Law,Total
H
P
P2
=2HX =
16
15
87. CS2 C S
6 gm 32 gm
CO2 C O
12 gm 32 gm
6 gm 16 gm
with fixed mass of C (= 6 gm), ratio of weight of S & O
O
S
w
w=
16
32= 2.
in SO2 S O
10 gm 10 gm
O
S
'w
'w=
10
10= 1.
O
S
w
w= 2 ×
O
S
'w
'w
Law of raciprocal proportions hold.
88. Volume of H2=
2
1× 22.4 = 11.2 L
89. d = 10.2 g/c.c.
V =3
4×
3
2
1
c.c.
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ACC_CHEMISTRY_Mole Concept 5 9
weight of ball = V × d
number of Fe atoms =3
4× ×
3
2
1
×
100
84×
56
1× N
A
= 4.8 × 1022 atoms
90. Under same condition of tempanol pressure equal volume of different gases contans equal molecular of
each gas.
91. Under same condition of temperature and pressure equal volume of different gases contains same
numberof molecules.
2Hn = nHe
atoms of H2 = 2 × atoms of He
92. In both cases
Al + 3H+ Al3+ +2
3H2
equal moles of H2 will be evolved in both cases.
93. X Y
mole ratio a : b
mole fraction (Let) x : 1 – x
therefore mole ratio b : a
mole fraction 1 – x : x
Case I
Mav = 16 x + 28 (1 – x) = 20
x =12
8=
3
2
Case II
Mav = 16 (1 – x) + 28 x
= 16 ×3
1+ 28
3
2=
3
72= 24 g/mol ]
94. 12C + 11 H2 +2
11O2 C12H22O11
moles12
84= 6
2
12= 6
4.22
56= 2.5
forlimiting12
6
11
6
2/11
5.2=
11
5
reagent Limittign reagent O2
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ACC_CHEMISTRY_Mole Concept 6 0
mole of sucrose produced =11
2× 2.5 =
11
5
mass of sucrose =11
5× 342 = 155.5
95. A4+ B2–
formula A B2
96. In B
with 32 part of X, mass of Y = 84
In C
with 32 part of X, mass of Y = 84 ×3
5= 140
with 16 part of X, mass of Y =2
140= 70 gm.
97. C2H5OH 2CO2
1 mole 2 mole
= 44 × 2 = 88 gm.
98. NH3 + O2 NO + H2O unbalanced
4NH3 + 5O2 4NO + 6H2O balanced
moles of NH3 (given) =17
8.6= 0.4
moles of O2 required =4
5× 0.4 = 0.5
99. weight of C–14 in 12 gm carbon = 12 ×100
2= 0.24 gm.
moles of C–14 =14
24.0
number of C–14 =14
24.0× 6.022 × 1023
= 1.032 × 10–22
100. Moles of H2SO4 left =98
1098 3– 23
20
1002.6
1001.3
= 0.5 × 10–3
101. dHg
= 13.6 g/cc
Let 'a' is edge length
volume = a3
mass of 1 mole = a3 × d × NA
= 200
a = 2.9 × 10–8 cm
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ACC_CHEMISTRY_Mole Concept 6 1
102. Let volume of one molecule of H2O = V
weight of 1 mole = V × d × NA = 18 gm.
V × 1 × NA = 18
V = 2310022.6
18
= 3 × 10–23 cm3
103. CO2 + 2NaOH Na2CO3 + H2O
moles of NaOH =40
12= 0.3
moles of CO2 =2
3.0= 0.15
C6H12O6 + 6O2 6O2 + 6H2O
moles of C6H12O6 =6
15.0
weight of glucos =6
15.0× 180 = 4.5 gm.
104. moles ofA= mole ofAO
M
35.1=
16M
88.1
M = 40.75 g/mol
105. moles of H2O = 5 × moles of (CuSO4 . 5H2O)
= 5 ×5.249
1000
mass of H2O = 5 ×5.249
1000× 18
= 360.7 gm.
106. CO + CO2 CO2
0.8 mole 0.8 mole
CO2 + 2 NaOH Na2CO3 + H2O
moles of NaOH required = 0.8 × 2 = 1.6
extra NaOH required = 1.6 –40
40= 0.6 mole
107. moles in container =44
4.4+
4.22
24.2= 0.2 moles
molecules = 0.2 × 6.022 × 1023 = 1.204 × 1023
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ACC_CHEMISTRY_Mole Concept 6 2
108. CaCO3 CaO + CO2
weight of CaCO3 in sample = 200 ×100
50= 100
moles of pure CaCO3 =100
100= 1
moles of CO2 = 1
volume (at NTP) = 22.4 L
109. F = 3.0 × 1024
noumber of formula unit ofAlF3 =3
F=
3
103 24
= 1.0 × 1024
110. 2 NaOH + H2SO4 Na2SO4 + H2O
mole 1 1/2
weight of H2SO4 (pure) required = 98 ×2
1= 49 gm
weight of H2SO4 (70 % pure) required = 49 ×70
100= 70 gm.
111. Weight of 1.425 L iso octane = 1425 × 0.8 = 1140 gm.
moles = 18128
1140
= 10
C8H18 + 12.5O2 8CO2 + 9H2O
moles of O2 required = 12.5 × 10 = 125 mole
112.8
1S
8+ O
2 SO
2(g)
weight of pure S = 200 ×100
80= 160 gm/
moles of S8=
832
160
moles of oxygen =832
160
× 8 = 5
volume of O2= 5 × 22.4 L
volume of air = 5 × 22.4 ×21
100= 533.3 L
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ACC_CHEMISTRY_Mole Concept 6 3
113. NO2 NOmoles x 1 – x
46x + 30 (1 – x) = 34
x =4
1
mole percentage = 25 %
114. CO2 + C 2COvolume of CO2 = 26 c.c. volume of CO = 52 c.c.
115. Weight of NaI = 3 ×100
5.0= 1.5 + 10–2
moles of NaI = 3 ×100
5.0= 10–4
number of I– ions = 10–4 × 6.022 × 1023
= 6.022 × 1019
116. C2H4O2 + 2O2 2CO2 + 2H2Oformaximumenrgymole ratio should be 1 : 2 let moles of C2H4O2 = x mole of O2 = 2x 60 x + 32 × 2x = 620
x = 5 moles of CO2 produced = 5 × 2 = 10 mass of CO2 = 10 × 44 = 440 g/mole
117. P4S3 + 8O2 P4O10 + 3SO2
weight 440 gm 384 gm
moles220
440= 2
32
384= 12
for limiting reagent =1
2
8
12 O2 is limiting reagent
mass of P4O10 produced =8
1× 12 × (31 × 4 + 10 × 16) = 426 gm
118. In M3O4 oxygen percentage is 27.6 %
with 27.6 gm oxygen weight of metal conbined = 72.4 gm
with 16 × 4 gm oxygen weight of metal combined =6.27
4.72× 16 × 4 = 168
Atomic weight of metal =3
168= 56 g/mol
formula of 2nd oxide
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ACC_CHEMISTRY_Mole Concept 6 4
M O
weight % 70 30
moles 1.25 1.875
mole ratio 1 : 1.5 2 : 3
formula M2O3
119. Ag2SO4 + 2NaCl 2 AgCl + Na2SO4moles 2 × 2 = 4 4 × 1 = 4 0after ppt 2 0 2
Agn = 2 × 2 = 4
24SO
n = 2 × 1 + 2 × 1 = 4
Nan = 2 × 2 = 4
total moles of ions = 12volume after mixing 4 + 2 = 6 L
total of ions =6
12= 2 M
120. [Na+] + [Ca2+] =100
60[Cl–]
i.e. Nan + 2Ca
n =100
60 Cln
no option match with requirement.
121. Ca(OH)2 + CO2 CaCO350 × 5 = 25 milimoles
25milimoles formed.CaCO2 + 2HCl CaCl2 + H2O + CO2
0.1 N = 0.1 M
2 ×3CaCOn = nHCl
2 × 25 = (0.1) × VV = 500 mL
EXERCISE-2
1. n =32
4+
2
2=
8
9
P =V
nRT=
1
2730821.08/9= 25.215 atm
2. M M(NO2)
2
M
12= 23214M
6.39
(POAC for metal)
M = 40 (error : use 39.6 at place of 14.8 g)
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ACC_CHEMISTRY_Mole Concept 6 5
3. BaCO3
BaO + CO2
3COn =3BaCOn = 16312137
85.9
= 0.05
2COV = 0.05 × 22.4 = 1.12 L
error : Volume of CO2at STP
4. M1
V1
+ M2
V2
= MV
2.5 × 1 + 3 × 0.5 = (M) × (5.5)
M = 0.73 M
5. Gram molecularmass : mass of one mole of molecule.
6. N2
+ 3H2
2NH3
30 30 0
100 % yield 20 0 20
in case of 50 % yield 20+2
1×10 0+
2
3×10 10 L
error : Option 4 10 L NH3, 25 L N
2, 15 LH
2
7. Na4
[Fe (CN)6]
number of Na atom in 2 mole
= 2 × 4 × 6 × 1023 = 48 × 1023
8. BaCl2+ H
2S BaS + 2 HCl
SH2n = n
BaS=
32137
69.1
= 0.01
SH2m = 0.01 × 34 = 0.34 g
9. H2O (18 g H
2O has 16 gm O)
weight of O =18
16× 72 = 64 kg
10. CaCO
number of proton in one mole (100 gm) = 20 + 6 + 8 × 3 = 50
number of protons in 10 gm =100
50× 10 × N
A= 5 × 6.022 × 1023
11. CuSO4
. 5 H2O
number of atoms = 21
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ACC_CHEMISTRY_Mole Concept 6 6
12. X =1612
270
= 5
13. O2
: N2
weight 1 : 4
mole32
1:
28
4
7 : 32
14. Ag2
CO3
2Ag + CO2
+2
1O
2
fromstiochiometry 276 gm 2 × 108 = 216 gm
with 2.76 gm 2.16 gm
15. Weight of one molecule C60
H12
= 60 × 12 + 122
= 842 amu
= 842 × 1.66 × 10–24 gm
= 1.4 × 10–2 gm
16. Moles of H2O =
18
34
17. PbO + 2HCl PCl2
6.5 3.2
molea168.2
5.6
= 0.029
5.36
2.3= 0.087
PbO is limiting reagent.
18. C3H
8+ 5O
2 3CO
2+ 4H
2O
with 1 L C3H
8, 5 L O
2is required.
19. C + O2
CO2
12 gm
= 1 mole 1 mole = 22.4 L
20. 2H2
+ O2
2H2O
weight 10 64
moles 5 2
O2is limitingreagent
moles of H2O formed = 2 × 2 = 4.
21. weight mole mole ratio
C 24 2 1
H 4 4 2
O 32 2 1
emperical formula = CH2O
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ACC_CHEMISTRY_Mole Concept 6 7
22. Number of molecules =4.22
1012.1 7× 6.02 × 1023 = 3.01 × 1012
23. mass of 1 mole e–
= 9.1 × 10–31 × 6.022 × 1023 kg
= 0.55 mg
24. M2O (weight of M = 56 × 2 = 112 gm
weight of O = 16 x)
112
x16=
8.2
8.24
x = 3
25. Number of atoms =17
25.4× 4 × 6.022 × 1023
= 6.022 × 1023
26. Number of moles =.W.M
.wtGiven
27. 1 molal solution
1 mole in 1000 gm of H2O
mole fraction =
18
10001
1
= 0.0177
EXERCISE-3
1. mass number of oxygen is 16.
2. Mass remains constant but moles can be changed.
3. 2 Fe2O
3+ 3C 4Fe + 3CO
2
4. We cann't say reaction will be always 100 % complete.
5. Molality of pure ethanol is lesser because of higher M.W. of ethanol then pure water.
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Good / Importantquestions
Exercise IV
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