Some Basic Concept of Chemistry - SelfStudys

JEE - CHEMISTRY

CLASSIFICATION OF MATTER

LAWS OF CHEMICAL COMBINATIONLaw of conservation of mass [Lavoisier]In a chemical change total mass remains conservedi.e. mass before the reaction is always equal to massafter the reaction.H

2 + 1/2 O

2 H

2O ()

(g) (g) 1 mole 1/2 mole 1 molemass before the reaction = 1 × 2 + 1/2 × 32 = 18 gmmass after the reaction = 1 × 18 = 18 gm

SOLVED EXAMPLE

Example-1

A 15.9g sample of sodium carbonate is added to a

solution of acetic acid weighing 20.0g. The two

substances react, releasing carbon dioxide gas to the

atmosphere. After reaction, the contents of the

reaction vessel weigh 29.3g. What is the mass of

carbon dioxide given off during the reaction?Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9

gm. From the conservation of mass, the final mass ofthe contents of the vessel should also be 35.9 gm.But it is only 29.3 gm. The difference is due to themass of released carbon dioxide gas.Hence, the mass of carbon dioxide gas released

= 35.9 – 29.3 = 6.6 gm

Law of constant composition [Proust]All chemical compounds are found to have constantcomposition irrespective of their method ofprepration or sources. In H

2O, Hydrogen & oxygen combine in 2 : 1 molar

ratio, this ratio remains constant whether it is Tap water,river water or seawater or produced by any chemicalreaction.

Example-2

The following are results of analysis of two samples ofthe same or two different compounds of phosphorusand chlorine. From these results, decide whether thetwo samples are from the same or different compounds.Also state the law, which will be obeyed by the givensamples.

Amount P Amount ClCompound A 1.156 gm3.971 gm Compound B1.542 gm 5.297 gm

Sol. The mass ratio of phosphorus and chlorine incompound A, m

P : m

Cl = 1.156:3.971 = 0.2911:1.000

The mass ratio of phosphorus and chlorine incompound B, m

P : m

Cl = 1.542:5.297 = 0.2911:1.000

As the mass ratio is same, both the compounds aresame and the samples obey the law of definiteproportion.

2Some Basic Conceptof Chemistry

Law of multiple proportions [Dalton]When one element combines with the other element toform two or more different compounds, the mass ofone element, which combines with a constant mass ofthe other bear a simple ratio to one another. Carbon is found to form two oxides which contain42.9% & 27.3% of carbon respectively show that thesefigures shows the law of multiple proportion.

First oxide Second oxideCarbon 42.9 % 27.3 %Oxygen 57.1 % 72.7%

GivenIn th first oxide, 57.1 parts by mass of oxygen combinewith 42.9 parts of carbon.

1 part of oxygen will combine with 42 9

571

.

. part of carbon

= 0.751Similarly in 2nd oxide

1 part of oxygen will combine with 27 3

727

.

. part of carbon

= 0.376The ratio of carbon that combine with the same mass ofoxygen = 0.751 : 0.376 = 2 : 1This is a simple whole no ratio this means above datashows the law of multiple proportion.

Example-3

2.5 ml of a gaseous hydrocarbon exactly requires12.5 ml oxygen for complete combustion and produces7.5 ml carbon dioxide and 10.0 ml water vapour. All thevolumes are measured at the same pressure andtemperature. Show that the data illustrates Gay Lussac’slaw of volume combination.

Sol. Vhydrocarbon

: Voxygen

: Vcarbon dioxide

: Vwater vapour

= 2.5 : 12.5 :7.5 : 10.0= 1 : 5 : 3 : 4 (simple ratio)Hence, the data is according to the law of volumecombination.

BASIC DEFINITIONS :Relative atomic mass :One of the most important concept come out fromDalton’s atomic theory was that of relative atomic massor relative atomic weight. This is done by expressingmass of one atom with respect to a fixed standard.Dalton used hydrogen as the standard (H = 1). Later onoxygen (O = 16) replaced hydrogen as the reference.Therefore relative atomic mass is given asOn hydrogen scale :Relative atomic mass (R.A.M)

= Mass of one atom of an element

mass of one hydrogen atom

On oxygen scale :

Relative atomic mass (R.A.M)

= atomoxygenoneofmass

161

elementanofatomoneofMass

The present standard unit which was adopted

internationally in 1961, is based on the mass of one

carbon-12 atom.

Relative atomic mass (R.A.M)

= atom12Coneofmass

121

elementanofatomoneofMass

Atomic mass unit (or amu) :

The atomic mass unit (amu) is equal to th

12

1

mass of

one atom of carbon-12 isotope.

1 amu = 12

1 × mass of one C-12 atom

~ mass of one nucleon in C-12 atom.

= 1.66 × 10–24 gm or 1.66 × 10–27 kg

Today, amu has been replaced by ‘u’ which is known as

unified mass

Atomic & molecular mass :

It is the mass of 1 atom of a substance it is expressed in

amu.

Atomic mass = R.A.M × 1 amu

Relative molecular mass

=atom12Coneofmass

121

cetansubstheofmoleculeoneofmass

Molecular mass = Relative molecular mass × 1 amu

Note

Relative atomic mass is nothing but the number ofnucleons present in the atom.

SOLVED EXAMPLE

Example-4

Find the relative atomic mass of ‘O’ atom and its atomic

mass.

Sol. The number of nucleons present in ‘O’ atom is 16.

relative atomic mass of ‘O’ atom = 16.

Atomic mass = R.A.M × 1 amu = 16 × 1 amu = 16 amu

MOLE CONCEPTMOLE

Mole is a chemical counting S unit and defined asfollows :A mole is the amount of a substance that contains asmany entities (atoms, molecules or other particles) asthere are atoms in exactly 0.012 kg (or 12 gm) of thecarbon-12 isotope.From mass spectrometer we found that there are6.023 × 1023 atoms present in 12 gm of C-12 isotope.The number of entities in 1 mol is so important that it isgiven a separate name and symbol known as Avogadroconstant denoted by N

A.

i.e. on the whole we can say that 1 mole is the collectionof 6.02 × 1023 entities. Here entities may represent atoms,ions, molecules or even pens, chair, paper etc alsoinclude in this but as this number (N

A) is very large

therefore it is used only for very small things.

How big is a mole ?

602,200,000,000,000,000,000,000

Amount of water inworld's oceans (litres)

Age of earth (seconds)Population of earth

Distance from earthto sun (centimeters)

Avogadro's number

NoteIn modern practice gram-atom and gram-molecule

are termed as mole.

Gram Atomic Mass :

The atomic mass of an element expressed in gram is

called gram atomic mass of the element.

or

It is also defined as mass of 6.02 × 1023 atoms.

or

It is also defined as the mass of one mole atoms.

For example for oxygen atom :

Atomic mass of ‘O’ atom = mass of one ‘O’ atom = 16

amu

gram atomic mass = mass of 6.02 × 1023 ‘O’ atoms = 16

amu × 6.02 × 1023

= 16 × 1.66 × 10–24 g × 6.02 ×1023 = 16 g

( 1.66 × 10–24 × 6.02 × 1023 ~ 1 )

Example-5

Total number of atoms of all elements present in 1 mole

of ammonium dichromate is ?

(A) 14 (B) 19

(C) 6 × 1023 (D) 114 × 1023

Ans. (D)Sol. (NH

4)

2Cr

2O

7 =19×6.02×1023 = 114 × 1023 atoms

Example-6

How many atoms of oxygen are their in 16 g oxygen.

Sol. Let x atoms of oxygen are present

So, 16 × 1.66 × 10–24 × x = 16 g

x = 2410x66.1

1 = N

A

Gram molecular mass :

The molecular mass of a substance expressed in gram

is called the gram-molecular mass of the substance.

or

It is also defined as mass of 6.02 × 1023 molecules

or

It is also defined as the mass of one mole molecules.

For example for ‘O2’ molecule :

Molecular mass of ‘O2’ molecule

= mass of one ‘O2’ molecule

= 2 × mass of one ‘O’ atom

= 2 × 16 amu

= 32 amu

gram molecular mass = mass of 6.02 × 1023 ‘O2’ molecules

= 32 amu × 6.02 × 1023

= 32 × 1.66 × 10–24 gm × 6.02 × 1023 = 32 gm

Gay-Lussac’s Law of Combining Volume :

According to him elements combine in a simple ratio of

atoms, gases combine in a simple ratio of their volumes

provided all measurements should be done at the same

temperature and pressure

H2 (g) + Cl

2 (g) 2HCl

1 vol 1 vol 2 vol

Avogadro’s hypothesis :

Equal volume of all gases have equal number of

molecules (not atoms) at same temperature and

pressure condition.

N.T.P. (Normal Temperature and Pressure)

At N.T.P. condition :

temperature = 0°C or 273 K, pressure = 1 atm = 760 mm

of Hg and volume of one mole of gas at NTP is found to

be experimentally equal to 22.4 litres which is known as

molar volume.

NoteMeasuring the volume is equivalent to counting

the number of molecules of the gas.

SOLVED EXAMPLE

Example-7

Calculate the volume in litres of 20 g hydrogen gas atNTP.

Sol. No. of moles of hydrogen gas = massMolecular

Mass

= gm2

gm20 = 10 mol

volume of hydrogen gas at NTP = 10 × 22.4 lt.

Y-map : Interconversion of mole - volume, mass and number ofparticles :

Mole

× 22.4 lt

22.4 lt

Volume at NTP

NA

× NA

Number

Mass

Note : Measuring the volume is equivalent to counting

the number of molecules of the gas.

Example-8

The number of atoms contained in 11.2 L of SO2 at

N.T.P. are -

(A) 3/2 x 6.02 x 1023 (B) 2 x 6.02 x 1023

(C) 6.02 x 1023 (D) 4 x 6.02 x 1023

Ans. (A)

Sol. 22.4 litre gas has = 1 mole

1 litre gas has = 4.22

1×11.2 =

2

1mole

= S + 2O = 3

= 2

3 × 6.02 × 1023

METHODS OF CALCULATIONS OF MOLE

(a) If no. of some species is given, then no. of moles

= Given no

NA

.

(b) If weight of a given species is given, then no of

moles = Given wt

Atomic wtfor atoms

.( ),

or = Given wt

Molecular wtfor molecules

.

.( )

(c) If volume of a gas is given along with its

temperature (T) and pressure (P) use n = PV

RT

where R = 0.0821 lit-atm/mol–K (when P is in atmosphere

and V is in litre.)

1 mole of any gas at NTP (0°C & 1 bar) occupies

22.7 litre.

1 mole of any gas at NTP (0°C & 1 atm) occupies

22.4 litre.

Atom : Atom is smallest particle which can not be

divided into its constituents.

RELATIONSHIP BETWEEN GRAM AND AMU

1 amu = 1

12 wt of one C - 12 atom.

for 1 mole C = 12 gm = 6.023 × 1023 atomswt of 6.023 × 1023 atoms = 12 gm

wt of 1 atom of C = 12

Ngm

A

(NA Avogadro’s number = 6.23 × 1023)

1 amu = 1

12 wt of one C - 12 atom

= 1

12

12

NA gm

1 amu = gmN

1

A

ELEMENTAL ANALYSIS

For n mole of a compound (C3H

7O

2)

Moles of C = 3n

Moles of H = 7n

Moles of O = 2n

SOLVED EXAMPLE

Example-9Find the wt of water present in 1.61 g of Na

2SO

4. 10H

2O

Sol. Moles of Na2SO

4. 10H

2O =

wt. in gram

molecular wt = 1.61

322= 0.005 molesMoles of water = 10 × moles of Na

2SO

4.10H

2O

= 10 × 0.005 = 0.05wt of water = 0.05 × 18 = 0.9 gm Ans.

PERCENTAGE FORMULAE COMPOSITION :% of element in compound

= atomic weight of element number of atom 100

total molecular weight of compound

Here we are going to find out the percentage of eachelement in the compound by knowing the molecularformula of compound.We know that according to law of definite proportionsany sample of a pure compound always possessconstant ratio with their combining elements.

SOLVED EXAMPLE

Example-10

Every molecule of ammonia always has formula NH3

irrespective of method of preparation or sources. i.e. 1mole of ammonia always contains 1 mol of N and 3 moleof H. In other words 17 gm of NH

3 always contains 14

gm of N and 3 gm of H. Now find out % of each elementin the compound.

Sol Mass % of N in NH3 =

3

3

Mass of N in 1 mol NH100

Mass of 1 mol of NH =

14 gm

17×100

= 82.35 %

Mass % of H in NH3 =

3

3

Mass of H is 1 mol NH100

Massof 1 mole of NH =

3

17×100 = 17.65 %

DENSITY :

(a) Absolute density (b) Relative density

Absolute density = Mass

volume

Relative density =density of substance

density of standard substance

Specific gravity = 2

density of substance

density of H O at 4 CVapour density : It is defined only for gas.

It is a density of gas with respect to H2 gas at same

temperature & pressure

V.D = 2H

d gas

d = PM RT

PM RTgas

H

/

/2

= M

Mgas

H2

= M

2

V.D = M

2

SOLVED EXAMPLE

Example-11

What is the V.D. of SO2 with respect to CH

4

Sol V.D. = 2

4

M.W. of SO

M.W. of CH

V.D = 16

64= 4

Example-12

7.5 litre of the particular gas at S.T.P. weighs 16 gram.

What is the V.D. of gas

Sol 7.5 litre = 16 gram

moles = M

16

4.22

5.7

M = 48 gram V.D = 2

48= 24

EMPIRICAL AND MOLECULAR FORMULA :

We have just seen that knowing the molecular formula

of the compound we can calculate percentage

composition of the elements. Conversely if we know

the percentage composition of the elements initially,

we can calculate the relative number of atoms of each

element in the molecules of the compound. This gives

us the empirical formula of the compound. Further if

the molecular mass is known then the molecular formula

can easily be determined.

The empirical formula of a compound is a chemical

formula showing the relative number of atoms in the

simplest ratio. An empirical formula represents the

simplest whole number ratio of various atoms present

in a compound.

The molecular formula gives the actual number of atoms

of each element in a molecule. The molecular formula

shows the exact number of different types of atoms

present in a molecule of a compound.

The molecular formula is an integral multiple of the

empirical formula.

i.e. molecular formula = empirical formula × n

where n = massformulaempirical

massformulamolecular

SOLVED EXAMPLE

Example-13

Acetylene and benzene both have the empirical formula

CH. The molecular masses of acetylene and benzene

are 26 and 78 respectively. Deduce their molecular

formulae.

Sol. Empirical Formula is CH

Step-1 The empirical formula of the compound is CH

Empirical formula mass = (1 × 12) + 1 = 13.

Molecular mass = 26

Step-2 To calculate the value of ‘n’

n = massformulaEmpirical

massMolecular =

13

26 = 2

Step-3 To calculate the molecular formula of the

Compound.

Molecular formula = n × (Empirical formula of the

compound)

= 2 × CH = C2 H

2

Thus the molecular formula is C2 H

2

Similarly for benzene

To calculate the value of ‘n’

n = massformulaEmpirical

massMolecular =

13

78 = 6

thus the molecular formula is 6 × CH = C6H

6

Example-14

An organic substance containing carbon, hydrogenand oxygen gave the following percentagecomposition.

C = 40.684% ; H = 5.085% and O = 54.228%The molecular weight of the compound is 118. Calculatethe molecular formula of the compound.

Sol. Step-1To calculate the empirical formula of the compound.

Element Symbol

Carbon C

Hydrogen H

Oxygen O

Percentage of element

40.687

5.085

54.228

At. massof element

12

1

16

Relative no.Percentage

At. massof atoms =

40.687

12= 3.390

5.085

1= 5.085

54.228

16= 3.389

Simplestatomic ratio

3.3903.389

=1

5.085

3.389=1.5

3.389

3.389=1

2

3

2

Empirical Formula is C2 H

3 O

2

Step-2 To calculate the empirical formula mass. The empirical formula of the compound is C

2 H

3 O

2 .

Step-3 To calculate the value of ‘n’

n = Molecular mass

Empirical formula mass = 118

59 = 2

Step-4 To calculate the molecular formula of the salt.Molecular formula = n × (Empirical formula) = 2 × C

2 H

3 O

2 = C

4 H

6 O

4

Thus the molecular formula is C4 H

6 O

4.

Chemical Reaction :It is the process in which two or more than twosubstances interact with each other where old bondsare broken and new bonds are formed.

9. STOICHIOMETRY BASED CONCEPT

(Problems Based on Chemical Reaction)

One of the most important aspects of a chemical equa-

tion is that when it is written in the balanced form, it

gives quantitative relationships between the various re-

actants and products in terms of moles, masses, mol-

ecules and volumes. This is called stoichiometry (Greek

word, meaning 'to measure an element'). For example, a

balanced chemical equation alongwith the quantitative

information conveyed by it is given below:

CaCO3

+ 2HClCaCl2

+ H2O + CO

2

1 Mole 2 Mole 1 Mole

1 Mole 1 Mole

40 + 12 + 3 × 16 2(1 + 35.5) 40 + 2 × 35.5

2 × 1 + 16 12 + 2 × 16

= 100 g = 73 g = 111 g

= 18 g = 44 g or 22.4 L at STP

Thus,

(i) 1 mole of calcium carbonate reacts with 2 moles of

hydrochloric acid to give 1 mole of calcium

chloride, 1 mole of water and 1 mole of carbon dioxide.

(ii) 100 g of calcium carbonate react with 73 g hydro-

chloric acid to give 111 g of calcium chloride, 18 g of

water and 44 g (or 22.4 litres at STP) of carbon dioxide.

Stoichiometry

N2

+ 3H2

2NH3

1 mole + 3 mole 2 mole

22.4 litre + 3 × 22.4 litre 2 × 22.4 litre (at STP)

1 litre + 3 litre 2 litre

1000 mL + 3000 mL 2000 mL

1 mL + 3 mL 2 mL

28 gm + 6 gm 34 g

(According to the law of conservation of mass)

Gram can not be represented by stoichiometry

The quantitative information conveyed by a chemical

equation helps in a number of calculations. The prob-

lems involving these calculations may be classified into

the following different types :-

(1) Mass - Mass Relationships i.e. mass of one of the

reactants or products is given and the mass of some

other reactant or product is to be calculated.

(2) Mass - Volume Relationships i.e. mass/volume of

one of the reactants or products is given and the vol-

ume/mass of the other is to be calculated.

(3) Volume - Volume Relationships i.e.volume of one of

the reactants or the products is given and the volume

of the other is to be calculated.

The general method of calculations for all the problems

of the above types consists of the following steps :-

(i) Write down the balanced chemical equation.

(ii) Write the relative number of moles or the relative

masses (gram atomic or molecular masses) of the reac-

tants and the products below their formula.

(iii) In case of a gaseous substance, write down 22.4

litres at STP below the formula in place of 1 mole

(iv) Apply unitary method to make the required calcu-

lations.

Quite often one of the reactants is present in larger

amount than the other as required according to the

balanced equation. The amount of the product formed

then depends upon the reactant which has reacted

completely. This reactant is called the limiting reactant.

The excess of the other is left unreacted.

Combustion reaction : (Problem based on combustion reactions)

: For balancing the combustion reaction : First of all

balance C atoms, Then balance H atom, Finally balance

Oxygen atom.

For Example : Combustion reaction of C2H

6 : C

2H

6 + O

2

CO2 + H

2O (skeleton equation)

balance C atoms C2H

6 + O

2 2CO

2 + H

2O

Now balance H atoms C2H

6 + O

2 2CO

2 +

3H2O

Now balance Oxygen atoms C2H

6 + 7

2O

2

2CO2 + 3H

2O

Type I Involving Mass-Mass Relationship

Example-15

How much iron can be theoritically obtained in thereduction of 1 kg of Fe

2O

3

Sol. Writing the balanced equation for the reaction.

w

weight 1000n mol

M 160

The equation shows that 2 mol of iron are obtainedfrom 1 mol of ferric oxide.

Hence, the obtained no. of moles of Fe = 2 1000

160

=12.5 mol = weight weight

Atomic weight 56

Weight of iron obtained = 12.5 × 56 g = 700 g

Example-16

What amount of silver chloride is formed by the actionof 5.850 g of sodium chloride on an excess of silvernitrate?

Sol. Writing the equation for the reaction

3 3

1 1 1 1

NaCl AgNO AgCl NaNO

w

weight 5.85n 0.1 mol

M 58.5

1 mol of AgCl is obtained with 1 mol of NaClHence, the number of moles of AgCl obtained with 0.1mol of NaCl = 0.1 mol

W

weightn =

M 0.1 mol w

weight weight= =

M 143.5

weight = 0.1 × 143.5 g = 14.35 g.

Type II Mass - Volume Relationship

Example-17At 100°C for complete combustion of 3g ethane therequired volume of O

2 & produced volume of CO

2 at

STP will be.

Sol.2 7 4 6

2C H2 64CO2(g) 6H O2 (g)

w

weightn

M =

3 1

30 10 = 0.1 mole

(a) Required moles of O2 =

70.1

2 = 0.35 mol

volume of O2 at STP = 0.35 × 22.4 = 7.84 litre

(b) Produced moles of CO2 =

4

2 × 0.1 = 0.2 mol

volume of CO2 at STP = 0.2 × 22.4 = 4.48 litre

Example-18

In the following reaction, if 10 g of H2, react with N

2.

What will be volume of NH3 at STP.

N2 + 3H

2 2NH

3

Sol. N2 + 3H

2 2NH

3 10 g

w

weight 10n

M 2 = 5 mol.

Produced moles of NH3 =

2 105 =

3 3 , Volume of

NH3 at STP =

1022.4

3 = 74.67 litre

Type III Volume-Volume Relationship

Example-19

At 100°C for complete combustion of 1.12 litre of butane(C

4H

10), the produced volume of H

2O

(g) & CO

2 at STP

will be.

Sol. C4H

10(g)+ 13

2O

2(g) 4CO

2(g) + 5H

2O

(g)

1.12 litreVolume of H

2O

(g) at STP = 5 × 1.12 = 5.6 litre

Volume of CO2(g)

at STP = 4 × 1.12 = 4.48 litre

Example-20

At 25°C for complete combustion of 5 mol propane

(C3H

8). The required volume of O

2 at STP will be.

Sol. For C3H

8 , the combustion reaction is

C3H

8(g) + 5O

2(g 3CO

2(g) + 4H

2O

(g)

5 mol

Required moles of O2 = 5 × 5 = 25 mol = V

22.4

volume of O2 gas at STP (V) = 25 × 22.4 = 560 litre

Example-21

3 litre of mixture of propane (C3H

8)

& butane (C

4H

10) on

complete combustion gives10 litre CO2. Find the

composition of mixture.

Sol. Let the volume of propane in the mixture = x litre,

The volume of butane in the mixture = (3 – x) litre

Now let us calculate the volume of CO2 evolved with

the help of Chemical equation.

Step I : Calculation of volume of CO2 from x litre of

propane.

C3H

8 + 5O

2 3CO

2+ 4H

2O

x litre 3x litreStep II : Calculation of volume of CO

2 from (3 – x) litre

of butane. The combustion equation for butane is:

C4H

10 + 13

2O

2 4CO

2+ 5H

2O

(3–x) litre 4(3–x) litreStep III : Calculation of composition of the mixture.Total volume of CO

2 formed in the step (I) and step (II)

= [3x + 4(3 – x)] litreBut the volume of CO

2 actually formed = 10 litre

3x + 4(3 – x) = 10or 3x + 12 – 4x = 10 or x = 2 litre volume of propane = x litre = 2 litre volume of butane = (3 – x) litre = (3 – 2) = 1 litre

LIMITING REAGENT (L.R.) CONCEPTLimiting Reagent (L.R.) : The reactant which iscompletely consumed in a reaction is called as L.R.

Ex 1 2 1 2 Stoichiometry A + 2B C + 2Dgiven 3 mol 9mol 3 – 3 = 0 mol 9 – 6 = 3 mol 3 mol 6 molL.R. = AFormula for checking L.R. =

given value ( may moles, volume, or molecules)

Stoichiometry Co-efficient

Least value indicate the L.R.

Ex. A B

31

3 92

4.5

3 < 4.5 So A is L.R.

Identification : More than 1 initial quantites of

reactants are given

Example-22

A + 5B C + 3D In this reaction which is a L.R.

Given 10 mol 10 mol

Sol. For A For B

1010

1

102

5

2 < 10 So B is L.R.

Example-23

H2(g)

+ 12

O2(g)

H2O

(g) ; In the above

reaction what is the volume of water vapour at STP.Given 4 g 32 g

Sol. 1 12

1

H2(g)

+ 12

O2(g)

H2O

(g)

4 g 32 g

For H2

For O2

4n

2 = 2mol

32n

32 =1mol

2

1=2 1

2

1=2 mol

Moles of H2O

(g) = 2 mol =

V

22.4

2 = 2 So Both H2 & O

2 are L.R.

volume of H2O

(g) at STP = 22.4 × 2 = 44.8 litre

Example-24

At NTP, In a container 100 mL N2 and 100 mL of H

2 are

mixed together. Then find out the produced volume of

NH3.

Sol. Balanced equation will be N2 + 3H

2 2NH

3.

Given 100mL 100mL

For determination of Limiting reagent. Now divided the

given quantities by staichiometry coefficients

100

1 = 100

100

3 = 33.3 (Limiting reagent)

In this reaction H2 is limiting reagent so reaction will

proceed according to H2.

According to stoichiometry from 3 mL of H2 produced

volume of NH3 = 2 mL

That is from 100 mL of H2 produced volume of NH

3 =

2

3 × 100 = 66.6 mL

PERCENTAGE YIELD :

The percentage yield of product =

actual yield

the theoretical maximum yield100

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.)

SOLVED EXAMPLE

Example-257.5 mL of a hydrocarbon gas was exploded with excessof oxygen. On cooling, it was found to have undergonea contraction of 15 mL. If the vapour density of thehydrocarbon is 14, determine its molecular formula.(C = 12, H = 1)

Sol. CxH

y + (x +

y

4) O

2 X CO

2 +

y

2 H

2O

7.5 mlon cooling the volume contraction = 15 mli.e. The volume of H

2O (g) = 15 ml

V.D. of hydrocarbon = 14Molecular wt. of C

xH

y = 28

12x + y = 28 ...(1)From reaction

7.5 y

2 = 15 y = 4

12 x + 4 = 2812x = 24 x = 2Hence Hydrocalbon is C

2H

4.

EQUIVALENT WEIGHT :

The equivalent weight of a substance is the number of

parts by weight of the substance that combine with or

displace directly or indirectly 1.008 parts by weight of

hydrogen or 8 parts by weight of oxygen or 35.5 parts

by weight of chlorine or 108 parts by weight of Ag.

Calculation of equivalent weight :

Equivalent weight Atomic weight

Valency factor

Equivalent weight of ionsformula weight of ion

Valency

Equivalent weight of ionic compound = equivalent

weight of cation + equivalent weight of anion

Ex. Equivalent weight of H2SO

4= Equivalent

weight of H++Equivalent weight of Anion(SO4

–2)

= 1 + 48 = 49

Equivalent weight of acid / base = Molecular weight

Basicity/ Acidicity

Equivalent weight of salt =

Molecular weight

Total charge on cation or anion

Ex. Na2SO

4 (salt) i.e. 2Na+ & SO

4–2

Total charge on cation or anion is 2

molecular weight of Na2SO

4 is = ( 2 × 23 + 32 + 16 × 4)

= 142

Equivalent weight of Na2SO

4 =

14271

2

Equivalent weight of an oxidizing or reducing agent

=Molecular weight of the substance

Number of electrons gain/ lost by one molecule

Concept of gram equivalent and law of chemicalequivalence:

Number of gram equivalent = (gram)W

E

(gram)W Valencefactor

M

= n × valence factor; whereAccording to it in a reaction equal gram equivalent ofreactant are reacts to give same number of gramequivalent of products.For a reactionaA + bB cC + dDNumber of gram equivalent of A = Number of gramequivalent of B = Number of gram equivalent of C =Number of gram equivalent of D

Methods for determination of the equivalent weight :

Hydrogen displacement method : This method is used

for those elements which can evolve hydrogen from

acids, i.e., active metals.equivalent weight of metal =

2

weight of metal1.008

weight of H gas (displaced)

Oxide formation method : A known mass of the element

is changed into oxide directly or indirectly. The mass

of oxide is noted.

Mass of oxygen = (Mass of oxide – Mass of element)

equivalent weight of element = weight of element

8weight of oxygen

Chloride formation method : A known mass of the

element is changed into chloride directly or indirectly.

The mass of the chloride is determined.equivalent weight of element

= weight of element

35.5weight of chlorine

Metal to metal displacement method : More active metal

can displace less active metal from its salt's solution.

The mass of the displaced metal bear the same ratio as

their equivalent weights.

1 1

2 2

m E

m E

Double decomposition method : this method is based

on the following points -

(a) The mass of the compound reacted and the mass of

product formed are in the ratio of their equivalent

masses.

(b) The equivalent mass of the compound

(electrovalent) is the sum of equivalent masses of its

radicals.

(c) The equivalent mass of a radical is equal to the

formula mass of the redical divided by its charge.

AB + CD AD (ppt.) + CB

Mass of AB Equivalent mass of AB

Mass of AD Equivalent mass of AD

Equivalent mass of A Equivalent mass of B

Equivalent mass of A Equivalent mass of D

Silver salt method : This method is used for findingthe equivalent weight of carbonic (organic) acids. Aknown mass of the RCOOAg is changed into Agthrough combusion. The mass of Ag is determined.

Equivalent weight ofRCOOAg

Equivalent weight of Ag

weight of RCOOAg

weight of Ag

equivalent weight of RCOOAg

= weight ofRCOOAg

108weight of Ag

By electrolysis :1 1

2 2

w E

w E

Where w1 & w

2 are deposited weight of metals at

electrodes and E1 and E

2 are equivalent weight

respectively.Methods for determination of Atomic weight - Atomic weight = equivalent weight × n

where n = valency Dulong and Petit's law - This law is applicable only

for solids (except Be, B, Si, C)

Atomic mass × specific heat (in calorie/gram) 6.4

or atomic mass (approximate) = 6.4

specific heat

Example-26

Specific heat of a metal is 0.031 cal per degree per gram,

and its equivalent weight is 103.6. Calculate the exact

atomic weight of the metal.

Sol. According to Dulong and Petit's law -

approximate atomic weight = 6.4

0.031 = 206.45

Valency of metal

Approximate atomic weight 206.45= 1.99 2

Equivalent weight 103.6

So, the exact atomic weight of the element =

Equivalent weight × valency = 103.6 × 2 = 207.2

Example-27

A chloride of an element contains 49.5% chlorine. Thespecific heat of the element is 0.064. Calculate theequivalent mass, valency and atomic mass of theelement.

Sol. Mass of chlorine in the metal chloride = 49.5Mass of metal = (100 – 49.5) = 50.5Equivalent weight of metal

weight of metal35.5

weight of chlorine 50.5

35.549.50

= 36.21

Now according to Dulong and Petit's law,Approximate

at. wt. of the metal 6.4

specific heat

6.4100

0.064

Approximate atomic weight 100Valency 2.7 3

Equivalent weight 36.21

Hence, exact atomic weight = 36.21 × 3 = 108.63Law of isomorphism : Isomorphous substances formcrystals which have same shape and size and can growin the saturated solution of each other.Examples of isomorphous compounds -(1) H

2SO

4 and K

2CrO

4

(2) ZnSO4.7H

2O and FeSO

4.7H

2O and MgSO

4.7H

2O

(3) KClO4 and KMnO

4

(4) K2SO

4.Al

2 (SO

4)

3.24H

2O and K

2SO

4.Cr

2(SO

4)

3.24H

2O

Conclusions -Masses of two elements that combine with same massof other elements in their respective compounds are inthe ratio of their atomic masses.Mass of one elements (A) that combines with a certain mass of other element

Mass of other element (B) that combines with the same mass of other element

= Atomic mass of A

Atomic mass of B

2. The valencies of the elements forming isomorphouscompounds are the same.Atomic mass from vapour density of a chloride -Required condition – chloride of element should bevapour.Required data - (i) Vapour density of chloride.(ii) Equivalent weight of element.

Let the valency of the element be x. The formula of itschloride will be MCl

x .

Molecular weight = Atomic weight of M + 35.5x

Atomic weight = Equivalent weight × valency

or A = E × x

Molecular weight = E × x + 35.5 x

or 2 × V.D. = x(E + 35.5)

or 2 V.D.

x=E+35.5

Example-28

The oxide of an element contains 67.67% of oxygen

and the vapour density of its volatile chloride is 79.

Calculate the atomic weight of the element.

Sol. Calculation of equivalent weight : weight of oxygen

= 67.67 g

weight of element = 100 – 67.67 = 32.33 g

67.67 g of oxygen combines with 32.33 g of element

8 g of oxygen combines with = 32.33 8

67.67

= 3.82 g

of element

Equivalent weight of the element = 3.82

Suppose M represents one atom of the element and x

is its valency. The molecular formula of the volatile

chloride would be MClx.

Formula weight of chloride = 3.82 × x + 35.5 x

= 39.32 x

But molecular weight of Chloride = 2 × V.D.

39.32 x = 2 × 79 x = 2 79

39.32

= 4

Now atomic weight = Equivalent weight × valency of

element = 3.82 × 4 = 15.28

Specific heat method : If P

V

C

C = is given, then

Case I. If = 5/3 = 1.66 Atomicity will be one

Case II. If = 7/5 = 1.4 Atomicity will be two

Case III. If = 4/3 = 1.33 Atomicity will be three

Atomic weight = Molecular weight

Atomicity

Example-29

Vapour density of a gas is 16. If the ratio of specific

heat at constant pressure and specific heat at constant

valume is 1.4. Then find out its atomic weight.

Sol. Given :P

V

C

C = 1.4 and vapour density = 16

We know that Molecular weight = 2 × vapour density

Molecular weight = 2 × 16 = 32

Here = 1.4 so atomicity will be 2.

Molecular weight 32Atomic weight 16

Atomicity 2

Methods for determination of Molecular weight :

Molecular weight = 2 × vapour density

Diffusion method (only for gases) :- According to

Graham's diffusion law

1

rate of diffusion of a gasMolecular weight or density

1 2

2 1

r M

r M

CONCENTRATION OF SOLUTIONConcentration of solution can be expressed in any ofthe following ways.(a) % by wt amount of solute dissolved in 100 gm ofsolution

4.9% H2SO

4 by wt.

100 gm of solution contains 4.9 gm of H2SO

4

(b) % by volume volume of solute dissolved in 100ml of solution

x% H2SO

4 by volume

100 ml of solution contains x ml H2SO

4

(c) % wt by volume wt. of solute present in 100 mlof solution(d) % volume by wt. volume of solute present in100 gm of solution.

CONCENTRATION TERMSMolarity (M) : No. of moles of solute present in 1000 mlof solution.

molarity (M) = (lit) solution of volume

solute of moles

M = l)solution(m of volume

solute of m.moles

Molality (m)No. of moles of solute present in 1000 gm of solvent

m = moles of solute

wt.of solvent in kg , m = m.moles of solute

wt.of solvent in gm

Normality (N)No of gm equivalents of solute present in 1000 ml ofsolution

N = gm equivalents of solute

volume of solution(lit) =

m. equivalent of solute

volume of solution in (ml)

Formality (F)The formality is the no. of gm -formula weights of theionic solute present in 1000 ml of solution.

F = (lit)solution of volume wtformula

gm in wt

Mole fractionThe mole fraction of a perticular component in a solutionis defined as the number of moles of that componentper mole of solution.If a solution has nA mole A & n

B mole of B.

mole fraction of A (XA) =

n

n nA

A B

mole fraction of B (XB) =

n

n nB

A BX

A + X

B = 1

Parts per million (ppm) : = solvent of Mass

solute of Mass × 106

610 solution of Mass

solute of Mass

SOLVED EXAMPLE

Example-30

0.2 mole of HCl and 0.1 mole of barium chloride were

dissolved in water to produce a 500-mL solution. The

molarity of the Cl– ions is -

(A) 0.06 M (B) 0.09 M

(C) 0.12 M (D) 0.80 MAns. (D)

Sol. HCl Cl0.2 mole

BaCl2 2 Cl–

2 × 0.1 = 0.2Total moles of Cl– = 0.4

M = vm

1000w

Molarity = 500

10004.0 = 0.8 4.0

m

w

RELATIONSHIP BETWEEN MOLARITY, MOLALITY &DENSITY OF SOLUTIONLet the molarity of solution be 'M', molality be 'm' andthe density of solution be d gm/m.Molarity implies that there are M moles of solute in1000 ml of solution wt of solution = density × volume= 1000 d gm wt of solute = MM

1

where M1 is the molecular wt of solute

wt of solvent = (1000d – MM1) gm

(1000d – MM1) gm of solvent contains M moles of

solute

1000 gm of solvent have = mole1000MM–d1000

M

1

= Molalityno. of moles of solute present in 1000 gm of solvent =

1000

1000 1

M

d MM– = Molality

on simplyfying

1000

M

m

1Md 1

Relationship between molality & mole fraction

consider a binary solution consisting of two

components A (Solute) and B (Solvent).

Let xA & x

B are the mole fraction of A & B respectively.

xA =

n

n nA

A B , xb =

n

n nB

A B

If molality of solution be m then

mn

mass of solventA 1000 =

A

B B

n1000

n M

where MB is the molecular wt of the solvent B

mx

x MA

B B

1000

molality = BM

1000

B of fraction mole

Aof fraction mole

m = mole fraction of solute 1000

mole fraction of solvent molecular wt. of solvent

SOLVED EXAMPLE

Example-31An aqueous solution is 1.33 molal in methanol.

Determine the mole fraction of methanol & H2O

Sol. molality = mole fraction of solute

1000mole fraction of solvent mol.wt of solvent

1.33= 1000Mx

x

BB

A ,

B

A

x

x

1000

1833.1

,

23 94

1000

.

x

xA

B

xA = 0.02394 x

B, x

A + x

B = 1 1.02394 x

B = 1

xB 1

102394. = 0.98, x

A = 0.02 Ans.

2nd Method : Let wt of solvent = 1000 gm,

molality = 1.33 = moles of solute

mole fraction of solute = solvent of molessolute of moles

solute of moles

,

18/100033.1

33.1

181000

+m

m

mole fraction of solute = 0.02

mole fraction of solvent = 1 – 0.02 = 0.98

Example-32225 gm of an aqueous solution contains 5 gm of urea.

What is the concentration of the solution in terms of

molality. (Mol. wt. of urea = 60)

Sol. Mass of urea = 5 gm

Molecular mass of urea = 60

Number of moles of urea = 60

5 = 0.083

Mass of solvent = (255 – 5) = 250 gm

Molality of the solution =

gram in solvent of Mass

solute of moles of Number × 1000

= 250

083.0 × 1000= 0.332.

Example-330.5 g of a substance is dissolved in 25 g of a solvent.Calculate the percentage amount of the substance inthe solution.

Sol. Mass of substance = 0.5 gMass of solvent = 25 g

percentage of the substance (w/w) = 100255.0

5.0

= 1.96

Example-34Find the relative atomic mass, atomic mass of thefollowing elements.(i) Na (ii) F (iii) H (iv) Ca (v) Ag

Sol. (i) 23, 23 amu (ii) 19, 19 amu

(iii) 1, 1.008 amu , (iv) 40, 40 amu,

(v) 108, 108 amu.

Example-35A sample of (C

2H

6) ethane has the same mass as 107

molecules of methane. How many C2H

6 molecules does

the sample contain ?

Sol. Moles of CH4 =

A

7

N

10

Mass of CH4 =

A

7

N

10 × 16 = mass of C

2H

6

So Moles of C2H

6 =

30N

1610

A

7

So No. of molecules of C2H

6=

7

A

10 16

N 30

× N

A= 5.34× 106.

Example-36From 160 g of SO

2 (g) sample, 1.2046 × 1024 molecules

of SO2 are removed then find out the volume of left

over SO2 (g) at NTP.

Sol. Given moles = 64

160 = 2.5.

Removed moles = 23

24

10023.6

102046.1

= 2.

so left moles = 0.5.

volume left at STP = 0.5 × 22.4 = 11.2 lit.

Example-3714 g of Nitrogen gas and 22 g of CO

2 gas are mixed

together. Find the volume of gaseous mixture at NTP.

Sol. Moles of N2 =

28

14 = 0.5.

moles of CO2 =

44

22 = 0.5.

so total moles = 0.5 + 0.5 = 1.

so vol. at STP = 1 × 22.4 = 22.4 lit.

Example-38Show that in the reaction N2 (g) + 3H2(g) 2NH3 (g),mass is conserved.

Sol. N2 (g) + 3H2(g) 2NH3 (g)moles before reaction 1 3 0moles after reaction 0 0 2Mass before reaction = mass of 1 mole N2(g) + mass of

3 mole H2(g)= 14 x 2 + 3 x 2 = 34 gmass after reaction = mass of 2 mole NH3= 2 x 17 = 34 g.

Example-39Calculate the mass in gm of 2NA molecules of CO2 -(A) 22 gm (B) 44 gm(C) 88 gm (D) None of these.

Sol. (C) NA molecules of CO2 has molecular mass = 44. 2NA molecules of CO2 has molecular mass = 44 × 2 = 88 gm.

Example-40How many carbon atoms are present in 0.35 mol ofC6H12O6 -(A) 6.023 × 1023 carbon atoms(B) 1.26 × 1023 carbon atoms(C) 1.26 × 1024 carbon atoms(D) 6.023 × 1024 carbon atoms

Sol. (C) 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

Example-41How many molecules are in 5.23 gm of glucose(C6H12O6) -(A) 1.65 × 1022 (B) 1.75 × 1022

(C) 1.75 × 1021 (D) None of theseSol. (B)

180 gm glucose has = NA molecules

5.23 gm glucose has = 180

10023.623.5 23

= 1.75 × 1022 molecules

Example-42What is the weight of 3.01 × 1023 molecules of ammonia-(A) 17 gm (B) 8.5 gm(C) 34 gm (D) None of these

Sol. (B) 6.023 × 1023 molecules of NH3 has weight = 17 gm 3.01 × 1023 molecules of NH3 has weight

= 23

23

10023.6

1001.317

= 8.50 gm

Example-43How many molecules are present in one ml of watervapours at NTP -(A) 1.69 × 1019 (B) 2.69 × 10–19

(C) 1.69 × 10–19 (D) 2.69 × 1019

Sol. (D) 22.4 litre water vapour at NTP has

= 6.023 × 1023 molecules 1 × 10–3 litre water vapours at NTP has

=236.023 10

22.4

× 10–3 = 2.69 × 10+19

Example-44How many years it would take to spend Avogadro'snumber of rupees at the rate of 1 million rupees in onesecond -(A) 19.098 × 1019 years (B) 19.098 years(C) 19.098 × 109 years (D) None of these

Sol. (C) 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

Example-45An atom of an element weighs 6.644 × 10–23 g. Calculateg atoms of element in 40 kg-(A) 10 gm atom (B) 100 gm atom(C) 1000 gm atom (D) 104 gm atom

Sol. (C) weight of 1 atom of element = 6.644 × 10–23 gm weight of 'N' atoms of element = 6.644 × 10–23 × 6.023 × 1023 = 40 gm 40 gm of element has 1 gm atom.

40 x 103 gm of element has 40

1040 3 ,

= 103 gm atom.

Example-46The density of O2 at NTP is 1.429g / litre. Calculate thestandard molar volume of gas-(A) 22.4 lit. (B) 11.2 lit(C) 33.6 lit (D) 5.6 lit.

Sol. (A) 1.429 gm of O2 gas occupies volume = 1 litre.

32 gm of O2 gas occupies = 32

1429. ,

= 22.4 litre/mol.

Example-47Which of the following will weigh maximum amount-(A) 40 g iron(B) 1.2 g atom of N

(C) 1 × 1023 atoms of carbon(D) 1.12 litre of O2 at NTP

Sol. (A)(A) Mass of iron = 40 g(B) Mass of 1.2 g atom of N = 14 × 1.2 = 16.8 gm

(D) Mass of 1×1023 atoms of C= 23

23

10023.6

10112

=1.99gm.

(D) Mass of 1.12 litre of O2 at NTP = 4.22

2.132 = 1.6 g

Example-48How many moles of potassium chlorate to be heated toproduce 11.2 litre oxygen -

(A) 1

2 mol (B)

1

3 mol

(C) 1

4 mol (D)

2

3 mol.

Sol. (B) 2 KClO3 2KCl + 3O2Mole for reaction 2 2 3 3 × 22.4 litre O2 is formed by 2 mol KClO3

11.2 litre O2 is formed by 4.223

2.112

= 1

3 mol KClO3

Example-49Calculate the weight of lime (CaO) obtained by heating200 kg of 95% pure lime stone (CaCO3).(A) 104.4 kg (B) 105.4 kg(C) 212.8 kg (D) 106.4 kg

Sol. (D) 100 kg impure sample has pure

CaCO3 = 95 kg 200 kg impure sample has pure CaCO3

= 100

20095 = 190 kg. CaCO3 CaO + CO2

100 kg CaCO3 gives CaO = 56 kg.

190 kg CaCO3 gives CaO = 100

19056 = 106.4 kg.

Example-50calculate the weight of one atom of Ag – (At. wt. of Ag

= 108)

(A) 17.93 × 10–23gm (B) 16.93 × 10–23 gm

(C) 17.93 × 1023 gm (D) 36 × 10–23 gm

Sol. (A)

NA atoms of Ag weigh 108 gm

1 atom of Ag weigh = A

108

N

= 2310023.6

108

= 17.93 × 10–23 gm.

Example-51In 5g atom of Ag (at. wt. = 108), calculate the no. ofatoms of Ag (N = NA) -(A) 1 N (B) 3N (C) 5 N (D) 7 N

Sol. (C) 1 gm atom of Ag has atoms = N 5 gm atom of Ag has atoms = 5N.

Example-52Calculate the number of atoms of each element presentin 122.5 g of KClO3.

Sol. No. of moles of KClO3 = 5.122

5.122= 1.

(mol. wt. of KClO3 = 122.5)From the formula KClO3, we know that 1 mole of KClO3

contains 1 mole of K atoms, 1 mole of Cl atoms and 3moles of O atoms.

23

23

23

10022.63Oofatomsof.no

10022.61Clofatomsof.no

10022.61Kofatomsof.no

Example-53The vapour density (hydrogen = 1) of a mixtureconsisting of NO2 and N2O4 is 38.3 at 26.7ºC. Calculatethe number of moles of NO2 in 100 g of the mixture.

Sol. Wt. of NO2 = x g.

obs. mol. wt. (wt./mole) = molestotal

gin.wt

.3.382

92

x100

46

x

100

(0.437 mole)

Example-54Calculate the following for 49 gm of H

2SO

4

(a) moles (b) Molecules (c) Total H atoms(d) Total O atoms(e) Total electrons

Sol. Molecular wt of H2SO

4 = 98

(a) moles = wt in gm 49

molecular wt 98 =

1

2mole

(b) Since 1 mole = 6.023 × 1023 molecules.

1mole

2 = 6.023 × 1023 ×

1

2 molecules = 3.011 × 1023

molecules(c) 1 molecule of H

2SO

4 Contains 2 H atom, so

3.011 × 1023 of H2SO

4 contain 2 × 3.011 × 1023 atoms =

6.023 × 1023 atoms(d) 1 molecules of H

2SO

4 contains 4 O atoms

3.011 × 1023 molecular of H2SO

4 contains = 4 × 3.011 ×

1023 = 12.044 × 1023

(e) 1 molecule of H2SO

4 contains 2H atoms + 1 S atom +

4 O atom this means 1 molecule of H2SO

4 Contains

(2 + 16 + 4 × 8) e–

So 3.011 × 1023 molecules have 3.011 × 1023 × 50electrons = 1.5055 × 1025 e–

Example-55Calculate the total ions & charge present in 4.2 gm ofN–3

Sol. mole = wt in gm

Ionic wt = 4 2

14

. = 0.3

total no of ions = 0.3 × NA ions

total charge = 0.3 NA × 3 × 1.6 × 10–19

= 0.3 × 6.023 × 1023 × 3 × 1.6 × 10–19 , = 8.67 × 104 C Ans.

Example-56Find the total number of iron atom present in 224 amuiron.

Sol. Since 56 amu = 1 atom

therefore 224 amu = 1

56 × 224 = 4 atom Ans.

Calculation of mole no. of entity, law of chemical combination,Molar volume of ideal gases at STP, Average molar massQ.1 A sample of pure carbon dioxide, irrespective of its

source contains 27.27% carbon and 72.73% oxygen.The data support(1) Law of constant composition(2) Law of conservation of mass(3) Law of reciprocal proportions(4) Law of multiple proportions

Q.2 The law of definite proportions is not applicable tonitrogen oxide because(1) Nitrogen atomic weight is not constant(2) Nitrogen molecular weight is variable(3) Nitrogen equivalent weight is variable(4) Oxygen atomic weight is variable

Q.3 1 amu is equal to

(1) 12of 12

1C (2) 16-Oof

14

1

(3) 2of 1 Hg (4) 231066.1 kg

Q.4 In chemical scale, the relative mass of the isotopic

mixture of oxygen atoms ),,( 181716 OOO is assumed to

be equal to(1) 16.002 (2) 16.00(3) 17.00 (4) 11.00

Q.5 1 mol of 4CH contains

(1) 231002.6 atoms of H

(2) 4 g atom of Hydrogen

(3) 231081.1 molecules of 4CH

(4) 3.0 g of carbon

Q.6 7.5 grams of a gas occupy 5.8 litres of volume at STPthe gas is(1) NO (2) N

2O (3) CO (4) CO

2

Q.7 The number of atoms in 4.25 g of 3NH is approximately

(1) 1 × 1023 (2) 2 × 1023

(3) 4 × 1023 (4) 6 × 1023

Q.8 One litre of a gas at STP weight 1.16 g it can possible be

(1) C2H

2(2) CO (3) O

2(4) CH

4

Q.9 The mass of a molecule of water is

(1) 26103 kg (2) 25103 kg

(3) 26105.1 kg (4) 26105.2 kg

Q.10 If NA

is Avogadro’s number then number of valence

electrons in 4.2 g of nitride ions )( 3N

(1) 2.4 AN (2) 4.2 AN

(3) AN6.1 (4) AN2.3

Q.11 The number of molecule at NTP in 1 ml of an ideal gaswill be

(1) 23106 (2) 191069.2

(3) 231069.2 (4) None of these

Q.12 Volume of a gas at STP is 71012.1 cc. Calculate thenumber of molecules in it

(1) 201001.3 (2) 121001.3

(3) 231001.3 (4) 241001.3

Q.13 g4.4 of an unknown gas occupies L24.2 of volumeat standard temperature and pressure. The gas may be(1) Carbon dioxide (2) Carbon monoxide(3) Oxygen (4) Sulphur dioxide

Q.14 The number of oxygen atoms in 4.4 g of 2CO is approx.

(1) 23102.1 (2) 22106

(3) 23106 (4) 231012

Q.15 The total number of protons in 10 g of calcium

carbonate is ( 230 10023.6 N )

(1) 24105057.1 (2) 24100478.2

(3) 24100115.3 (4) 24100956.4

Q.16 Number of molecules in 100 ml of each of

232 and , CONHO at STP are

(1) In the order 322 NHOCO

(2) In the order 223 COONH (3) The same

(4) 223 OCONH

EXERCISE-I

Q.17 The number of water molecules in 1 litre of water is

(1) 18 (2) 100018

(3) AN (4) AN55.55

Q.18 2g of oxygen contains number of atoms equal to that in

(1) g5.0 of hydrogen (2) g4 of sulphur

(3) g7 of nitrogen (4) g3.2 of sodium

Empirical Formula, % Composition of a given compound by mass,

% By mole, Minimum molecular mass determination.

Q.19 Caffeine has a molecular weight of 194. If it contains

28.9% by mass of nitrogen, number of atoms of nitrogen

in one molecule of caffeine is

(1) 4 (2) 6 (3) 2 (4) 3

Q.20 The percentage of oxygen in NaOH is

(1) 40 (2) 60 (3) 8 (4) 10

Q.21 A compound (60 g) on analysis gave C = 24 g, H = 4 g,

O = 32 g. Its empirical formula is

(1) 222 OHC (2) OHC 22

(3) 22OCH (4) OCH 2

Q.22 What is the % of OH 2 in OHCNSFe 23 3.)(

(1) 45 (2) 30 (3) 19 (4) 25

Q.23 A hydrocarbon contains 86% carbon, 488ml of the

hydrocarbon weight 1.68 g at STP. Then the

hydrocarbon is an

(1) Alkane (2) Alkene

(3) Alkyne (4) Arene

Q.24 The simplest formula of a compound containing 50%

of element X (atomic mass 10) and 50% of element Y

(atomic mass 20) is

(1) XY (2) YX 2 (3) 3XY (4) 32 YX

Stoichiometry, Equation based calculations

Q.25 How much of NaOH is required to neutralise 1500

3cm of 0.1 HClN (Na = 23)

(1) 40 g (2) 4 g (3) 6 g (4) 60 g

Q.26 In the reaction,

)(6)(4)(5)(4 223 gOHgNOgOgNH , When 1

mole of ammonia and 1 mole of 2O are made to react to

completion

(1) 1.0 mole of OH 2 is produced

(2) 1.0 mole of NO will be produced

(3) All the oxygen will be consumed

(4) All the ammonia will be consumed

Q.27 Haemoglobin contains 0.33% of iron by weight. The

molecular weight of haemoglobin is approximately

67200. The number of iron atoms (At. wt. of Fe = 56)

present in one molecule of haemoglobin is

(1) 6 (2) 1 (3) 4 (4) 2

Q.28 2H evolved at STP on complete reaction of 27 g of

Aluminium with excess of aqueous NaOH would be

(1) 22.4 (2) 44.8

(3) 67.2 (4) 33.6 litres

Q.29 What is the concentration of nitrate ions if equal

volumes of 0.1 M 3AgNO and 0.1 M NaCl are mixed

together

(1) 0.1 N (2) 0.2 M

(3) 0.05 M (4) 0.25 M

Q.30 g12 of Mg (at. mass 24) will react completely with acid

to give

(1) One mole of 2H

(2) 1/2 mole of 2H

(3) 2/3 mole of 2O

(4) Both 1/2 mol of 2H and 1/2 mol of 2O

Q.31 100 g CaCO3 reacts with 1litre 1 N HCl. On completion

of reaction how much weight of 2CO will be obtain

(1) 5.5 g (2) 11 g (3) 22 g (4) 33 g

Concentration term

Q.32 What is the normality of a 1 M solution of 43POH

(1) 0.5 N (2) 1.0 N (3) 2.0 N (4) 3.0 N

Q.10 Mole fraction of ethyl alcohol in aqueous ethyl alcohol(C

2H

5OH) solution is 0.25. Hence percentage of ethyl

alcohol by weight is :(1) 54% (2) 25% (3) 75% (4) 46%

Q.11 74 gm of sample on complete combustion gives 132 gmCO

2 and 54 gm of H

2O. The molecular formula of the

compound may be(1) C

5H

12(2) C

4H

10O

(3) C3H

6O

2(4) C

3H

7O

2

Q.12 Weight of oxygen in Fe2O

3 and FeO is in the simple

ratio for the same amount of iron is :(1) 3 : 2 (2) 1 : 2(3) 2 : 1 (4) 3 : 1

Q.13 A person needs on average of 2.0 mg of riboflavin(vitamin B

2) per day. How many gm of butter should be

taken by the person per day if it is the only source ofriboflavin? Butter contains 5.5 microgram riboflavin pergm.(1) 363.6 gm (2) 2.75 mg(3) 11 gm (4) 19.8 gm

Q.14 The oxide of a metal contains 30% oxygen by weight. Ifthe atomic ratio of metal and oxygen is 2 : 3, determinethe atomic weight of metal.(1) 12 (2) 56 (3) 27 (4) 52

Q.15 When a mixture of 10 mole of SO2 , 15 mole of O

2 was

passed over catalyst , 8 mole of SO3 was formed. How

many mole of SO2and O

2 did not enter into combination?

(1) 2 moles of SO2, 11 moles of O

2

(2) 3 moles of SO2, 11.5 moles of O

2


2


2

Q.16 C6H

5OH(g) + O

2(g) CO

2(g) + H

2O(l)

Magnitude of volume change if 30 ml of C6H

5OH (g) is

burnt with excess amount of oxygen, is(1) 30 ml (2) 60 ml (3) 20 ml (4) 10 ml

Q.17 Mass of sucrose C12

H22

O11

produced by mixing 84 gmof carbon, 12 gm of hydrogen and 56 lit. O

2 at 1 atm &

273 K according to given reaction, is C(s) + H2(g) +

O2(g) C

12H

22O

11(s)

(1) 138.5 (2) 155.5(3) 172.5 (4) 199.5

Q.1 A sample of Calcium phosphate Ca3(PO

4)

2 contains 8

mol of O atoms. The number of mole of Ca atoms in thesample is :(1) 4 (2) 1.5 (3) 3 (4) 8

Q.2 Ratio of masses of H2SO

4 and Al

2 (SO

4)

3 each

containing 32 grams of S is __________.(1) 0.86 (2) 1.72(3) 0.43 (4) 2.15

Q.3 Which has maximum number of atoms of oxygen(1) 10 ml H

2O(l)

(2) 0.1 mole of V2O

5

(3) 12 gm O3(g)

(4) 12.044 × 1022 molecules of CO2

Q.4 Mass of one atom of the element A is 3.9854 × 10–23.How many atoms are contained in 1g of the element A?(1) 2.509 × 1022 (2) 6.022 × 1023

(3) 12.044 × 1023 (4) None

Q.5 The number of atoms present in 0.5 g-atoms of nitrogenis same as the atoms in(1) 12 g of C(2) 32 g of S(3) 8 g of oxygen (4) 24g of Mg

Q.6 How many moles of magnesium phosphateMg3(PO4)2 will 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.7 64 g of an organic compound has 24 g carbon and 8 ghydrogen and the rest is oxygen. The empirical formulaof the compound is :(1) CH4O (2) CH2O (3) C2H4O (4) None

Q.8 Two elements X (atomic mass=75) and Y (atomicmass=16) combine to give a compound having 75.8%of X. The formula of the compound is:(1) X

2Y

3(2) X

2Y (3) X

2Y

2(4) XY

Q.9 A definite amount of gaseous hydrocarbon was burntwith just sufficient amount of O

2. The volume of all

reactants was 600 ml, after the explosion the volume ofthe products [CO

2(g) and H

2O(g)] was found to be 700

ml under the similar conditions. The molecular formulaof the compound is :(1) C

3H

8(2) C

3H

6(3) C

3H

4(4) C

4H

10

EXERCISE-II

Q.18 What volume (in ml) of 0.2 M H2SO4 solution should be

mixed with the 40 ml of 0.1 M NaOH solution such that

the resulting solution has the concentration of H2SO4

as 55

6M.

(1) 70 (2) 45 (3) 30 (4) 58

Q.19 For the reaction

2x + 3y + 4z 5w

Initially if 1 mol of x, 3 mol of y and 4 mol of z is taken.

If 1.25 mol of w is obtained then % yield of this reaction

is

(1) 50% (2) 60% (3) 70% (4) 40%

Q.20 If 10 g of Ag reacts with 1 g of sulphur , the amount of

Ag2S formed will be :

(1) 7.75 g (2) 0.775 g

(3) 11 g (4) 10 g

Q.21 A solution of A (MM = 20) and B (MM = 10), [Mole

fraction XB = 0.6] having density 0.7 gm/ml then molarity

and molality of B in this solution will be

________________ and ______________

respectively.

(1) 30M,75m (2) 40M,75m

(3) 30M,65m (4) 50M,55m

Q.22 125 ml of 8% w/w NaOH solution sp. gravity = 1 is

added to 125 ml of 10 % w/v HCl solution. The nature

of resultant solution would be _________

(1) Acidic (2) Basic

(3) Neutral (4) None

Q.23 36.5 % (w/w) HCl has density equal to 1.20 g mL–1. The

molarity (M) and molality (m), respectively, are

(1) 15.7, 15.7 (2) 12, 12

(3) 15.7, 12 (4) 12, 15.7

Q.24 An aqueous solution of ethanol has density 1.025 g/

mL and it is 2M. What is the molality of this solution ?

(1) 1.79 (2) 2.143

(3) 1.951 (4) None of these.

Q.25 500 mL of a glucose solution contains 6.02 × 1022

molecules. The concentration of the solution is(1) 0.1 M (2) 1.0 M (3) 0.2 M (4) 2.0 M

Q.26 Equal moles of H2O and NaCl are present in a solution.

Hence, molality of NaCl solution is :(1) 0.55 (2) 55.5 (3) 1.00 (4) 0.18

Q.27 Decreasing order of mass of pure NaOH in each of the

aqueous solution.

(I) 50 g of 40% (W/W) NaOH

(II) 50 ml of 50% (W/V) NaOH (dsol

= 1.2 g/ml).

(III) 50 g of 15 M NaOH (dsol

= 1 g/ml).

(1) I, II, III (2) III, II, I

(3) II, III, I (4) III = II = I.

Q.28 Mole fraction of A in H2O is 0.2. The molality of A in

H2O is :

(1) 13.9 (2) 15.5 (3) 14.5 (4) 16.8

Q.29 What is the molarity of H2SO4 solution that has a

density of 1.84 g/cc and contains 98% by mass of H2SO4?

(Given atomic mass of S = 32)

(1) 4.18 M (2) 8.14 M

(3) 18.4 M (4) 18 M

Q.30 The molarity of the solution containing 2.8% (W/V)

solution of KOH is : (Given atomic mass of K = 39 ) is:

(1) 0.1 M (2) 0.5 M

(3) 0.2 M (4) 1 M

Q.31 A solution of FeCl3 is

M

30 its molarity for Cl– ion will be

:

(1) M

90(2)

M

30(3)

M

10(4)

M

5

Q.32 If 500 ml of 1 M solution of glucose is mixed with 500 ml

of 1 M solution of glucose final molarity of solution

will be:

(1) 1 M (2) 0.5 M

(3) 2 M (4) 1.5 M

Q.33 What volume of a 0.8 M solution contains 100 milli

moles of the solute?

(1) 100 mL (2) 125 mL

(3) 500 mL (4) 62.5 mL

Q.34 The molarity of Cl¯ in an aqueous solution which was

(w/V) 2% NaCl, 4% CaCl2 and 6% NH

4Cl will be

(1) 0.342 (2) 0.721

(3) 1.12 (4) 2.18

Q.35 2M of 100 mL Na2 SO

4 is mixed with 3M of 100 mL NaCl

solution and 1M of 200 mL CaCl2 solution. Then the

ratio of the concentration of cation and anion.

(1) 1/2 (2) 2

(3) 1.5 (4) 1

Which of the following is/are correct.(A) The HCl is in excess.(B) 117.0 g of NaCl is formed.(C) The volume of CO

2 produced at NTP is 22.4 L.

(D) None of these

Q.6 A sample of a mixture of CaCl2 and NaCl weighing 4.44

g was treated to precipitate all the Ca as CaCO3, which

was then heated and quantitatively converted to 1.12gof CaO. (At . wt. Ca = 40, Na = 23, Cl = 35.5)(A) Mixture contains 50% NaCl(B) Mixture contains 60% CaCl

2

(C) Mass of CaCl2 is 2.22 g

(D) Mass of CaCl2 1.11 g

Q.7 A + B A3B

2 (unbalanced)

A3B

2 + C A

3B

2C

2 (unbalanced)

Above two reactions are carried out by taking 3 moleseach of A and B and one mole of C. Then which optionis/are correct?(A) 1 mole of A

3B

2C

2 is formed

(B) 21 mole of AA3B

2C

2 is formed

(C) 1/2 mole of A3B

2 is formed

(D) 21 mole of AA3B

2 is left finally

Q.8 21.2 g sample of impure Na2CO

3 is dissolved and reacted

with a solution of CaCl2, the weight of precipitate of

CaCO3 is 10.0 g. Which of the following statements is/

are correct ?(A) The % purity of Na

2CO

3 is 50%

(B) The percentage purity of Na2CO

3 is 60%

(C) The number of moles of Na2CO

3 = CaCO

3 = 0.1 mol.

(D) The number of moles of NaCl formed is 0.1 mol.

Q.9 A mixture of C3H

8 and O

2 having total volume 100 ml in

an Eudiometry tube is sparked & it is observed that acontraction of 45 mL is observed what can be thecomposition of reacting mixture.(A) 15 mL C

3H

8 & 85 mL O

2

(B) 25 mL C3H

8 & 75 mL O

2

(C) 45 mL C3H

8 & 55 mL O

2

(D) 55 mL C3H

8 & 45 mL O

2

MCQ/COMPREHENSION/MATCHING/NUMERICALQ.1 Which is/are correct statements about 1.7 g of NH

3 :

(A) It contain 0.3 mol H – atom(B) it contain 2.408 1023 atoms(C) Mass % of hydrogen is 17.65%(D) It contains 0.3 mol N-atom

Q.2 If 27 g of Carbon is mixed with 88 g of Oxygen and isallowed to burn to produce CO

2 , then :

(A) Oxygen is the limiting reagent.(B) Volume of CO

2 gas produced at NTP is 50.4 L.

(C) C and O combine in mass ratio 3 : 8.(D) Volume of unreacted O

2 at STP is 11.2 L.

Q.3 + AgNO

3 (Excess) Silver salt

Ag (metal)

If 0.5 mole of silver salt is taken and weight of residueobtained is 216 g. (Ag = 108 g/mol).Then which the following is correct :(A) n = 4(B) n = 2(C) M.wt. of silver salt is 718 g/mol(D) M.wt. of silver salt is 388 g/mol

Q.4 For the following reactionN2 + 3H2 2NH3

Identify the compositions which will produce sameamount of NH3.(A) 140 g N2 & 35 g H2

(B) 18 g H2 & 52 g N2

(C) Total 20 moles of mixture having N2 & H2 present instoichiometric ratio (No limiting reagent)

(D)136gm of mixture having mass fraction of H2 = 6

34

Q.5 For the following reaction : Na2CO

3 + 2HCl

2NaCl + CO2 + H

2O

106.0 g of Na2CO

3 reacts with 109.5 g of HCl.

EXERCISE-III

Q.10 The incorrect statement(s) regarding 2M MgCl2

aqueous solution is/are (dsolution = 1.09 gm/ml)

(A) Molality of Cl¯ is 4.44 m

(B) Mole fraction of MgCl2 is exactly 0.035

(C) The conc. of MgCl2 is 19% w/v

(D) The conc. of MgCl2 is 19 × 104 ppm

Q.11 Solutions containing 23 g HCOOH is/are :

(A) 46 g of 70%

v

w HCOOH (d

solution = 1.40 g/mL)

(B) 50 g of 10 M HCOOH (dsolution

= 1 g/mL)

(C) 50 g of 25%

w

w HCOOH

(D) 46 g of 5 M HCOOH (dsolution

= 1 g/mL)

Q.12 Which of the following solutions contains same molar

concentration ?

(A) 166 g KI per liter solution

(B) 33.0 g (NH4)

2 SO

4 in 200 mL solution

(C) 25.0 g CuSO4.5H

2O in 100mL solution

(D) 27.0 mg Al3+ per mL solution

Q.13 If 100 ml of 1M H2SO

4 solution is mixed with 100 ml of

9.8%(w/w) H2SO

4 solution (d = 1 g/mL) then :

(A) concentration of solution remains same

(B) volume of solution become 200 mL

(C) mass of H2SO

4 in the solution is 98 g

(D) mass of H2SO

4 in the solution is 19.6 g

Comprehension # 1 (Q. 14 to 16)

According to the Avogadro’s law, equal number of

moles of gases occupy the same volume at identical

condition of temperature and pressure. Even if we have

a mixture of non-reacting gases then Avogadro’s law is

still obeyed by assuming mixture as a new gas.

Now let us assume air to consist of 80% by volume of

Nitrogen (N2) and 20% by volume of oxygen (O

2). If air

is taken at STP then its 1 mol would occupy 22.4 L. 1

mol of air would contain 0.8 mol of N2 and 0.2 mol of O

2

hence the mole fractions of N2 and O

2 are given by

8.0X2N , 2.0X

2O

Q.14 Volume occupied by air at NTP containing exactly 11.2

g of Nitrogen :

(A) 22.4 L (B) 8.96 L

(C) 11.2 L (D) 2.24 L

Q.15 If air is treated as a solution of O2 and N

2 then % w/w of

oxygen is :

(A) 9

10(B)

9

200

(C) 9

700(D)

9

350

Q.16 Density of air at NTP is :

(A) 1 g/L (B) 7

9 g/L

(C) 7

2 g/L (D) can’t be determined

Comprehension # 2 (Q. 17 to 19)

A chemist decided to determine the molecular formula

of an unknown compound. He collects following

informations :

(I) Compounds contains 2 : 1 'H' to 'O' atoms(number of

atoms).

(II) Compounds has 40% C by mass

(III) Approximate molecular mass of the compound is

178 g

(IV) Compound contains C, H and O only.

Q.17 What is the % by mass of oxygen in the compound

(A) 53.33% (B) 88.88%

(C) 33.33% (D) None of these

Q.18 What is the empirical formula of the compound

(A) CH3O (B) CH

2O

(C) C2H

2O (D) CH

3O

2

Q.19 Which of the following could be molecular formula of

compound

(A) C6H

6O

6(B) C

6H

12O

6

(C) C6H

14O

12(D) C

6H

14O

6

Q.20 One type of artificial diamond (commonly called YAG for yttrium aluminium garnet) can be represented by the formula

Y3Al5O12.[Y = 89, Al =27]

Column I Column II

Element Weight percentage

(A) Y (P) 22.73%

(B) Al (Q) 32.32%

(C) O (R) 44.95%

Q.21 Column-I Column-II

(A) 100 mL of 0.2 M AlCl3 solution + 400 ml (p) Total concentration of cation(s) = 0.12 M

of 0.1 M HCl solution

(B) 50 mL of 0.4 M KCl + 50 ml H2O (q) [SO42–] = 0.06 M

(C) 30 mL of 0.2 M K2SO4 + 70 ml H2O (r) [SO42–] = 2.5 M

(D) 200 mL 24.5% (w/v) H2SO4 (s) [Cl¯] = 0.2 M

Q.22 Column - I Column - II

(A) A gaseous organic compound containing (p) One mole of compound contains 4NA atoms of Hydrogen.

C = 52.17%, H = 13.04% & O = 34.78%

(by weight) having molar mass 46 g/mol.

(B) 0.3 g of an organic compound containing (q) The empirical formula of the compound is same as

C, H and O on combustion y ields 0.44 g its molecule formula.

of CO2 and 0.18 g of H

2O, with two O

atoms per molecule.

(C) A hydrocarbon containing C = 42.857% (r) Combustion products of one mole of compound

and H = 57.143% (by mole) containing 3C contains larger number of moles of CO2 than that of

molecule. atoms per H2O.

(D) A hydrocarbon containing 10.5 g carbon (s) CO2 gas produced by the combustion of 0.25 mole

per gram of hydrogen having vapour of compound occupies a volume of 11.2 L at NTP.

density 46.

NUMERICAL BASED QUESTIONSQ.23 Find the oxidation state of osmium (Os) in OsO

4

Q.24 Find the Oxidation number of nitrogen in (NH4)

2SO

4

Q.25 Molarity of H2SO

4 is 18 M. Its density is 1.8 g/cm3,

calculate molality

Q.26 Find the percentage of oxygen in NaOH.

Q.27 Find the oxidation number of cobalt in K3[Co(NO

2)

6]

Q.28 Calculate the mass of NaOH is required to neutralise

1500 cm3 of 0.1 N HCl (Na = 23)

Q.29 One mole of potassium chlorate (KClO3) is thermally

decomposed and excess of aluminium is burnt in thegaseous product. How many mol of aluminium oxide(Al

2O

3) are formed ?

Q.30 100 g CaCO3 reacts with 1litre 1 N HCl. On completion

of reaction how much weight of CO2 will be obtained ?

Q.31 2M of 100 ml Na2 SO

4 is mixed with 3M of 100 ml NaCl

solution and 1M of 200 ml CaCl2 solution. Then the

ratio of the concentration of cation and anion.

Q.32 What is the quantity of water that should be added to16 g. methanol to make the mole fraction of methanol as0.25.

Q.8 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mlof sodium hydroxide solution. The amount of NaOH in50 mL of the given sodium hydroxide solution is

[JEE Main - 2019 (January)](1) 40 g (2) 10 g (3) 20 g (4) 80 g

Q.9 A solution of sodium sulphate contains 92 g of Na+

ions per kilogram of water. The Molality of Na+ ions inthat solution in mol kg–1 is

[JEE Main - 2019 (January)](1) 12 (2) 4 (3) 8 (4) 16

Q.10 The hardness of water sample (in terms of equivalentsof CaCO

3) containing 10–3 M CaSO

4 is

(molar mass of CaSO4 = 136 g mol–1)

[JEE Main - 2019 (January)](1) 10 ppm (2) 50 ppm(3) 90 ppm (4) 100 ppm

Q.11 Water sample with BOD vales of 4 ppm and 18 ppmrespectively are [JEE Main - 2019 (January)](1) Clean and clean(2)Highly polluted and clean(3) Clean and highly polluted(4) Highly polluted and highly polluted.

Q.12 The amount of sugar (C12

H22

O11

) required to prepare 2L of its 0.1 M aqueous solution is:

[JEE Main - 2019 (January)](1) 136.8 g (2) 17.1 g(3) 68.4g (4) 34.2 g

Q.13 8 g of NaOH is dissolved in 18 g of H2O. Mole fraction

of NaOH in solution and molality (in mol kg–1) of thesolution respectively are :

[JEE Main - 2019 (January)](1) 0.2, 22.20 (2) 0.2, 11.11(3) 0.167, 11.11 (2) 0.167, 22.20

Q.14 The volume strength of 1 M H2O

2 is : (Molar mass of

H2O

2 = 34 g mol–1) [JEE Main - 2019 (January)]

(1) 5.6 (2) 16.8(3) 11.35 (4) 22.4

Q.15 An open vessel at 27ºC is heated until two fifth of theair (assumed as an ideal gas) in it has escaped from thevessel. Assuming that the volume of the vessel remainsconstant, the temperature at which the vessel has beenheated is : [JEE Main - 2019 (January)](1) 500º C (2) 500 K(3) 750º C (4) 750 K

JEE-MAINPREVIOUS YEAR'S

Q.1 1 gram of a carbonate (M2CO

3) on treatment with excess

HCl produces 0.01186 mole of CO2, the molar mass of

M2CO

3 in g mol–1 is [JEE Main-2017]

(1) 1186 (2) 84.3 (3) 1186 (4) 11.86

Q.2 The most abundant elements by mass in the body of ahealthy human adult are :Oxygen (61.4%) : Carbon (22.9%), Hydrogen (10.0%),and Nitrogen (2.6%). The weight which a 75kg personwould gain if all 1H atoms are replaced by 2H atoms is

[JEE Main-2017](1) 15 kg (2) 37.5 kg(3) 7.5 kg (4) 10 kg

Q.3 The ratio of mass percent of C and H of an organiccompound (C

XH

YO

Z) is 6 : 1. If one molecule of the

above compound (CXH

YO

Z) contains half as much

oxygen as required to burn one molecule of compoundC

XH

Y completely to CO

2 and H

2O. The empirical formula

of compound CXH

YO

Z is : [JEE Main-2018]

(1) 2 4C H O (2) 3 4 2C H O

(3) 2 4 3C H O (4) 3 6 3C H OQ.4 The concentration of dissolved oxygen (DO) is cold

water can go upto: [JEE Main - 2019 (January)](1) 14 ppm (2) 8 ppm(3) 10 ppm (4) 16 ppm

Q.5 A 10 mg effervescent tablet containing sodiumbicarbonate and oxalic acid releases 0.25 ml of CO

2 at T

= 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0

L under such condition, what is the percentage ofsodium bicarbonate in each tablet? [Molar mass ofNaHCO

3 = 84 g mol–1]

[JEE Main - 2019 (January)](1) 0.84 (2) 33.6 (3) 16.8 (4) 8.4

Q.6 Water filled in two glasses A and B have BOD values of10 and 20 respectively. The correct statement regardingthem is [JEE Main - 2019 (January)](1) B is more polluted than A(2) A is suitable for drinking, whereas B is not(3) Both A and B ar suitable for drinking(4) A is more polluted than B

Q.7 For the following reaction the mass of water producedfrom 445 g of C

57H

110O

6 is 2C

57H

110O

6 (s) + 163 O

2(g) 114 CO

2 (g) + 110 H

2O (l)

[JEE Main - 2019 (January)](1) 490 g (2) 445 g (3) 495 g (4) 890 g

EXERCISE-IV

Q.16 100 mL of a water sample contains 0.81 g of calciumbicarbonate and 0.73 g of magnesium bicarbonate. Thehardness of this water sample expressed in terms ofequivalents of CaCO3 is : (molar mass of calciumbicarbonate is 162 g mol–1 and magnesium bicarbonateis 146 gmol–1) [JEE Main 2019](1) 1,000 ppm (2) 10,000 ppm(3) 100 ppm (4) 5,000 ppm

Q.17 For a reaction,N

2(g) + 3H

2(g) 2NH

3(g) ;

identify dihydrogen (H2) as a limiting reagent in the

following reaction mixtures. [JEE Main 2019](1) 14g of N

2 + 4g of H

2

(2) 28g of N2 + 6g of H

2

(3) 56g of N2 + 10g of H

2

(4) 35g of N2 + 8g of H

2

Q.18 The percentage composition of carbon by mole inmethane is : [JEE Main 2019](1) 80% (2) 25%(3) 75% (4) 20%

Q.19 The maximum prescribed concentration of copper indrinking water is : [JEE Main 2019](1) 5 ppm (2) 0.5 ppm(3) 0.05 ppm (4) 3 ppm

Q.20 0.27 g of a long chain fatty acid was dissolved in 100cm3

of hexane. 10 mL of this solution was added dropwiseto the surface of water in a round watch glass. Hexaneevaporates and a monolayer is formed. The distancefrom edge to centre of the watch glass is 10 cm. What isthe height of the monolayer ? [JEE Main 2019][Density of fatty acid = 0.9 g cm–3 , = 3](1) 10–8 m (2) 10–6 m(3) 10–4 m (4) 10–2 m

Q.21 The strength of 11.2 volume solution of H2O

2 is : [Given

that molar mass of H = 1 g mol–1 and O = 16 g mol–1][JEE Main 2019]

(1) 13.6% (2) 3.4%(3) 34% (4) 1.7%

Q.22 25 g of an unknown hydrocarbon upon burningproduces 88 g of CO

2 and 9 of H

2O. This unknown

hydrocarbon contains. [JEE Main 2019](1) 20g of carbon and 5 g of hydrogen(2) 24g of carbon and 1 g of hydrogen(3) 18g of carbon and 7 g of hydrogen(4) 22g og carbon and 3 g of hydrogen

Q.23 10 mL of 1mM surfactant solution forms a monolayer

covering 0.24 cm2 on a polar substrate. if the polar head

is approximated as cube,what is its edge length?

[JEE Main 2019]

(1) 2.0 pm (2) 2.0 nm

(3) 1.0 pm (4) 0.1 nm

Q.24 What would be the molality of 20% (mass/mass)

aqueous solution of KI? [JEE Main 2019]

(molar mass of KI = 166 g mol–1)

(1) 1.08 (2) 1.48

(3) 1.51 (4) 1.35

Q.25 5 moles of AB2 weight 125 × 10–3 kg and 10 moles of

A2B

2 weigh 300 × 10–3 kg. The molar mass of AMA and

molar mass of BMB in kg mol–1 are :

[JEE Main 2019]

(1) MA = 50 × 10–3 and M

B = 25 × 10–3

(2) MA = 25 × 10–3 and M

B = 50 × 10–3

(3) MA = 5 × 10–3 and M

B = 10 × 10–3

(4) MA = 10 × 10–3 and M

B = 5 × 10–3

Q.26 The mole fraction of a solvent in aqueous solution of a

solute is 0.8. The molality (in mol kg–1) of the aqueous

solution is [JEE Main 2019]

(1) 13.88 × 10–1 (2) 13.88 × 10–2

(3) 13.88 (4) 13.88 × 10–3

Q.27 At 300 K and 1 atmospheric pressure, 10 mL of a

hydrocarbon required 55 mL of O2 for complete

combustion and 40 mL of CO2 is formed. The formula

of the hydrocarbon is : [JEE Main 2019]

(1) C4H8 (2) C4H7Cl (3) C4H10 (4) C4H6

Q.28 Consider the following statements [JEE Main 2019]

(i) The pH of a mixture containing 400 mL of 0.1 M

H2SO4 and 400 mL of 0.1 M NaOH will be approximately

1.3.

(ii) Ionic product of water is temperature dependent.

(iii) A monobasic acid with Ka = 10–5 has a pH = 5. The

degree of dissociation of this acid is 50%.

(iv) The Le Chatelier's principle is not applicable to

common-ion effect.

the correct statement are :

(1) (i), (ii) and (iv) (2) (i), (ii) and (iii)

(3) (i) and (ii) (4) (ii) and (iii)

Q.29 The minimum amount of O2(g) consumed per gram of

reactant is for the reaction : [JEE Main 2019]

(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31,

C = 12, H = 1)

(1) C3H

8(g) + 5 O

2(g ) 3 CO

2(g) + 4 H

2O(l)

(2) P4(s) + 5 O

2(g) P

4O

10(s)

(3) 4 Fe(s) + 3 O2(g) 2Fe

2O

3(s)

(4) 2 Mg(s) + O2(g) 2 MgO(s)

Q.30 Amongst the following statements, that which was not

proposed by Dalton was :[JEE Main-2020 (January)]

(1) When gases combine or reproduced in a chemical

reaction they do so in a simple ratio by volume

provided all gases are at the same T & P.

(2) Matter consists of indivisible atoms

(3) Chemical reactions involve reorganization of atoms.

These are neither created nor destroyed in a

chemical reaction.

(4) All the atoms of a given element have identical

properties including identical mass. Atoms of

different elements differ in mass.

Q.31 The volume (in mL) of 0.125 M AgNO3 required to

quantitatively precipitate chloride ions in 0.3 g of

[Co(NH3)6]Cl3 is . [JEE Main-2020 (January)]M[Co(NH3)6]Cl3=267.46 g/molMAgNO3 = 169.87 g/mol

Q.32 NaClO3 is used, even in spacecrafts, to produces O2.

The daily consumption of pure O2 by a person is 492L

at 1 atm, 300 K. How much amount of NaClO3, in grams,

is required to produce O2 for the daily consumption of

a person at 1 atm, 300 K> .

[JEE Main-2020 (January)]

NaClO3(s) + Fe(s) O2(g) + NaCl(s) + FeO(s)

R = 0.082 L atm mol–1 K–1

Q.33 The hardness of a water sample containing 10–3 M

MgSO4 expressed as CaCO

3 equivalents (in ppm) is

___________. (molar mass of MgSO4 is 120.37 g/mol)

[JEE Main-2020 (January)]

Q.34 Complex A has a composition of H12

O6Cl

3Cr. If the

complex on treatment with conc. H2SO

4 loses 13.5% of

its original mass, the correct molecular formula of A is

[Given : Atomic mass of Cr = 52 amu and Cl = 35 amu]

(1) [Cr(H2O)

6]Cl

3

(2) [Cr(H2O)

4Cl

3]Cl.2H

2O

(3) [Cr(H2O)

3Cl

3].3H

2O

(4) [Cr(H2O)

4Cl]Cl

2.H

2O

Q.35 6.023 × 1022 molecules are present in 10 g of a substance‘x’. The molarity of a solution containing5 g of substance ‘x’ in 2 L solution is ______× 10–3

Q.36 The strengths of 5.6 volume hydrogen peroxide (ofdensity 1 g/mL) in terms of mass percentage andmolarity (M), respectively, are :(Take molar mass of hydrogen peroxide as 34 g/ mol)(1) 1.7 and 0.5 (2) 0.85 and 0.5(3) 1.7 and 0.25 (4) 0.85 and 0.25

Q.37 The mole fraction of glucose (C6H

12O

6) in an aqueous

binary solution is 0.1. The mass percentage of water init, to the nearest integer, is ________

Q.38 The volume strength of 8.9 M H2O

2 solution calculated

at 273 K and 1 atm is _____.(R = 0.0821 L atm K–1 mol–1)(Rounded off to the nearest integer)

Q.39 The mass of ammonia in grams produced when 2.8 kgof dinitrogen quantitatively reacts with 1 kg ofdihydrogen is _______.

[JEE Main-2020 (September)]

Q.40 The minimum number of moles of O2 required for

complete combustion of 1 mole of propane and 2 molesof butane is _______.[JEE Main-2020 (September)]

Q.41 The average molar mass of chlorine is 35.5 g mol–1. Theratio of 35Cl to 37Cl in naturally occuring chlorine isclose to [JEE Main-2020 (September)](1) 1 : 1 (2) 2 : 1(3) 3 : 1 (4) 4 : 1

Q.42 A solution of two components containing n1 moles of

the 1st component and n2 moles of the 2nd component is

prepared. M1 and M

2 are the molecular weights of

component 1 and 2 respectively. If d is the density ofthe solution in g mL–1, C

2 is the molarity and x

2 is the

mole fraction of the 2nd component, then C2 can be

expressed as [JEE Main-2020 (September)]

(1) 22

1 2 2 1

1000xC

M x (M M )

(2) 22

1 2 2 1

1000d xC

M x (M M )

(3) 22

2 2 2 1

d xC

M x (M M )

(4) 22

2 2 2 1

d xC

M x (M M )

JEE-ADVANCEDPREVIOUS YEAR'SQ.1 Reaction of Br

2 with Na

2CO

3 in aqueous solution gives

sodium bromide and sodium bromate with evolution ofCO

2 gas. The number of sodium bromide molecules

involved in the balanced chemical equation is[JEE-2012]

Q.2 Dissolving 120 g of urea (mol. wt. 60) in 1000 g of watergave a solution of density 1.15 g/mL. The molarity ofthe solution is : [JEE-2012](A) 1.78 M (B) 2.00 M(C) 2.05 M (D) 2.22 M

Q.3 29.2% (w/w) HCl stock solution has a density of 1.25 gmL–1. The molecular weight of HCl is 36.5 g mol–1. Thevolume (mL) of stock solution required to prepare a 200mL solution of 0.4 M HCl is : [JEE-2012]

Q.4 A compound H2X with molar weight of 80 g isdissolved in a solvent having density of 0.4 g ml–1.Assuming no change in volume upon dissolution,the molality of a 3.2 molar solution is

[JEE Advanced-2014]

Q.5 The mole fraction of a solute in a solution is 0.1. At 298

K, molarity of this solution is the same as its molality.

Density of this solution at 298 K is 2.0 gcm–3. The ratio

of molecular weights of the solute and solvent,

solute

solvent

MW

MW

, is [JEE Advanced-2016]

Q.6 In neutral or faintly alkaline solution, 8 moles of

permanganate anion quantitatively oxidize

thiosulphate anions to produce X moles of a sulphur

containing product. The magnitude of X is

[JEE Advanced-2016]

Q.7 The ammonia prepared by treating ammonium sulphate

with calcium hydroxide is completely used by

NiCl2.6H

2O to form a stable coordination compound.

Assume that both the reactions are 100% complete. If

1584 g of ammonium sulphate and 952g of NiCl2.6H

2O

are used in the preparation, the combined weight (in

grams) of gypsum and the nickel-ammonia coordination

compound thus produced is___.

(Atomic weights in g mol–1 : H = 1, N = 14, O = 16, S = 32,

Cl = 35.5, Ca = 40, Ni = 59)

[JEE Advanced-2018]

Q.8 The mole fraction of urea in an aqueous urea solution

containing 900 g of water is 0.05. If the density of the

solution is 1.2 g cm–3, the molarity of urea solution is

___ [JEE Advanced - 2019]

(Given data : Molar masses of urea and water are 60 g

mol–1 and 18 g mol, respectively)

Q.9 Aluminium reacts with sulfuric acid to form aluminium

sulfate and hydrogen. What is the volume of hydrogen

gas in liters (L) produced at 300 K and 1.0 atm pressure,

when 5.4 g of aluminium and 50.0 mL of 5.0 M sulfuric

acid are combined for the reaction ?

(Use molar mass of aluminium as 27.0 g mol–1, R = 0.082

atm L mol–1 K–1) [JEE(Advanced) - 2020]

ANSWER KEY

EXERCISE-IQ.1 (1) Q.2 (3) Q.3 (1) Q.4 (2) Q.5 (2) Q.6 (1) Q.7 (4) Q.8 (1) Q.9 (1) Q.10 (1)Q.11 (2) Q.12 (2) Q.13 (1) Q.14 (1) Q.15 (3) Q.16 (3) Q.17 (4) Q.18 (2) Q.19 (1) Q.20 (1)Q.21 (4) Q.22 (3) Q.23 (2) Q.24 (2) Q.25 (3) Q.26 (3) Q.27 (3) Q.28 (4) Q.29 (3) Q.30 (2)Q.31 (3) Q.32 (4)

EXERCISE-IIQ.1 (3) Q.2 (1) Q.3 (3) Q.4 (1) Q.5 (3) Q.6 (2) Q.7(1) Q.8 (1) Q.9 (1) Q.10 (4)Q.11 (3) Q.12 (1) Q.13 (1) Q.14 (2) Q.15 (1) Q.16 (2) Q.17 (2) Q.18 (1) Q.19 (1) Q.20 (1)Q.21 (1) Q.22 (1) Q.23 (4) Q.24 (2) Q.25 (3) Q.26 (2) Q.27 (2) Q.28 (1) Q.29 (3) Q.30 (2)Q.31 (3) Q.32 (1) Q.33 (2) Q.34 (4) Q.35 (4)

EXERCISE-IIIMCQ/COMPREHENSION/MATCHING/NUMERICALQ.1 (A,B,C) Q.2 (B,C,D) Q.3 (A,C) Q.4 (A,C) Q.5 (A,B,C)

Q.6 (A,C) Q.7 (B,D) Q.8 (A,C) Q.9 (A, B) Q.10 (B,D)

Q.11 (A,B) Q.12 (A,C,D) Q.13 (A,B,D) Q.14 (C) Q.15 (B)

Q.16 (B) Q.17 (A) Q.18 (B) Q.19 (B)

Q.20 (A) R, (B) P, (C) Q

Q.21 (A - p,s); (B - s); (C - p,q); (D - r)

Q.22 (A - q,s); (B - p, s); (C - p, q, r); (D - q, r)

Q.23 [+ 8] Q.24 [– 3] Q.25 [500] Q.26 [40] Q.27 [+ 3]

Q.28 [6g] Q.29 [1] Q.30 [22g] Q.31 [1] Q.32 [27g]

EXERCISE-IVJEE-MAINPREVIOUS YEAR'SQ.1 (2) Q.2 (3) Q.3 (3) Q.4 (3) Q.5 (1) Q.6 (1) Q.7 (3) Q.8 (Bonus) Q.9 (3) Q.10 (4)Q.11 (3) Q.12 (3) Q.13 (3) Q.14 (3) Q.15 (3) Q.16 (2) Q.17 (3) Q.18 (4) Q.19 (4) Q.20 (2)Q.21 (2) Q.22 (2) Q.23 (1) Q.24 (3) Q.25 3) Q.26 (3) Q.27 (4) Q.28 (2) Q.29 (3) Q.30 (1)Q.31 [26.80 to 27.00] Q.32[2120.00 to 2140.00] Q.33 [100.00 to 100.00] Q.34 (2) Q.35 [25.00] Q.36 (1)Q.37 [47] Q.38 [100] Q.39 [3400] Q.40 [18] Q.41 (3) Q.42 (2)

JEE-ADVANCEDPREVIOUS YEAR'SQ.1 [5] Q.2 (C) Q.3 [8 mL.] Q.4 [8] Q.5 [9] Q.6 [6] Q.7 [29.92] Q.8 [2.98 or 2.99] Q.9 [6.15]

Q.1 (1)

Q.2 (3)

Q.3 (1)

Q.4 (2)

Q.5 (2) 1 Mole of 4CH contains 4 mole of hydrogen atom i.e.

4g atom of hydrogen.

Q.6 (1) 5.8L of gas has mass gm5.7

22.4L “ “ “ = 96.284.228.5

5.7

So molecular weight = 29So, molecular formula of compound is NO

Q.7 (4) 17gm 3NH contains 23106 molecules of 3NH

4.25gm 3NH contains = 25.417

106 23

No. of atoms 417

25.4106 23

23106 .

Q.8 (1) 1L of gas at S.T.P. weight 1.16g 22.4 L of gas at S.T.P. weight 16.14.22

26984.25 This molecular weight indicates that given compound

is 22 HC .

Q.9 (1) 23106 molecules has mass gm18

1 molecules has mass 23106

18

gm23103

kg26103 .

Q.10 (1) 14 gm 3N ions have AN8 valence electrons

4.2gm of 3N ions have AA N

N4.2

14

2.48

Q.11 (2) 22400 ml at NTP has 2310023.6 molecule

1 ml at NTP has =22400

10023.6 23

= 23100002688.0 191069.2 .

Q.12 (2) 22400cc of gas at STP has 23106 molecules

1.12 710 of gas at STP has

22400

1012.1106 723 1214 1031003. .

Q.13 (1) 2.24L of gas has mass = 4.4gm

22.4L of gas has mass 444.2224.2

4.4

So given gas is 2CO because 2CO has molecular mass

= 44.

Q.14 (1) 44g of CO2 has 231062 atoms of oxygen

4.4g of CO2 has = 4.4

44

1012 23

23102.1 atoms.

Q.15 (3) 100gm 233 10023.6 CaCO molecules

10gm 3CaCO = 10100

10023.6 23

2210023.6 molecule

1 molecule of 3CaCO = 50 protons

2210023.6 molecule of 3CaCO 2210023.650

24100115.3

Q.16 (3) According to avogadro’s hypothesis equal volumesof all gases under similar conditions of temperatureand pressure contains equal no. of molecules.

Q.17 (4)V

Md (d = density, M= mass, V =volume)

Since d = 1So, VM 18gm = 18ml18ml = N

A molecules (N

A = avogadro’s no.)

1000ml 100018

AN = 55.555 N

A.

EXERCISE-I

EXERCISE Solution

Q.18 (2) 2gm of oxygen contains atom8

1

16

2 mole

also 4g of sulphur 8

1

32

4 mole.

Q.19 (1) 100gm caffeine has 28.9gm nitrogen

194gm caffeine has = gm06.56194100

9.28

No. of atoms in caffeine 414

06.56 .

Q.20 (1) 40gm NaOH contains 16gm of oxygen

100gm of NaOH contains 10040

16 =40% oxygen.

Q.21 (4) gmC 24 , gmH 4 , gmO 32

So, Molecular formula 242 OHC

So, Empirical formula OCH 2

(Simplest formula).

Q.22 (3) In OHCNSFe 23 3.)(

% of %19100284

1832

OH .

Q.23 (2)Element At.wt. Mole Ratio EmpiricalformulaC =86% 12 7.1 1 CH

2

H =14% 1 14 2 Belongs to alkene

Q.24 (2)Element % (1) At.wt.(2) a/b RatioX 50 10 5 2Y 50 20 2.5 1

Simplest formula YX 2

Q.25 (3) Eq.wt.

1000)W(

V

gmN

1500ml of 0.1N HCl = 150ml (N)

40150

1000)(W1

gm

, W (gm) gm61000

40150

.

Q.26 (3)xxxxtt

t

OHNOONH gggg

645141

00110

6454 )(2)()(2)(3

Oxygen is limiting reagent

So, 2.05

1X all oxygen consumed

Left 2.02.0413 NH .

Q.27 (3) 100gm Hb contain = 0.33gm Fe

67200gm Hb = Fegm100

33.067200

gm atom of Fe = 456

33.0672

.

Q.28 (4)L

gmHNaAlONaOHAlOH

6.334.222

3

2227

2 2

3

Q.29 (3) 0.1 M 3AgNO will react with 0.1 M NaCl to form 0.1 M

3NaNO . But as the volume is doubled, conc. of

MNO 05.02

1.03

Q.30 (2) 22 HMg

2

1

24

12

gm

gmn mole of 2H

Q.31 (3) OHCOCaClHClCaCOgNg

244

222100

3 2

100 g 3CaCO with 2 N HCl gives 44 g 2CO


Q.32 (4) 43 POH is tribasic so 3133 MN .

EXERCISE-II

Q.1 (3) In Ca3(PO

4)

2

mole of Ca atom

mole of O atom = 3

8

Mole of ‘O’ atom = 3

8 (mole of Ca atom)

Mole of ‘Ca’ atom = 3

Q.2 (1) H2SO

4A

2(SO

4)

3 32 3 × 32

so total molecular

mass = 9831

(A2(SO

4)

3)

31

× 342

114

11498

= 0.86

Q.3 (3) (1) n = 18

110 = 0.55

(2) n = 0.1 × 5 = 0.5

(3) n = 4812

× 3 = 0.75

(4) n = NAN

= 0.2 × 2 = 0.4

Q.4 (1) No. of atoms = 9854.31023

= 2.509 × 10+22

Q.5 (3) (1) n = 1212

= 1 (2) n = 168

= 05

(3) n = 3232

= 1 (4) n = 2424

= 1

Q.6 (2) Moles of Mg3(PO4)2 = 25.08

1 = 3.125 × 10–2

Q.7 (1) C H OMass 24 8 32

Moles12

24

1

816

32

Ratio 2 8 2Simple integer ratio 1 4 1Hence empirical formula is CH4O

Q.8 (1) X Y

75.8

75

24.2

16

1.01 1.5 × 22 3

Q.9 (1) CxH

y +

4yx O

2 XCO

2 + OH2y

2 (g)

(g) (g)

700600

2yx

4yx1

x + 7 = 4y5

by option (1)

Q.10 (4) Mole fraction of H2O = 1 – 0.25 = 0.75

OHOHHC

OHHC

252

52

XX

x

= OHOHHC

OHHC

252

52

nn

n

or wt. %

= 1875.04625.0

4625.0

× 100 = 46%.

Q.11 (3) CO2 = 132 g =

44132

mole = 3 mole

H2O = 54 g =

1854

mole = 3 mole

C atoms = 3 moleH atoms = 6 mole

by option 3

Q.12 (1) Fe2O

3 =

163562

= 37

FeO = 1656

= 27

Fe2O

3 : FeO =

37

× 27

= 3 : 2

Q.13 (1) amount of butter = 6

3

105.5

102

= 363.6 gm

Q.14 (2) M2O

3 0.30 × (2M + 48) = 48

0.6 M = 0.7 × 48M = 7 × 8 = 56

Q.15 (1) 2 SO2 + O

2 2 SO

3

Initial mole 10 15 0Final mole (10 – 2x) (15 – x) 2x Given 2x = 8 x = 4 Mole of SO

2 left = 10 – 2 × 4 = 2

Mole of O2 left = 15 – 4 = 11

Q.16 (2) C6H

5OH + 7O

2 6CO2 + 3H

2O()

30 ml6 × 30 = 180 ml of CO

2 is produced

Volume used initially= 30 + 210 = 240

(for C6H

5OH) (for O

2)

change in volume = 240 – 180 = 60 ml

Q.17 (2) C = 84/12 = 7 mole

H2 = 12 g = 6 mole

O2 = 56/22.4 = 5/2 mole

12C + 11H2 + 11/2 O

2 C12

H22

O11

L.R. = O2

11/2 mole O2 produce 1 mole sucrose

5/2 mole O2 will for 5/11 mole sucrose

mass of sucrose = 5/11 × (mol. mass)= 5/11 × 342= 155.45 g

Q.18 (1) Let volume is V mlH2SO4 + 2 NaOH Na2SO4 + 2H2O

mmole 0.2 V 40 x 0.1

m. moles of H2SO4 remains = 0.2 V – 2

1.040

40V2

1.0x40V2.0

=

55

6V = 70 mL

Q.19 (1) 1 Mol of x will give = 2

5 = 2.5 mol

But % yield = 1005.2

25.1 = 50%

Q.20 (1) Explanation : 2 Ag + S Ag2 S

2 × 108 g of Ag reacts with 32 g of sulphur

10 g of Ag reacts with 216

32 × 10 =

216

320 > 1 g

It means ‘S’ is limiting reagent32 g of S reacts to form 216 + 32 = 248 g of Ag

2S

1 g of S reacts to form = 32

248 = 7.75 g

Alternately

neq

of Ag = 108

10 = 0.0925 n

eq of S =

16

1 = 0.0625

(neq

= number of equivalents)Since n

eq of S is less than n

eq of Ag

0.0625 eq of Ag will react with 0.0625 eq of S toform 0.0625 eq of Ag

2S

Hence , amount of Ag2S = n

eq× Eq. wt. of Ag

2S

= 0.0626 × 124 = 7.75 g

Q.21 (1) Molality = AA

B

MXX × 1000

mB = 75 m

m = 1

M 1000d 1000 M M

M = 30

Q.22 (1) NaOH =

8125ml 1

10040

mole

HCl =

10125

10036.5

= 0.34 mole

HCl > NaOHAcidic

Q.23 (4) M = 2Mw

d10weightby% =

5.36

2.1105.36 = 12

M

m = )5.36100(5.36

10005.36

= 5.63

1000 = 15.7 m

Q.24 (2) 1000 mL solution contain 2 mole of ethanol or 1000 ×1.025 g solution contain 2 mole of ethanolwt. of solvent = 1000 × 1.025 – 2 × 46

m = 462025.11000

2

× 1000

m = 933

2 × 1000 = 2.143

Q.25 (3) Molarity = 23

22

1002.6

1002.6

2/1

1 = 0.2

Q.26 (2) Let, OH2n = n

NaCl = n

m = )kg(solventof.wt

soluteofMole=

18n

n

× 1000

= 18

1×1000 = 55.55 m.

Q.27 (2)

Q.28 (1) Mole fraction of A i.e. XA = molesTotal

nA

So XH2O

= molesTotal

n OH2

Now OH

A

OH

A

22n

n

X

X

and molality = 18X

1000X

18n

1000n

OH

A

OH

A

22

= 188.0

10002.0

= 13.9 Ans.

Q.29 (3) Molarity = Gmm

1.84 10 98 = 18.4 M

{ M = soluteofmass.Mol

10)d()w/w(% } (d in g/ml.)

Q.30 (2) Weight of KOH = 2.8 gramVolume of solution = 100 ml

M = 100 56

1000 2.8

= 56

28 = 0.5 M

Q.31 (3) Molarity of Cl– = 3 (molarity of FeCl3) = 3

30

M =

10

M.

Q.32 (1) M1V

1 + M

2V

2

= MR [V

1 + V

2]

1 × 500 + 1 × 500= M

R [500 + 500]

MR = 1

Q.33 (2) Mole = M V100 10–3 = 0.8 VV = 0.125

Q.34 (4) Moles of Cl– in 100 ml of solution

= 111

4

5.58

2 × 2 +

5.53

6 = 0.2184

Molarity of Cl– = 100

2184.0 × 1000 = 2.184.

Q.35 (4) Conc. of cation = 400

200300400

Conc. of anion = 400

400300200

Ratio of the conc. = 1

EXERCISE-III

Q.1 (A,B,C)Mole of NH

3 = 1.7 = 0.1

Mole H atom = 0.3Total atoms = 0.4 6.02 1023 = 2.408 1023

% H = 17

13 100 = 17.65%

Q.2 (B,C,D)

C + O2 CO

2

Mass 27 88

Moles 12

27

32

88

C is limiting reagent

Moles of CO2 produced = moles of C =

12

27 = 2.25

Volume of CO2 at STP = 2.25 22.4 = 50.4 L

Ratio of C and O in CO2 = 12 : 32 = 3 : 8

Moles of unreacted O2 = 2.75 – 2.25 = 0.5

Volume of unreacted O2 at STP = 0.5 22.4 = 11.2 L

Q.3 (A,C)

0.5 x n = 108

216 = mol of Ag

n = 4M.wt = 58 + [165]n g/mol = 718 g/mol

Q.4 (A,C)

Convert all the wt. in mole and use limiting reagentconcept find out the mole produced of NH3.

In (A) & (C) it comes equal to 10 moles

Q.5 (A,B,C)(Mw of Na

2CO

3 = 106, Mw of HCl = 36.5, Mw of NaCl =

58.5)

Moles of Na2CO

3 =

106

106 = 1.0 mol

Moles of HCl = 5.36

5.109 = 3.0 mol

(A) Since for 1 mol of Na2CO

3, 2 mol of HCl is required.

So, HCl is in excess (3 – 2) = 1.0 molTherefore, Na

2CO

3 is the limiting quantity.

(B) Weight of NaCl formed = (1.0 mol Na2CO

3)

32CONamol

NaClmol2

NaClmol

NaClg5.58 = 1 × 58.5 = 117.0 g NaCl

(C) 1 mol of Na2CO

3 = 1 mol of CO

2 = 22.4 L at NTP

Q.6 (A,C)

CaCl2 CaCO

3 CaO

56

12.1 = 0.02 mole CaO

Moles of CaCl2 = 0.02 Mole

Mass of CaCl2 = 0.02 × 111 = 2.22 g

% of CaCl2 =

44.4

22.2 × 100 = 50 %

Q.7 (B,D)

3A + 2B A3B

2

Initial mole 3 3 0Final mole 0 3–2 1

A3B

2 + 2C A

3B

2C

2

Initial mole 1 1 0

Final mole 1 – 2

1 0

2

1

Q.8 (A,C) Mw of CaCO3 = 100, Mw of Na

2CO

3 = 106

Mw of HNO3 = 63 g mol–1

Na2CO

3 + CaCO

3 CaCO

3 + 2NaCl

(a) moles of CaCO3 =

100

10 = 0.1 mol

moles of Na2CO

3 = moles of CaCO

3 2 × moles of NaCl

Weight of Na2CO

3 = 0.1 × 106 = 10.6 g

% purity Na2CO

3 =

2.21

6.10 = 100 = 50%

(b) wrong(c) correct(D) moles of NaCl = 2 × 0.1 = 0.2 mol

Q.9 (A, B)

Q.10 (B,D)

Molality of Cl¯ = 44.419009.11000

210002

Q.11 (A,B)

Q.12 (A,C,D)[Mw of KI, (NH

4)

2SO

4, CuSO

4, CuSO

4.5H

2O and Al3+,

respectively, are, 166, 132, 160, 250 and 27 g mol–1]

Q.13 (A,B,D)

(A) Molarity of second solution is = M

xd10 = 1 M

(B) Volume = 100 + 100 = 200 mL

(D) Mass of H2SO

4 =

1000

1200 × 98 = 19.6 g.

Q.14 (C) 11.2 g of N2 28

2.11 = 0.4 mole

air = 0.5 mole 0.5 × 22.4 = 11.2 Ltr air

Q.15 (B) 1 mole of air 0.8 mole of N2 = 0.8 × 28 g N

2

0.2 mole of O2 = 0.2 ×32 g O

2

% w/w O2 =

22

2

NO

O

ww

100w

=

288.0322.0

100322.0

= 22.2%

Q.16 (B) Density of air at NTP1 mole of air = 0.8 mole N

2 + 0.2 mole O

2

= 0.8 × 28 + 0.2 × 32 = 28.8 g = 22.4 Ltr volume.

D = V

m=

4.22

8.22= 1.2857 g/L

Q.17 (A)

Q.18 (B)

Q.19 (B)

Q.20 (A) R, (B) P, (C) Q

% of Y = 1612275389

389

× 100

= 594

100267 = 44.95%

% Al = 100594

275

= 22.73

% O = 100594

1612

= 32.32%

Q.21 (A - p,s); (B - s); (C - p,q); (D - r)

(A) Molarity of cation = 21

2211

VV

VMVM

= 500

4001.01002.0 =

5

6.0 = 0.12

Molarity of Cl— = 500

4001.0100)2.0(3 =

5

4.06.0

= 0.2

(B) Molarity of cation = 100

04.050 = 0.2

Molarity of Cl— = 100

0504.0 = 0.2

(C) Molarity of cation = 100

030)2.0(2 = 0.12

Molarity of SO42– =

100

2.030= 0.06

(D) 24.5 g H2SO

4 in 100 mL solution

Molarity = 1.0

984.25

= 2.5

Concentration of cation = 2 × 2.5 MConcentration of SO

42– = 2.5 M.

Q.22 (A - q,s); (B - p, s); (C - p, q, r); (D - q, r)

(A) C : H : O = 12

17.51 :

1

04.13 :

16

78.34 = 4 : 12 : 2 or

2 : 6 : 1 Empirical formula = C

2H

6O & molar mass = 46 g/mol

Mol formula = C2H

6O

C2H

6O + 3O

2 2CO

2 + 3H

2O

1 mole 44.8 L at STP0.25 mole (11.2 L at STP)(B) Mass of C in organic compound = mass of C in CO

2

= 44

44.0 12 = 0.12 g

Mass of H in organic compound = Mass of H in H2O

= 18

18.0 2 = 0.02 g

Mass of O in organic compound = 0.3 – (0.12 + 0.02)= 0.16 g

C : H : O = 12

12.0 :

1

02.0 :

16

16.0 = 0.01 : 0.02 : 0.01

= 1 : 2 : 1 Empirical formula = CH

2O, but it contains 2 O atom

per molecule Molecular formula = C

2H

4O

2

1 mole of C2H

4O

2 contains 4 N

A hydrogen atoms.

C2H

4O

2 + 2O

2 2CO

2 + 2H

2O

1 mole 44.8 L0.25 mole 11.2 L(C) C : H = 42.857 : 57.143

= 3 : x (given)On solving, x = 4 molecular formula = C

3H

4

1 mole of C3 H

4 contains 4N

A hydrogen atoms.

Empirical formula is same as molecular formula

C3H

4 + 4O

2 3CO

2 + 2H

2O

2COn > OH2n

(D) C: H = 12

5.10 :

1

1 =

8

7 : 1 = 7 : 8 Empirical

formula = C7H

8

Mol wt. = 2 VD = 2 46 = 92 Mol formula = Empirical formula = C

7H

8

C7H

8 + 9O

2 7CO

2 + 4H

2O

2COn > OH2

n

NUMERICAL VALUE BASED

Q.23 [+ 8]

x + 4 (–2) = 0x = +8

Q.24 [– 3]x + 4 (+1) = +1x = –3

Q.25 [500]

M = )solution.lit1(

soluteofmoles18

Mass of solution = 1000 × 1.8 = 1800 g.Mass of solute = 18 × 98 = 1764.Mass of solvent = (1800 – 1764) g = 36 g.

Hence molality = solventofmass

soluteofmoles × 1000 =

36

18 × 1000

= 500.

Q.26 [40] 40gm NaOH contains 16gm of oxygen

100gm of NaOH contains 10040

16 =40% oxygen.

Q.27 [3][Co(NO

2)

6]3–

x + 6 (–1) = – 3x = +3

Q.28 [6]

Eq.wt.

1000)W(

V

gmN

1500ml of 0.1N HCl = 150ml (N)

40150

1000)(W1

gm

, W (gm) gm61000

40150

.

Q.29 [1]2Al + KClO

3 Al

2O

3 + KCl

(excess) 1 mole (KClO3 is L.R.) 1 mole (KClO

3 L.R)

From mole-mole analysis

1

n3KClO

= 1

n32OAl

32OAln = 1 mole

Q.30 [22g]

OHCOCaClHClCaCOgNg

244

222100

3 2



Q.31 [1]

Conc. of cation = 400

200300400

conc. of anion = 400

400300200

ratio of the conc. = 1

Q.32 [27g]Let wg water in added to 16 g CH3OH

molality = 32W

100016

= W

500

W

500 =

BA

A

m)x1(

1000x

= 1875.0

100025.0

W = 27 gm.

EXERCISE-IV

JEE-MAINPREVIOUS YEAR'S

Q.1 (2) Given chemical eqn

M2CO

3 + 2CHl 2MCl + H

2O + CO

2

1 gm 0.01186 mol from the balanced chemical eqn

1

M = 0.01186 M = 84.3 gm/mol

Q.2 (3) Mass in body of a healthy human adult has:Oxygen = 61.4%, carbon = 22.9 %Hydrogen = 10.0% and Nitrogen = 2.6%Total weight of person = 75kg

Mass due to 1H is = 75 × 10

100 = 7.5 kg

1H atoms are replaced by 2H atoms.So mass gain by person = 7.5 kg

Q.3 (3)12x 6

y 1

x y z2x y for C H O

x y 2 2g 2 gy y

C H g x O xCO H O4 2

no. of oxygen atom in x y zC H O z

no. of oxygen atom required for x yC H

combustion is y y

x 2 2x4 2

So 1 y

z 2x2 2

yz x

4

2x 3xz x

4 2

3xx : 2x :

2

2x : 4x : 3x2 : 4 : 3Hence C

2H

4O

3

Q.4 (3) In cold water, dissolved oxygen can reach aconcentration upto 10 ppm.

Q.5 (1) Let NaHCO3 = x gm

Then, H2C

2O

4 = (10 – x) gm

3NaHCOn =

x

1682 NaHCO

3 Na

2CO

3 + H

2O + CO

2

2COn =

x

168

2 2 4H C O

10 xn

90

H

2C

2O

4 H

2O + CO

2 + CO

2CO

10 xn

90

Total CO2 =

x 10 x 0.25

168 90 25

On solving ‘x’ % = x

100 10x10

Q.6 (1) More polluted water has high biological oxygendemand.

Q.7 (3) 2C57

H110

O6 (s) + 163 O

2 (g) 114 CO

2 (g) + 110 H

2O (l)

57 110 6Molesof C H O

22Molesof H O

110

2massof H O445

890 182 110

Mass of H2O = 495 g

Q.8 (4)Eq. of (COOH)2 = Eq. of NaOH50 × 0.5 × 2 = 25 × M × 1

Mass of NaOH in 50 mL =50 2

40 4g1000

Q.9 (3) Molality of w 1000

Na 2M W

(Na

2SO

4 contains

two Na+ ions)

92 10002 8

23 1000

Q.10 (4) As 1L solution have 10–3 mol CaSO4

Eq. of CaSO4 = eq. of CaCO

3

ln 1L solution

4 3CaCO CaCO v.fn v.f. n

3

3CacO10 2 n 2

3

3CaCOn 10 mol in 1L

3

3CaCOw 100 10 g in 1Lsolution

3hardness in terms of CaCO

33

CaCO 6 6

Total

w 100 10 g10 10 100ppm

w 1000g

Q.11 (3) Clean water has BOD value of less than 5 ppm andhighly polluted water has a BOD value of 17 ppm ormore.

Q.12 (3) 12 22 11C H On0.1

2

12 22 11C H On 0.2

12 22 11C H OWt 0.2 342 68.4

Q.13 (3) Moles of NaOH = 8

0.240

Moles of H2O =

181

18

Moles fraction of NaOH = 0.2

0.1671.2

Molality = 8 1000

11.1140 18

Q.14 (3) Volume strength = 11.35 × M = 11.35 (STP)

Q.15 (3) n1T

1 = n

2T

2

n × 300 = 22n

n T5

2 23

300 T T 500K5

Q.16 (2) eq 3 eq 3 2 eq 3 2n .CaCO N Ca(HCO ) n Mg(HCO )

Or, W 0.81 0.73

2 2 2100 162 146

w 1.0

61.0Hardmess 10 10000ppm

100

Correct option : (2)

Q.17 (3) N2(g) + 3H

2(g) 2NH

3(g)

(1) 0.5 mol 2 mol (LR)(2) 1 mol 3 mol (completion)(3) 2 mol 5 mol (LR)(4) 1.25 mol 4 mol (LR) Ans. (3)

Q.18 (4) CH4

% by mole of carbon 1molatom

1005molatom

= 20%

Q.19 (4) The maximum prescribed concentration of Cu in drink-ing water is 3 ppm.

Q.20 (2) Radius of watchglass = 10 cm surface area = r2 = 3 × (10cm)2

= 300 cm2

mass of fatty acid in 10 ml solution

10 0.270.027gm

100

volume of fatty acid 20.027g

0.03cm0.9g / ml

Height volumeof fattyacid

surfacearea of watch glass

3

2

0.03cm

300cm 0.0001 cm = 10–6 m

Q.21 (2) Volume strength = 11.2 × molarity = 11.2 molarity = 1 M strength = 34 g/L

% w/w 34

100 3.4%1000

Q.22 (2) x y 2 2 2

Y YC H X O – xCO H O

4 2

25

M

25 y 25X

M 2 M2 0.5

25

C x 2M25

H y 1M

2y y yC H 24 gmC ygmHor

24 :1ratiobymass

Q.23 (1) Millimoles = 10 × 10–3 = 10–2

Moles = 10–5

No. of molecules = 6 × 1023 × 10 –5 = 6 ×10+8

surface area occupied by one molecule

–18 218

0.240.04 10 cm

6 10

–20 24 10 a 10a 2 10 cm pm

Q.24 (3)W

% 20W

100 gm solution has 20 gm KI80 gm solvent has 20 gm KI

2020 1000166M 1.506 1.51 mol / kg

80 166 801000

Q.25 (3) 5[MA + 2MB] = 125MA + 2MB = 25 ......(1)2MA + 2MB = 30 ......(2)from eq. (1) & (2)MA = 5MB = 10

Q.26 (3) Xsolvent = 0.8

If nT = 1nSolvent = 0.8

2molality 13.88

0.8 18

1000

Q.27 (4) CxHy + y

x4

O2 xCO2 + y

2H2O

10 10y

x4

10x

By given data, 10y

x4

= 55 .......(1)

10x = 40 .......(2) x = 4, y = 6 C4H6

Q.28 (2)(1) H2SO4 + 2NaOH Na2SO4 + 2H2O 400×.1=40 400×.1=40 20 0

[H+] = 20 2

800

=

1

20 pH = – log

1

20

pH = 1.3 so (1) is correct

(2) log 2

1

Kw

Kw

= 1 2

H 1 1

2.303R T T

so ionic product of water is temp. dependent hence (2)is correct.(3) Ka = 10–5, pH = 5 [H+] = 10–5

Ka =

2c

(1 )

Ka =

[H ].

(1 )

10–5 =

510 .

(1 )

1 – = =

1

2 = 50%

so (3) is correct.(4) Le-chatelier's principle is applicable to common-Ioneffect so option (4) is wrong correct answer (2)

Q.29 (3) C3H

8(g ) + 5O

2(g) 3CO

2(g) + 4H

2O ( )

Each lg of C3H

8 requires 3.63 g of O

2

P4(s) + 5O

2g P

4O

10(s)

Each lg of P4 requires 1.29 g of O

2

4Fe(s) + 3O2(g) 2Fe

2O

3(s)

Each lg of Fe requires 0.428 g of O2

2Mg(s) + O2(g) 2MgO(s)

Each lg of Mg requires 0.66 g of O2 therefore least

amount of O2 is required in option (3).

Q.30 (1) Refer Notes

Q.31 [26.80 to 27.00][ML6]Cl3 + 3AgNO3 3AgCl 0.3g v ml, 0.125 M

30.33 0.125 V 10

267.46 or,,

0.3 3 1000V 26.92ml

267.46 0.125

Q.32 [2120.00 to 2140.00]mol of NaClO3 = mol of O2

mol of 2

PV 1 492O 20mol

RT 0.082 300

mass of NaClO3 = 20 × 106.5 = 2130 g

Q.33 [100.00 ]10–3 molar MgSO

4 10–3 moles of MgSO

4 present in 1

L solutions.

3 4CaCO MgSOn n

3

36

(in termof CaCO )10 100

ppm 101000

3(in termof CaCO )ppm 100 ppm

Q.34 (2) 13.5% of H12

O6Cl

3Cr is equal to 36(approx).

Around two moles of water are are lost during heating. Formula of complex could be [Cr(H

2O)

4Cl

3]Cl.2H

2O

Q.35 [25.00]Mass of 6.023 × 1022 molecules of a substance= 10 gMass of 6.023 × 1023 molecules of the substance= 100 g

Molar mass of the substance = 5

100 2 = 2.5 × 10–2

= 25 ×10–3

Q.36 (1) Volume strength = 5.6V.S = 11.2 × Molarity

Molarity (M) = V.S. 5.6

11.2 11.2 = 0.5

0.5 M 0.5 mole in 1L

Mass percentage = 0.5 34

1001000 d

= 17 100

1.71000 1

Q.37 [47]

Q.38 [100]Molarity of H

2O

2 solution = 8.9 M

Volume strength of H2O

2 solution

= 8.9 × 11.2 =100 V

Q.39 [3400]

N2 (g) + 3H

2(g) 2NH

3(g)

Number of moles of N2 =

32.8 10100

28

Number of moles of H2 =

1000500

2

Number of moles of NH3 produced = 200

Mass of NH3 produced = 200 × 17 = 3400 gm

Q.40 [18]C

3H

8(g) + 5O

2(g) 3CO

2(g) + 4H

2O(l)

C4H

10(g) +

13

2O

2 (g) 4CO

2(g) + 5H

2O(l)

No. of moles of O2 required to oxidise 1 mole of propane

and 2 moles of butane = 5 + 2 ×13

2

= 18

Q.41 (3) Average molar mass = 35 3 37 1

35.54

35Cl : 37Cl = 3 : 1

Q.42 (2) To express C2 in terms of mole fraction x

2

1st component 2nd componentmole n

1n

2

m.w M2

M2

mass n1M

1n

2M

2

mass of solution = n1M

1 + n

2M

2

mole fraction x2 = 2

1 2

n

n n

n1 = 2 2

2

n (1 x )

x

Mass of solution = n1M

1 + n

2 M

2

= 2 1 22 2

2

n M (1 x )n M

x

= 2

2

n

x [M2x

2 – x

2 M

1 + M

1]

Volume of solution

= 2 2 2 2 1 1

2

n [M x x M M ]Litre

1000dx

C2 =

2 2

2 2 2 2 1 1

1000n dx

n [M x x M M ]

22

1 2 2 1

1000dxC

M x (M M )

JEE-ADVANCEDPREVIOUS YEAR'SQ.1 [5] The balance chemical equation is

3Br2 + 3Na

2CO

3 5NaBr + NaBrO

3 + 3CO

2

Q.2 (C) Mole = 60

120= 2

mass of solution = 1120 g

V = 100015.1

1120

= 115

112L

M = 112

1152 = 2.05 mol/litre

Q.3 [8 mL.]29.2% (w/w) HCl has density = 1.25 g/mlNow, mole of HCl required in 0.4 M HCl = 0.4 × 0.2 mole= 0.08 moleif v mol of orginal HCl solution is takenthen volume of solution = 1.25 v

mass of HCl = (1.25 v × 0.292)

mole of HCl = 5.36

292.0v25.1 = 0.08

so, v = mol25.129.0

08.05.36

= 8 mL

Q.4 [8] Given 3.2 M solution moles of solute = 3.2 molConsider 1 L Solution. volume of solvent = 1 LP

solvent = 0.4 g.mL–1

msolvent

= P × V = 400 g

molality = kg4.0

mol2.3 = 8 molal

Q.5 [9] Given, molality = MolarityAnd assuming no volume change in forming solutionDensity of solvent = 1 gml–1

And density of solution (given) = 2 gml–1

Implies, solvent and solute are present in equalqualities

1

0.11 1

solutesolute

solvent solute

M

M M

9solute

solvent

M

m

Q.6 [6] 8KMnO4 + 3Na

2S

2O

3 + H

2O

8MnO2 + 3Na

2SO

4 + 3K

2SO

4 + 2KOH

No. of sulphur containing products is 3 + 3 = 6Q.7 [2992]

4 4 4 2 32 2gypsum(M 172) 24mole1584g12mol12mol

NH SO Ca OH CaSO .2H O 2NH

2 2 3 3 2 26952g 4mol 24mol M 232

4mol

NiCl .6H O 6NH Ni NH Cl 6H O

Total mass = 12 × 172 + 4 × 232 = 2992 g

Q.8 [2.98 or 2.99]

Xurea

= 0.05 = n

n 5019n = 50n = 2.6315

Vsol

= (2.6315 60 900)

1.2

= 881.5789 ml

Molarity = 2.6315 1000

881.5789

= 2.9849

Molarity = 2.98MQ.9 [6.15]

2Al+3H2SO

4 Al

2(SO

4)

3+3H

2

Moles of Al takes 5.4

0.227

moles of H2SO

4 taken

50 5.00.25

1000

As 2 4

0.2 0.25,H SO

2 3 is limiting reagent

Now, moles of H2 formed

30.25 0.25

3

Volume of H2 gas formed =

nRT

P

0.25 0.082 3006.15L

1

Some Basic Concept of Chemistry - SelfStudys

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