Chapter : 01

Fundamentals Of Chemistry

Self – Assessment Exercise 1.3

Vinegar is 5% Acetic Acid. It contains two Carbon atoms, four Hydrogen

atoms and two Oxygen atoms. Write its empirical and molecular Formula.

Empirical Formula of Vinegar = _________

Molecular Formula of Vinegar = _________

Caffeine (C8H10N4O2) is found in tea and coffee. Write the empirical Formula

for Caffeine.

Empirical Formula gives the simplest whole number ratio of atoms of each

element.

Empirical Formula for Caffeine = ____________

Self – Assessment Exercise 1.4

Potassium Chlorate (KClO3) is used commonly for the Laboratory

preparation of Oxygen gas. Calculate its formula mass.

Solution:

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Formula mass of KClO3= ________________

When baking soda NaHCO3 is heated carbon dioxide is released, which is

responsible for the rising of cookies and bread. Determine the formula masses

of baking soda and carbon dioxide.

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Following compounds are used as fertilizers. Determine the formula masses.

(a) Urea ,(NH2)2 CO

Solution:

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(b) Ammonia nitrate, NH4NO3

Solution:

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Self- Assessment Exercise 1.5 Explain why?

1. An oxide ion Oˉ² has -2 charge.

An oxide atom has 8 protons in the nucleus and 8 electrons revolving

around its nucleus. Thus its nucleus has a total charge of +8. Oxygen

gains 2 electrons to complete its valence shell electronic configuration.

Around the nucleus, in the ion are 10 electrons, with a total charge of -

10.So, the charge on oxide ion is:

Total charge on nucleus =+8

Charge on electron= -10

Charge on ion= +8-10

Charge on ion= -2

2. Magnesium ion, Mg+2 has +2 charge.

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3.Sulphide ion, S-2 has -2 charge.

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Self- Assessment Exercise 1.6

Identify ions, molecular ions and free radicals from the following specie.

CN- ,He+2 , N-3, CH+4 , CN∙

Ions Molecular ions Free radicals

Self -Assessment Exercise 1.8

Calculate the mass of one mole of:

(a) Copper

Given Data:

Number of mole of copper (cu) = 1 mole

To Find:

Mass of Cu = ?

Formula used:

Solution:

Result:

b) Iodine

Given data:

No of moles of iodine =1 mole

Molar mass of iodine =2 127

=254 g/mole

To Find:

Mass of iodine =?

Formula Used:

Solution:

Result:

c) Potassium:

Given Data:

No of mole of potassium (K)= 1 mole

Molar mass of K =1 x 39

=39g/mole

To find:

Solution:

Result:

d) Oxygen

Given data:

No of mole of oxygen O2 = I mole

Molar mass of O2 = 2 x 16

= 32g/mole

To find:

Mass of O2 =?

Formula used:

Solution:

Result:

Self – Assessment Exercise 1.9

The molecular formula of a compound used for bleaching hair is H2O2.

Calculate

a) Mass of this compound that would contain 2.5 moles

b) No of moles of this compound that would exactly weigh 30 g.

Given data:

Number of moles ofH2O2 =2.5 moles

Molar mass of H2O2= 2x1+2x16

=2+32

=34 g /mole

To find:

Mass of H2O2=?

Formula used:

Solution:

Result:

b) Data Given:

Mass of H2O2= 30g

Molar Mass = 34g/mol

To Find:

No of moles of H2O2 =?

Formula used:

Solution:

Result:

2: A spoon of table salt, NaCl contains 12.5grams of this salt. Calculate the

number of moles it contains.

Data Given:

Mass of NaCl = 12.5g

Molar mass of NaCl = 1×23+1×35.5

= 23+35.5

= 58.5g/mol

To find:

Number of moles =?

Formula used:

Solution:

Result:

3: Before the digestive system x-rayed, people are required to swallow

suspensions of barium sulphate (BaSO4). Calculate mass of one mole of

BaSO4

Given data:

No. of mole of BaSO4 = 1 mole

Molar mass of BaSO4 = 137+32+4×16

= 233g/mol

To find:

Mass of BaSO4=?

Formula Used:

Solution:

Result:

Self- Assessment Exercise 1.10

1: Aspirin is a compound that contains carbon, hydrogen and oxygen. It is

used as a painkiller. An aspirin tablet contains 1.25×1030 molecules. How

many moles of this compound are present in the tablet?

Given data:

No. of molecules= 1.25×1030 molecules

Avogadro’s no =NA = 6.022×1023

To find:

Number of moles (n) =?

Formula used:

Solution:

Result:

2: A method used to prevent rusting in ships and underground pipelines

involving connecting the ion to a block of a more active metal such as

magnesium. This method is called cathodic protection. How many moles of

magnesium are present in 1 billion (1×109) atoms of magnesium?

Data given:

No. of atoms = 1×109 atoms

Avogadro’s number = 6.022×1023

To find:

Number of moles=?

Formula used:

Solution:

Result:

Review Questions

(vi) What is the number of molecules in 9.0g of steam?

Data Given:

Mass of steam (H2O) = 9g

Molar mass of steam (H2O) = 2 x 1 + 16

=18g / mol

Avogadro’s number = NA = 6.022 x 1023

To Find:

Number of molecules =?

Formula Used:

Solution:

Results:

Q.6 Calculate the number of moles of each substance in samples with the

following masses:

a) 2.4g of He

Data Given:

Mass of Helium (He) = 2.4g

Molar mass of He = 1 x 4 = 4g /mol

To Find:

Number of moles =?

Formula Used:

Solution:

Result:

b) 250mg of Carbon:

Data Given:

Mass of Carbon = 250mg = 250 / 1000 = 0.25g

Molar mass of Carbon = 12g / mol

To Find:

No. of moles =?

Formula Used:

Solution:

Result:

c) 15g of Sodium Chloride

Data Given:

Mass of NaCl = 15g

Molar Mass of NaCl = 23 + 35.5

= 58.5g / moles

To Find:

No. of moles =?

Formula Used:

Solution:

Result:

d) 40g of Sulphur

Data Given:

Mass of Sulphur (S) = 40g

Molar mass of Sulphur = 32g / mol

To Find:

No. of moles =?

Formula Used:

Solution:

Result:

e) 1.5kg of MgO

Data Given:

Mass of MgO = 1.5 kg = 1.5 x 1000 = 1500 g

Molar mass of MgO = 24 + 16 = 40 g/mol

To Find:

No. of Moles =?

Formula Used:

Solution:

Result:

Q.7:Calculate the mass in grams of each of the following samples.

a) 1.2 moles of K

Data Given:

No. of moles of K = 1.2 moles

Molar Mass of K = 39 g / mol

To Find:

Mass of K =?

Formula Used:

Solution:

Result:

b) 75 moles of H2

Data Given:

No. of moles of H2 75 moles

Molar mass of H2 = 2 x 1

=2g / mol

To Find:

Solution:

Result:

(c) 0.25 Moles of steam

Given data:

No. of moles of steam (H2O) =0.25 moles

Molar mass of H2O = 2x1+16

=18g

To Find:

Mass of steam=?

Formula used:

Solution:

Result:

(d) 1.05 moles of CuSO4.5H2O

Given data:

No of moles of CuSO4.5H2O = 1.05 moles

Molar mass = 63.5+32+4x16+5(2x1+16)

=249.5g

To find:

Mass of CuSO4.5H2O =?

Formula used:

Solution:

Result:

(e) 0.15 moles of H2SO4

Given data:

No of moles of H2SO4 = 0.15 moles

Molar mass of H2SO4 = 2x1+32+4+16

= 98g

To find:

Mass of H2SO4 =?

Formula used:

Solution:

Result:

Q.8- Calculate the number of molecules present in each of the

following samples.

(a) 2.5 Moles of carbon dioxide

Given data:

No of moles = 2.5 moles

Avogadro’s number = 6.022x1023

To find:

No of molecules =?

Formula used:

Solution:

Result:

(b) 3.4 moles of ammonia NH3

Given data:

No of ammonia NH3 = 3.4

Avogadro’s no = NA = 6.022x1023

To find:

No of molecules =?

Formula used:

Solution:

Result:

(c) 1.09 moles of benzene, C6H6

Given data:

No of moles = 1.09

Avogadro’s no = 6.022x1023

To find:

No of molecules =?

Formula used:

Solution:

Result:

(d) 0.01 moles of acetic acid CH3COOH

Given data:

No of moles = 0.01 moles

Avogadro’s no = 6.022x1023

To find:

No of molecules =?

Formula used:

Solution:

Result:

Q.14: Calculate the number of atoms in each of following samples.

(a) 3.4 moles of Nitrogen atoms

Data Given:

No. of moles of N-atom = 3.4 moles

Avogadro’s number, NA=6.022×1023

To Find: No. of atoms = ?

Formula Used:

Solution:

Result:

(b) 23g Na

Data Given:

Mass of Na = 23g

Molar Mass of Na = 23g / mol

To Find: No. of atoms =?

Formula used:

Solution:

Result:

(c) 5g of H atoms

Data Given:

Mass of H- atoms = 5 kg

Molar mass of H- atoms = 1 g / mol

NA =6.022 ×1023

To Find: No. of atoms = ?

Formula used:

Solution:

Result:

Self assessment No 1

Draw Bohr’s Model for the following atoms indicating the

location for electrons, protons and neutrons

⚫ Carbon Atomic No 6, Mass No 12

Proton no

Neutron no

⚫ Chlorine Atomic no 17 Mass no .35

Proton no

Neutron no

⚫ Self assessment No 3

Write the complete electronic configuration for the following

elements

1. 13Al

2. 14Si

3. 15P

4. 16S

5. 17Cl

6. 18Ar

⚫ Self assessment No 4

Write the electronic configuration of for the following

isotopes

A) 146C

B) 3517Cl

C) 3717Cl

Review Question

1) Distinguish between shell and sub shell

SShell SuSSSubshell 1. The electron in an atom revolves

aroud the nucleus in one of the

circular paths called shell or orbit.

Each shell has a fixed energy. So each

shell is also called an energy level.

1. A shell or energy level is sub divided

into subshell or sub energy level.

2.Each shell is described by its “n”

value

2. ‘n’ value of shell is placed before the

symbol of a subshell.

3- n=1 , it is K- shell

n= 2, it is L-shell

n= 3, it is m- shell

n=4, it is n- shell

3-n=1, for K shell. It has only one

subshell 1s

n=2, for L shell, it has two subshells.2s

2p

n=3, for M shell it has three subshells 3s,

3p and 3d.

n=4 for N- shell it has four subshells

4s,4p,4d, 4f

S subshell can accommodate maximum

2 electrons.

P subshell can accommodate 6

electrons.

d subshell can accommodate maximum

10 electrons

F subshell can accommodate maximum

14 electrons.

4. Asthe value of n increases the

distance of electrons from the

nucleus and energy and energy of the

shell increases.

4. The increasing order of energy of the

sub shells belonging to different shells

is

1s<2s< 3s<3p<4s<3d…….

2- An atom is electricall neutral why?

Ans

3-How may subshells are there in N shell?

4 Five notation for subshells in subshells of M- shell.

Ans

5 List the subshells of M shell in order of increasing energy

Ans

6 Can you identify an atom without knowing number of neutrons in

it.

Ans

5. Draw Bohr’s model for the following atoms indicating the

location of electrons protons and neutrons

a) Potassium Atomic no 19 Mass no 39

No of protons

No of neutron

No of electrons

b) Silicon Atomic no 14 mass no 28

c) Argon Atomic no 18 mass no 40

⚫ Write electronic configuration for the following elements

No of protons

No of neutron

No of electrons

No of protons

No of neutron

No of electrons

No of protons

No of neutron

No of electrons

2814Si 1s2 2s2,2p6, 3s2 ,3p2

2412Mg

2713Al

4018Ar

⚫ Explain how Bohr’s atomic theory differed from Rutherford’s

atomic theory

Bohrs

Rutherford

⚫ Describe the presence of subshell in a shell

Ans.

⚫ The atomic no of an element is 23 and its mass no is 56

a-How many protons and electrons does of an atom of this

element have?

b-How many neutrons does this atom have?

THINK TANK

Q- M-24 is a radioactive isotope used to diagnose restricted

blood circulation, for example in legs. How many electrons,

proton and neutrons are there in this isotope Valence shell

electronic configuration of M is 3s1

R- Two isotopes of are 3517Cl and 37

17Cl. How do these

isotopes differ? How are they alike?

Q- How many electrons can be placed in all of the sub shells

in the n=2 shell?

Q - Mass number of an atom indicates total no. Of protons

and neutrons in the nucleus. Can you find an atom without

any electron?

Self assessment exercises

Self assessment exercise 3.1

In which period and group following elements are present in

the periodic table

ELEMENT PERIOD GROUP

A Mg 3 IIA

B Ne

C Si

D B

Identify the group and the period of the following elements

on the basis of electronic configuration.

A) 2814Si 1s2, 2s2, 2p6 ,3s2, 3p2

K L M

So, the Valence shell is M

As, n= 3 Si is present in 3rd period. Total no of electrons in

the valence subshell are 2 +2 =4, it is I group IV A.

B) 199F

C) 3216S

D) 4018Ar

SELF ASSESSMENT 3.4

1- Obtain the valence shell configuration of Al and S from

their position in the periodic table

A) Al

Period no. =n value of shell

Group no. =number of Valence electrons

Period no. Of aluminum is 3

As n=3 therefore valence shell is M so valence electron

will be present in 3s, 3p subshell

The group no is 3, so there are three electrons in the valence

shell.

Two electrons will fill 3s subshell and the remaining in 3p

subshell. Thus, the valence shell electronic configuration in

3s23p1

B) 3216S

2- Find out the position of Ne ( atomic no 10) and Cl ( atomic

no 17) in the periodic table

⚫ Ne

Electronic Configuration of Ne = 1s2 2s2, 2p6

Valence shell has configuration 2s2, 2p6

period no 2

Group no 2+6= 8

Therefore Neon is present in second period of group VIIIA

⚫ Cl

SELF ASSESSMENT EXERCISE 3.6

Using the periodic table Choose the element whose atom you

expect to have smaller atomic radius in each of the following

pairs

A) O or S

O will have smaller atomic radius as compared to S

REASON

By using the electronic configuration we know

8O= 1s2, 2s2, 2p4 16S= 1s2, 2s2, 2p6, 3s2,3p4

K L

n=2 its Valence shell is L n=3 its Valence shell

is M

As in O the Valence shell is L which is smaller than M, it

means it have smaller radius.

B) O or F

Self Assessment No 3.7

Which atom has smaller ionization energy?

a) B or N

B will have smaller ionization energy

REASON

As we move from left to right in a period, the shielding effect

remains constant but progressively nuclear charge increases.

A strong force of attraction between the nucleus and the

valence electron increases. This leads to increase in ionization

energy from left to right in a period. As Boron will come

earlier in a period so it have smaller energy.

b) Be or Mg

C) C or Si

i. Write the valence shell electronic configuration of an

element present in the 3 period and Group III A.

ii. Write two ways in which isotopes of an element differ

iii. Which atom has higher shielding effect ,Li or Na?

iv. Explain why Na has higher ionization energy?

v. Alkali metals belong to S-block in the periodic table, why?

⚫ Arrange the elements in each of the following groups in

order of increasing ionization energy

A) Li, Na, K

B) Cl, Br,I

⚫ Arrange the elements in each of the following element of

decreasing shielding effect

A) Li, Na , K

B- Cl, Br,I

C) Cl, Br

⚫ Specify which of the following elements you would expect

to have to have the greatest electron affinity

S,P,Cl

⚫ Write the valence electronic shell configuration For the

following groups

Metals Group Electronic

Configuration

Alkali metals 1st ns1

Alkaline earth

metals

Halogens

Nobel gases

⚫ Write types of elements have the highest ionization

energies and what type of elements the lowest ionization

energies. Argue

ANS

⚫ Two atom have electronic configuration 1s2, 2s2, 2p6 and

1s2, 2s2, 2p6, 3s1 the ionization energy of one is

2080KJ/mol. Match each ionization energy with one of

the given electronic configuration Give reason for your

choice

ANS.

⚫ Use the second member of each group from group IA ,

IIA to judge the member of valence electron in an atom

of the element is the same as its group number.

ANS

⚫ Argue in what region of the periodic table you will find

elements

A- High ionization energy

B- Low ionization energy

Find the number of electron in valence shell of the following atoms using the

periodic table

Silicon

Sulphur

Bromine

Argon

Potassium

Nitrogen

1. Describe the formation of cations for the following metal atoms.

a. Li(atomic no.3)

b. Al(atomic no.13)

a. Li(atomic no:3)

Li (1s2 2s1) Li+ (1s2)

Li belongs to group IA to the periodic table. It has 1 electron in the

valence shell. Li atom loses its valence electron and is left with a duplet.

-e’

+ 1e’

b. Al (atomic no.13)

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2. Represent the formation of cations for the following metal atoms using

electron dot structures.

a. K b. Ca

+ 1e’

1. Describe the formation of anions by the following non-metals.

a. Sulphur (atomic no. 16)

b. Chlorine (atomic no. 17)

S (1S2 2S2 2P6 3S2 3P4) + 2e’ (1S2 2S2 2P6 3S2 3P6)

Sulphur belongs to VIA of the periodic table. It has 6 electrons in the valence shell.

S atom requires 2 electrons to complete octet.

+ 2e’

-2

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2. Represent the formation of anions by the following non-metal using

electron dot structures.

a. N b. P c. Br d. H

+ 3e’

+ 1e’

-1

1. For each of the following pairs of atoms, use electron dot and electron

cross structures to write the equation for the information ionic

compound.

a. Mg & O b. AI & CI

Mg atom has two electrons in valence shell. It losses 2 electrons to form Mg+2 ion.

O atom has six electron in valence shell. It gains 2 electrons to form O-2 ion. For

each Mg+2 ion, one O-2 ion is required to form MgO.

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+

1. Recognize the following compounds as having ionic bounds:

a. KCI b. AICI3 c. MgF2

K is a metal while Cl is a non-metal. K atom has 1e’ in valence shell. It

losses 1e’ to form K+ ion while Cl atom has 7e’ in valence shell. It gains 1e’ to

form Cl- ions. K by losing 1e’ & Cl by gaining 1e’ achieve nearest nobel gas

configuration. Therefore, KCl is an ionic compound.

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• Draw electron cross and dot structures for the following molecules:

a. NH3 that is used to manufacture urea.

b. CCI4 a dry cleansing agent.

c. SiCI4 used to make smoke screen.

d. H2S a poisonous gas.

• N atom has 5e’ in valence shell and each H atom has 1e’ in valence shell.

So, N atom needs 3 electrons to complete its octet. Each H needs 1 electron

to complete its duplet.

• N is the central atom and will form 3 single bond with H atoms.

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• Si atom has 4e’ in valence shell and each Cl atom has 7e’ in valence shell. So,

Si atom needs 4 electrons to complete its octet. Each Cl needs 1 electron to

complete its octet.

• Si is the central atom and will form 4 single covalent bond with Cl atoms.

or

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1. Find the number of valence electrons in the following atoms using the

periodic table:

a. Boron b. Neon c. Rubidium d. Barium e. Arsenic

Boron

Neon

Rubidium

Barium

Arsenic

2. Represent the formation of captions for the following metal atoms using

electron dot structures.

a. AI b. Sr c. Ba

3. Describe the information of anions for the following non-metal atoms:

a. P b. Br c. H

4. Represent the formation of action for the following metal atoms using

electron dot structures.

a. Mg b. Li c. Be

5. For each of the following pairs of atoms, use electron dot and electron

cross structures to write the equation for the formation of ionic

compound.

a. K & CI b. Ca & S c. AI & N

6. Recognize the following compounds as having ionic bonds.

a. MgCI2 b. KBr c. Nal

7. An atom of an element has atomic number 9 and mass number 19.

a. State the number of protons and neutrons in the nucleus of his atom.

b. State the number of electron in this atom.

c. Show with electron cross dot diagrams, the information of ionic in the

reaction of this atom and sodium atom.

Chapter 5

Physical States of Matter

Self Assessment Exercises 5.1:

(1) A student obtained following data in an experiment at 200C.

P (atm) V(dm3)

0.350 0.707

0.551 0.450

0.762 0.325

0.951 0.261

1.210 0.205

Solution:

P (atm) V(dm3) PxV atm.dm3

Result:

The above table shows that an inverse relationship exists between

pressure and volume and product of P and V must be constant if

student carefully collect the data.

P α 1/V

P = 1/V constant

PV=constant

The product of P and V is constant according to Boyle’s law.

(2) Ammonia gas is used as refrigerant 0.474 atm. pressure is

required to change 2000cm3 sample of ammonia initially at 1.0 atm

to 4.22 dm3 at constant temperature. Show that this data satisfies

Boyle’s law.

Data: P1= 1.0dm3

V1 = 4.22 dm3

P2 = 0.474 atm

V2 = 4.22 dm3

Solution:

Result: _____________________________________________________________________

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Self Assessment Exercises 5.2:

(1) A chemist obtained data shown in table in an experiment at 1atm.

Temp (0C) Volume (cm3) Temp (0C) Volume (cm3)

25 117.5 35 121.3

30 119.4 40 123.3

Solution:

Temp (0C ) Volume (cm3) Temp (K) V/T (cm3/K)

Result:

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(2) A bacterial culture isolated from sewage produces 36.4 cm3 of

methane gas at 270C and 760mm Hg. This gas occupies

33.124cm3 at 00C and same pressure. Explain volume –

temperature relationship from this data.

Data:

Solution:

According to Charles’s law

Result:

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(3) A perfect elastic balloon filled with helium gas has a volume

of1.25 x103 dm3 at 1.00 atm and 250C on ascending to a certain

altitude where temperature is 150C the volume of ballon becomes

1.208x103 dm3 .Show that this data satisfies the Charles’s law.

Data:

Solution:

Result:

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Self Assessment Exercises 5.3:

Give reason:

(1) When you put nail polish remover on your palm, you feel a

sensation of coldness.

When we put nail polish remover on our palm, we feel a sensation

of coldness because due to due to evaporation high kinetic energy

molecules vaporize and temperature of remaining molecules falls down.

To compensate this deficiency of energy, the molecules of liquid

remover absorb energy from the surroundings. As a result the

temperature of surroundings decreases and we feel cooling.

(3) Wet clothes dry quickly in summer than in winter.

Wet clothes dry quickly in summer than in winter because the rate

of evaporation is directly proportional to temperature. It increase with

the increase in temperature because of increase in kinetic energy of

molecules .As in summer , the temperature of the surroundings is high

so water molecules present in wet clothes absorb heat from the

surroundings and evaporate easily.

Self Assessment Exercises 5.5:

(1) The boiling point of water on the top of Mount Everest is 700C

while at Murree 980C.Explain this difference.

Boiling point is the temperature at which vapour pressure of a

liquid becomes equal to the external or atmospheric pressure. At Mount

Everest at about 8850m above sea level, atmospheric pressure is only

34kPa.So water boils at this height above sea level, when its vapour

pressure is 34kPa at 700 C.

At Murree the atmospheric pressure is less than 1atm, so vapour

pressure becomes equal to atmospheric pressure at 980C, so water boils

at 980 C.

(2)If you try to cook an egg in boiling water while camping at an

elevation of 0.5 km in the mountain, you will find that it takes

longer than it does at home .Explain why?

At an elevation of 0.5 km in the mountain, the atmospheric

pressure is less than 1 atm, so water boils at low temperature. It means

boiling water has less heat energy so if we try to cook an egg in boiling

water it takes longer time than it does at home.

Self Assessment Exercises 5.6:

Sodium chloride, an ionic compound has a high melting point of

8010C.Wherease molecular solid such as ice has relatively melting

point of 00C .Explain this difference.

Sodium chloride ,an ionic solid has very strong electrostatic force

of attraction between sodium and chloride ions and thus has got a very

high melting point 8010C.Wherease ice is a molecular solid and its

molecules are attracting each other by weak intermolecular forces

,hence it melts at a very low temperature i.e 00C.

Review Questions

Q2: Give short answers.

(iv) Is evaporation a cooling process?

Yes, evaporation is a cooling process. When the high kinetic

energy molecules vaporize from the surface, the temperature of the

remaining molecules falls down. To compensate this deficiency of

energy, the molecules of liquid absorb heat energy from the

surrounding, as a result the temperature of the surrounding decreases

and we feel cooling effect. For example, when we put a drop of alcohol

on our palm, the alcohol evaporates and we feel cooling effect.

(v) Can you make water boil at 700C?

Yes we can you make water boil at 700C by decreasing

atmospheric pressure e.g at mount Everest at about 8850m above sea

level atmospheric pressure is only 34kPa.So water boils at this height

above sea level ,whenits vapour pressure is only 34 kPa at 700C.

(vi)Express the pressure 400mmHg in kPa.

Q9: Plots of vapour pressure versus temperature of four liquids are

given in the following table.

a) Find the boiling point of each liquid when the atmospheric

Pressure is 1atm.

Boiling point of:

• Chloroform: ______________

• Ethanol: ___________________

• Water: ____________________

• Ethanoic acid: ____________

b) At what temperature ethanol will boil when the atmospheric

Pressure is 51kPa.

_________________________________________________________

_________________________________________________________

c) How can you make the water to boil at 800 C?

_________________________________________________________

d) At what temperature chloroform will boil when the external

pressure is 50kPa?

_________________________________________________________

_________________________________________________________

e) Can you boil chloroform at 00C?

_________________________________________________________

f) Predict the boiling point of chloroform at 600mmHg.

_________________________________________________________

_________________________________________________________

Q12: The air in a perfectly elastic balloon occupies 885cm3, during

the fall when the temperature is 200C. During the winter, the

temperature on a particular day is -100C, the balloon occupies

794.39cm3.If the pressure remains constant show that given data

proves the volume-temperature relation according to Charles’s law.

Data:

Solution:

Result:

_________________________________________________________

_________________________________________________________

Q13:In the past the gas volume was used as a way to measure temp

devices called gas thermometer. An experimenter obtains following

data from a gas thermometer.

Volume (dm3) Temp( 0C )

2.7 0

3.7 100

5.7 300

Show that gas thermometer obtains result according to Charles’s

law.

Solution:

Volume (dm3) Temp0C Temp (K) V/T (dm3/K)

2.7 0

3.7 100

5.7 300

Result:

_________________________________________________________

_________________________________________________________

Q14:In automobiles engine the gaseous fuel air mixture enters the

cylinder and is compressed by a moving piston before it is ignited.

If the initial cylinder volume is 990cm3 .The fuel air mixtures

initially has a pressure of 1.0atm and final pressure 11.0atm .Do

you think this change occurs according to the Boyle’s law.

Data:

P1= P2 =

V1= V2 =

Solution: According to Boyle’s law

Result:

_________________________________________________________

_________________________________________________________

Q15: A sample of neon that is used in a neon sign has a volume of

1500cm3 at a pressure of 636 torr. The volume of the gas after it

is pumped into the glass tube of the sign is1213.74cm3. When it

shows a pressure of 786 torr. Show that this data obey Boyle’s

law.

Data: V1 :____________ V2 :______________

P1:____________ P2: ______________

Solution: According to Boyle’s law

Result:

_________________________________________________________

Think Tank

(1)Following table shows data collected from an experimenter by a

Student.

Volume (dm3) Pressure (mmHg)

400 353.5

320 442

240 589

200 707

Do you think that the student collected data carefully or carelessly?

Defend.

Solution:

Volume (dm3) Pressure (mmHg) PV (mmHg .dm3)

Result:

_________________________________________________________

_________________________________________________________

_________________________________________________________

Q2: The water level in an aquarium decrease slowly even though the

Tank does not leak. Interpret it.

The water level in an aquarium decreases slowly because the liquid

molecules constantly change into gas or vapours and this process is

called evaporation.

Q3: Illustrate types of attractive forces you expect the molecules of

HF and HCl.

There are two types of attractive forces present between the

molecules of HF and HCl.

➢ Covalent bonding is present in HF and HCl. They mutually share

their molecules as these compounds are made of non-metals

only.

➢ Intermolecular forces are present between molecules of HF and

HCl due to electro negativity difference.

Q4: Name two substances that are solids at 250C.Name two

substances that are liquid at 250C.

➢ Solid substances: (i)_____________ (ii)_______________

➢ Liquid substances: (i)_____________ (ii)_______________

Q5: Identify the process occurring in each of the followings.

(a)Mothballs slowly disappear:_____________________

(b)A cold windshield become covered with ice when struck by rain

Drops: ________________

Q6: An autoclave is used to sterilize surgical equipment. It is far

more effective to produce steam by autoclave than steam

produced from boiling water in the open atmosphere because

it generates steam at a pressure of two atmosphere. Argue why

an autoclave is such an efficient sterilization device.

The pressure inside an autoclave is 2atm which is higher than

normal atmospheric pressure, so more heat is required to boil

water in an autoclave. The high heat content kills germs more

efficiently than steam produced from boiling water in the

open atmosphere.

Q7: The following table shows the melting points and boiling points

of four substances.

Substances Melting point(0C) Boiling point(0C)

A -123 -79

B -17 58

C 52 305

D -6 120

(a)Write the physical state of each substance at room temperature

and at 1atm.

A:_____________ B:____________ C:___________

D:_____________

(b) Choose substances that exist as a liquid for the longest range of

temperature.

_________________________________________________________

_________________________________________________________

(c) Interpret what will happen to the substance B when it is heated

from 00C to 1000C.

_________________________________________________________

_________________________________________________________

(d)) Interpret what will happen to the substance D when it is cooled

from 1000C to -100C.

_________________________________________________________

_________________________________________________________

Structure of carbon allotropes

Diamond:

Graphite:

Bucky Ball:

Structure of Phosphorous allotropes

White phosphorous:

Red phosphorous:

CHAPTER # 6

SOLUTIONS

SELF ASSESSMENT EXCERCISES

6.1. The maximum amount of sodium acetate that dissolves in 100g of water at 0°C is 119g

and 170g at 100°C.

(a) If you add 170g of sodium acetate in 100g of water at 0°C, how much will dissolve?

Ans.: It will dissolve 119g of sodium acetate in 100g of water at 0°C.

(b) Is the solution saturated, unsaturated or supersaturated?

Ans.: It is a saturated solution.

(c) If the solution is heated to 100°C, is the solution now saturated, unsaturated or

supersaturated?

Ans.: It will be supersaturated solution as it will dissolve all (170g) of solute in it at 100°C.

(d) If the solution is cooled back to 0°C and no crystals appear .Is the solution now

saturated, unsaturated or supersaturated?

Ans.: It will be a supersaturated solution.

6.2.: What are the physical states of solute and solvent in each of the following solutions?

Also identify type of solution.

Ans.:

No. Solute Solvent Type of Solution

a.

b.

c.

d.

e.

f.

g.

6.3. 1. Write four ways to express percentage of solutions.

Ans.: Four ways are:

a) Mass by mass percent c) Volume by mass percent

b) Mass by volume percent d) Volume by volume percent

2. A saline solution is administered intravenously to a person suffering from severe

dehydration. This is labeled as 0.85% m/v of NaCl. What does it mean?

Ans. 0.85% m/v solution of NaCl means 0.85g of NaCl in 99.15 cm3 of water to make 100 cm3

of solution.

6.4. Potassium Chlorate (KClO3) is a white solid. It is used in making matches and dyes.

Calculate the molarity of the solution that contains:

a) 1.5 moles of this compound dissolved in 250 cm3 of solution.

No. of moles of solute = 1.5 moles

Volume of solution = 250 cm3 = 250/1000 = 0.25 dm3

Molarity = No. of moles of solute

dm3 of solution

M = 1.5 moles

0.25 dm3

M = 6M

b) 75g of this compound dissolved to produce 1.25 dm3 of solution.

Mass of solute = 75g

Volume of solution = 1.25dm3

Molarity =?

Molar mass of KClO3 = 39+ 35.5 + 3(16)

= 122.5g/mol

Molarity = Mass of solute x 1

Molar mass of solute dm3 of solution

.

(c). What is the molarity of a 50cm3 sample of potassium chlorate solution that yields 0.25g

residue after evaporation of water.

Vol. of soln.= 50cm3

Mass of solute =0.25g

Molarity = ?

Molar mass of KClO3 = 39 + 35.5 + (16 x 3) = 122.5 gmol-1

6.5. 1. Sodium hydroxide solutions are used to neutralize acids and in the preparation of

soups and rayon. If you dissolve 25g 0f NaOH to make 1dm3 of solution, what is the

molarity of this solution?

Mass of solute =25g

Vol. of soln. =1dm 3

Molar Mass of NaOH = 23+16+1

=40g/mol

Molarity =?

2. A solution of NaOH has concentration 1.2M. Calculate the mass of NaOH in g/dm3in this

solution.

Molarity = 1.2 M

Vol. of soln.= 1dm3

Mass of NaOH = ? Molar mass of NaOH = 23+16+1= 40 gmol-1

Molarity = mass of NaOH x 1 .

Molar mass of NaOH dm3 of soln.

Mass of NaOH = Molarity x molar mass of NaOH x dm3 of soln.

= 1.2M x 40g/mol x 1dm3

Mass of NaOH =48 g

3. A solution is prepared by dissolving 10g of hemoglobin in enough water to make up 1dm3

in volume. Calculate molarity of this solution. Molar mass of hemoglobin is 6.51x104 gmol-1.

Mass of hemoglobin = 10g

Vol. of soln. =1dm3

Molar mass of hemoglobin= 6.51x 104g/mol

Molarity = ?

Molarity = mass of hemoglobin x 1 .

Molar mass of hemoglobin dm3 of soln.

6.6 1. How can you prepare 500cm3 of 0.2M KMnO4 Soln.?

Vol. of soln. =500cm3

Molar mass of KMnO4= 39+55+(16x4)

= 158g/mol

Molarity = mass of KMnO4 x 1000 .

Molar mass of KMnO4 cm3 of solution

Mass of KMnO4 = Molarity x molar mass of KMnO4 x cm3 of soln.

1000

= 0.2M x 158g/mol x500cm3

1000

Mass of KMnO4 =15.8g

a) Weight out 15.8g of KMnO4.

b) Add this solid into a beaker, add some water to dissolve it.

c) Transfer this solution to 500cm3 volumetric flask and add more water.

d) Keep adding water until the volume of the solution rises to the etched line and mix the

solution. This is desired solution.

2. How can you prepare 25cm3 of 0.25m solution of CuSO4.5H2O (Blue Vitriol)?

Vol. of solution =25cm3

Molarity = 0.25m

Molar Mass of CuSO4 .5H2O= 63.5 + 32 + (16×4) + 5(2+16)

=159.5 +90

= 249.5g/mol

a) Weight out 1.55g of CuSO4.5H2O.

b) Add this solid into a beaker; add some water to dissolve it.

c) Transfer this solution to 25cm3 volumetric flask and add more water.

d) Keep adding water until the volume of solution rises to the etched line and mix the

solution. This is desired soln.

6.7 (1). A stock of hydrochloric acid is 12.1M. How many cm3 of this solution should you

use to prepare 500cm3 of 0.1m HCl?

M1=12.1M

V1=?

M2=0.1M

V2=500cm3

Given HCl = Desired HCl

M1V1= M2V2

V1 =M2V2

M1

= 0.1M× 500cm3

12.1 M

V1= 4.13 cm3

Transfer 4.13cm3 of 12.1 M HCl to 500 cm3 volumetric flask and dilute it by adding water up to

the mark and mix. Resulting solution is 0.1M HCl.

2) K2Cr2O7 is a red- orange compound. It is a strong oxidizing agent and is used in the

estimation of iron content in ores. A stock solution is 2.5M K2Cr2O7. How many cm3 of this

solution you need to dilute to make 50cm3 of 0.05M K2Cr2O.

M1= 2.5 M

V1 =?

M2= 0.05M

V2= 50cm3

Given K2Cr2O7 = Desired K2Cr2O7

Transfer 1cm3 of 2.5M K2Cr2O7 to 50cm3 volumetric flasks and dilute it by adding water up to

the mark and mix. Resulting solution is 0.05M K2Cr2O7.

3) Commercial acetic acid is 17.8 molar. How can you convert this into 0.1M acetic acid?

Commercial volumetric flask has volume = 1000cm3

M1=17.8 M

V1=?

M2= 0.1M

V2= 1000cm3

Given acetic acid = Desired acetic acid

Transfer 5.26 cm³ of 17.8 M acetic to 1000cm³ volumetric flasks and dilute it by adding water up

to the mark and mix .resulting solution is 0.1M acetic acid.

6.8. Sodium chloride and glucose both are soluble in water. But the solubility of NaCl is

greater than glucose .Explain why?

Ans.: Glucose and water molecules are polar in nature. Water molecules form H-bonding with

one another as well as with glucose molecules .This means water and glucose molecules have

similar structures and intermolecular forces.

On the other hand NaCl is an ionic compound. When its crystal is added in water it breaks into

ions Na+ and Cl- , the negative end of water molecules are attracted to Na+ and negative end is

attracted to Cl- . These attractive forces are strong enough to overcome the strong attraction that

exist between ions of NaCl, thus NaCl readily dissolves in water.

The attractive forces between Na+ and Cl- ions with water are strong as compared to the

intermolecular forces (H-bonding) present between water and glucose molecules. Hence

solubility of NaCl is greater than glucose.

REVIEW QUESTIONS

2. ( i) differentiate between saturated and unsaturated solution

Saturated Solution Unsaturated Solution

The solution which cannot dissolve more

solute at a particular temperature is called a

saturated solution.

The solution which can dissolve of the

solute at a given temp. is called an

unsaturated solution.

Take 100g of water in a beaker add sugar in

it , go on adding sugar till it starts settling

down at the bottom of the beaker at a

particular temperature . this is saturated

solution

Take 100g water in a beaker. Add a

teaspoon sugar in it. Stir it, the sugar will

dissolve in it. This is unsaturated solution.

It has excess solute settled at the bottom of

beaker.

It does not have any solute at the bottom of

the beaker.

(iii) Can you call colloid a solution?

Ans.: Colloids are heterogeneous mixtures of tiny particles but they looks homogeneous like

solutions. In colloid dispersed phase is like solute in solutions and dispersion medium is like

solvent in solutions.

4. Explain why CHȝOH is soluble in water but C6H6 is not.

Ans.: Methanol readily dissolves in water. Water molecules are polar. Two H-atoms bounded to

an O- atom are slightly positively charged, and O-atom has a slight negative charge. Water

molecules form H-bonds with one another. Methanol molecules are also polar and exhibit H-

bonding. This means water and methanol molecules have similar structures and intermolecular

forces. They can form of H-bonds with each other and are miscible to each other, while C6 H6 is

non-polar in nature, the attraction between water molecules and benzene is very weak, So, it is

insoluble in water.

5- How can you prepare 250cm3 of 0.5M MgSO4 from a stock of 2.5M MgSO4?

M1=2.5M

V1=?

M2=250cm3

V2= 250cm3

Given MgSO4=Req. MgSO4

Transfer 50cm3 of 2.5M to 250cm3 volumetric flask and dilute it by adding water up to the mark

and mix. Resulting solution is 0.5M MgSO4.

6-Copy and complete the following table for aqueous solution of NaOH.

Mass of Solute Moles of Solute Vol. of Solution Molarity

20g A 500cm3 B

C 0.25 D 0.25

E F 200cm3 0.1

(B) =? (Molarity)

M= Given mass of solute x 1000

Molar mass of solute cm3of Sol.

M. mass of NaOH = 23+16+1 = 40g/mol

M = 20g x 1000

40g/mol 500cm3

Molarity = 1M

(A) =? (moles of Solute)

No. of moles of solute = Molarity x dm3 of soln.

‘

Moles =

(D) =? (Vol. of Soln.)

dm3 of solution= no. of moles of solute

Molarity

Vol. of soln. =

(C) = ? (Mass of solute)

Molarity = Given mass of solute x 1 .

M. mass of solute dm3 of soln.

Given mass =

‘ Mass of solute=

(F)= ? ( moles of solute)

Req. vol. of soln. = 200 cm3 = 200 =0.2 dm3

1000

No. of moles = M x dm3 of solution

No. of moles = 0.02 moles

(E)= ? ( mass of solute)

M= Given mass of solute x 1 .

M. mass of solute dm3 of soln.

Given mass=

Mass of solute=

Results:

A = 0.5moles B = 1M C = 10g

D = 1dm3/1000cm3 E = 0.8g F = 0.02 moles

8. What is the molarity of a solution prepared by dissolving 1.254g HCl gas into enough

water to make 30cm3 of solution?

Data:

Molarity=?

Mass of HCl=1.254g

M. Mass of HCl=1+35.5=376.5 gmol-1

Volume of solution=30cm3

Formula used:

Molarity=?

Solution:

M= Mass of HCl x 1000 .

M. Mass of HCl cm3 of solution

9. Formalin is an aqueous solution of formaldehyde (HCHO) used as a preservative for

biological specimens. A biologist wants to prepare 1 dm3 of 11.5M formalin .What mass of

formaldehyde he requires?

Data:

Vol. of solution=1dm3

Molarity of solution=11.5M

M. Mass of formaldehyde=1+12+1+16=3-0g mol-1

Mass of formaldehyde=?

Formula used:

Molarity=Mass of solute x 1……

M. Mass of solute dm3 of soln.

Solution:

10. A solution of Ca (OH)2 is prepared by dissolving 5.2mg of Ca(OH)2 to a total vol. of

1000cm3.Calculate the molarity of this solution?

Data:

Mass of Ca(OH)2=5.2mg

Vol. of soln.=1000cm3

M. mass of Ca(OH)2=40+(16+1)x2=74 g mol-1

Molarity=?

Formula used:

Molarity= mass of Ca(OH)2 x 1000 .

M. mass of Ca(OH)2 cm3 of soln.

Solution:

11. Calculate no. of moles of solute present in 1.25cm3 of 0.5 M of H3PO4 of solution.

Data:

Vol. of solution=1.25cm3=1.25=0.001254dm3

1000

Molarity of solution=-0.54M

Moles of H3PO4=?

Formula applied:

Molarity= Moles of solute

dm3 of solution

Solution:

12. Calculate the new molarity when 100cm3 of water is added to 100cm3 of 0.5M HCl.

M1=-0.5M Given HCl=Desired HCl

V1=100cm3 M1V1=M2V2

M2=? M1V1=M2

V2

V2=100+100=200cm3

THINK TANK

1. 100g of solid solute is placed in 100g of water at 20°C and all of it dissolves. Then another

4.0g of the solute is added at 20°C and all of it dissolves

a. Is the first solution saturated, unsaturated or supersaturated? Interpret

Ans.: The first solution is unsaturated because it dissolves another 4.0g of solute in it, at 20°C.

b. Is it Possible to tell from this information that the final solution is unsaturated or

Saturated? Agree

Ans.: Yes, the solution is unsaturated because it has the capacity to dissolve more of the change.

2. Suggest a Convenient way to change.

a. A Saturated Solution to an unsaturated solution

Ans.: By adding more solvent to the solution to the solution until all of the solute dissolves.

b. An unsaturated solution to a saturated solution

Ans.: By adding more solute to the solution gradually until extra amount of solute settles at the

bottom.

3. Knowing the molarity of a solution is more meaningful then knowing whether a solution

is dilute or concentrated defends this statement.

Ans.: Molarity of a solution is concerned with the exact amount of solute in a solvent while

dilute or concentrated solution do not include exact amount of the solute. Thus, molarity of a

solution is more meaningful then concentration of solution.

4. Design an experiment to determine the solubility of table sugar in water at room

temperature.

Ans.: First of all, prepare a saturated solution of table sugar in 100g of water at room

temperature. Take this solution in a reweighed china dish. Place china on a burner and heat it

slowly till water evaporates completely. Cool china dish and weigh it. Calculate the mass of table

sugar present in it.

5. Design an experiment to prepare 10% mass by volume solution of CuSO4.5H2O

(Nelathota).

Ans.: Mass by volume percent solution of CuSO4.5H2O is the mass of the solute dissolved per

100 parts by volume of a solution. So, 10⁒ m/v CuSO4, 5H2O solution means that 10g

CuSO4.5H2O in 90 cm³ water to make 100 cm³ of solution.

6. Which solution is more dilute 50 cm³ of 0.2 NaOH or 100 cm³ of 0.1M NaOH? Justify.

Ans.: First of all find the no. of moles of moles of both solutions.

Volume = 50cm³ = 50/1000 = 0.05dm³

M = 0.2M

No. of moles = ?

No. of moles = M × dm³ of soln.

= 0.2 × 0.05

= 0.01 moles

Calculate mass of NaOH:

Mass of solute = No. of moles x M. mass

Volume = 100cm³ = 100/1000 = 0.1dm³

M = 0.1M

No. of moles = ?

No. of moles = M × dm³ of soln.

= 0.1 × 0.1

= 0.01 moles

Calculate mass of NaOH:

Mass of solute = No. of moles x M. mass

Both Solutions are having same concentration, as amount of solute is equal.

7. Which solution is more concentration 100cm³ of 0.1M HCl or 100cm³ 0.1M NaOH?

Justify.

Find out the no. of moles of both solutions.

HCl

Volume = 100cm3 = 100/1000=0.1dm3

M=0.1M

No. of moles = M x dm3 of solution

NaOH

Volume = 100cm3 =100/1000=0.1dm3

M=0.1m

No. of moles = M x dm3 of solution

Calculate the mass of both the solution

m. mass of HCl = 36.5 g/mol

Mass of solute = no of moles x M. Mass

m. mass of NaOH = 40 g/mol

Mass of solute = no of moles x M. Mass

Hence 100cm3 of 0.1M NaOH will be more concentrated than 100cm3 of 0.1M HCl.

8. Benzene is a common organic solvent. Its use is now restricted because this can cause

cancer. The recommended limit of expose to benzene is 0.32 mg per dm3 of air. Calculate

the molarity of this solution.

Data:

Molar Mass of C6H6= 0.32 mg= 0.32/1000=0.00032g

Molar Mass of C6H6= (12x6) + (1x6) =78gmol-1

Volume= 1dm3

Formula applied:

Molarity = Mass ofC6H6 x 1 .

M. Mass ofC6H6 dm3 of soln.

Solution:

9. A patient in a hospital is often administered an intravenous (IV) drip containing an

aqueous solution. This aq. Soln. contains 0.85% (mass by volume) of sodium chloride or

5% (mass by volume) of glucose. Calculate the molarity both the solutions.

NaCl Glucose

0.85% m/v of NaCl means:

0.85g of NaCl per 100cm3 of solution.

M. Mass of NaCl= 23+35.5

= 58.5gmol-1

5% m/v of glucose means:

5g of glucose per 100cm3 of soln.

M. Mass of glucose 6= (12x6) + (1x12) +(16x6)

=180gmol-1

Molarity =Mass of solute x 1000 .

M. Mass of solute cm3 of soln.

M=

M =

10. 100cm3 of NaOH solution was heated to complete dryness, 1.5g residue left behind.

What is the molarity of the solution?

Data:

Volume of solution= 100cm3

Mass of NaOH= 1.5g

M. Mass of NaOH= 23+16+1=40gmol-1

Formula Applied:-

Molarity = Mass of solute x 1000 .

M. Mass of solute cm3 of soln.

Solution:

Chapter: 07 ELECTROCHEMISTRY

Self-Assessment Exercise 7.1

Identify elements undergoing oxidation and reduction in the following reactions.

a. N2 + 3H2 → 2NH3

N in N2 gains H-atom, thus N atoms undergo reduction. At the same time

H-atom combine with N-atom and forms NH3, thus H-atom undergoes

oxidation.

b. 2H2 + O2 → 2H2O

___________________________________________________________________________

___________________________________________________________________________

c. Fe2O3 + 3CO → 2Fe + 3CO2

___________________________________________________________________________________________

___________________________________________________________

d. 4Al + 3O2 → 2Al2O3

____________________________________________________________________________________________

___________________________________________________________

Self-Assessment Exercise 7.2

In the following reactions, identify which element is oxidized and which element is

reduced.

a. Mg + S → MgS

Group II A metals form M+2 cations, and that group VI A form X-2 anions.

This means in this reaction each Mg atom loses two electrons to form Mg+2.

So it is oxidized. Each S atom gains two electrons to form S-2. So S is

reduced.

b. 4Na + O2 → 2Na2O

_____________________________________________________________

_____________________________________________________________

c. 2Al + 3Cl2 → 2AlCl3

_____________________________________________________________

_____________________________________________________________

___________________________________________________________

d. Mg + H2 → MgH2

____________________________________________________________

_____________________________________________________________

Self-Assessment Exercise 7.3

Determine the oxidation number of N in NO2.

a. Suppose the oxidation number of N=X

b. There are 2 oxygen atoms, therefore overall oxidation state for oxygen is

2(-2) = -4

X+(-4)= 0

c. The sum of oxidation number must be zero.

X-4=0

X=+4

So, oxidation number of N in NO2 is +4.

Determine the oxidation number of N in HNO3.

__________________________________________________________________

__________________________________________________________________

• ____________________________________________________________

Determine the oxidation number of S in SO2.

__________________________________________________________________

__________________________________________________________________

• _____________________________________________________________

Determine the oxidation number of S in H2SO4.

__________________________________________________________________

__________________________________________________________________

__________________________________________________________________

• _____________________________________________________________

Self-Assessment Exercise 7.4

The torch cell discharges electricity because of an oxidation- reduction

reaction that takes place between zinc and manganese dioxide.

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3

1. Identify the oxidizing and reducing agents in the following reactions.

2. Zn is being oxidized because there is an increase in its oxidation state.

3. Therefore zinc is reducing agent.

4. The reactant manganese is being reduced in MnO2 because the oxidation

number of Mn decreases, so MnO2 is an oxidizing agent.

2. Identify the oxidizing and reducing agents in the following reactions.

a. 2S + Cl2 → S2Cl2

__________________________________________________________________________________________

__________________________________________________________________________________________

__________________________________________________________________________________________

__________________________________________________________________________________________

b. 2Na + Br2 → 2NaBr

__________________________________________________________________________________________

__________________________________________________________________________________________

__________________________________________________________________________________________

__________________________________________________________________________________________

c. H2 + S → H2S

__________________________________________________________________________________________

__________________________________________________________________________________________

__________________________________________________________________________________________

__________________________________________________________________________________________

Self-Assessment Exercise 7.5

Sketch an electrolytic cell for the electrolysis of fused KCl.

Self-Assessment Exercise 7.6

Sketch a voltaic cell labeling the cathode, the anode and the direction of flow

of the electrons. Use the following chemicals:

Silver, Zinc, Silver Nitratre (AgNO3) & Zinc sulphate (ZnSO4).

REVIEW QUESTIONS

Q. 2 Give short answers.

ii. What is the oxidation number of Cr in chromic acid (H2CrO4).

• There are 2 H-atoms, therefore, overall oxidation state for H is 2(+1) =+2

• There are 4 oxygen atoms, therefore overall oxidation state for oxygen is

4(-2) = -8

• Suppose oxidation for Cr is X. The sum of oxidation number must be zero.

• ___________________________________________________________

iii. Identify reducing agent in the following reaction.

CuO + H2 → Cu + H2O

____________________________________________________________

____________________________________________________________

____________________________________________________________

____________________________________________________________

vi. Explain one example from daily life which involves oxidation-

reduction reaction.

Following reaction occurs when we burn sui gas

CH4 + 2O2 → 2H2O + heat

Since C in CH4 loses H-atom and combine with oxygen atoms, thus C

atom under oxidation at the same time O-atoms combine with H-

atoms thus O-atoms undergo reduction.

Q. 13 (c) Distinguish between electrolytic and voltaic cell.

Electrolytic Cell Voltaic Cell

• An electro-chemical cell that

uses electrical energy to

derive a chemical reaction is

called an electrolytic cell.

• An electro-chemical cell that

converts chemical energy

into electricity is called

voltaic or galvanic cell.

• Non-spontaneous redox

reaction takes place in

electrolytic cell.

• Spontaneous redox reaction

takes place in voltaic cell.

• Electricity is required to

carry out chemical reactions.

• Chemical reaction takes

place in order to generate

electricity.

• Downs Cell

• Nelson Cell

• Galvanic Cell

• Battery

Q. 14 State the substances which are oxidized or reduced. Give

reasons for your answer.

a. N2 + 3H2 → 2NH3

• N-atoms combine with H-atoms to from NH3, thus N-atoms

undergo reduction as reduction is the gain of hydrogen or gain of

electrons.

• Hydrogen is oxidized because oxidation is the loss of electrons.

b. CO2 + 2Mg → 2MgO + C

________________________________________________________

___________________________________________________________

c. Mg + H2O → MgO + H2

d. H2S + Cl2 → 2HCl + S

__________________________________________________________

__________________________________________________________

_________________________________________________________

e. 2NH3 + 3CuO → 3Cu + N2 +3H2O

__________________________________________________________

__________________________________________________________

__________________________________________________________

Q. 15 (b) Find the oxidation state of Nitrogen in the following

compounds.

i. NO2

• Suppose the oxidation number of N = X

• There is one N-atom, therefore overall oxidation state for N is X.

• The oxidation state for O is -2.

• The sum of Oxidation number must be zero.

• ____________________________________________________

ii. N2O

• Suppose the oxidation number of N = X

• There are two N-atoms, therefore overall oxidation state for N is

2X.

• The oxidation state for O is -2.

• The sum of Oxidation number must be zero.

• ____________________________________________________

iii. N2O3

• Suppose the oxidation number of N = X

• There are two N-atoms, therefore overall oxidation state for N is

2X.

• The oxidation state for O is -2.Since there are 3 O-atoms so,

overall oxidation state of O is -6.

• The sum of Oxidation number must be zero.

• ____________________________________________________

iv. HNO3

• Suppose the oxidation number of N = X

• The oxidation state for H = +1.

• There are two N-atoms, therefore overall oxidation state for N is

2X.

• The oxidation state for O is -2.Since there are 3 O-atoms so,

overall oxidation state of O is -6.

• The sum of Oxidation number must be zero.

• ____________________________________________________

Q. 16 Find the oxidation state for S in the following compounds.

a. H2S

• Suppose the oxidation number of S = X

• The oxidation state of H is +1. Since there are two H-atoms,

therefore overall oxidation state for H is +2.

• The sum of Oxidation number must be zero.

• ____________________________________________________________________________

b. H2SO3

• Suppose the oxidation number of S = X

• The oxidation state of H is +1. Since there are two H-atoms,

therefore overall oxidation state for H is +2.

• The oxidation state of oxygen in O3 is -6.

• The sum of Oxidation number must be zero.

• ____________________________________________________________________________

c. Na2S2O3

• Suppose the oxidation number of S = X

• The oxidation state of Na in Na2 is +2.

• The oxidation state of oxygen in O3 is -6.

• The sum of Oxidation number must be zero.

• ____________________________________________________________________________

Q. 17 (b) Identifying the oxidizing and reducing agents in the following

reaction.

a. H2S + Cl2 → 2HCl + S

__________________________________________________________

__________________________________________________________

_________________________________________________________

b. 2FeCl2 + Cl2 → 2FeCl3

__________________________________________________________

__________________________________________________________

_________________________________________________________

c. 2KI + Cl2 → 2KCl + I2

__________________________________________________________

__________________________________________________________

_________________________________________________________

d. Mg + 2HCl → MgCl2 + H2

__________________________________________________________

__________________________________________________________

_________________________________________________________

Q. 18 Hydrogen peroxide reacts with silver oxide and lead (II) sulphide

according t the following equations.

i. H2O2 + Ag2O → 2Ag + H2O + O2

ii. 4H2O2 + PbS → PbSO4 + 4H2O

Is hydrogen peroxide an oxidizing or reducing agent in these

reactions? Argue

i. H2O2 + Ag2O → 2Ag + H2O + O2

• Oxygen is being oxidized because there is an increase in its

oxidation state.

• Silver is reduced because there is a decrease in its oxidation

state.

• The reactant H2O2 contains oxygen that is being oxidized so

H2O2 is reducing agent.

• The reactant Ag2O contains Ag that is being reduced so Ag2O

is an oxidizing agent.

ii. 4H2O2 + PbS → PbSO4 + 4H2O

• Oxygen is being reduced because there is a decrease in its

oxidation state.

• Sulphur is oxidized because there is an increase in its oxidation

state.

• The reactant H2O2 contains oxygen that is being reduced so

H2O2 is oxidizing agent.

• The reactant PbS contains Ag that is being oxidized so PbS is a

reducing agent.

THINK TANK Q. 1: What materials do you need to electroplate copper onto an iron

nail? Construct a diagram showing how these materials should be

arranged.

Q. 5: Would you expect corrosion to occur more rapidly in a desert or

in a rainforest? Justify you stand.

Corrosion occurs more rapidly in rain forest because corrosion

depends upon the amount of moisture in the atmosphere as there is more

moisture in the rainforest.

Chapter 8

Chemical reactivity Self-Assessment exercises

8.1. (1) In each of the following pairs identify, which is more

metallic?

(a) Be , B

Be is more metallic because both elements are present in the

same period i.e 2nd and electro positivity decreases from left

to right in the period so Be lies before B in the second period

so it is more metallic than B .

(b) Si , Al

________________________________________________________

(b) K, Li

(2)Using the periodic table, rank each set of elements in order of

decreasing metallic character.

(a) Na, Li, K

__________________________________________

(b) Al, Na , Mg

__________________________________________

(3)Which is more basic Li2O or Na2O? Argue.

___________________________________________________________

___________________________________________________________

8.2: Identify the position of Mg and Ca in periodic table.

Electronic configuration of Mg =

Valence shell electronic configuration =

Period number=

Group number =

Mg is present in _______ period of group ___________.

Electronic configuration of Ca =

Valence shell electronic configuration =

Period number=

Group number =

Ca is present in ________period of group __________.

Review Questions

Q2 (ii) Which element is more metallic Mg or Al? Explain.

Mg and Al belongs to same period .Electropositive character of

element decreases across period and Mg comes after Al in same

period so Mg is more metallic than Al.

(iv) Arrange the followings in order of increasing acidic strength.

HF, HI ,HBr , HCl

_______________________________________________________

(v) Can , F2 oxidize all the halide ions to free halogens?

Yes, F2 oxidize all the halide ions to free halogens

Oxidizing power of halogens is the highest.Due to relative

strength as oxidizing agent , it is possible for a free halogen to

oxidize the ion of halogen next to it in the group.

Q3: Arrange the following oxides in order of decreasing basic

Character.

BeO ,CaO , MgO , SrO

__________________________________________________________

Q4: Rank the each set of elements in order of increasing metallic

Character.

(a) Al , Na , Mg

________________________________________________________

(b) Na , Li , K

________________________________________________________

Q7. What is a half reaction?

Daniel cell consists of two half cell joined in series.Oxidation

half reaction occurs at anode and reduction half reaction occurs

at cathode .Such oxidation and reduction reactions are called

half cell reactions.

Think Tank

Q1.How do electrons participate in half reaction?

Oxidation half Reaction occurs at anode and reduction half

reaction occurs at cathode .Such oxidation and reduction reactions are

called half cell reaction. Electrons flow from anode through the external

wire to cathode.

Q5. An iron bar is silver plated which metal would be anode and

which is cathode?

Silver will act as anode and iron will act as cathode.

Q7: Compare and contrast alkali and alkaline earth metals.

Alkali metals

➢ The group I-A elements

except hydrogen are known

as alkali metals.

➢ These elements have general

electronic configuration ns1

in their valence shell.

➢ Alkali metals lose one

electron to complete their

outermost shell.

➢ Alkali metals are greater in

size than alkaline earth

metals.

➢ Alkali metals have low value

of ionization energy than

alkaline earth metals.

Alkalineearth metals

➢ The group II-A elements are

known as alkali metals.

➢ These elements have general

electronic configuration ns2

in their valence shell.

➢ Alkaline earth metals lose

two electrons to complete

their outermost shell.

➢ Alkaline earth metals are

smaller in size than alkali

metals.

➢ Alkaline earth metals have

high value of ionization

energy than alkaline earth

metals.

Q: Choose three metals that would not be suitable for coinage.

Elaborate your answer.

➢ Sodium would not be suitable for coinage as it is extremely

reactive metal.It readily reacts with water to form sodium

hydroxide and liberates hydrogen gas. This reaction is highly

exothermic. For this reason Na is not suitable for coinage.

➢ Iron would not be suitable for coinage because it reacts with

oxygen and moisture in the atmosphere and gets corrode.

➢ Calcium would not be suitable for coinage because reacts with

oxygen at room temperature.

Ca+ O2 → 2CaO