Chapter : 01 Fundamentals Of Chemistry
Transcript of Chapter : 01 Fundamentals Of Chemistry
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
____________________________________________________________
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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
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c. Mg + H2O → MgO + H2
d. H2S + Cl2 → 2HCl + S
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e. 2NH3 + 3CuO → 3Cu + N2 +3H2O
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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
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b. 2FeCl2 + Cl2 → 2FeCl3
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c. 2KI + Cl2 → 2KCl + I2
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d. Mg + 2HCl → MgCl2 + H2
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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
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(b) K, Li
(2)Using the periodic table, rank each set of elements in order of
decreasing metallic character.
(a) Na, Li, K
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(b) Al, Na , Mg
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(3)Which is more basic Li2O or Na2O? Argue.
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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
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(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
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Q4: Rank the each set of elements in order of increasing metallic
Character.
(a) Al , Na , Mg
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(b) Na , Li , K
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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