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Wednesday 7 September 2016
Assignments Class 12 (Chapter 1 to 3)
ASSIGNMENTS
CHAPTER 1- SOLID
STATE
Level
one
Q.1. Why is glass considered
a super cooled liquid?
Q.2 .Why do solids have a
definite volume?
Q.3. Classify the following
as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic
Acid, Teflon, potassium
nitrate, cellophane, polyvinyl chloride, fiber glass, copper.
Q.4. what type of solids are
electrical conductors, malleable and ductile?
Q.5. Name the parameters that
characterize a unit cell.
Q.6. What is meant by the
term 'coordination number'?
Q.7. What is the coordination
number of atoms:
a) In a cubic close-packed
structure?
b) In a body-centred cubic
structure?
Level Two
Q.1. Solid A is a very hard
electrical insulator in solid as well as in molten state and melts at
Extremely high temperature. What type
of solid is it?
Q.2. Ionic solids conduct
electricity in molten state but not in solid state. Explain.
Q.3. Classify the following
as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid,
Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiber glass, copper.
Q.4. Define the term
amorphous. Give a few examples of amorphous solids.
Level
three
Q.5. What makes a glass
different from a solid such as quartz? Under what conditions quartz could be converted into glass?
CHAPTER 2-SOLUTIONS
Level One
Q1.What is meant by
semi-molar and deci-molar solutions?
Q2.What will be the mole
fraction of water in ethanol solution containing equal number of moles of water
and ethanol?
Q3.State Roult’s Law in its
general form in reference to solutions?
Q4.How Is colligative
property of the solution when a solute in a solution undergoes:
(i)association
(ii)dissociation ?
Q5.Give one practical
application of depression of freezing point in automobiles.
Q6.Mention one large scale use of the
phenomenon called “Reverse Osmosis”.
Q7.Which one will have
greater boiling point 0.1 M NaCl or 0.1 BaCl2 solutions
in water?
Q8.What will happen to the
vapour pressure of water, if a tablespoon of sugar is added to it?
Q9.Why does the use of
pressure cooker reduce the cooking time?
Q10. The dissolution of
ammonium chloride in water is an endothermic process .What is the effect of
temperature on its solubility?
Q11. Why is liquid ammonia
bottle first cooled in ice before opening it?
LevelTwo
Q1. Out of three solutions
(i) 1M cane sugar solution (ii) 1M NaCl solution (iii) 1M BaCl2 solution,
which one will have
(a)highest boiling point (b)highest freezing point (c)lowest osmotic pressure?
Q2.(a)For which of the
following, the van Hoff factor is not greater than 1?
NaNO3 , BaCl2 , K3Fe(CN)6 , C6H12O6
(b)If glycerine and methyl
alcohol are sold at the same price per kg, which one would be preferred for
use, as an anti-freeze in the car radiators?
Q3.Define Osmotic Pressure.
Arrange the following solutions in increasing order of their osmotic pressure.
urea,sucrose,NaCl,glucose.(if 1g substance is added).
Q4.What is the effect of the
following changes on the osmotic pressure of a solution containing a non-volatile
solute:
(i)Solvent is added to the
solution
(ii)Solute is added to the
solution
(iii)Temperature is raised ?
Q5.Carbon Tetrachloride and
Water are immiscible whereas, ethanol and water are miscible in all proportions.
Co-relate this behavior with molecular structures of these compounds.
Q6.Give reason, why at higher
altitudes, people suffer from a disease called Anoxia.
In this disease, they become
weak and cannot think clearly.
Level Three
Q1.Why do aquatic species
feel more comfortable in lakes in winters than in summers?
Q2. Why is a person suffering
from high blood pressure is adviced to take minimum quantity of
common salt?
Q3. If Kf for water is 1.86˚C/m,
explain why 1 m NaCl in water does not have a freezing point equal
to : (i) -1.86˚C (ii) -3.72˚C
Q4.Derive a relationship
between relative lowering of the vapour pressure and the mole fraction of the
volatile liquid.
Q5. Why dissolution of some
solid compounds is exothermic while that of some others is endothermic?
CHAPTER 3 ELCTROCHEMISTRY
Level One
Q 1 What is the effect of
temperature on molar conductivity?
Q 2 What are the units of
molar conductivity?
Q 3 Name the factors on which
emf of a cell depends.
Q 4 What is the emf of the
cell when the cell reaction attains equilibrium?
Q 5 What is the electrolyte
used in dry cell?
Q 6 How is the cell constant
calculated from conductance value?
Q 7 What flows in the
internal circuit of a galvanic cell?
Q 8 Define electrochemical
series.
Level Two
Q 1 How can you increase
reduction potential of an electrode for the reaction :
Mn+ aq + ne− ⟶ M(s) ?
Q 2 What happens when a piece
of copper is added to (a) an aq solution of FeSO4
(b) aq solution of FeCl3
Q 3 Define corrosion. Write
chemical formula of rust.
Q 4 State Kohlrausch Law. How
do you determine molar conductivity of weak electrolyte at infinite dilution?
Q 5 What is fuel cell? Write
its electrode reaction.
Q 6 Electrolysis OF KBr give
bromine at anode, but KF does not give F2 .
Level Three
Q 1 Write three differences
between potential difference and emf.
Q 2 How many grams of sodium
will be deposited at cathode by passing 10 A current for 10 s
through the electrolysis of
NaCl.
Q 3 The resistance of a
conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ξ©. What is the
cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1.
Q 4 Calculate maximum work
done that can be obtained from the following cell under standard
conditions at 250C
ππ π | ππ2+ ∥ πΆπ’2+ ππ |πΆ(π )
πΈ0πΆπ’2+/πΆπ’= 0.34 π£, πΈ0ππ2+//ππ= −0.76 π£
ASSIGNMENTS
(Answers)
CHAPTER
1- SOLID STATE
Level one
Ans.1. Glass is an amorphous
solids, it has tendency to flow but very slowly. This is the cause that glass is
considered as super cooled liquid.
Ans.2. The intermolecular
force of attraction make the particles of solid closely packed and force them
To only oscillate at their
fixed positions. These give solids a definite volume.
Ans.3.Polyurethane, Teflon,
cellophane, polyvinyl chloride, fiber glass – Amorphous solids
Naphthalene, benzoic acid,
potassium nitrate, copper – Crystalline solids.
Ans.4. Metallic solids are
conductor of electricity, malleable and ductile.
Ans.5.Unit cells are
characterize on six parameters – dimensions along three edges and three angles between their edges, i.e. a, b, c which are edges and Ξ±, Ξ² and Ξ³ which are angles between the edges.
Ans.6. Coordination number: -
it is the total number of surrounding atoms of any constituent particle present
in the crystal lattice.
Ans.7. The coordination
number of atoms
a) In a cubic close-packed
structure is 12
b) In a body-centred cubic
structure is 8
Level Two
Ans.1.Given
solid ‘A’ is a covalent solids, such as diamond.
Ans.2. Ionic solids conduct
electricity because of movement of their ions. In solid state ions present in ionic
solids do not move hence do not conduct electricity while in molten state ions
can move and thus conduct electricity.
Ans.3.Polyurethane, Teflon,
cellophane, polyvinyl chloride, fiber glass – Amorphous solids
Naphthalene, benzoic acid,
potassium nitrate, copper – Crystalline solids.
Ans.4.
Amorphous
solids exhibits following properties: -
• Their
constituent particles are of irregular shapes and have short range order.
• They
are isotropic in nature and melt over a range of temperature. Because of which
they
are also known as pseudo solids or
super cooled liquids.
• When
cut with a sharp-edged tool, they cut into two pieces with irregular surfaces.
• They do
not have definite heat of fusion.
• Examples
- glass, rubber, plastic etc.
Level three
Ans.5.
The
arrangement of the constituent particles is different in glass & quartz. In
glass, the
Constituent
particles have short range order whereas in quartz, the constituent particles have both long range as well as
short range orders. We can convert Quartz into glass by heating and then cooling it rapidly.
CHAPTER
2-SOLUTIONS
Level One
Ans.1.Semi molar means
M/2,i.e.,half mole of the solute is present in 1 liter of the solution.
Dec molar means M/10,i.e.,one-tenth mole of
the solute is present in 1 liter of the solution.
Ans.2. 0.5
Ans.3. The relative lowering of the vapour pressure
is equal to the mole fraction of the solute.
Ans.4. (i) Value of
colligative property decreases.
(ii)Value of colligative property
increases.
Ans.5. The use of anti-freeze
(ethylene glycol) in car radiators.
Ans.6. In the Desalination of
Sea water.
Ans.7. 0.1 M BaCl2.
Ans.8. The vapour pressure of
water will decrease as the addition of a non-volatile solute lowers the vapour pressure of the solvent.
Ans.9.At higher pressure over
the liquid (due to increase in pressure of cooker ,the liquid boils at higher
temperature. Therefore, cooking occurs faster.
Ans.10. Since the dissolution
of NH4Cl in water
is an endothermic process, its solubility increases with the rise in
temperature (Le-Chatelier principle)
Ans.11. At room temperature
the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure
decreases. Therefore, the liquid ammonia will not splash out.
Level Two
Ans.1. (a) highest boiling
point -1M BaCl2 solution
(b) Highest freezing point 1M cane
sugar solution
(c) Lowest osmotic pressure -1M cane
sugar solution
Ans.2. (a) C6H12O6
(b) Methyl alcohol
Ans.3. Osmotic Pressure is
defined as the hydrostatic pressure built up on the solution which, just, stops
the flow of solvent.
Sucrose < Glucose < Urea
< NaCl
Ans.4. (i) Decreases
(ii)Increases
(iii) Increases
Ans.5. CCl4 does
not form hydrogen bond with water, while ethyl alcohol forms hydrogen bond with
water.
Ans.6. At higher altitudes,
the partial pressure of oxygen is less than that at the ground level. This leads
to low concentration of oxygen in the blood and the tissues of the people
living in the higher altitudes. As a result of low oxygen in blood, the people
become weak and are unable to think clearly. These are the symptoms of the
condition, known as Anoxia.
Level
Three
Ans.1. Aquatic species
require dissolved O2 for breathing. As solubility of gases decreases with
increase in temperature, less
oxygen is available in summer than winter. Hence, they feel more
comfortable in winters (low
temperature) when the solubility of O2 is
higher.
Ans.2. Osmotic pressure is
directly proportional to the concentration of the solutes. Our body fluids
contain a number of solutes.
If a person takes more salt, the concentration of Na+ and Cl- ions
entering the body fluid ,
raises the concentration of the solutes. As a result, osmotic pressure
increases which may rupture
the blood cells.
Ans.3.
(i) There are two moles if
ions per mol of NaCl. Therefore, ΞTf cannot
be equal to 1.86 and freezing point cannot be equal to 0 - 1.86 = -1.86.
(ii) ΞTf cannot
be equal to 2×1.86 = 3.72 and freezing point = -3.72 because there are strong
interionic attractions for 1m
solution so that the degree of ionization is not 100% at the freezing
point . The value of I is
somewhat less than 2.
Ans.4. According to Roult’s Law , (po -ps )/ po =x2 (mole
fraction of the non-volatile solute)
Mole fraction of the volatile
liquid, x1 =1- x2 or x2 =1- x1 therefore,
(po
-ps )/ po=1- x1 or x1=1- (po -ps) / po = ps/ po
Ans.5. If solute-solvent
interactions are stronger than the solute- solute or solvent- solvent
interactions, energy
is released and the
dissolution is exothermic. And if solute-solvent interactions are weaker than
the solutesolute
or solvent- solvent
interactions, energy is absorbed and the dissolution is endothermic.
CHAPTER
3 ELCTROCHEMISTRY
Level One
Ans.1. Molar conductivity of
an electrolyte increases with increase in temperature.
Ans.2. cm2ohm-1mol-1 or S Cm2mol-1
Ans.3. nature of reactants,
concentration of solution, temperature, pressure of gas.
Ans.4. zero
Ans.5. a paste of NH4Cl, MnO2 and
Carbon
Ans.6. cell
constant= Specific conductance/ Observed conductance
Ans.8.The arrangement of
various electrodes in the decreasing or increasing order of their standard reduction
potential.
Level
Two
Ans.1. (i) Increase in
concentration of Mn+ ions in solution.
(ii) By increasing the temperature.
Ans.2. (a) nothing will
happen when the piece of copper is added to FeSO4 solution
because reduction potential of copper is more than reduction potential of iron.
(b) Copper will dissolve in
aq. Solution of FeCl3 because reduction potential of iron is more than
reduction potential of copper.
Ans.3. Corrosion is a process of determination of metal as a result of its reaction with air and water and surroundings. It is due to formation of sulphide, oxide, carbonate and hydroxide. Formula of rust is Fe2O3.xH2O
Ans.4.Molar conductivity of
strong electrolytes at infinite dilution is the sum of two values one
depending upon cation and
other on the anion. Using Kohlraush law we can determine molar
conductivity of weak
electrolyte at infinite dilution.
Ans.5. The device which
converts chemical energy of fuel into electrical energy is called fuel cell.
2π»2 + 4ππ»− ⟶ 4π»2π + 4π
2π»2π + 2π»2 + 4π ⟶ 4ππ»−
__________________________
2π»2 + π2 ⟶ 22π»2π
Ans.6. Oxidation takes place
at anode. Now higher the oxidation potential easier to oxidise. Oxidation
potential of π΅π−, π»2π, πΉ− are in
the following order.
π΅π− > π»2π > πΉ−
Therefore in aq. Solution of
KBr gives Br2 at anode, but KF give oxygen O2 instead
of F2.
Level
Three
Ans.1. E.M.F POTENTIAL
DIFFERENCE
(i)Potential difference of
two electrodes when no current is flowing through
circuit. Potential difference
between electrode in closed circuit.
(ii)It is maximum voltage
obtained from a cell. It is less than maximum voltage obtained
from cell.
(iii)It is responsible for
steady flow of current. It is not responsible for steady flow of
current.
Ans.2. m = zct
= 23×10×10/96500
= 23/965 g sodium
= 0.024 g sodium (approx.)
Ans.3. Conductivity = 1/R ×
cell constant
cell constant = R × conductivity
= 1500 ×
0.146 × 10-3 cm-1
Ans.4 0ππππ = πΈ0πΆπ’ 2+πΆπ’− πΈ0πΆπ’ 2+πΆπ’
= 0.34 – ( – 0.76)
= 0.34 + 0.76
= 1.10 v
△ πΊ0 = −ππΉπΈ0ππππ
= −2 × 96500 × 1.10 = −212300 π½
= −212.3 ππ½
Tuesday 30 August 2016
Solid State Question Bank
QUESTION BANK OF SOLID STATE
LEVEL 1
1.
How does
amorphous silica differ from quartz?
2.
Why glass
is called supper cooled liquids?
3.
Some of
the very old glass objects appear slightly milky instead of being transparent
why?
4.
What is
anisotropy?
5.
What is
the coordination number of atoms?a) in fcc structure b) in bcc structure.
6.
How many
lattice points are there in one cell of a) fcc b) bcc c) simple cubic
7.
What are
the co‐ordination numbers of octahedral voids and tetrahedral
voids?.
8.
Why common
salt is sometimes yellow instead of being of being pure white?
9.
A compound
is formed by two elements X and Y. The element Y forms ccp and atoms of X
occupyoctahedral voids. What is formula of the compound?
LEVEL 2
1.
Explain
how electrical neutrality is maintained in compounds showing Frenkel and
Schottkydefect.
2.
Calculate
the number of atoms in a cubic unit cell having one atom on each corner and
twoatoms on each body diagonal.
3.
Gold
crystallizes in an FCC unit cell. What is the length of a side of the cell(r=0.144mm)?
4. Classify each of the following as
either a p‐type or n‐type semi‐conductor.
a) Ge doped with In (b) B doped with
Si
5.
In terms
of band theory what is the difference between a conductor, an insulator and a
semiconductor?
6.
The
electrical conductivity of a metal decreases with rise in temperature while
that of a semiconductorincreases. Explain.
7.
What type
of substances would make better permanent magnets, ferromagnetic
orferromagnetic, why?
8.
Zinc oxide
is white but it turns yellow on heating. Explain?
9.
In
compound atoms of element Y forms ccp lattice and those of element X occupy
2/3rd oftetrahedral voids. What is the formula of the compound?
10.
AgCl is
doped with 10–2mol% of CdCl2, find the concentration of
cationvacancies.
Level Three Hots
Q-1 CsCl have bcc
arrangement and its unit cell edge length is 400 pm.
Calculate interionic distance.
Q-2 Analysis show
that Ni Oxide has formula Ni 0.98 O 1.8 What fractions
of Ni exist as Ni2+ and Ni3+ ions?
Q-3 on terms of
band theory what is the difference
(i) between
conductor and insulators.
(ii) between
conductor and semi-conductors.
Q-4. KF has NaCl structure calculate the edge
length of its unit cell if its
dencity is 2.48 cm-3.
molar mass of KF = 58.1
g mol-1,
NA = 6.02 x 1023 mol-1.
Q5. In the cubic crystal of CSCl d = 3.97g/cm3.
the eight corners are
occupied by Cl- with Cs-
at centre and vice versa .Calculate the distance
between Cs+ and Cl-
ions. What is the radius ratio of two
ions-. (atmass of Cs =
132.9 and Cl = 35.46.
VALUE-BASED QUESTIONS
Q.1 A goldsmith under training was asked
by his boss to make bangles of gold. He used a bar of 24carat for this purpose
, The bangles made by him were so soft that they could be easily bent. The boss
told him not to use pure gold but rather 22 carat gold. The bangles were now
quite hard and could not be bent on applying some force?
(a)
What was the mistake commited by goldsmith?
(b)
Why did boss ask him to use 22 carat gold?
(c)
In what way is 22 carat gold better than 24 carat gold?
(d)
What is the value associated with this?
Q.2
A student
working in the laboratory passed electric current through a rod of copper as
well as solid copper sulphate crystals . To his surprise, the current could pass
through copper rod connected in a cell and not through the crystals .
(a)
was his observation correct?
(b)
what was the reson behind this observation?
(c)
how is this behaviour helpful in explaning conducting of electrolytes?
Q.3 In India , there is shortage of
electricity . people have to face big cuts particulary when the summer is in
peak . Industry has to suffer a lot . conventional sources of electricity such
as hydel power, thermal power and even nuclear power etc.fail to meet the ever
growing demand of electricity . now a days , there is emphasis on the use of solar energy . photovoltaic material can
convert solar energy into electrical energy.
(a)
what are the drawbacks of traditional source of energy?
(b)
how is solar energy better than conventional source of energy
(c)
why is solar power not very popular in india?
(d)
what is the value associated with the use of solar energy?
Q.4
Daimond
and graphite are two crystalline allotropic forms of carbon . however , they
differ in most of their physical characteristis. for example , diamond is very
hard while graphite is quite soft. Daimond is a poor conductor of e.lectricity
. on the other hand , graphite is a good conductor . Answer the following
questions.
(a)why
is diamond hard while graphite is soft?
(b)why
is diamond a poor conductor of
electricity while graphite is a good conductor?
(c)
why are certain diamonds very shining?
(d)
what is the value associated with the diamonds?
Q.5 Sudanshu made a model of
the unit cell of diamond. It resembled the unit cell of ZnS. If the unit cell
of ZnS has 4 units of ZnS per unit cell. It has the same packing efficiency as
ZnS. But diamond is the hardest known substance.
(a)What is the number of atoms of carbon per unit cell of diamond?
(b)What is the value that Sudanshu can derive from these facts?
(c)What is the value that Sudanshu can derive from these facts?
QUESTION
BANK - SOLID STATE (ANSWERS)
LEVEL 1.
Ans1. In amorphous silica, SiO4
tetrahedral are randomly joined to each other whereas in quartz they
arelinked in a regular manner.
Ans2.Schottky defect.
Ans3. It has tendency to flow like
liquid.
Ans4. Due to crystallization
Ans5. Physical properties show
different values when measured along different directions in crystalline solids
Ans6. a) 12 b) 8
Ans7. a) 14 b) 9 c) 8
Ans8. 6 and 4
respectively
Ans9. Due to the presence of electrons in some lattice sites
in place of anions these sites act as F‐centers.These electrons when excited
impart color to the crystal.
Ans10. No. of Y atoms be N No. of
octahedral voids N
No. of X atoms be =N Formula XY
LEVEL 2.
Ans1. In compound showing Frenkel
defect, ions just get displaced within the lattice. While incompounds showing
Schottky defect, equal number of anions and Cations are removed fromthe
lattice. Thus, electrical neutrality is maintained in both cases.
Ans2. 8 corner *1/8 atom per unit cell = 1atom
There are four body diagonals in a
cubic unit cell and each has two body centre atoms.
So 4*2=8 atoms therefore total
number of atoms per unit cell =1+8=9
Ans3. r=0.144nm
a=2*√2r
=2*1.414*0.144nm
=0.407nm
Ans4. (a) Ge is group 14 elements
and In is group 13 element. Therefore, an electron deficit holeiscreated. Thus
semi‐conductor is p‐type.
(b) Since B is group 13 element and
Si is group 14 elements, there will be a free electron, thus it is n type semi
conductor.
Ans5. The energy gap between the
valence band and conduction band in an insulator is very large while in a
conductor, the energy gap is very small or there is overlapping between valence
band andconductionband.
Ans6. In metals with increase of
temperature, the kernels start vibrating and thus offer resistance to the flow
of electrons. Hence conductivity decreases. In case of semi‐conductors, with increase oftemperature, more electrons can
shift from valence band to conduction band. Hence conductivity increases.
Ans7.. Ferromagnetic substances make
better permanent magnets. This is because the metal ions of a ferromagnetic
substance are grouped into small regions called domains. Each domain acts as
tiny magnet and get oriented in the direction of magnetic field in which it is
placed. This persists even in the absence of magnetic field.
Ans8. When ZnO is heated it looses
electron and which occupies the
interstitial sites . This electron absorbs the radiation from the
visible region and transmitted radiations and reflects the colour.
Ans9.
No. of Y atoms per unit cell in ccp
lattice=4
No. of tetrahedral voids= 2*4=8
No. of tetrahedral voids occupied by
X= 2/3*8=16/3
Therefore formula of the compound =X16/3
Y4
=X16 Y12
=X4 Y3
Ans10. Doped atom = 10–2mol%
of CdCl2
= 10–2 x 6.022 X 1023 atoms.
One crystal of CdCl2
create one cation vacancy when doped with AgCl
No of cation vacancies = 6.022 X 1021
LEVEL 3.
Ans1 Face Diagonal = √a2
+a2 = √ 2 .a
Body Diagonal = √(√2a)2 + a2
Body Diagonal = √3 . a .
Body Diagonal= √3 x 400 = 1.73x400
Body Diagonal = 692
Interior Distance = ½
x Body Diagonal = ½ x 692 = 346pm.
Ans 2. Ni 2+
ions present the ratio of Ni : O is 1:1
i.e, 100Ni2+ ion should
have 100 O2-. Now replace 3Ni2+ ions
2Ni3+ ions are required it
means presence of two Ni3+ would reduce one No. of Ni out of 100
numbers.In Ni 0.98 O 1.00 ratio 98:100. It means two Ni ions are lesser
than oxide ions ( out of 100 atoms) Then four Ni3+
ions are present
and 96 Ni2+ are present for each 100 oxide ions.
Thus % of Ni2+ is
96% and Ni3+ is 4%.
Ans3.
(i) in conductor the
energy bands are very close overlapped by the
conduction bands. Due to this reason e- can flow
very easily from
valency band to conductor band under an electric field,
thereby
showing conductivity.
ii) In semiconductor the gap between valence band and conduction bond is small which make some of the e- enable to jump from valences band to conduction band. Therefore conductance is observed with the increace in temp. more e- jump. Hence the electrical conductivity of semiconductor increase with the increase in temperature.
Ans4. For Nacl
d = 2xM/a3 x NA
a3 = 2 x M /d x NA
a3 = 4 x 581/2.48 x 6.02 x 10-3
= 155.659 x 10-24
a = 5.379 x 10-8 cm
= 537.9 p.m.
Ans5. In one unit cell there is one CsCl 1/8 x 8 Cl- + 1 Cs +
d = z x M/a3 x Noa3 = 2 x M / d x No
a3 = volume , Z = 1m = 132.9 + 35.46 = 168.36
f = 3.97 g cm-3No = 6.02 x 1023.
a3= 1 x 168.36/6.02 x 1023 x 3.97 = 7.04 x 10-23 cm3
a = (7.04 x 10-23)1/3a = 4.13 x 10-8 cm = 413 pm.
Hence the diagonals = √3 x 413 = 715.34 pm.
As the structure with Cs+ at centre and Cl- at corner
2r+ + 2r- = 715.34 or
r+ + r- = 715.34 / 2 = 357.62 pm
We assume 2Cl- touch with each other than length of cell is
2r- = 413 pm
r- = 206.5 pm
r+ =357.67 - 206 .50 =151.17 pm
r+/r- = 151.17/206.50 = 0.73
VALUE-BASED ANSWERS
Ans.1 (a) ornaments are not made from 24 carat gold which is completely pure.
(b) 22 carat gold contains traces of copper or silver as impurity.
(c) It is better than puregold since it is comparatively hard, malleable and ductile.
Infact, 22 carat is an alloy.
(d) The boss guided the junior worker about the technique for making gold jewellery.As stated above , pure gold is extremely soft and is also brittle . That is why bangles could not be made out of that. the information is extremley valuable for all new comers in the field.
Ans.2 (a) yes , his observation was correct.
(b) Copper metal is conducting by the movement of electrons while copper sulphate crystals are expected to be conducting by the movement of ions. Since ions do not move in the solid state therfore , copper sulphate is non-conducting in the solid state.
(c) Electrolytes are either taken in molten state or aqueous solution in the electrolyte cells(e.g. electrolysis,electroplating).They conduct due to the movement of ions.
Ans.3 (a) Traditional source of electricity ;both thermal and nuclear energy lead to emissionof smoke,fly ash and extremely harmful radiations.
(b) Sun is the only source of solar energy of energy in a country like India where the span of summer is very long solar energy is very good option .
(c) Solar power is not very popular in India mainly due to lack of awareness.
(d) Use of solar energy in houese,hotelsand in small industrial unitscan solve our electricity problems.
Ans.4 (a) Structure of diamond is very compact on the other hand graphite has a layer structure.
(b) In diamond all the four valence electrons of carbon are involved in bond formation in graphite three valence electrons of carbon are involved while one electron is delocalised to conduct electricity.
(c) Diamond has a high refractive index as a result certain diamonds produce very high total internal refraction responsible for bluish tinge that is why these are very shiny and expensive.
Ans.5 a) The number of atoms of Carbon per unit cell is 8 in diamond.
b) The C—C bond is very strong in diamond (due to small size of Carbon) unlike the Zn—S bond in ZnS.
c) Though from the same background ie with the same structure the property can be different, thus, with a little effort, we can do same things differently and bring about major change
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