R.E. SOCIETY’S ABHYANKAR KULKARNI JR. COLLEGE, … · R.E. SOCIETY’S ABHYANKAR KULKARNI JR....

R.E. SOCIETY’S

ABHYANKAR KULKARNI JR. COLLEGE, RATNAGIRI.

XII SCIENCE WORK FROM HOME - 2020 CHEMISTRY

Date: 20/05/2020 Page 1 of 16

Answers

A] Multiple choice questions [1 Mark]

1. a) KBr & NaCl 2. d) unit cell 3. a) polymorphous

4. b) covalent 5. c) face centred cubic 6. b) covalent

7. a) NaCl 8. a) glass 9. c) CaCl2

10. d) any one of the above three

11. a) CH4 12. b) Cu

13. d) solubility in polar solvent

14. b) covalent 15. b) graphite

16. d) a and b both 17. c) glass 18. c) K2SO4

19. c) 7 20. b) 2 21. d) 3

22. b) silica 23. b) tetragonal 24. c) solid

25. d) Ice 26. d) barium chloride 27. b) Diamond

28. c) Covalent

B] Answer in one sentence. [1Mark]

1. 14 2. Covalent solid 3. Covalent 4. Molecular solid

5. Polymorphism 6. Orthorhombic 7. Triclinic

8. Anisotropy:- The ability of crystalline solids to change their physical properties when measured in different directions is called Anisotropy.

9. Isomorphous:- Refer page No. 1

10. Isomorphism :-Refer page No. 1

11. Polymorphous :- Refer page No. 2

12. Polymorphism:- Refer page No. 2

13. Unit cell :-Refer page No. 3

C] Answer the following [2Marks / 3 Marks /4 Marks]

1) Classification of solids:-Refer page No. 1

2) i) Crystalline solids :- Refer page No. 1 ii) Amorphous solids :- Refer page No. 1

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3) Properties of crystalline solids:- Refer page No. 1

4) Why does a crystalline solid has a sharp melting point?

Ans:- a) there is regularity and periodicity in the arrangement of constituent particles.

b) long range arrangement of constituent particles.

c) interparticle forces are identical , hence thermal energy needed to melt the solid is same.

5) Explain, ‘Amorphous solids do not have sharp melting points’.

Ans:- a) the constituent particles are randomly arranged.

b) The order is only short range

c) hence thermal energy needed to melt the solid is not uniform.

6) Distinguish between crystalline solids and amorphous solids.

Crystalline solids Amorphous solids

1. there is regularity and periodicity in the arrangement of constituent particles.

1. the constituent particles are randomly arranged.

2. long range arrangement of constituent particles.

2. The order is only short range.

3. Crystalline solids are anisotropic. 3. Amorphous solids are isotropic .

4. Crystalline solids have sharp melting points.

4. They do not have sharp melting point.

5. These are true solids. 5. These are pseudo solids or super cooled liquids.

7) Types of crystalline solids :-Refer page No. 2

8) i) Molecular solid:- Refer page No. 2

ii) Ionic solids :- Refer page No. 2

iii) metallic solids:- Refer page No. 2

iv) Covalent solids or network solids:- Refer page No. 2

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9) Classification of the solids into different types:

Amorphous solids Plastic

Covalent solids P4 molecule, S8 molecule ,Tetra phosphorous decoxide, Graphite, Diamond, Silicon, SiC,

Molecular solids Iodine molecule, Ice, SO2

Ionic solids Ammonium phosphate, NaCl, CaF2, CaCO3, and ZnS.

Metallic solids Rubidium, Brass

10) Number of atoms per unit cell in the following crystal structure.

a) Simple cubic :-Refer page No. 4 b) Body centred cubic :-Refer page No. 4 c) Face centred cubic :-Refer page No. 4 d) Body centred tetragonal :-Refer page No. 4

11) Relation between molar mass, ‘ρ’ of the substance and ‘a’ :-Refer page No. 4

12) Number of atoms in the following unit cells:

i) scc :- 1 ii) bcc :-2 iii) fcc:- 4

D] Numerical problems: - [1Mark / 3 Marks]

1) Copper crystallises into a fcc structure and the unit cell has edge length 3.61 x 108

cm. Calculate the density of Cu. Atomic mass of copper is 63.5 g mol-1.

Formula :- ρ = 𝑛 𝑥 𝑀

𝐚𝟑𝑁𝐴 n = 4

ρ = 8.97 g cm -3

2) Silver crystallises into a fcc structure. If density of silver is 10.51 g cm -3. Calculate the volume of unit cell. Atomic mass of silver is 108 g mol-1.

Formula :- V = 𝑛 𝑥 𝑀

ρ𝑁𝐴 n = 4

V = 6.824 cm3

3) A metal crystallises into cubic faces namely face centred (fcc) and body centred (bcc), whose unit cell edge lengths are 3.5 Å and 3.0 Å respectively. Find the ratio of the densities of fcc and bcc.

Formula :- Density of fcc unit cell

Density of bcc unit cell =

𝜌𝑓𝑐𝑐

𝜌𝑏𝑐𝑐 =

𝑛𝑓𝑐𝑐 𝑥 𝑀𝑁𝐴𝑎

3 𝑓𝑐𝑐⁄

𝑛𝑏𝑐𝑐 𝑥 𝑀𝑁𝐴𝑎

3 𝑏𝑐𝑐⁄ =

𝑛𝑓𝑐𝑐

𝑛𝑏𝑐𝑐 (

𝑎𝑓𝑐𝑐

𝑎𝑏𝑐𝑐)3 = 1.26

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4) Unit cell of iron crystal has edge length of 288 pm and density of 7.86 g cm-3. Determine the type of crystal lattice. Atomic mass of Iron is 56 g mol-1.

a = 288 pm = 2.88 x 10-8 cm

Formula :- 𝑛 = ρ 𝑥 𝐚𝟑𝑁𝐴

𝑀 = 2.01 ≅ 2

i.e. type of crystal lattice is bcc.

5) Calculate the number of atoms present in 2 gm of crystal which has fcc crystal lattice having edge length of 100 pm and density 10 g cm -3.

Formula :- volume of unit cell = a3

Mass of unit cell = Density x volume = 1 x 10-23 g

Since 1 x 10-23 g crystal contains 4 atoms

∴ 2 g crystal contains = ? atoms

Number of atoms = 8 x 1023

6) A cubic unit cell of a crystal consists of atoms of A and B elements. Atoms of A occupy corners of the unit cell while one B atom is present at the body centre. Determine the formula of the crystalline compound.

Ans :- Since 1/8th atom is contributed at each corner & there are 8 corners in unit cell

Therefore number of atoms of A = 1/8 x 8 = 1

There is one atom of B at body centre ∴ formula of compound is AB

7) A cubic unit cell has atoms A at the corners while the atoms B at the face of centres. Find the formula of the compound.

Ans :- AB3

8) A cubic unit cell has atoms A at the corners while the atoms B at the face of centres and C at the body centre. Find the formula of the compound.

Ans :- AB3C

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(Solid State)

§§ Co-ordination Number :- The number of neighbouring spheres that touch any given sphere is its co-ordination number .

§§ Packing of particles in crystal lattice:-

Constituent particles (considered as hard sphere) of a crystalline solid are closed packed .

Larger the co-ordination number , the closer are the spheres to each other.

Closed packed structures:-

a) Closed packing in one dimension :- It results by arranging the spheres to touch each other in a row. See fig.1.3 (a)

b) Closed packing in two dimension :- It results by stacking the two rows together such that they are in contact with each other .

There are two ways,

i) Square close packing :- Spheres are aligned vertically and horizontally as in fig.1.3 (b)

Since 1st and 2nd rows are same, they are labelled as ‘A’. Hence this arrangement is called AAAA...type two dimensional arrangement.

In this arrangement, every sphere touches four neighbouring spheres, hence co-ordination number is 4.

A square is obtained by joining the centres of these four closest neighbours. Therefore it is called square close packing in two dimension.

ii) Hexagonal close packing :- In this arrangement , spheres of 2nd row fit in depressions of the 1st row i.e. staggered arrangement results. If the 1st row is ‘A’ type , the 2nd row being different is ‘B’ type.

In staggered manner , the spheres in 3rd row are aligned with the spheres of 1st row, hence 3rd row is ‘A’ type . Similarly the spheres in 4th row are aligned with the

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spheres of 2nd row , hence 4th row is ‘B’ type . Hence resulting two dimensional arrangement is ‘ABAB.......’ type.

In this arrangement , every sphere touches 6 neighbouring spheres, hence co-ordination number is 6. A regular hexagon is obtained by joining the centres of these six closest neighbours. Therefore it is called Hexagonal close packing in two dimension. See Fig.1.3 (c)

In this arrangement free spaces i.e. voids are triangular in shape. Hexagonal close packing is more efficient than square close packing because co-ordination number in hexagonal close packing is higher than in square close packing. The free space in hexagonal close packing is also less than square close packing .

c) Closed packing in three dimensions :-

Stacking of two dimensional layers give rise to three dimensional crystal structure.

i) Stacking of square close pack layers :- The structure that results on stacking square closed packed layers is simple cubic (sc). Its unit cell is primitive cubic unit cell.

In this arrangement spheres of all the layers are perfectly aligned vertically and horizontally. Hence all the layers are alike and are labelled as ‘A’ layers. This arrangement of layers is described as ‘AAAA....’ type three dimensional.

Polonium is the only metal that crystallises in simple cubic closed packed structure.

ii) Stacking of two hexagonal close packed layers :-

Spheres of 2nd layer (labelled as ‘B’) are placed in the depression of the 1st layer, labelled as ‘A’. All triangular voids of the 1st layer are not covered by the spheres of 2nd layer.

The triangular voids that are covered by spheres of 2nd layer generate tetrahedral voids. (A tetrahedral void is surrounded by four spheres.

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On joining the centres of these four spheres a tetrahedron is formed.) The remaining triangular voids of the 1st layer are covered by triangular voids of 2nd layer results in octahedral void surrounded by six spheres .

iii) Placing 3rd hexagonal close packed layer :- There are two ways,

a) To align the spheres of 3rd layer with the spheres of 1st layer, resulting pattern is ABABAB... and hexagonal close packed (hcp) arrangement. Fig.1.8 (a)

Metals such as Mg, Zn have hexagonal close packed crystal structure.

b) To cover the octahedral voids by spheres of 3rd layer called as ‘C’ layer , in such a way that the spheres of 3rd layer do not align with the spheres of 2nd and 1st layer. Spheres of 4th layer get aligned with the spheres of 1st layer hence 4th layer is called ‘A’ layer. The resulting pattern is ABCABC.... and arrangement is cubic close packed (ccp) this is same as fcc. Fig.1.8 (b)

Metals such as Cu , Ag have cubic close packed (ccp or fcc ) crystal structure.

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§§ Number of voids per atom in hcp and ccp :-

There are two tetrahedral voids associated with each atom while there is one octahedral void per atom.

§§ Packing efficiency :- The magnitude of Packing efficiency gives a measure of how tightly particles are packed together.

Packing efficiency is the fraction or percentage of the total space occupied by the spheres (particles).

Packing efficiency = 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒄𝒄𝒖𝒑𝒊𝒆𝒅 𝒃𝒚 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔 𝒊𝒏 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍

𝒕𝒐𝒕𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍 𝑿 𝟏𝟎𝟎

Packing efficiency of metal crystal in simple cubic lattice :-

Step i) Radius of sphere :- In simple cubic unit cell , particles (spheres ) are at the corners and touch each other along the edge as shown in figure

a = 2r

∴ r = 𝑎

2 eq-1

where r = radius of atom and

a = edge length of unit cell

Step ii) Volume of sphere :-

Volume of sphere = 4

3𝜋𝑟3

From equation -1 ⇒ Volume of sphere = 4

3𝜋(

𝑎

2)3 =

𝝅 𝒂𝟑

𝟔

Step iii) Total volume of particles :-

Since simple cubic unit cell contains only one particle ,

∴ volume occupied by particle in unit cell = 𝝅 𝒂𝟑

𝟔

Step iv) Packing efficiency :-

Packing efficiency = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑐𝑐𝑢𝑝𝑖𝑒𝑑 𝑏𝑦 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑖𝑛 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙

𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑋 100

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∴ Packing efficiency =

𝜋 𝑎3

6

𝑎3 x 100 =

100𝜋

6 =

100 𝑋 3.142

6 = 52.36 %

Thus in simple cubic lattice 52.36 % of total space is occupied by particles and 47.64 % is empty space i.e. void volume.

Packing efficiency of metal crystal in body centred cubic lattice :-

Step i) Radius of sphere :- In bcc unit cell , particles occupy the corners and in addition one particle is at the centre of the cube. The particle at the centre of the cube touches two corner particles along the diagonal of the cube as shown in figure,

To obtain radius of the particle (sphere) , Pythagorus therom is applied ,

For triangle FED , ∠FED = 90 o

∴(FD)2 = (FE)2 + (ED)2 = a2 + a2 = 2a2 …….eq-1 ....... FE = ED = a

For triangle AFD , ∠ADF = 90 o

∴ (AF)2 = (AD)2 + (FD)2 …….eq-2

Substituting eq-1 into eq-2

∴ (AF)2 = a2 + 2a2 = 3a2 ....... AD = a

∴ AF = √3 a …….eq-3

From figure we can write

AF = 4r …….eq-4

From eq-3 and eq-4 we can write

4r = √3 a

∴ r = √𝟑

𝟒 a …….eq-5

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3𝜋𝑟3

From eq -5


3𝜋(

√3

4 a)3


3𝜋 X

(√3)3

64 𝑎3

Volume of sphere = √𝟑 𝝅𝒂𝟑

𝟏𝟔

Step iii) Total volume of particles :- Since bcc unit cell contains two particles ,

∴ volume occupied by particles in unit cell = √𝟑 𝝅𝒂𝟑

𝟏𝟔 𝑋 2 =

√𝟑 𝝅𝒂𝟑

𝟖




∴ Packing efficiency =

√3 𝜋𝑎3

8

𝑎3 x 100 =

√3 𝜋𝑎3

8𝑎3 X 100

= √3 𝑋100 𝑋 3.142

8 = 68 %

Thus in bcc 68 % of total space is occupied by particles and 32 % is empty space i.e. void volume .

Packing efficiency of metal crystal in face centred cubic lattice ( ccp or hcp lattice) :-

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Step i) Radius of sphere :- The corner particles are assumed to touch the particle at the centre of face ABCD as in figure,

The triangle ABC is right angled with ∠ ABC = 90 o.

According to Pythagoras theorem,

(AC)2 = (AB)2 + (BC)2 = a2 + a2 = 2a2 ....... AB = BC = a

Hence, AC = √2 a eq-1

From figure, AC = 4r eq-2

From eq-1 and eq-2,

4r = √2 a ⇒ ∴ r = √𝟐

𝟒 a =

𝑎

2√2 eq-3



3𝜋𝑟3

From eq -3, ⇒ Volume of sphere = 4

3𝜋(

𝑎

2√2 )3

From eq -1, ⇒ Volume of sphere = 4

3𝜋 𝑎3 X (

1

(2√2) )3

∴ Volume of sphere =𝝅𝒂𝟑

𝟏𝟐√𝟐

Step iii) Total volume of particles :- Since fcc unit cell contains four particles ,

∴ volume occupied by particles in unit cell = 𝜋𝑎3

12√2 𝑋 4 =

𝝅𝒂𝟑

𝟑√𝟐




∴ Packing efficiency =𝜋𝑎3

3√2 𝑎3 x 100 =

𝜋

3√2 X 100 = 74 %

Thus in fcc/ccp/hcp crystal lattice 74 % of total space is occupied by particles and 26 % is empty space i.e. void volume .

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§§ Number of particles in ‘x’ g metal :- It can be calculated as,

We have, ⇒ M = ρ 𝐚𝟑𝑁𝐴

𝑛 eq-1

We know that, Since Molar mass ‘M’, contains 𝑁𝐴 particles

∴ x g metal contains 𝑥𝑁𝐴

𝑀 particles. From eq-1

Number of particles in ‘x’ g metal = 𝑥𝑁𝐴

𝛒𝐚𝟑𝑁𝐴𝑛⁄

= 𝒙𝒏

𝛒𝐚𝟑

§§ Number of unit cells in ‘x’ g metal :-

Since ‘n’ particles corresponds to 1 unit cell

∴ 𝑥𝑛

𝛒𝐚𝟑 Particles corresponds to

𝑥𝑛

𝛒𝐚𝟑 x

1

𝑛 unit cells

∴ Number of unit cells in ‘x’ g metal = 𝒙

𝛒𝐚𝟑

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§§ Summary

Unit cell /Lattice

Relation between

a & r

Volume of one particle

Total volume occupied by particles in unit cell

Co-ordination number of atoms Packing efficiency

sc r = a/2 = 0.5 a 𝜋 𝑎3

6= 0.5237a3

𝜋 𝑎3

6= 0.5237a3

6: four in the same layer , one directly above & one directly below

52.4%

bcc r = √3

4 a

=0.433a

√3 𝜋𝑎3

16= 0.34a3

√3 𝜋𝑎3

8= 0.68a3

8: four in the layer below & four in the layer above.

68%

fcc/ccp r = √2

4 a =

𝑎

2√2

= 0.3535a

𝜋𝑎3

12√2 = 0.185 a3

𝜋𝑎3

3√2 = 0.74 a3 12: six in its own layer , three

above & three below 74%

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Question Bank

A ] Multiple choice questions [1 Mark]

1) In crystal lattice formed by bcc unit cell the void volume is.....

a) 68% b) 74% c) 32% d) 26%

2) The co-ordination number of atoms in fcc crystal lattice is....

a) 2 b) 4 c) 6 d) 8

3) A compound forms hcp structure. Number of octahedral voids and tetrahedral voids in 0.5 mole of substance is respectively ......

a) 3.011 x 1023,6.022 x 1023 b) 6.022 x 1023 , 3.011 x 1023

c) 4.011 x 1023, 2.022 x 1023 d) 6.011 x 1023, 12.022 x 1023

4) Pb has fcc structure with edge length of unit cell 495 pm . Radius of Pb atom is...

a) 205 pm b) 185 pm c) 260 pm d) 175 pm

5) Two metals A and B crystallise in bcc and fcc type structures. The ratio of number of atoms of A and B present in unit cell of their crystal is .....

a) 0.5 b)1 c) 2 d) 4

6) Potassium crystallises in bcc structure . Hence co-ordination number of it is...

a) 4 b) 6 c) 8 d) 12

7) The ratio of closed packed atoms to octahedral holes in hexagonal close packing is..

a) 1:1 b) 1:2 c) 2:1 d) 1:3

8) The number of octahedral sites per sphere in fcc structure is...

a) 1 b) 2 c) 3 d) 4

9) The number of tetrahedral sites per sphere in ccp structure is...

a) 1 b) 2 c) 3 d) 4

10) The ratio of Packing efficiency in scc , bcc and fcc crystalline structure is ....

a) 1:1.2:1.3 b) 1:1.12:1.23 c) 1:1.3:1.4 d) 1:1.25:1.38

11) The correct sequence of the Atomic layers in cubic close packing is....

a) ABABA b) ABACABAC c) ABCABC d) AABBAABB

12) The Packing efficiency in the unit cell of fcc crystal is.....

a) 68% b) 74% c) 52.4% d) 47.6%

13) If ‘r’ is the radius of an atom in bcc unit cell having edge length ‘a’ then, …..

a) r = √3

4 a b) r =

4

√3 a c) r = 2 √2 a d) r = √

3

4 a

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14) The close packed structure has N number of atoms. The number of tetrahedral voids and octahedral voids are...

a) N , N b) 4N , 3N c) 2N , N d) 2N , 2N

15) If all the lattice points in ccp structure namely corners ,faces and edge centres and body centre are occupied by atoms then total number of atoms in the unit cell will be...

a) 8 b) 12 c) 14 d) 16

16) An ionic solid crystallises in bcc structure . If the ionic radii of cation and anion are 0.84 Å and 1.07 Å, the length of body diagonal is.....

a) 1.91 Å b) 2.75 Å c) 3.82 Å d) 2.32 Å

17) Mg crystal has ..... structure ‘

a) simple cubic b) bcc c) hcp d) tetragonal

18) There are two tetrahedral holes per anions in...

a) ccp & hcp b) bcc & fcc

c) bcc & hcp d) simple cubic cell and tetrahedral

19) The radii of Na+ and Cl- ions are 95 pm and 181 pm respectively .The edge length of NaCl unit cell is...

a) 276 pm b) 138 pm c) 552 pm d) 45 pm

20) The fraction of the total volume occupied by the atoms present in a simple cube is..

a) 𝜋

4 b)

𝜋

6 c)

𝜋

3√2 d)

𝜋

4√2

B ] Answer in one sentence.[1Mark]

1) What is the ratio of close packed atoms to tetrahedral holes in cubic packing ?

2) What is the packing fraction for a simple cubic structure ?

3) What is the percentage of space occupied in face centred cubic structure?

4) What is the co-ordination number of Na+, if NaCl crystallises in ccp structure?

5) What is the radius of atom of the element , if a body centred cubic unit cell of an element has edge length 1.3 Å.

6) In which crystal system the Packing efficiency is maximum ?

7) How many total number of different primitive unit cells are possible?

8) What is packing efficiency?

9) What is coordination number ?

10) Sketch a tetrahedral void .

11) A solid is hard, brittle and electrically non-conductor but on melting conducts electricity. What type of solid is it?

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C] Answer the following. [2M/3M]

1) Mention two properties that are common to both hcp and ccp lattices.

2) Distinguish between tetrahedral voids and octahedral voids.

3) Distinguish between Hexagonal close packing and square close packing.

4) Explain AAAA and ABAB type of three dimensional packing.

5) Write packing fraction, Packing efficiency and void space for the following

Crystal structure: a) scc b) bcc c) fcc d) hcp

6) How many tetrahedral void and octahedral void are present in a closed packed structure?

7) What is a relation between edge length of the unit cell and radius ‘r’ of atom in the following? a) scc b) bcc c) fcc

8) Calculate Packing efficiency in the following crystals:

a) scc crystal b) bcc structure c) fcc structure

9) How are the spheres arranged in 1st layer of simple cubic close-packed structures? How are the successive layers of spheres placed above this layer?

D] Numerical problems :-

1) The unit cell of metallic silver is fcc. If radius of Ag atom is 144.4 pm . Calculate edge length of unit cell, volume of Ag atom , volume of a unit cell that is occupied by Ag atoms and the percent of empty space.

2) A compound is formed by two elements A and B . The atoms of elements B forms ccp structure . The atoms of A occupy 1/3rd of tetrahedral voids . What is the formula of the compound ?

3) Niobium forms bcc structure . The density of Niobium is 8.55 g/cm3 and edge length is 330.6 pm. How many atoms and unit cells are present in 0.25 g of Niobium?

4) A compound forms hcp structure. What is the number of octahedral voids, tetrahedral voids and total voids formed in 0.8 mol of it?

5) The density of Iridium is 22.4 g/cm3 the unit cell of it is fcc and molar mass is 192.2 g/mol. Calculate radius of Iridium atom .

6) Aluminium crystallizes in cubic close packed structure with unit cell edge length of 353.6 pm. What is radius of Al atom? How many unit cells are there in 1.00 cm3 of Aluminium.

𝐖𝐎𝐑𝐊 𝐅𝐑𝐎𝐌 𝐇𝐎𝐌𝐄 & 𝑆𝑇𝐴𝑌 𝑆𝐴𝐹𝐸

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R.E. SOCIETY’S ABHYANKAR KULKARNI JR. COLLEGE, … · R.E. SOCIETY’S ABHYANKAR KULKARNI JR....

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