Solid State Test

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Solid State Test - 59

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Question 1
1 / -0

Fraction of empty space in ABAB type arrangement in 3D is :

A
$$0.74$$

B
$$0.26$$

C
$$0.68$$

D
$$0.32$$

Solution

Percentage of packing fraction of ABAB (HCP) arrangement in 3D $$= 74\%$$
Percentage of vacant space $$= 26\% $$
Fraction of vacant space $$=0.26$$
Question 2
1 / -0

The pyknometric density of sodium chloride crystal is $$2.165\times 10^{3}\ kg\ m^{-3}$$ while its X-ray density is $$2.178\times 10^{3} kg\ m^{-3}$$. The fraction of unoccupied sites in sodium chloride crystal is:

A
$$5.96\times 10^{-3}$$

B
$$5.96$$

C
$$5.96\times 10^{-2}$$

D
$$5.96\times 10^{-1}$$

Solution

Therefore volume unoccupied site in sodium chloride $$=5.96 \times 10^{-3}$$
Question 3
1 / -0

Statement 1: Increasing temperature increases the density of point defects.
Statement 2: The process of formation of point defects in solids is endothermic and has $$\Delta S > 0$$.

A
Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is false

D
Statement-1 is false, Statement-2 is true

Solution

The process of formation of point defects in solids is endothermic.
Hence, $$\Delta H > 0$$ and also, $$\Delta S > 0$$.
When temperature is increased, $$\Delta G = \Delta H-T\Delta S < 0$$. The process of formation of defects become spontaneous. Hence, more and more defects are created per unit volume.
Thus, increasing temperature increases the density of point defects.
Question 4
1 / -0

Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is $$124$$ pm, the density (in g cm$$^{-3}$$) of iron in both these structures are, respectively :

A
$$7.887,\ 8.59$$

B
$$8.59,\ 7.887$$

C
$$14.287,\ 16.248$$

D
$$16.248,\ 14.287$$

Solution

For bcc unit cell, the edge length, $$\displaystyle a = \frac {4r}{\sqrt{3}} = \frac {4 \times 124}{\sqrt{3}} = 286.4$$ pm
The number of formula units per unit cell $$\displaystyle Z=2 $$
Density $$\displaystyle = \frac {\text {Z} \times \text {Molar mass}}{\text {Avogadro's number } \times a^3} $$
Density $$\displaystyle = \frac {2 \times 55.85}{6.023 \times 10^{23} \times (286.4 \times 10^{-10})^3} = 7.887gcm^{-3}$$
For fcc unit cell, the edge length, $$\displaystyle a = \frac {4r}{\sqrt{2}} = \frac {4 \times 124}{\sqrt{2}} = 350.8$$ pm
The number of formula units per unit cell $$\displaystyle Z=4 $$
Density $$\displaystyle = \frac {\text {Z} \times \text {Molar mass}}{\text {Avogadro's number } \times a^3} $$
Density $$\displaystyle = \frac {4 \times 55.85}{6.023 \times 10^{23} \times (350.8 \times 10^{-10})^3} = 8.59gcm^{-3}$$
Question 5
1 / -0

An element crystallizes in a structure having FCC unit cell of an edge length equal to $$200$$ pm. Calculate the density (in g cm$$^{-3}$$) if $$200$$ g of this element contains $$24\times10^{23}$$ atoms.

A
$$41.67$$

B
$$83.34$$

C
$$119.43$$

D
None of the above

Solution

$$ M = \dfrac{N_A \times 200}{24 \times 10^{23}} = 50 $$ g
For fcc $$z = 4 $$
$$ d = \dfrac{ zM}{N_A \times V }$$
$$ d = \dfrac{ 4 \times 50}{6.023 \times 10^{23} \times (200 \times 10^{-10})^3 }$$
$$ d = \dfrac{ 200}{6.023 \times 8 \times 10^{-1} }$$
$$ d = 41.67 $$ $$gcm^{-3} $$
Question 6
1 / -0

Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is $$407$$ pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold $$=197$$ amu.

A
$$19.4\:g/cm^{3}$$, $$143.9$$ pm

B
$$38.8\:g/cm^{3}$$, $$ 143.9$$ pm

C
$$19.4\:g/cm^{3}$$, $$287.8$$ pm

D
None of the above

Solution

For an FCC crystal, atoms are in contact along the diagonal of the unit cell.
$$4r= \sqrt{2}a$$, where $$a$$ is the edge length of unit cell and $$r$$ is the radius of atom.
Substituting values in the above expression, we get
$$r= \dfrac{ \sqrt{2} \times 407}{4} =143.9$$ pm
Number of atoms per unit cell (FCC) $$=4$$
Mass of $$1$$ unit cell $$=197 \times 4=788\:amu=1.66 \times 10^{-27} \times 788\:kg=1.30808 \times 10^{-24}\:kg$$
Volume of $$1$$ unit cell $$= (407 \times 10^{-12})^{3}\:m^{3}=6.741 \times 10^{-29}\:m^{3} $$ $$=2.824 \times 10^{-28}\:m^{3}$$
$$\text{Density}= \dfrac{1.30808 \times 10^{-24}}{6.741 \times 10^{-29}}=19404.83 \:kg/m^{3}=19.4\:g/cm^{3}$$
Question 7
1 / -0

The face centred cubic cell of platinum has a length of $$0.392 nm$$. Calculate the density of platinum $$(g/cm^3): $$

(Atomic weight : Pt=195)

A
20.9

B
20.4

C
19.6

D
21

Solution

$$Density, d = \dfrac{Z{\times} M}{N_A{\times}a^3}$$

Here, Z = No. of atoms in fcc crystal = 4

$$N_a =$$ Avagadro No. $$= 6\times 10^{23} mol^{-1}$$, $$a=0.392\ nm = 0.392 \times 10^{-7} cm $$

$$M = 195\ g/mol$$

$$ d = \dfrac{4{\times} 195}{N_a{\times}{(0.392\times 10^{-7})}^3}$$

$$ d = 21\dfrac{g}{cm^3}$$

Option D is correct.
Question 8
1 / -0

The distance between an octahedral and tetrahedral void in fcc lattice would be

A
$${\sqrt 3a}$$

B
$$\displaystyle\frac{\sqrt{3a}}{2}$$

C
$$\displaystyle\frac{\sqrt{3a}}{3}$$

D
$$\displaystyle\frac{\sqrt{3}a}{4}$$

Solution

Distance between an octahedral and tetrahedral void in fcc lattice is $$\dfrac{\text{body diagonal}}{4}$$ $$=\dfrac{\sqrt { 3 } a}{4}$$
Question 9
1 / -0

An element X (atomic weight $$= 24$$ amu) forms a face-centred cubic lattice. If the edge length of the lattice is $${4\times10^{- 8}}$$ cm and the observed density is $$2.40 \times 10^{3}$$ kg m$$^{-3}$$, then the percentage occupancy of lattice points by element X is: (use $$ {N_{A}= 6\times 10^{23}}$$)

A
$$96$$

B
$$98$$

C
$$99$$

D
none of the above

Solution

As we know, density $$(d)= \dfrac{nM}{VN_A}$$
Here, $$a$$ is the edge length of the unit cell, $$M$$ is the molecular mass and $$V$$ is the volume of an unit cell.
$$d = \dfrac{4\times24}{(4\times10^{-8})^3\times6\times10^{23}} = 2.5$$ g/cc
The observed density is $$2.4$$ g/cc.
So, the percentage occupancy of lattice points by element X is $$\dfrac{2.4\times100}{2.5} = 96\%$$
Question 10
1 / -0

For orthorhombic system axial ratios are $$a\neq b\neq c$$ and the axial angles are

A
$$\alpha = \beta = \gamma \neq 90^o$$

B
$$\alpha = \beta = \gamma = 90^o$$

C
$$\alpha = \gamma= 90^o , \beta \neq 90^o$$

D
$$\alpha \neq \beta \neq \gamma = 90^o$$

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Solid State Test - 59

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Solid State Test - 59

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SHARING IS CARING

Question 1

$$0.74$$

$$0.26$$

$$0.68$$

$$0.32$$

Solution

Question 2

$$5.96\times 10^{-3}$$

$$5.96$$

$$5.96\times 10^{-2}$$

$$5.96\times 10^{-1}$$

Solution

Question 3

Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1

Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation for Statement-1

Statement-1 is true, Statement-2 is false

Statement-1 is false, Statement-2 is true

Solution

Question 4

$$7.887,\ 8.59$$

$$8.59,\ 7.887$$

$$14.287,\ 16.248$$

$$16.248,\ 14.287$$

Solution

Question 5

$$41.67$$

$$83.34$$

$$119.43$$

None of the above

Solution

Question 6

$$19.4\:g/cm^{3}$$, $$143.9$$ pm

$$38.8\:g/cm^{3}$$, $$ 143.9$$ pm

$$19.4\:g/cm^{3}$$, $$287.8$$ pm

None of the above

Solution

Question 7

20.9

20.4

19.6

21

Solution

Question 8

$${\sqrt 3a}$$

$$\displaystyle\frac{\sqrt{3a}}{2}$$

$$\displaystyle\frac{\sqrt{3a}}{3}$$

$$\displaystyle\frac{\sqrt{3}a}{4}$$

Solution

Question 9

$$96$$

$$98$$

$$99$$

none of the above

Solution

Question 10

$$\alpha = \beta = \gamma \neq 90^o$$

$$\alpha = \beta = \gamma = 90^o$$

$$\alpha = \gamma= 90^o , \beta \neq 90^o$$

$$\alpha \neq \beta \neq \gamma = 90^o$$

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