Solid State Test

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Solid State Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

The number of unit cells in 58.5 g of $$NaCl$$ is approximately:

A
$$6 \times 10^{20}$$

B
$$1.5 \times 10^{23}$$

C
$$9 \times 10^{23}$$

D
None of these

Solution

Atomic mass of sodium chloride $$= 23 + 35.5 = 58.5$$ g
this means 23 g has no. of toms $$=6.022 \times 10^{23}$$
therefore 58.5 g has no. of atoms $$= 6.022 \times 10^{23}\times 58.5/23$$
no. of atoms in a unit cell of fcc structure=4
therefore no. of unit cell which contains no. of atoms $$= 6.022 \times 10^{23} \times 58.5/23\times 4 = 1.5 \times 6\times 10^{23}$$
Question 2
1 / -0

How many units cells are there in $$ 1.00 g $$ cube shaped ideal crystal of $$ AB (Mw = 60 ) $$ which has a $$NaCl$$ type lattice?

A
$$ 6.02 \times 10^{23} $$

B
$$ 1.00 \times 20^{22} $$

C
$$ 2.50 \times 10^{21} $$

D
$$6.02 \times 10^{24} $$

Solution

$$NaCl$$ has Rock salt type lattice, so Numbers of atoms per unit cell = 4
That means 4 $$NaCl$$ farms = 1 unit cell
So in $$AB$$ (mw = 60) $$= 4 AB$$ forms one unit cell.
Now 1.0 gram of $$AB$$ has Numbers of moles = $$\dfrac{Given mass}{molar mass}$$
= $$\dfrac{1}{60}$$ mole
1 mole of $$AB =$$ $$6.022 \times 10^{23}$$
$$\dfrac{1}{60} $$ mole of $$AB = \dfrac{1}{60} \times 6.022 \times 10^{23}$$
= $$1.004 \times 10^{22} AB$$
$$4 AB$$ forms = 1 unit all
$$1 AB$$ forms = $$\dfrac{1}{4}$$ unit all
$$1.004 \times 10^{22} $$AB form = $$\dfrac{1}{4} \times 1.004 \times 10^{22}$$
$$9.50 \times 10^{21}$$
$$0.250 \times 10^{22}$$
$$2.50 \times 10^{21}$$
So, the answer is $$2.50 \times 10^{21}$$
Question 3
1 / -0

The density of crystalline sodium chloride is 5.85 g $${cm}^{-3}$$. What is the edge length of the unit cell?

A
$$4.04 \times 10^{-9}$$ cm

B
$$1.32 \times 10^{-14}$$ cm

C
$$7.8 \times 10^{-23}$$ cm

D
$$9.6 \times 10^{-24}$$ cm

Solution

Density $$(d) = 5.85\ g\ {cm}^{-3}$$

Molar Mass of Sodium Chloride $$(m) = 58.5 g/mol$$

$$N_A = 6.022 \times 10^{23}$$

Unit Cell Type is $$FCC,$$ $$Z = 4$$

We know,
$$d = \cfrac{(Z \times M)}{(N_A \times a^3)}$$

$$=> a^3 = \cfrac{(Z \times m)}{(N_A \times d)}$$

$$=> a^3 = \cfrac{(4 \times 58.5)}{(6.022 \times 10^{23} \times 5.85)}$$

$$=> a^3 = 0.066 \times 10^{-24}$$

$$=> a = 0.404 \times 10^{-8}$$

So, edge length of the unit cell of NaCl is $$0.404 \times 10^{-8}\ cm.$$
Question 4
1 / -0

Certain quantity of metal is crystallised $$60\%$$ in ccp structure $$40\%$$ in hcp structure. What is the ratio of number of unit cells in ccp and hcp?

A
$$3:2$$

B
$$1:1$$

C
$$9:4$$

D
$$4:9$$

Solution

Let number of moles of metal crystallized= $$n$$
Number of moles crystallized in $$ccp=n \times \cfrac {60}{100}=0.6n$$
Number of moles crystallized in $$hcp= n \times \cfrac {40}{100}=0.4n$$
as $$1$$ unit of $$ccp$$ contains= $$4$$ atoms of metal or $$\cfrac {4}{N_a}$$ moles of metal
Thus $$0.6n$$ moles contains= $$\cfrac {0.6 n \times N_a}{4}$$ unit cell of $$ccp$$
Similarly, $$1$$ unit cell of $$hcp$$ contains= $$6$$ atoms of metal or $$\cfrac {6}{N_a}$$ moles of metal.
Thus $$0.4n$$ moles contains= $$\cfrac {0.4n \times N_a}{6}$$ unit cell of $$hcp$$
Thus ratio of number of unit cell of $$ccp$$ and $$hcp$$ is
$$R=\cfrac {\text{unit cell of ccp}}{\text {unit cell of hcp}}$$
$$=\cfrac {\cfrac {0.6 \times n \times N_a}{4}}{\cfrac {0.4n \times N_a}{6}}=\cfrac {9}{4}$$
So, the ratio is $$9:4$$
Question 5
1 / -0

Salt Ab has a zinc blende structure. The radius of $$A^{2+}$$ and $$B^{2-}$$ ions are $$0.7\mathring{A}$$ and $$1.8\mathring{A}$$ respectively. The edge length of AB unit cell is:

A
$$2.5 \mathring{A}$$

B
$$5.0.9\mathring{A}$$

C
$$5 \mathring{A}$$

D
$$5.77 \mathring{A}$$

Solution

In ZnS like structure
$$r_A^+ + r_B^- = \dfrac{a\sqrt{3}}{4} $$
$$a= 5.77 A^0$$
Question 6
1 / -0

Packing efficiency of body centred unit cell is:

A
85%

B
68%

C
33.33%

D
45%

Solution

Total number of atoms per unit cell in bcc= $$2$$
From $$\Delta EFD$$,
$$\Rightarrow b^2= a^2+a^2$$
$$\Rightarrow b= \sqrt {2} a$$
$$\Delta AFD$$,
$$\Rightarrow c^2= a^2+b^2=a^2+2a^2=3a^2$$
$$\Rightarrow c= \sqrt {3}a$$
But $$c=4r$$ ($$r$$= radius of sphere)
$$\therefore \sqrt {3}a=4r$$
$$\Rightarrow a=\cfrac {4r}{\sqrt {3}}$$
Total number of atoms per unit cell in bcc= $$2$$
Packing efficiency= $$\cfrac {\text {Volume occupied by 2 spheres in unit cell }\times 100}{\text{Total volume of unit cell}}$$
$$=\cfrac {2 \times \cfrac {4}{3}\Pi r^3\times 100}{a^3}$$
$$=\cfrac {2 \times \cfrac {4}{3}\Pi r^3 \times 100}{\left[\cfrac {4}{\sqrt {3}}r\right]^3}=68$$ %

$$\therefore$$ Packing efficiency of bcc $$=$$ $$68$$ %
Question 7
1 / -0

The density of solid argon (Ar = 40 g / mol) is 1.68 g/mL at 40 K. If the argon atom is assumed to be a sphere of radius 1.50 x $$10^{-8}$$ cm, what % of solid Ar is apparently empty space? (use $$N_A = 6 \times10^{23}$$)

A
35.64

B
64.36

C
74%

D
none of these

Solution

Volume of one atom of $$Ar =\frac{4}{3}πr^3$$
Also, number of atoms in 1.65 g/mL $$= \frac{1.65}{40}\times 6.023 \times {10}^{23}$$
∴ Total volume of all atoms of Ar in solid state $$=\frac{4}{3}πr^3 \times \frac{1.65}{40}\times 6.023 \times {10}^{23}$$
$$ =\frac{4}{3}\times 3.14\times (1.54\times 10^{-8})^3\times \frac{1.65}{40}\times 6.023 \times {10}^{23} =0.380 cm^3$$
Volume of solid argon $$= 1 cm^3$$
∴% Empty space$$ =\frac{1−0.380}{1}\times 100 = 62$$%
Question 8
1 / -0

$$A_2B$$ molecules (molar mass =259.8 g/mol) crystallises in a hexagonal lattice as shown in figure. The lattice constants were $$a = 5 \mathring{A}$$ and $$ b= 8 \mathring{A}$$. If density of crystal is 5 $$g/cm^3$$ then how many molecules are contained in given unit cell? (use $$N_A= 6 \times 10^{23}$$)

A
6

B
4

C
3

D
2

Solution

Volume of unit cell$$= a^2\sin 60^o\times b$$
$$=173.2\times 10^{-24}$$
$$\Rightarrow a=5\overset {o}{A}$$
Mass of unit cell= $$173.2\times 10^{-24}\times 5\times 6 \times 10^{23}= 519.6 g$$
Number of molecules presenr in given cell= $$\cfrac {519.6}{259.8}=2$$ .
Question 9
1 / -0

The compound $${ MX }_{ 4 }$$ is tetrahedral . The number of $$X-M-X$$ angles in the compound is:

A
three

B
four

C
five

D
six

Solution

$$MX_4$$ is tetrahedral, therefore total number of angles $$XMX$$ is six.
Question 10
1 / -0

Which forms a crystal of NaCl?

A
$$NaCl\ molecules$$

B
$$Na^{+}$$ and $$Cl^{-}ions$$

C
$$Na\ and\ Cl\ atoms$$

D
$$None\ of\ these$$

Solution

NaCl crystal structure is formed by $$Na^+$$ cation and $$Cl^-$$ anion.
$$Cl^-$$ forms a FCC lattice and $$Na^+$$ occupies all the octahedral voids.
It is also called as Rock Salt type structure.

Hecne, the correct option is B

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Solid State Test - 48

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Solid State Test - 48

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Question 1

$$6 \times 10^{20}$$

$$1.5 \times 10^{23}$$

$$9 \times 10^{23}$$

None of these

Solution

Question 2

$$ 6.02 \times 10^{23} $$

$$ 1.00 \times 20^{22} $$

$$ 2.50 \times 10^{21} $$

$$6.02 \times 10^{24} $$

Solution

Question 3

$$4.04 \times 10^{-9}$$ cm

$$1.32 \times 10^{-14}$$ cm

$$7.8 \times 10^{-23}$$ cm

$$9.6 \times 10^{-24}$$ cm

Solution

Question 4

$$3:2$$

$$1:1$$

$$9:4$$

$$4:9$$

Solution

Question 5

$$2.5 \mathring{A}$$

$$5.0.9\mathring{A}$$

$$5 \mathring{A}$$

$$5.77 \mathring{A}$$

Solution

Question 6

85%

68%

33.33%

45%

Solution

Question 7

35.64

64.36

74%

none of these

Solution

Question 8

6

4

3

2

Solution

Question 9

three

four

five

six

Solution

Question 10

$$NaCl\ molecules$$

$$Na^{+}$$ and $$Cl^{-}ions$$

$$Na\ and\ Cl\ atoms$$

$$None\ of\ these$$

Solution

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