Solid State Test

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Solid State Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Stacking (a pile) of square close packed layer gives rise to:

A
BCC structure

B
FCC strncture

C
SC structure

D
none of the above
Question 2
1 / -0

MgO has NaCl type structure and its unit cell edge length is $$421$$ pm. Its density (in g cm$$^{-3}$$) will be:

A
$$3.54$$

B
$$1.46$$

C
$$2.14$$

D
None of the above

Solution

The expression for density is as follows:
$$\rho =\dfrac{ZM}{N_{0}a^{3}}$$
As MgO has the same structure as NaCl, we can say that MgO has FCC structure.
So, putting values accordingly, we get
$$\rho =\dfrac { 4\times 40 }{ (6.022\times { 10 }^{ 23 })\times (4.21\times { 10 }^{ -8 })^{ 3 } } =3.54$$ g cm$$^{-3}$$
Question 3
1 / -0

Of the three cubic lattices, the one that has the largest mass of empty space is the :

A
simple cubic

B
ccp

C
bcc

D
both simple cubic and bcc

Solution

Amount pf empty space per unit volume in
simple cubic = 100 - 52 = 48%
closed cubic packing = 100 - 74 = 26%
body centred cubic = 100 - 68 = 32%
Question 4
1 / -0

When a viscous liquid is cooled rapidly, _________ solid is formed.

A
eutectic

B
liquid crystal

C
super-cooled

D
amorphous
Question 5
1 / -0

If the radius of $$B ^-$$ ion is $$0.180$$ nm, the size (in nm) of the cation that can fit in the tetrahedral void is:

A
$$0.125$$

B
$$0.04$$

C
$$0.08$$

D
$$0.180$$

Solution

For tetrahedral void, we have $$\dfrac {r_+ }{r_-}=0.225$$

$$\implies r_+=0.225\times r_-= 0.225 \times 0.180 =0.04$$ nm

Hence, the size of cation that can fit in the tetrahedral void is $$0.04$$ nm.
Question 6
1 / -0

If a diamond has a face-centered cubic structure with two atoms per lattice point and a unit cell of the edge of $$3.569\:\mathring{A}$$. The density (in g cm$$^{-3}$$) of the diamond is :

A
$$3.569$$

B
$$2.569$$

C
$$1.899$$

D
$$7.946$$

Solution

Diamond is a carbon allotrope. We can use density formula to calculate the density as follows:

Density $$=\rho =\dfrac { ZM }{ N_{ 0 }a^{ 3 } } $$

Substituting values in the above expression, we get $$\rho=\dfrac { 2\times 12 }{ (6.022\times { 10 }^{ 23 })\times (3.569\times { 10 }^{ -8 })^{ 3 } } $$

On solving the above equation, we get $$\rho$$ as $$3.569$$ g cm$$^{-3}$$.

Hence, the correct option is $$\text{A}$$
Question 7
1 / -0

Lithium borohydride $$(LiBH_4) $$ crystallizes in an orthorhombic system with $$4$$ molecules per unit cell. The unit cell dimensions are $$a = 6.8$$ $$\mathring {A } $$, $$b = 4.4$$ $$\mathring {A} $$ and $$c = 7.2$$ $$\mathring {A} $$. If the molar mass of the compound is $$21.76$$ g/mol, the density (in g cm$$^{-3}$$) of the compound will be:

A
$$0.67$$

B
$$0.33$$

C
$$1.72$$

D
none of the above

Solution

Density $$ = \dfrac{zM}{N_A \times (abc \times 10^{-24})}$$

Density $$ = \dfrac{4 \times 21.76}{6.023 \times 10^{23} \times (6.8 \times 4.4 \times 7.2 \times 10^{-24})}$$

Density $$= 0.67 \text{ gcm}^{-3}$$

Hence, the correct option is $$\text{A}$$
Question 8
1 / -0

If the effective radius of iron (Fe) atom is $$141.4$$ pm which has rock salt type structure, then its density (in g cm$$^{-3}$$) will be:
[$$Fe = 56$$ amu]

A
$$5.81$$

B
$$2.63$$

C
$$1.08$$

D
None of the above

Solution

In rock salt type structure its form fcc type unit cell where edge legth, $$a = 2\sqrt 2r $$

$$\implies a \approx 400 \text{ pm} $$

Density, $$d = \dfrac{zM}{N_A \times (a)^3}$$

For FCC structure, $$z=4$$

Now,

Density, $$d = \dfrac{4 \times 56}{6.02 \times 10^{23} \times (400 \times 10^{-10})^3} \\$$

Density, $$d =5.81 \text{ g cm}^{-3}$$

Hence, the correct option is $$\text{A}.$$
Question 9
1 / -0

The total fraction of space occupied by voids in simple cubic structure is:

A
$$0.58$$

B
$$0.25$$

C
$$0.48$$

D
$$0.86$$

Solution

Since packing fraction of a simple cubic structure is 0.52, thus, empty(void) space is 0.48.
Question 10
1 / -0

The compound CuCl has zinc blende structure and the edge length of its unit cell is $$500$$ pm, its density (in g cm$$^{-3}$$) is :
(Given that the atomic weight of $$Cu$$ is $$63.5$$ amu and of $$Cl $$ is $$35.5$$ amu.)

A
$$5.24$$

B
$$1.14$$

C
$$2.99$$

D
None of the above

Solution

density $$ = \dfrac{zM}{N_a \times (a \times 10^{-10})^3}$$
density $$ = \dfrac{4 \times 99}{6 \times 10^{23} \times (500 \times 10^{-10})^3}$$
density $$= 5.24 \text{ gcm}^{-3}$$

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Solid State Test - 27

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Solid State Test - 27

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SHARING IS CARING

Question 1

BCC structure

FCC strncture

SC structure

none of the above

Question 2

$$3.54$$

$$1.46$$

$$2.14$$

None of the above

Solution

Question 3

simple cubic

ccp

bcc

both simple cubic and bcc

Solution

Question 4

eutectic

liquid crystal

super-cooled

amorphous

Question 5

$$0.125$$

$$0.04$$

$$0.08$$

$$0.180$$

Solution

Question 6

$$3.569$$

$$2.569$$

$$1.899$$

$$7.946$$

Solution

Question 7

$$0.67$$

$$0.33$$

$$1.72$$

none of the above

Solution

Question 8

$$5.81$$

$$2.63$$

$$1.08$$

None of the above

Solution

Question 9

$$0.58$$

$$0.25$$

$$0.48$$

$$0.86$$

Solution

Question 10

$$5.24$$

$$1.14$$

$$2.99$$

None of the above

Solution

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