Solid State Test

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Solid State Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

Sodium crystallizes in the cubic lattice and the edge of the unit cell is $$ 430 pm$$. Calculate no of atoms in the unit cell.
[atomic mass of sodium = $$ 23.0$$ amu.
density of sodium = $$ 0.9623$$ g $$cm^{-3}$$.]

A
4

B
3

C
2

D
5

Solution

$$\rho=0.9623$$ $$gcm^{-3}$$
$$M_{Na}=23$$ $$amu$$
$$a=4.3 \mathring A$$
$$Z=?$$
$$\rho=\cfrac{Z\times M_A}{N_A\times a^3}$$
$$\implies 0.9623=\cfrac{Z\times 23}{6.023\times 10^{23}\times 79.5\times 10^{-24}}$$
$$\implies Z=2$$
Question 2
1 / -0

If the ratio of the coordination number of $$A$$ to that of $$B$$ is $$x:y$$, then the ratio of number of atoms of $$A$$ to that of $$B $$ in the unit cell is:

A
$$x:y$$

B
$$y:x$$

C
$${ x }^{ 2 }:y$$

D
$$y:{ x }^{ 2 }$$

Solution

Coordination number of A = Number of atoms of B
Coordination number of B = Number of atoms of A
So, The Formula is $${A}_{y}{B}_{x}$$ and the ratio of number of atoms is y : x
So option B is correct
Question 3
1 / -0

Silver crystallizes in fcc structure with edge length of unit cell, $$408.7pm$$. Hence density of silver is $$(Ag = 108\ g\ mol^{-1})$$

A
$$10.5\ g \ cm^{-3}$$

B
$$9.82\ g \ cm^{-3}$$

C
$$12.06\ g \ cm^{-3}$$

D
$$8.86\ g \ cm^{-3}$$

Solution

Density $$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$$
$$\displaystyle Z=4 \ atoms=$$ number of atoms in fcc unit cell
$$\displaystyle M=108 \ g/mol =$$ atomic mass of silver
$$\displaystyle N_0=6.023 \times 10^{23} \ atoms/mol=$$ Avogadro's number
$$\displaystyle a=408.7 \ pm = 408.7 \times 10^{-10} \ cm = $$ edge length of unit cell
$$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$$
$$\displaystyle \rho = \dfrac {4 \ atoms \times 108 \ g/mol}{6.023 \times 10^{23} \ atoms/mol \times (408.7 \times 10^{-10} \ cm)^3}$$
$$\displaystyle \rho = 10.5 \ g/cm^3$$
Question 4
1 / -0

In chromium chloride $$(CrCl_3)$$, $$Cl^-$$ ions have cubic close packed arrangement and $$Cr^{3+}$$ ions are present in the octahedral holes. The fraction of the total number of holes occupied is:

A
$$1/3$$

B
$$1/6$$

C
$$1/9$$

D
$$1/12$$

Solution

In CCP, No. of Tetrahedral holes $$=8$$
No. of Octahedral holes $$=4$$
Total $$=12$$ holes
$$Cr^{+3}$$ ions are in octahedral holes
Fraction of occupied holes $$=\cfrac{4}{12}=\cfrac{1}{3}$$
Question 5
1 / -0

Copper crystallizes in fcc structure and the edge length of the unit cell is $$3.61A^o$$. Hence the density of copper is:
(atomic mass of $$Cu = 63.5\ g \ mol^{-1}$$)

A
$$5.17\ g \ cm^{-3}$$

B
$$0.72\ g \ cm^{-3}$$

C
$$7.09\ g \ cm^{-3}$$

D
$$8.97\ g \ cm^{-3}$$

Solution

The density $$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$$
$$\displaystyle Z=4 \ atoms =$$ number of Cu atoms in one fcc unit cell.
$$\displaystyle M=63.5 \ g/mol = $$ atomic mass of copper.
$$\displaystyle N_0 = 6.023 \times 10^{23} \ atoms/mole=$$ Avogadro's number.
$$\displaystyle a=3.61 \ A^o =3.61 \times 10^{-8} \ cm = $$ edge length of the unit cell.
The density $$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$$
The density $$\displaystyle \rho = \dfrac {4 \ atoms \times 63.5 \ g/mol }{6.023 \times 10^{23} \ atoms/mole \times (3.61 \times 10^{-8} \ cm)^3}$$
The density $$\displaystyle \rho =8.97 \ g \ cm^{-3}$$
Question 6
1 / -0

The atomic radius of metal crystallizing in fcc structure true is $$1.25A^o$$. Hence the volume occupied by atoms in cell crystal is:

A
$$4.078\times 10^{-22}cm^3$$

B
$$6.09\times 10^{-23}cm^3$$

C
$$3.27\times 10^{-23}cm^3$$

D
$$2.68\times 10^{-22}cm^3$$

Solution

For $$FCC$$ unit cell
$$ 4r = \sqrt2\times a$$ or $$a = 2\sqrt2\times r = 3.5A^o$$
Volume $$= a^3 = (3.5\times 10^{-10})^3 = 4.078\times 10^{-22}cm^3$$
Question 7
1 / -0

The crystal system possess by $${K}_{2}{Cr}_{2}{O}_{7}$$ is:

A
Tetragonal

B
Cube

C
Hexagonal

D
Triclinic

Solution

The crystal system possessed by $$K_2Cr_2O_7$$ is Triclinic. It has been determined by single-crystal $$X-ray$$ diffraction.
Question 8
1 / -0

A metal $$M$$ is crystallised in $$F.C.C$$ lattice. The number of unit cells in it having $$2.4\times 10^{24}$$ atoms are:

A
$$6.023\times 10^{23}$$

B
$$\frac{6.023\times 10^{23}}{2}$$

C
$$2\times 6.023\times 10^{23}$$

D
$$4\times 6.023\times 10^{23}$$

Solution

In FCC lattice $$Z_M=4$$.
$$4$$ atoms of $$M$$ is present in a unit cell of $$M$$.
Number of unit cell $$=\cfrac{2.4}{4}\times 10^{24}$$ atom $$=6\times 10^{23}$$ atom $$\approx N_A$$ (Avogadro's number)
Question 9
1 / -0

The rank of atoms in the hexagonal until cell is:

A
4

B
3

C
5

D
6

Solution

Rank of atom is another term for effective number of atoms for a unit cell, i.e., $$Z_{effective}$$
Therefore, For HCP structure rank is $$6$$.
Question 10
1 / -0

In fcc, the neighbouring number of atoms for any lattice points is:

A
6

B
8

C
12

D
14

Solution

There are 12 nearest neighbors to a given lattice point in a FCC lattice. We say that the coordination number of a lattice point in the FCC lattice is 12. The distance of the nearest lattice point in terms of the lattice parameter (i.e, the edge length of the cubic unit cell) is $$\dfrac{a}{\sqrt{2}}$$

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Solid State Test - 45

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Solid State Test - 45

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Question 1

4

3

2

5

Solution

Question 2

$$x:y$$

$$y:x$$

$${ x }^{ 2 }:y$$

$$y:{ x }^{ 2 }$$

Solution

Question 3

$$10.5\ g \ cm^{-3}$$

$$9.82\ g \ cm^{-3}$$

$$12.06\ g \ cm^{-3}$$

$$8.86\ g \ cm^{-3}$$

Solution

Question 4

$$1/3$$

$$1/6$$

$$1/9$$

$$1/12$$

Solution

Question 5

$$5.17\ g \ cm^{-3}$$

$$0.72\ g \ cm^{-3}$$

$$7.09\ g \ cm^{-3}$$

$$8.97\ g \ cm^{-3}$$

Solution

Question 6

$$4.078\times 10^{-22}cm^3$$

$$6.09\times 10^{-23}cm^3$$

$$3.27\times 10^{-23}cm^3$$

$$2.68\times 10^{-22}cm^3$$

Solution

Question 7

Tetragonal

Cube

Hexagonal

Triclinic

Solution

Question 8

$$6.023\times 10^{23}$$

$$\frac{6.023\times 10^{23}}{2}$$

$$2\times 6.023\times 10^{23}$$

$$4\times 6.023\times 10^{23}$$

Solution

Question 9

4

3

5

6

Solution

Question 10

6

8

12

14

Solution

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