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#### Example – 1:

- Silver crystallizes in face centred cubic structure. The edge length of a unit cell is found to be 408.7 pm. Calculate the density of silver.(Ag = 108 g mol
^{-1}) **Solution:****Given:**The edge length of the unit cell = a = 408.7 pm = 408.7 x 10^{-10}cm, Atomic mass of silver = M = 108 g mol^{-1}, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = face centred cubic**To Find:**Density of silver =?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

#### Example – 2:

- Copper crystallizes in face centred cubic structure. The edge length of a unit cell is found to be 3.61 x 10
^{-8}cm. Calculate the density of copper if molar mass of copper is 63.5 g mol^{-1} **Solution:****Given:**The edge length of the unit cell = a = 3.61 x 10^{-8}cm, molar mass of copper = M = 63.5 g mol^{-1}, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = face centred cubic**To Find:**Density of copper =?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

#### Example – 3:

- Determine the density of caesium chloride which crystallizes in bcc type structure with an edge length of 412.1 pm. The atomic mass of Cs and Cl are 133 and 35.5 respectively.
**Solution:****Given:**The edge length of the unit cell = a = 412.1 pm = 412.1 x 10^{-10}cm, atomic mass of Cs = 133, atomic mass of Cl = 35.5, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = bcc**To Find:**Density of caesium chloride =?

Volume of unit cell = a^{3} = (4.121 x 10^{-8} cm)^{ 3} = (4.121)^{ 3 }x 10^{-24} cm^{3}.

Mass of one atom of Cs = Atomic mass of Cs / Avogadro’s number

Mass of one atom of Cs = 133 g mol^{-1} / 6.022 x 10^{23} mol^{-1}

Mass of one atom of Cs = 22.09 x 10^{-23} g

Mass of one atom of Cl = Atomic mass of Cl / Avogadro’s number

Mass of one atom of Cl = 35.5 g mol^{-1} / 6.022 x 10^{23} mol^{-1}

Mass of one atom of Cl = 5.894 x 10^{-23} g

Caesium chloride has a body centred cubic structure such that caesium atom is at centre

and eight chlorine atom at the eight corners of the cube.

Thus there is 1/8 x 8 = 1atom of Cl and 1 atom of Cs.

Thus mass of unit cell of caesium chloride = 22.09 x 10^{-23} g + 5.894 x 10^{-23} g

Thus mass of unit cell of caesium chloride = 27.98 x 10^{-23} g

Density of caesium chloride = mass of unit cell/volume of unit cell

Density of caesium chloride = 27.98 x 10^{-23} g /(4.121)^{ 3 }x 10^{-24} cm^{3}.

Density of caesium chloride = 3.997 g/cm^{3} = 4 g/cm^{3}

#### Example – 4:

- A metal crystallises as fcc structure and the unit cell has a length of edge 3.72 x 10
^{-8}cm. Calculate the density of the metal. Given the atomic mass of metal as 68.5 g mol^{-1}. **Solution:****Given:**The edge length of the unit cell = a = 3.72 x 10^{-8}cm, atomic mass of copper = M = 68.5 g/mol, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = face centred cubic**To Find:**Density of metal = ?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

#### Example – 5:

- Sodium crystallises in bcc structure. If the atomic radius of sodium is 186 pm. Find a) edge length b) volume of the unit cell and c) density of sodium crystal. Atomic mass of sodium is 23 g mol
^{-1}. **Solution:****Given:**Atomic radius of sodium = r = 186 pm = 186 x 10^{-10}cm = 1.86 x 10^{-8}cm, atomic mass of copper = M = 23 g/mol, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = bcc**To Find:**Edge length = a = ? Volume of unit cell =? Density of sodium =?

Volume of unit cell = a^{3} = (4.30 x 10^{-8} cm)^{ 3} = (4.3)^{ 3 }x 10^{-24} cm^{3 }= 79.4 x 10^{-24} cm^{3}

The density of sodium is 0.962 g cm^{-3}.

**Example – 6:**

- copper crystallises in fcc type unit cell. The edge length of a unit cell is 360.8 pm. The density of metallic copper is 8.92 g cm
^{-3}. Determine atomic mass of copper. **Solution:****Given:**The edge length of the unit cell = a = 360.8 pm = 360.8 x 10^{-10}cm = 3.608 x 10^{-8}cm, Density of copper = 8.92 g cm^{-3}, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = fcc**To Find:**the atomic mass of copper =?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Atomic mass of copper is 63.07 g mol^{-1}.

#### Example – 7:

- Silver crystallises in fcc type unit cell. The edge length of a unit cell is 4.07 10
^{-8}cm. The density of metallic silver is 10.5 g cm^{-3}. Determine atomic mass of silver. **Solution:****Given:**The edge length of the unit cell = a = 4.07 x 10^{-8}cm, Density of silverr = 10.5 g cm^{-3}, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = fcc**To Find:**Atomic mass of silver = ?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Atomic mass of silver is 106.6 g mol^{-1}

#### Example – 8:

- Face centred cubic crystal lattice of copper has a density of 8.966 g cm
^{-3}. Calculate the volume of the unit cell. Given the molar mass of copper 63.5 g/mol. - Solution:
- Given: Molar mass of copper = 63.5 g mol
^{-1}, Density of copper = 8.966 g cm^{-3}, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = fcc - To Find: Volume of unit cell =?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

The volume of the unit cell is 4.704 x 10^{-23} cm^{3}

#### Example – 9:

- Silver crystallizes as fcc structure. If the density of silver is 10.51 g cm
^{-3}. calculate the volume of a unit cell. Given: molar mass of silver 108 g/mol. **Solution:****Given:**Molar mass of silver = 108 g mol^{-1}, Density of silver = 10.51 g cm^{-3}, Avogadro’s number N = 6.022 x 10^{23}mol^{-1}. Type of crystal structure = fcc**To Find:**Volume of unit cell =?

The number of atoms in the unit cell of a face centred cubic structure is n = 4

The volume of the unit cell is 6,825 x 10^{-23} cm^{3}

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