Question

Total number of voids present in 38 mg of osmium metal which crystallizes as hcp lattice is? (Atomic mass of Os is 190 amu, Na = 6 x 1023)​

Answer 1

Answer:

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Answer 2

Concept:

A hexagonal close packing arrangement has one additional atom in the center and is made up of alternating layers of spheres or atoms stacked in a hexagon shape. Between these two hexagonal layers is a layer of atoms that is triangular in shape and whose atoms fill the tetrahedral gaps left by the top and bottom levels.

Given:

The mass of osmium metal = 38 mg = 0.038 g

The atomic mass of osmium = 190 amu = 190 g/mol

Avogadro's number = 6 x $10^{23}$

Find:

The total number of voids present in 38 mg of osmium metal that crystallizes as hcp lattice is?

Solution:

We are aware that one hcp lattice has a total of 6 atoms.

• In addition, the number of octahedral voids in one hcp lattice is equal to the number of atoms in one hcp lattice = 6

• The number of tetrahedral voids in one hcp lattice = 2(number of atoms in one at hcp lattice) = 2 * 6 = 12

• As a result, the total number of voids in hcp lattice = 6 + 12 = 18

Currently, we know that the number of moles can be calculated as:

      Number of moles of osmium = $\frac{Given mass}{Molar mass}$ = $\frac{0.038 g}{190 g/mol}$ = 0.0002 mol

We know that,

• 1 mole = 6.022 * $10^{23}$ atoms

• So, 0.0002 mol = 0.0002 mol * 6.022 * $10^{23}$ atoms

Thus,

• The number of unit cells = $\frac{total number of atoms}{number of atoms in HCP lattice}$ = $\frac{1.2 * 10^{20}}{6}$ = $2 * 10^{19}$

• The total number of voids in 38 mg osmium metal = number of units cells × voids in one unit cell
• The total number of voids in 38 mg of osmium metal = $2 * 10^{19} * 18 = 3.6 * 10^{20}$  

Hence, the total number of voids present in 38 mg of osmium metal which crystallizes as hcp lattice is $3.6 * 10^{20}$.

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