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 × 1023)

Answer 1

Answer:

3.6 x 10^20

Explanation:

this is the answer that i got

Answer 2

Answer:

The total number of voids in $38mg$ of osmium metal is $3.6 \times 10^{20}$.

Explanation:

Given,

The mass of osmium metal which crystallizes in a hcp lattice = $38mg$ = $0.038g$

The molar mass of osmium = $190u$ = $190g/mol$

The total number of voids in $38mg$ of osmium metal =?

As we know,

• The total number of atoms in one hcp lattice = $6$

Also,

• The number of octahedral voids in one hcp lattice = number of atoms in one hcp lattice = $6$
• The number of tetrahedral voids in one hcp lattice = 2( number of atoms in one hcp lattice ) = $2 \times 6$ = $12$

Therefore, the total number of voids in one hcp lattice = $6+12$ = $18$.

Now, as we know,

• The number of moles = $\frac{Given mass}{Molar mass}$ = $\frac{0.038}{190}$ = $0.0002mol$

Also,

• 1 mole = $6.022 \times 10^{23}$ atoms
• $0.0002mol$ = $0.0002mol\times 6.022 \times 10^{23}$ = $1.2 \times 10^{20}$ atoms.

Now,

• Number of units cells = $\frac{Total number of atoms}{Number of atoms in one hcp lattice}$ = $\frac{1.2 \times 10^{20}}{6}$ = $2 \times 10^{19}$

Thus,

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