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:

I don't know

Explanation:

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Answer 2

Total number of voids present in 38 mg of osmium metal which crystallizes as hcp lattice is  $3.6$ x $10^{20}$

Explanation:

• For an FCC crystal, atoms are in contact along the diagonal of the unit cell
• Given $38 mg$ of osmium metal
• Atomic mass of Os is $190$
• Na =$6$ x $10^{23}$
• The relation between edge length (a) and radius of atom (r) for FCC lattice is √$(2a)$= $4r$ .
• where a is the edge length of unit cell and r is the radius of atom.
• Substituting values in the above expression, we get
• r=$143.9 pm$
• Number of atoms per unit cell (FCC) =4
• Mass of 1 unit cell =$190$×$4$=$780$amu
• $1.66$×$10^{-24}$ ×$780$kg =$1.30808$×$10^{-24}$kg
• Volume of $1$ unit cell =($407$×$10^{-12}$ )$m3$ =$32.824$×$10^{-28}$ $m3$
• Density=$19.4g/cm3$

• Total number of voids present in 38 mg of osmium metal which crystallizes as hcp lattice is  $3.6$ x $10^{20}$