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Total number of voids present in 38 mg of osmium metal which crystallizes as hcp lattice is? (Atomic mass of Os is 190 amu, N A = 6 × 10 23 )

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)

Grade:12

1 Answers

Priyanshu Gujjar
101 Points
one month ago

For an FCC crystal, atoms are in contact along the diagonal of the unit cell.
 4r=2a, 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=42×407=143.9 pm
Number of atoms per unit cell (FCC) =4
Mass of 1 unit cell =197×4=788amu=1.66×1027×788kg=1.30808×1024kg
Volume of 1 unit cell =(407×1012)3m3=6.741×1029m3 =2.824×1028m3
Density=6.741×10291.30808×1024=19404.83kg/m3=19.4g/cm3

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