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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, 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)

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
94 Points
2 years ago

For an FCC crystal, atoms are in contact along the diagonal of the unit cell.
 4r=2​a, 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=1.66×10−27×788kg=1.30808×10−24kg
Volume of 1 unit cell =(407×10−12)3m3=6.741×10−29m3 =2.824×10−28m3
Density=6.741×10−291.30808×10−24​=19404.83kg/m3=19.4g/cm3

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