Answer in Brief
Sum
Gold occurs as face centred cube and has a density of 19.30 kg dm-3. Calculate atomic radius of gold (Molar mass of Au = 197)
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Solution
Unit cell of FCC `=1/8xx8+6xx1/2`
=4 atoms
Mass of unit cell of FCC `=4xx197/(6.022xx10^23)`
=130.85 x 10-23g
`rho=19.3` g/cm-3
Volume of unit cell=`(130.85xx10^-23)/19.3`
=6.78x10-23 cm3
a3=6.78x10-23
a is edge of unit cell
`a= root(3)(6.78xx10^(-23)`
=4.08x10-8 cm
For FCC
`a=sqrt8.r`
`r=a/sqrt8=(4.08xx10^(-8))/sqrt8`
r=1.44x10-8 cm
r=144 pm
Concept: Number of Atoms in a Unit Cell
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