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^{-23}g

`rho=19.3` g/cm^{-3}

Volume of unit cell=`(130.85xx10^-23)/19.3`

=6.78x10^{-23} cm^{3}

a^{3}=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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