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Question
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 `= 1/8 xx 8 + 6 xx 1/2`
= 4 atoms
Mass of unit cell of FCC `= 4 xx 197/(6.022 xx 10^23)`
= 130.85 × 10−23 g
`rho = 19.3` g/cm−3
Volume of unit cell = `(130.85xx10^-23)/19.3`
= 6.78 × 10−23 cm3
a3 = 6.78 × 10−23
a is edge of unit cell
`a = root(3)(6.78 xx 10^(-23)`
= 4.08 × 10−8 cm
For FCC
`a = sqrt8.r`
`r = a/sqrt8`
= `(4.08 xx 10^(-8))/sqrt8`
= 1.44 × 10−8 cm
= 144 pm
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