Question

# Naturally occurring gold crystallises in face centred cubic structure and has a density of $$19.3$$ g cm$$^{-3}$$. Find atomic radius of gold. (Au$$=197$$g mol$$^{-1}$$)

Solution

## density of Gold = $$19.3 g/cm^3$$density = $$\dfrac{z M}{9^3 . NA}$$$$19.3 = \dfrac{4 \times 197}{9^3. 6.022 \times 10^{23}}$$$$9^3 = \dfrac{4 \times 197}{6.022 \times 10^{23} \times 19.3}$$ value of $$z = 4$$$$9^3 = 6.78 \times 10^{-23} = 4.077 \times 10^{-8}$$$$\sqrt{2} a = 4^r$$$$r = \dfrac{\sqrt{2} 9}{4} = \dfrac{\sqrt{2} \times 4.077 \times 10^{-8}}{4}$$$$= 1.44 \times 10^{-8} cm$$Chemistry

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