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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 $$z = 4$$
$$9^3 = 6.78 \times 10^{-23} = 4.077 \times 10^{-8}$$
$$\sqrt{2} a = 4^r$$
$$r = \dfrac{\sqrt{2} 9}{4} = \dfrac{\sqrt{2} \times 4.077 \times 10^{-8}}{4}$$
$$= 1.44 \times 10^{-8} cm$$

1060687_878006_ans_c0a25b768e8f478a90fd511911f1d95a.png

Chemistry

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