Gold has a close packed structure which can be viewed as spheres occupying 0.74 of the total volume. — Solid State Chemistry Question
Question
Gold has a close packed structure which can be viewed as spheres occupying 0.74 of the total volume. If the density of gold is $19.7 \text{ g/ml}$, the atomic radius of gold (in pm) is $[\text{Au} = 197, (1.43)^3 \times 4\pi = 37]$
💡 Solution & Explanation
Volume of 1 mole of Au $= \frac{197 \text{ g}}{19.7 \text{ g/ml}} = 10 \text{ cm}^3$. Volume occupied by atoms in 1 mole $= 0.74 \times 10 = 7.4 \text{ cm}^3$. Volume of 1 Au atom $= \frac{7.4}{6.022 \times 10^{23}} \approx 12.3 \times 10^{-24} \text{ cm}^3$. Thus, $\frac{4}{3}\pi r^3 = 12.33 \times 10^{-24} \implies 4\pi r^3 = 37 \times 10^{-24}$. Given $(1.43)^3 \times 4\pi = 37$, $r^3 = (1.43 \times 10^{-8})^3 \implies r = 1.43 \times 10^{-8} \text{ cm} = 143 \text{ pm}$. Therefore, correct answer is 0143.