The density of graphite is 2.4 g/ml and the spacing between the layers is found by X-ray diffraction — Solid State Chemistry Question
Question
The density of graphite is 2.4 g/ml and the spacing between the layers is found by X-ray diffraction to be $2\sqrt{3} \text{ \AA}$. If the carbon–carbon distance in the molecular layers is $x \text{ \AA}$, then the value of $1.08 \times x^2$ is
💡 Solution & Explanation
The area of the primitive rhombus base of graphite is $\frac{3\sqrt{3}}{2} x^2 \text{ \AA}^2$, containing 2 carbon atoms. The volume for one layer spacing ($2\sqrt{3} \text{ \AA}$) is $\frac{3\sqrt{3}}{2} x^2 \times 2\sqrt{3} = 9x^2 \text{ \AA}^3 = 9x^2 \times 10^{-24} \text{ cm}^3$. Density $d = \frac{2 \times 12}{N_A \times V} \implies 2.4 = \frac{24}{6.022 \times 10^{23} \times 9x^2 \times 10^{-24}} \implies x^2 = \frac{24}{2.4 \times 6.022 \times 0.9} \approx 1.845$. Thus, $1.08 \times x^2 = 1.08 \times 1.845 \approx 1.99 \approx 2$. Therefore, correct answer is 2.