What is the approximate density in g/cm³ of titanium metal if the atomic radius of a titanium atom is 0.145 nm and the atoms form a body-centered cubic unit cell?

Question

Quizplus · Accepted Answer

In a body-centered cubic (bcc) unit cell, there are 2 atoms per unit cell. The density ($\rho$) can be calculated using the formula $\rho = \frac{Z \times M}{a^3 \times N_A}$, where $Z$ is the number of atoms per unit cell (2 for bcc), $M$ is the molar mass of titanium ($47.867 \, \text{g/mol}$), $a$ is the edge length of the unit cell, and $N_A$ is Avogadro's number ($6.022 \times 10^{23} \, \text{mol}^{-1}$). The edge length ($a$) can be found from the atomic radius ($r$) since for a bcc structure, $a = 4r/\sqrt{3}$. Given $r = 0.145 \, \text{nm} = 0.145 \times 10^{-7} \, \text{cm}$, we find $a$ and then calculate the density. The calculated density is approximately $4.5 \, \text{g/cm}^3$ to $5 \, \text{g/cm}^3$, making option C the closest answer.

What Is the Approximate Density in G/cm3 of Titanium Metal

What Is the Approximate Density in G/cm³ of Titanium Metal