Question

What is the volume of unit cell of a metal (at. mass $25\text{ g mol}^{-1}$) having BCC structure and density $3\text{ g cm}^{-3}$ ?

• A. $3.64 \times 10^{-23}\text{ cm}^3$
• B. $1.56 \times 10^{-24}\text{ cm}^3$
• C. $2.76 \times 10^{-23}\text{ cm}^3$
• D. $1.88 \times 10^{-24}\text{ cm}^3$

Hint:

Always remember the $Z$ values for the three standard cubic lattices: Simple Cubic (SC) $\rightarrow Z=1$, Body-Centered Cubic (BCC) $\rightarrow Z=2$, Face-Centered Cubic (FCC) $\rightarrow Z=4$.

Answer 1

The Correct Option is C

Step 1: Start from the density formula.
$$d = \frac{Z M}{N_A V} \Rightarrow V = \frac{Z M}{N_A d}$$

Step 2: Set the values.
For BCC, $Z = 2$. Also $M = 25$, $d = 3$, and $N_A = 6.022 \times 10^{23}$.

Step 3: Compute.
$$V = \frac{2 \times 25}{6.022 \times 10^{23} \times 3} \approx 2.76 \times 10^{-23}\ \text{cm}^3$$
\[ \boxed{2.76 \times 10^{-23}\ \text{cm}^3} \]