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: Understanding the Question:
We need to calculate the volume of a single unit cell ($V$) given the metal's molar mass, its crystal structure type, and its bulk density.

Step 2: Key Formula or Approach:
The theoretical density ($d$) of a crystal lattice is given by the formula:
$$d = \frac{Z \times M}{N_A \times V}$$
Where $Z$ is the number of atoms per unit cell, $M$ is the molar mass (g/mol), $N_A$ is Avogadro's number ($6.022 \times 10^{23}\text{ mol}^{-1}$), and $V$ is the volume of the unit cell ($\text{cm}^3$).
Rearranging to solve for Volume ($V$):
$$V = \frac{Z \times M}{N_A \times d}$$

Step 3: Detailed Explanation:
First, identify the values for the variables:
$M = 25\text{ g mol}^{-1}$
$d = 3\text{ g cm}^{-3}$
For a Body-Centered Cubic (BCC) structure, the number of atoms per unit cell is $Z = 2$.
Substitute these values into the rearranged formula:
$$V = \frac{2 \times 25}{6.022 \times 10^{23} \times 3}$$
$$V = \frac{50}{18.066 \times 10^{23}}$$
$$V = \frac{50}{18.066} \times 10^{-23}\text{ cm}^3$$
$$V \approx 2.767 \times 10^{-23}\text{ cm}^3$$
Rounding to two decimal places to match the options gives $2.76 \times 10^{-23}\text{ cm}^3$.

Step 4: Final Answer:
The volume of the unit cell is $2.76 \times 10^{-23}\text{ cm}^3$, corresponding to option (C).