Question

What is molar mass of metal with BCC structure having density $10\ \text{g}\ \text{cm}^{-3}$ and edge length 200 pm?

• A. $90.2\ \text{g}\ \text{mol}^{-1}$
• B. $24.1\ \text{g}\ \text{mol}^{-1}$
• C. $48.0\ \text{g}\ \text{mol}^{-1}$
• D. $60.5\ \text{g}\ \text{mol}^{-1}$

Hint:

Always convert picometers (pm) to centimeters (cm) before cubing when the density is in $\text{g/cm}^3$. The standard conversion factor is $1\ \text{pm} = 10^{-10}\ \text{cm}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must calculate the molar mass ($M$) of a metallic crystal given its unit cell structure (BCC), edge length ($a$), and mass density ($\rho$).

Step 2: Key Formula or Approach:
The density of a crystal unit cell is defined by the formula:
$$\rho = \frac{Z \times M}{a^3 \times N_A}$$ Rearranging to solve for molar mass ($M$):
$$M = \frac{\rho \times a^3 \times N_A}{Z}$$

Step 3: Detailed Explanation:
For a Body-Centered Cubic (BCC) lattice, the number of atoms per unit cell is $Z = 2$.
Density $\rho = 10\ \text{g}\ \text{cm}^{-3}$.
Edge length $a = 200\ \text{pm} = 2 \times 10^{-8}\ \text{cm}$.
Avogadro's number $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.
Substitute these values into the rearranged formula:
$$M = \frac{10 \times (2 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}$$ $$M = \frac{10 \times (8 \times 10^{-24}) \times 6.022 \times 10^{23}}{2}$$ $$M = \frac{80 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = 40 \times 6.022 \times 10^{-1}$$ $$M = 4 \times 6.022 = 24.088 \approx 24.1\ \text{g}\ \text{mol}^{-1}$$

Step 4: Final Answer:
The calculated molar mass is $24.1\ \text{g}\ \text{mol}^{-1}$, matching option (B).