Question

The density of $\beta$-Fe is 7.6 g/cm$^3$. It crystallizes in a cubic lattice with $a = 290$ pm. 
What is the value of $Z$? ($Fe = 56$ g/mol, $N_A = 6.022 \times 10^{23}$ mol$^{-1}$)

• A. $2$
• B. $1$
• C. $4$
• D. $6$

Hint:

- The number of atoms per unit cell $Z$ helps determine the crystal structure: - $Z = 1$ for simple cubic, - $Z = 2$ for BCC (body-centered cubic), - $Z = 4$ for FCC (face-centered cubic).

Answer 1

The Correct Option is A

Step 1: Use the density formula for a unit cell $${Density} = \frac{Z \cdot M}{N_A \cdot a^3}$$ Where: - $Z$ = number of atoms per unit cell - $M$ = molar mass of Fe = 56 g/mol - $N_A$ = Avogadro’s number = $6.022 \times 10^{23}$ - $a = 290$ pm = $290 \times 10^{-10}$ cm - Given density $\rho = 7.6$ g/cm³ 
Step 2: Calculate the unit cell volume $$a^3 = (290 \times 10^{-10})^3 = 2.44 \times 10^{-23} { cm}^3$$ Step 3: Solve for $Z$ $$Z = \frac{\rho \cdot N_A \cdot a^3}{M}$$ $$Z = \frac{(7.6) \times (6.022 \times 10^{23}) \times (2.44 \times 10^{-23})}{56}$$ $$Z = \frac{1.116 \times 10^1}{56} = 2$$ Thus, $Z = 2$, indicating a body-centered cubic (BCC) structure.