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\(^{-1}\))

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- 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).
Updated On: May 22, 2025
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The Correct Option is A

Solution and Explanation

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.

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