Question

An element with molar mass \( 2.7 \times 10^{-2} \) kg mol\(^{-1}\) forms a cubic unit cell with an edge length of 405 pm. If its density is \( 2.7 \times 10^3 \) kg m\(^{-3}\), the number of atoms present in one unit cell is: (Given: \( N_A = 6.023 \times 10^{23} \) mol\(^{-1}\))

• A. 2
• B. 4
• C. 6
• D. 12
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Hint:

For FCC unit cells, \( Z = 4 \), while for BCC and simple cubic, \( Z = 2 \) and \( Z = 1 \), respectively.

Answer 1

The Correct Option is B

Step 1: Using the Density Formula for a Unit Cell The number of atoms per unit cell (\( Z \)) can be calculated using the formula: \[ Z = \frac{\rho \times N_A \times a^3}{M} \] where: - \( \rho = 2.7 \times 10^3 \) kg m\(^{-3}\) (density) - \( M = 2.7 \times 10^{-2} \) kg mol\(^{-1}\) (molar mass) - \( a = 405 \) pm = \( 405 \times 10^{-12} \) m (edge length of the unit cell) - \( N_A = 6.023 \times 10^{23} \) mol\(^{-1}\) (Avogadro’s number) \bigskip Step 2: Calculating the Volume of the Unit Cell \[ a^3 = (405 \times 10^{-12})^3 \] \[ = 6.64 \times 10^{-29} \text{ m}^3 \] \bigskip Step 3: Substituting Values into the Formula \[ Z = \frac{(2.7 \times 10^3) \times (6.023 \times 10^{23}) \times (6.64 \times 10^{-29})}{2.7 \times 10^{-2}} \] \[ Z = \frac{(1.08 \times 10^{-2}) \times (6.023 \times 10^{23})}{2.7 \times 10^{-2}} \] \[ Z = \frac{6.5 \times 10^{21}}{2.7 \times 10^{-2}} \] \[ Z = 4 \] Thus, the number of atoms per unit cell is: \[ \mathbf{4} \] \bigskip