Question

X-ray diffraction studies show that a metal 'M' crystallises in an fcc unit cell with edge length of 400 pm. If the density of the metal is \(7.5 \times 10^3 { kg/m}^{-3}\), the number of unit cells present in 0.015 kg of it is:

• A. \(6.6 \times 10^{23}\)
• B. \(6.25 \times 10^{22}\)
• C. \(3.125 \times 10^{22}\)
• D. \(3.125 \times 10^{23}\)

Hint:

Ensure that calculations involving small scales such as picometers and molecular masses are done with precision to avoid significant rounding errors in final outcomes.

Answer 1

The Correct Option is C

Given:

• Crystal structure: fcc (Face-Centered Cubic)
• Edge length, \(a = 400 \, \text{pm} = 400 \times 10^{-12} \, \text{m} = 4 \times 10^{-10} \, \text{m}\)
• Density, \(\rho = 7.5 \times 10^{3} \, \text{kg/m}^3\)
• Mass of metal sample, \(m = 0.015 \, \text{kg}\)

Step 1: Calculate the volume of one unit cell (\(V_{\text{cell}}\)): \[ V_{\text{cell}} = a^3 = (4 \times 10^{-10} \, \text{m})^3 \] \[ = 4^3 \times (10^{-10})^3 \, \text{m}^3 \] \[ = 64 \times 10^{-30} \, \text{m}^3 \] Step 2: Calculate the mass of one unit cell (\(m_{\text{cell}}\)) using the density formula: \( \rho = \frac{m_{\text{cell}}}{V_{\text{cell}}} \). \[ m_{\text{cell}} = \rho \times V_{\text{cell}} \] \[ = 7.5 \times 10^{3} \, \text{kg/m}^3 \times 64 \times 10^{-30} \, \text{m}^3 \] \[ = (7.5 \times 64) \times (10^{3} \times 10^{-30}) \, \text{kg} \] \[ = 480 \times 10^{-27} \, \text{kg} \] \[ = 4.8 \times 10^{-25} \, \text{kg} \] Step 3: Calculate the number of unit cells present in 0.015 kg of metal: \[ \text{Number of unit cells} = \frac{\text{Total mass of metal}}{\text{Mass of one unit cell}} \] \[ = \frac{m}{m_{\text{cell}}} = \frac{0.015 \, \text{kg}}{4.8 \times 10^{-25} \, \text{kg}} \] \[ = \frac{0.015}{4.8} \times 10^{25} \] \[ = \frac{15 \times 10^{-3}}{48} \times 10^{25} \] \[ = \frac{15}{48} \times 10^{22} \] \[ = \frac{5}{16} \times 10^{23} \] \[ = 0.3125 \times 10^{23} \] \[ = 3.125 \times 10^{22} \] The number of unit cells present in 0.015 kg of the metal is \(3.125 \times 10^{22}\). 
Correct Answer: (3) \(3.125 \times 10^{22}\)