Question

An element with a molar mass 27g mol\(^{-1}\) forms a cubic unit cell. Calculate the number of atoms present in a unit cell if the density of metal is 2.7g cm\(^{-3}\).
\([a^3\times N_A = 40\) cm\(^3\) mol\(^{-1}]\)

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(6\)

Hint:

Use density = Z x M / (a^3 x N_A) and the given value of a^3 x N_A to find Z.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The density of a crystal is the mass of the atoms in one unit cell divided by the volume of that cell. From that we can count how many atoms \(Z\) the cell holds.

Step 2: Key Formula or Approach:
\[ d = \frac{Z \times M}{a^3 \times N_A} \quad \Rightarrow \quad Z = \frac{d \times (a^3 N_A)}{M} \]

Step 3: Detailed Explanation:
Given \(d = 2.7\text{ g cm}^{-3}\), \(M = 27\text{ g mol}^{-1}\) and \(a^3 N_A = 40\text{ cm}^3\text{ mol}^{-1}\).
\[ Z = \frac{2.7 \times 40}{27} = \frac{108}{27} = 4 \]
Four atoms per cubic unit cell means the structure is face-centred cubic (8 corners x 1/8 + 6 faces x 1/2 = 4).

Step 4: Why the other options are wrong.
\(Z = 1\) is simple cubic, \(Z = 2\) is body-centred cubic. With \(Z = 1\) the density would be only 0.675 g/cm\(^3\) and with \(Z = 2\) only 1.35 g/cm\(^3\), not 2.7. \(Z = 6\) is not a common cubic arrangement.

Final Answer:
Each unit cell holds 4 atoms, option (C). \[ \boxed{Z = 4} \]