Question

An element crystallises in fcc unit cell with cell edge length of \(3.608\times 10^{-8}\) cm, the density of element is \(8.92\text{ gcm}^{-3}\). Calculate the atomic mass of element (\(\text{N}_A = 6.022\times 10^{23}\)).

• A. \(60\) g/mol
• B. \(65\) g/mol
• C. \(63\) g/mol
• D. \(108\) g/mol

Hint:

Use M = rho a^3 N_A / Z with Z = 4 for fcc.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The density of a cubic crystal is linked to its unit cell edge and the atoms per cell. For fcc the number of atoms per unit cell is \(Z = 4\).

Step 2: Key Formula:
\[ \rho = \frac{Z\,M}{a^3 N_A} \quad\Rightarrow\quad M = \frac{\rho\,a^3 N_A}{Z} \]

Step 3: Compute the cell volume:
\[ a^3 = (3.608\times 10^{-8})^3 = 46.97\times 10^{-24} = 4.697\times 10^{-23}\ \text{cm}^3 \]

Step 4: Substitute:
\[ M = \frac{8.92\times 4.697\times 10^{-23}\times 6.022\times 10^{23}}{4} \]
The numerator is \(8.92\times 4.697 = 41.90\), and \(41.90\times 6.022 = 252.3\). Dividing by 4 gives
\[ M \approx 63.1\ \text{g/mol} \]

Step 5: Choose:
The closest option is 63 g/mol, option (C). The element is copper. If one wrongly used \(Z = 2\) or \(Z=1\) the result would be 126 or 252, which are not offered.

Final Answer:
The atomic mass is about 63 g/mol, option (C). \[ \boxed{63\ \text{g/mol}} \]