Question

The crystal structure of an element has fcc lattice. If the edge length of the crystal is \(4\ \text{\AA}\), what is the atomic weight \((g\ mol^{-1})\) of the element, if the density of the crystal is \(11.21\ g\ cm^{-3}\)? \[ (N_A=6.023\times10^{23}\ mol^{-1}) \]

• A. 63.5
• B. 85.5
• C. 108.0
• D. 197.0

Hint:

For crystal density problems, always use \[ \rho=\frac{ZM}{N_Aa^3} \] where \(Z=4\) for fcc, \(Z=2\) for bcc and \(Z=1\) for simple cubic lattices.

Answer 1

The Correct Option is C

Step 1: Use the density relation for a crystal lattice.
The density of a crystal is given by \[ \rho=\frac{ZM}{N_Aa^3} \] where \[ \rho=11.21\ g\,cm^{-3} \] \[ Z=4 \quad \text{(for fcc lattice)} \] \[ M=\text{atomic mass} \] \[ N_A=6.023\times10^{23}\ mol^{-1} \] \[ a=4\ \text{\AA} \]

Step 2: Convert edge length into centimeters.
Since \[ 1\ \text{\AA}=10^{-8}\ cm \] \[ a=4\times10^{-8}\ cm \] Therefore, \[ a^3=(4\times10^{-8})^3 \] \[ =64\times10^{-24} \] \[ =6.4\times10^{-23}\ cm^3 \]

Step 3: Substitute the values into the density formula.
\[ 11.21 = \frac{4M} {(6.023\times10^{23})(6.4\times10^{-23})} \] First calculate the denominator: \[ (6.023\times10^{23})(6.4\times10^{-23}) = 38.5472 \] Hence, \[ 11.21=\frac{4M}{38.5472} \]

Step 4: Solve for \(M\).
\[ 4M = 11.21\times38.5472 \] \[ 4M \approx432 \] \[ M = \frac{432}{4} \] \[ M = 108 \]

Step 5: Final conclusion.
Therefore, the atomic weight of the element is \[ \boxed{108.0\ g\ mol^{-1}} \] Hence, option (3) is correct.