Question

An element crystallising in fcc lattice has a density of \(8.92\,\text{g cm}^{-3}\) and edge length of \(3.61\times 10^{-8}\,\text{cm}\). What is the atomic weight of element?
\((N=6.022\times 10^{23}\,\text{mol}^{-1})\)

• A. \(126.356\,u\)
• B. \(63.178\,u\)
• C. \(31.589\,u\)
• D. \(47.383\,u\)

Hint:

For cubic crystals, \[ \rho=\frac{ZM}{a^3N_A} \] Remember: \[ Z=4 \text{ for fcc lattice} \]

Answer 1

The Correct Option is B

Step 1: Use the density formula for unit cell.
Density of a crystal is given by \[ \rho=\frac{ZM}{a^3N_A} \] where \[ \rho=\text{density} \] \[ Z=\text{number of atoms per unit cell} \] \[ M=\text{molar mass} \] \[ a=\text{edge length} \] \[ N_A=\text{Avogadro number} \]

Step 2: Identify the values.
For fcc lattice, \[ Z=4 \] Given, \[ \rho=8.92\,\text{g cm}^{-3} \] \[ a=3.61\times 10^{-8}\,\text{cm} \] \[ N_A=6.022\times 10^{23}\,\text{mol}^{-1} \]

Step 3: Rearrange the formula for molar mass.
\[ M=\frac{\rho a^3N_A}{Z} \] Substituting the values, \[ M= \frac{ (8.92)(3.61\times 10^{-8})^3(6.022\times 10^{23}) }{4} \]

Step 4: Calculate \(a^3\).
\[ (3.61\times 10^{-8})^3 = 3.61^3\times 10^{-24} \] \[ =47.045\times 10^{-24} \] \[ =4.7045\times 10^{-23}\,\text{cm}^3 \] Now, \[ M= \frac{ (8.92)(4.7045\times 10^{-23})(6.022\times 10^{23}) }{4} \] \[ M= \frac{ 252.712 }{4} \] \[ M\approx 63.178 \] Thus, \[ M=63.178\,u \]

Step 5: Final conclusion.
Therefore, the atomic weight of the element is \[ \boxed{63.178\,u} \]