Question

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

• A. 126.356 u
• B. 63.178 u
• C. 31.589 u
• D. 47.383 u

Hint:

In an FCC lattice, there are 4 atoms per unit cell. The density formula \(\rho = \frac{Z \cdot M}{a^3 \cdot N}\) is key to finding the atomic weight.

Answer 1

The Correct Option is B

For an FCC lattice, the number of atoms per unit cell (\(Z\)) is 4. The density (\(\rho\)) of a crystal is given by the formula: \[ \rho = \frac{Z \cdot M}{a^3 \cdot N} \] where \(M\) is the atomic weight, \(a\) is the edge length, and \(N\) is Avogadro's number. Rearranging for \(M\): \[ M = \frac{\rho \cdot a^3 \cdot N}{Z} \] Given: \(\rho = 8.92 \, \text{g cm}^{-3}\), \(a = 3.61 \times 10^{-8} \, \text{cm}\), \(N = 6.022 \times 10^{23} \, \text{mol}^{-1}\), and \(Z = 4\). First, calculate \(a^3\): \[ a^3 = (3.61 \times 10^{-8})^3 = 4.7045 \times 10^{-23} \, \text{cm}^3 \] Now, substitute the values: \[ M = \frac{8.92 \times 4.7045 \times 10^{-23} \times 6.022 \times 10^{23}}{4} = \frac{8.92 \times 4.7045 \times 6.022}{4} \] \[ 8.92 \times 4.7045 = 41.96414 \] \[ 41.96414 \times 6.022 = 252.668451 \] \[ M = \frac{252.668451}{4} = 63.16711275 \, \text{g mol}^{-1} \approx 63.17 \, \text{u} \] Comparing with the options, the closest value is 63.178 u. Thus, the correct answer is 63.178 u.