Solid StatehardNUMERICAL

An unknown element crystallises in an FCC lattice. Its theoretical density is given as and the unit Solid State Chemistry Question

Question

An unknown element crystallises in an FCC lattice. Its theoretical density is given as $10\text{ g cm}^{-3}$ and the unit cell edge length is exactly $400\text{ pm}$. What is the atomic mass of this element in $\text{g mol}^{-1}$? (Use $N_A = 6.0 \times 10^{23}\text{ mol}^{-1}$)

Answer: 96

💡 Solution & Explanation

Using the density formula: $M = \frac{d \times N_A \times a^3}{Z}$. We know $Z=4$ for FCC, $d = 10\text{ g cm}^{-3}$, and $a = 400\text{ pm} = 4 \times 10^{-8}\text{ cm}$. $M = \frac{10 \times 6.0 \times 10^{23} \times (4 \times 10^{-8})^3}{4} = \frac{10 \times 6.0 \times 10^{23} \times 64 \times 10^{-24}}{4} = \frac{384}{4} = 96\text{ g mol}^{-1}$.

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