An element A crystallises in an FCC structure. Exactly of this element contains atoms. If the measur — Solid State Chemistry Question
Question
An element A crystallises in an FCC structure. Exactly $200\text{ g}$ of this element contains $4.12 \times 10^{24}$ atoms. If the measured density of the crystal is $7.2\text{ g cm}^{-3}$, calculate the edge length ($a$) of the unit cell in $\text{pm}$.
💡 Solution & Explanation
Mass of 1 atom = $\frac{200}{4.12 \times 10^{24}} = 4.854 \times 10^{-23}\text{ g}$. For FCC, $Z=4$, so mass of one unit cell = $4 \times 4.854 \times 10^{-23} = 19.41 \times 10^{-23}\text{ g}$. Using density $d = \frac{\text{Mass}}{\text{Volume}}$, Volume $a^3 = \frac{19.41 \times 10^{-23}}{7.2} = 2.696 \times 10^{-23}\text{ cm}^3 = 26.96 \times 10^{-24}\text{ cm}^3$. $a = (27 \times 10^{-24})^{1/3} = 3 \times 10^{-8}\text{ cm} = 300\text{ pm}$.