Chemistry - Solid State Question with Solution | TestHub

Edge length of the unit cell, $a=400\text{pm}=4\times 10^{-8}\text{cm}$

Volume of the unit cell, $V=a^3=(4\times 10^{-8}\text{cm}{)}^3=64\times 10^{-24}{\text{cm}}^3$

Mass of one unit cell = Volume $\times$ Density = $64\times 10^{-24}{\text{cm}}^3\times 12.0{\text{g/cm}}^3=768\times 10^{-24}\text{g}$

Number of unit cells in $0.6\text{g}$ of metal = $\frac{0.6}{768\times 10^{-24}}=\frac{600}{768}\times 10^{21}=0.78125\times 10^{21}$

Since the metal forms an fcc unit cell, the number of atoms per unit cell is $4$.

Total number of atoms = $4\times 0.78125\times 10^{21}=3.125\times 10^{21}$

Answer: $3.125\times 10^{21}$