Question

Gold has a face-centered cubic arrangement with a unit cell edge length of 4.08 Å . How many moles of gold fit in a gold nanoparticle sheet with a length of 88.5 nm , a width of 26.8 nm , and a thickness of 13.1 nm ?

Answer 1

Volume of 1 FCC unit cell = (4.08 x 10^-10m)³/unit cell = 6.792 x 10^-29 m³/unit cell

1 unit cell of FCC has 4 atoms

we have total volume = 88.5 x 10^-9 m x 26.8 x 10^-9m x 13.1 x 10^-9m = 3.107 x 10^-23 m³

no. of unit cells in this volume = 3.107 x 10^-23 m³ / 6.792 x 10^-30 m³/unit cell = 4.57 x 10⁵ unit cell

total no. of atoms = 4 x no. of unit cells = 4 x 4.57 x 10⁵ = 1.828 x 10⁶ atoms

1 mol = 6.023 x 10²³ atoms

1.828 x10⁶ atoms = 1.828 x 10⁶ atoms / 6.023 x 10²³ = 3.035 x 10^-18 mol Au atoms