Question

The radius of Pt atom in an fcc structure is 0.1386 nm. Pt has atomic mass of 195.09 g/mol. Calculate the density of this fcc structure. The Avogadro’s number NAis 6.022 x 1023 per mole

Answer 1

21.51g/cm³

Step-by-step explanation:

To answer this question we need to know that, in 1 unit FCC cell:

Edge length = √8 * R

Volume = 8√8 * R³

And there are 4 atoms per unit cell

The mass of the 4 atoms of the cell, in grams, is:

4 atom * (1mol / 6.022x10²³atom) * (195.09g / mol) = 1.2958x10⁻²¹g

Volume in cm³:

0.1386nm * (1x10⁻⁷cm / 1nm) = 1.386x10⁻⁸cm

Volume = 8√8 * (1.386x10⁻⁸cm)³

Volume = 6.02455x10⁻²³cm³

And density, the ratio of mass and volume, is:

1.2958x10⁻²¹g / 6.02455x10⁻²³cm³ =

21.51g/cm³