A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10⁻¹⁰ m). The density of the metal is 7.87 g/cm³. What is the mass of an atom of this metal? (1 m = 10¹² pm).

The mass of an atom of this metal is 9.26 × 10⁻²³ grams

Given that:

The density of a unit cell can be computed as:

where;

∴

Making the molar mass (M) the subject, we have:

where;

Now, the molar mass (M) can be estimated as:

Finally, the mass of an atom of this metal is:

mass (m) of an atom of this metal = 9.26 × 10⁻²³ grams

Learn more about body-centered cubic structure here:

link

Therefore the mass of an atom of this metal is 55.74 g/mol.

Explanation:

Given, edge length = 2.866

                                =2.866×10⁻¹⁰ m.

Density of the metal = 7.87 g/ cm³

Body centered cubic has 2 atom per unit cell.

Atomic mass: Atomic mass is mass of the component per atom.

Density: Density is the ratio of mass to volume.

mass = number of atom × atomic mass = Z×M

Volume =( edge length)³× Avogadro Number =

z=  number of atom =2

M=atomic mass=?

a=edge length=2.866×10⁻¹⁰ m=

=Avogadro Number =6.023×10²³

g/mol

⇒M = 55.74 g/mol

Therefore the mass of an atom of this metal is 55.74 g/mol.

thats how its supposed to be