Question

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).

Answer 1

Answer:

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

Explanation:

Given, edge length = 2.866 $\AA$

                                =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.

$Density=\frac{mass}{volume}$

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

Volume =( edge length)³× Avogadro Number = $a^3\times N_A$

z=  number of atom =2

M=atomic mass=?

a=edge length=2.866×10⁻¹⁰ m=$=2.866\times 10^{-10} \times 10^2 cm$

$N_A$ =Avogadro Number =6.023×10²³

$Desity(d)=\frac{Z\times M}{a^3\times N_A}$

$\Rightarrow M=\frac{d \times a^3\times N_A}{Z}$

$\Rightarrow M=\frac{7.87 g/cm^3 \times( 2.866\times 10^{-10} \times 10^2 cm)^3 \times 6.023\times 10^{23}/mol}{2}$

$\Rightarrow M = 557.94\times 10^{-30}\times 10^{29}$ g/mol

⇒M = 55.74 g/mol

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