Question

An imaginary element with BCC structure and has an atomic radius of 0.17 nm, with a molar mass of 56.08 g/mol. What is the density of this element in g/cc? hint: you will need Avogadro's number and you will need to convert the given radius to cm.

Answer 1

Answer: The density of the given element is $3.07g/cm^3$

Explanation:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = 0.17 nm

a = edge length = ?

Putting values in above equation, we get:

$0.17=\frac{\sqrt{3}\times a}{4}\\\\a=\frac{0.17\times 4}{\sqrt{3}}=0.393nm$

To calculate the density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal = 56.08 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = $0.393nm=3.93\times 10^{-8}cm$    (Conversion factor:  $1cm=10^{7}nm$  )

Putting values in above equation, we get:

$\rho=\frac{2\times 56.08}{6.022\times 10^{23}\times (3.93\times 10^{-8})^3}\\\\\rho=3.07g/cm^3$

Hence, the density of the given element is $3.07g/cm^3$