Question

Phosphorus crystallizes in several different forms, one of which is a simple cube with an edge length of 238 pm. What is the density of phosphorus in the simple cubic crystal form?

Answer 1

Answer: The density of phosphorus is $3.81g/cm^3$

Explanation:

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

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

where,

$\rho$ = density

Z = number of atom in unit cell = 1  (CCP)

M = atomic mass of phosphorus = 31 g/mol

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

a = edge length of unit cell = $238pm=238\times 10^{-10}cm$    (Conversion factor:  $1cm=10^{10}pm$  )

Putting values in above equation, we get:

$\rho=\frac{1\times 31}{6.022\times 10^{23}\times (238\times 10^{-10})^3}\\\\\rho=3.81g/cm^3$

Hence, the density of phosphorus is $3.81g/cm^3$