Question

An element of atomic mass 107.9 g/mol has FCC unit cell.The edge length of the unit cell is 408.6 pm. Calculate the density of the unit cell.(Given NA =6.022×10^23mol.)

Answer 1

Answer:

4.88×10^13 g/mol

Explanation:

Apply formula

density=( no. of atoms in unitcell× atomic mass) ÷(NA×a^3)

here a is edge length

for fcc lattice no. of atoms=4

Answer 2

The density of given element is $10.51g/cm^3$

Explanation:

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 = 4  (FCC)

M = atomic mass of metal = 107.9 g/mol

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

a = edge length of unit cell = $408.6pm=408.6\times 10^{-10}cm$    (Conversion factor:  $1cm=10^{10}pm$  )

Putting values in above equation, we get:

$\rho=\frac{4\times 107.9}{6.022\times 10^{23}\times (408.6\times 10^{-10})^3}\\\\\rho=10.51g/cm^3$

Learn more about density of crystal lattice:

https://brainly.com/question/14592830

https://brainly.com/question/14698291

#learnwithbrainly