Calculate number of atoms per unit cell of an element having molar mass 92.0 g mol − 1 and density 8.6 g cm − 3 forming cubic unit cell structure. [ a 3 × N A = 21.5 cm 3 mol − 1 ]

Question

Calculate number of atoms per unit cell of an element having molar mass 92.0 g mol − 1 and density 8.6 g cm − 3 forming cubic unit cell structure. [ a 3 × N A ​ = 21.5 cm 3 mol − 1 ]

TestHub · Accepted Answer

The density of a cubic unit cell is given by the formula:

$d = \frac{Z \times M}{a ^{3} \times N _{A}}$

Where:
$d$ is the density of the unit cell
$Z$ is the number of atoms per unit cell
$M$ is the molar mass
$a^{3}$ is the volume of the unit cell
$N_{A}$ is Avogadro's number

Rearranging the formula to solve for $Z$ :

$Z = \frac{d \times ( a ^{3} \times N _{A} )}{M}$

Substituting the given values ( $d = 8.6 g cm^{- 3}$ , $a^{3} \times N_{A} = 21.5 cm^{3} mol^{- 1}$ , $M = 92.0 g mol^{- 1}$ ):

$Z = \frac{8.6 \times 21.5}{92.0}$

$Z = \frac{184.9}{92.0}$

$Z \approx 2$

The number of atoms per unit cell is $2$ , which corresponds to a body-centered cubic (bcc) structure.

Answer: 2

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