Question

The density of iron crystal is 8.54-gram cm^–3. If the edge length of the unit cell is 2.8 A° and atomic mass is 56 gram mol^–1, find the number of atoms in the unit cell.  (Given: Avogadro's number = 6.022x10²³,1A° = 1x10^-8 cm)

Answer 1

Given a = 2.8Å = ${\displaystyle 2.8\times {10}^{-8}}$ cm

Molar Mass = 56 gm

Density = 8.54 g cm^-3

Volume = (a)³

${\displaystyle ={(2.8\times {10}^{-8})}^3}$

${\displaystyle =21.95\times {10}^{-24}c{m}^3}$

Density of unit cell = ${\displaystyle \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}}$

∴ Mass of unit cell = density x volume

${\displaystyle =21.95\times {10}^{-24}\times 8.54}$

${\displaystyle =187.47\times {10}^{-24}}$

56 gm of Fe contains 6.022 x 10²³ atoms

${\displaystyle 187.47\times {10}^{-24}}$ gm Fe contain = ${\displaystyle \frac{6.022\times {10}^{23}\times 187.47\times {10}^{-24}}{56}}$

= ${\displaystyle 20.15\times {10}^{-1}}$

= 2.01

= 2

Number of atoms in unit cell is 2.