Unit cell edge length = 400 pm = 400 x 10-10 cm
Volume of the unit cell = (400 x 10-10 cm)3 = 6.4 x 10-23 cm3
Mass of the unit cell = (No. of atoms in the unit cell) x (Mass of each atom)
No. of atoms in the bcc unit cell = \(8 \times \frac 18 + 1 = 2 \) atoms
Mass of one atom = Atomic mass / Avogadro number
= \(\frac{100 g / mol}{6.023 \times 10^{23} atoms/ mol}\)
\(\therefore\) Mass of the unit cell = \(\mathrm{\frac {2\ atoms / unit \ cell \ \times \ 100g/mol}{6.023 \times 10^{23}\ atoms / mol}}\)
\(\therefore\) Density of the unit cell = \(\mathrm {\frac {Mass \ of \ the \ cell}{Vol. \ of \ the\ unit \ cell}}\)
\(= \frac {2 \ atoms \times 100g / mol}{6.023 \times 10^{23} atoms / mol \times 6.4 \times 10^{23} cm^{3}}\)
\(= 5.188 g \ cm^{-3}\)
Calculation of the number of unit cells in 10 g of A
Volume of 10 g of A = \(\frac {Mass}{Density}\)
\(= \frac {10 g}{5.188 g \ cm^{-3}}\)
\(= 1.9275 \) cm3
No. of units in 1.9275 cm3 (10 g) of the substance = Vol. of 10 g of substance / Vol. of one unit cell
\(= \frac{1.9275 \ cm^3}{6.4 \times 10^{-23} cm^{3}}\)
\(= 3 \times 10^{22}\) unit cells
Calculation of the number of atoms in 10 g of A
Total number of atoms in 10 g of A = Atoms/unit cell x No. of unit cells in 10 g of A
= 2 atoms/unit cell x 3 x 1022 unit cells
= 6 x 1022 atoms
