The Solid State

 Solution:
We have given that
Atomic mass of the metal = 50g
bcc unit cell, Z = 2
Density of metal = 5.96 g/cm³
Therefore, volume of the unit,


$$\begin{gathered} {\mathrm{a}}^3 = \frac{\mathrm{M}\times \mathrm{Z}}{{\mathrm{N}}_{\mathrm{A}}\times \mathrm{d}} \\ = \frac{50\times 2}{6.02\times {10}^{23}\times 5.96} \\ = 2.787 \times {10}^{-23} {\mathrm{cm}}^3 \end{gathered}$$

 Solution:
We have given that

Unit cell length,
a = 0.40806 nm  = 4.086 x 10^-10 m

  If fcc lattice the number of atoms per unit cell,
i.e. Z = 4

        M for Ag = 107.88 g mol<sup>-1

                      $=1.0788\times {10}^{-1} \mathrm{kg} {\mathrm{mol}}^{-1}$

              ${\mathrm{N}}_{\mathrm{A}} = 6.023 \times {10}^{23}$</sup>

Solution:
We have given that
Edge length of the unit cell
= 200 pm = 200 x 10^–10 cm

Vol. of the unit cell
= (200 x 10^–10 cm)³ = 8 x 10^–24 cm³
Therefore, volume of the substance
= Vol. of unit cell x Vol. of 1 unit cell.

Since the element has a fcc unit cell, number of atoms per unit cell = 4.
Total number of atoms = Atoms/unit cell x number of unit cells.
                                     = 24 x 10²³atoms = 4

atoms/unit cell x No. of unit cells.

∴   No. of unit cells = $\frac{24\times {10}^{23} \mathrm{atoms}}{4 \mathrm{atoms}/\mathrm{unit} \mathrm{cell}}$

                               $= 6\times {10}^{23} \mathrm{unit} \mathrm{cells}.$

∴  Volume of the substance
                   $6\times {10}^{23} \mathrm{unit} \mathrm{cell} \times 8\times {10}^{-24 }{\mathrm{cm}}^3/\mathrm{unit} \mathrm{cell} = 4.8 {\mathrm{cm}}^3$


Now,     $$\begin{gathered} \mathrm{Density} = \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ = \frac{200\mathrm{g}}{4.8 {\mathrm{cm}}^3} =4.17 \mathrm{g} {\mathrm{cm}}^{-3}. \end{gathered}$$

Solution:
we have given that 
Density, d = 6.23 g/cm³
              a = 400 pm
              M = 60g/mol^-1
 
Volume  = (a³)$=(400{)}^3 = (400\times {10}^{-10} \mathrm{cm}{)}^3$

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

$\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}=\frac{6.023\times {10}^{23} {\mathrm{mol}}^{-1}}{60 \mathrm{g}/\mathrm{mol}}$

$$\begin{gathered} \mathrm{Z} = \frac{6.023 \mathrm{g} {\mathrm{cm}}^{-3}\times (400{)}^3\times {10}^{-30} {\mathrm{cm}}^3\times 6.023\times {10}^{23} {\mathrm{mol}}^{-1}}{60 \mathrm{g} {\mathrm{mol}}^{-1}} \\ = \frac{6.023\times 64\times 6.023}{600} \\ = \frac{2401.49}{600}=4. \end{gathered}$$

Hence, the type of cubic unit cell is FCC

$\mathrm{Radius} = \frac{\mathrm{a}}{\mathrm{a}\sqrt{2}}=\frac{400}{2\sqrt{2}}=\frac{200}{\sqrt{2}}$

              $= \frac{200}{1.414}=141.4 \mathrm{pm}.$ 

So radius of element is 141.4 pm