157. Caesium chloride crystallises as a body centred cubic lattice and has a density of 4.0 g cm^–3.  Calculate the length of the edge of the unit cell of caesium chloride crystal.
         [Molar Mass of CsCl = 168.5 g mol^–1, N_A = 6.02 x 10²³ mol^–1]

Solution;
we have given that    

   $\mathrm{Z} =1, \mathrm{\rho } = 4 \mathrm{g} {\mathrm{cm}}^{-3},$
        $$\begin{gathered} \mathrm{M} = 168.5 {\mathrm{mol}}^{-1}, \\ {\mathrm{N}}_{\mathrm{A}} = 6.02\times {10}^{23}{\mathrm{mol}}^{-1} \\ \mathrm{\rho } = \frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3\times \mathrm{N}} \end{gathered}$$

or          $$\begin{gathered} {\mathrm{a}}^3 = \frac{\mathrm{Z}\times \mathrm{M}}{\mathrm{\rho }\times {\mathrm{N}}_{\mathrm{A}}} \\ = \frac{1\times 168.5\mathrm{g} {\mathrm{mol}}^{-1}}{4\times 6.02\times {10}^{23}} \\ = \frac{168.5\times {10}^{-23}}{24.08} \end{gathered}$$
$$\begin{gathered} \log {\mathrm{a}}^3 = \log 168.5 + \log {10}^{-23} - \log 24.08 \\ 3 \log \mathrm{a} = 2.2266 - 23.000-1.3816 \\ 3 \log \mathrm{a} = -22.1550+2-2 \\ 3 \log \mathrm{a} = \overline{24}+(2.1550) \\ \mathrm{a} = \mathrm{Antilog} \overline{8}.7183 = 5.228\times {10}^{-8}\mathrm{cm} \\ = 5.228\times {10}^{-8}\times {10}^{10}\mathrm{pm} = 522.8 \mathrm{pm}. \end{gathered}$$