Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

As sphere are touching each other

Therefore a = 2r

No. of spheres per unit cell = 1/8 × 8 = 1 

Volume of the sphere = 4/3 πr3 

Volume of the cube = a3= (2r)3 = 8r3

∴ Fraction of the space occupied = 1/3πr3 / 8r3 = 0.524

∴ % occupied = 52.4 %

we have 
edge length =200Pm
volume of the unit = ( 200 x10^-10 cm)³
                         = 8 x10^-24 cm^-3

$$\begin{gathered} \mathrm{In} \mathrm{a} \mathrm{FCC} \mathrm{unit} \mathrm{cell} \mathrm{there} \mathrm{are} \mathrm{four} \mathrm{atoms} \mathrm{per} \mathrm{unit} \mathrm{cell}. \\ \mathrm{therefore} \mathrm{mass} \mathrm{of} \mathrm{unit} \mathrm{cell} =\frac{200 \mathrm{x} 4}{24 \mathrm{x}{10}^{23}} =33.3 \mathrm{x}{10}^{-23}\mathrm{g} \\ \mathrm{Density} =\frac{\mathrm{mass} \mathrm{of} \mathrm{unit} \mathrm{cell} }{\mathrm{volume} \mathrm{of} \mathrm{unit} \mathrm{cell}} =\frac{33.3 \mathrm{x}{10}^{-23}\mathrm{g}}{8 \mathrm{x}{10}^{-24}{\mathrm{cm}}^3 } =41.6 \mathrm{g} {\mathrm{cm}}^{-3} \end{gathered}$$

density = 41.6 g cm^-3.

Formula mass of CuCl =63.5 +35.5

                                      =  99.0

The number of formula units per cell of ZnS is 4. It has face centred cubic structure.

$$\begin{gathered} we have \\ Z =4 \\ density \left(d\right) =3.4 g C{m}^{-3} \\ {N}_0 (Avogadro\text{'}s number) =6.023 x{10}^{23} \\ Mass =99.0 gram \\ Therefore u\sin g formula \\ density =\frac{Mass x Z}{unit cell volume(Cm{)}^3 x{N}_{avod.no.}} \\ d =\frac{M x Z}{a^3 x N_0} \\ {a}^3 =\frac{99 x4}{3.4 x6.023 x{10}^{23}} \\ {a}^3 =1.932 x{10}^{-22}c{m}^3 \\ unit cell length a= (1.932 x{10}^{-22}{)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ takinig \log both side \\ \log a =\frac{1}{3}\log 1.932 x{10}^{-22} \\ =\frac{1}{3}(0.2860-22) =\frac{1}{3}(-21.7140) \\ = -7.238 =0.762 x{10}^{-8} \\ taking anti\log \\ a =anti\log 0.762 x{10}^{-8} =5.758 x{10}^{-8} cm \\ = 5.758 x {10}^{-8}cm \end{gathered}$$

we have given,
Mass= 209 g
Number of  atom per unit cell =1 (Simple cubic)
density =91.5 g m^-3
N_A =6.023 x10²³
edge length of unit cell =?
 
By applying formula


$$\begin{gathered} \mathrm{density} =\frac{ \mathrm{Mass} \mathrm{x} \mathrm{Number} \mathrm{of} \mathrm{atom} \mathrm{per} \mathrm{unit} \mathrm{cell}}{\mathrm{volume} \mathrm{of} \mathrm{unit} \mathrm{cell} \mathrm{x} \mathrm{avogadro}\text{'}\mathrm{s} \mathrm{number}} \\ \mathrm{d}= \frac{\mathrm{M} \mathrm{x} \mathrm{Z}}{{\mathrm{a}}^3 \mathrm{x} {\mathrm{N}}_{\mathrm{A}}} \\ 91.5 =\frac{ 209 \mathrm{x} 1}{{\mathrm{a}}^3 \mathrm{x} 6.023 \mathrm{x} {10}^{23}} \\ {\mathrm{a}}^3 =\frac{209 \mathrm{x} 1}{91.5 \mathrm{x} 6.023 \mathrm{x} {10}^{23}} \end{gathered}$$

a= 15.59 x10^-8 cm

we know that

$\mathrm{d} =\frac{\mathrm{ZM}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}$

as we have given 
d= 7.80 g cm^–3.
M=56 g mol^–1
a= 290 pm or a³ =2.43 x10^-23
putting all value in above equation we get,

$$\begin{gathered} \mathrm{Z} =\frac{{\mathrm{a}}^3 \mathrm{x} \mathrm{d} {\mathrm{xN}}_{\mathrm{A}}}{\mathrm{M}} \\ \mathrm{Z}=\frac{ 2.43 \mathrm{x}{10}^{-23} \mathrm{x} 7.80 \mathrm{x}6.022 \mathrm{x}{10}^{23}}{56} =2.03 \end{gathered}$$

hence it belong to bcc crystal lattice.

b) Zinc oxide is white in colour at room temperature. On heating it loses electron and turns yellow in colour. 

$\mathrm{ZnO}\stackrel{∆}{\to } {\mathrm{Zn}}^{2+} +\frac{1}{2}{\mathrm{O}}_2 +2{\mathrm{e}}^{-}$ 

The electron liberate after heating, can act as charge carrier and thus on heating zinc oxide it exhibit electrical conductivity.