The nearest neighbour Ag atoms in the silver crystal are 2.87 x 10^–10 m apart. What is the density of silver? Silver crystallises in fcc form.

we have given :
mass of silver = 107.87 g
Z(FCC)          = 4
distance between nearest neighbour Ag atoms = 2.87 x10^-10

$$\begin{gathered} \frac{\mathrm{face} \mathrm{diagonal} }{2} =2.87 x{10}^{-10} m \\ \frac{\sqrt{2a}}{2} =2.87 x{10}^{-10} m \\ a= 2.87 x{10}^{-10} x \sqrt{2} \\ a= 2.87 x{10}^{-10} x 1.414 \\ a= 4.05 x {10}^{-10} \\ now u\sin g formula \\ d=\frac{ Z x M}{a^3 x N_A} \\ d =\frac{4 x 107.87}{(4.05 x 10{)}^3 x 6.02 x{10}^{23}} \\ d= 10.84 g c{m}^{-3} \end{gathered}$$

we have give that aluminium forms Fcc cubic crystal.
density = 2.7g cm^-3
Mass of aluminium = 27g
number of unit cell =4 (FCC)
we have find edge length
Thus using formula

$$\begin{gathered} \mathrm{d} =\frac{\mathrm{Z} \mathrm{xM}}{{\mathrm{a}}^3 \mathrm{x} {\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{putting} \mathrm{the} \mathrm{value} \mathrm{in} \mathrm{this} \mathrm{equation} \\ 2.7 =\frac{4 \mathrm{x}27}{6.023 \mathrm{x} {10}^{23} \mathrm{x} {\mathrm{a}}^3} \\ {\mathrm{a}}^3 = \frac{4 \mathrm{x} 27 }{6.023 \mathrm{x} 10{ }^{23} \mathrm{x} 2.7} \\ \mathrm{a} =4.05 \mathrm{x} {10}^{-8} \end{gathered}$$

length of the unit cell edge = 400Pm =400x10^-10 cm
 volume of the unit cell  = ( 400 x10^-10cm)³
                                 = 6.4 x 10^-23 cm^-3

as the element A forms a body centred  cubic lattice so no of atom per units cell is 2

z= 2 atoms unit cell

atomic mass of the element  =100 g/mol
density of element is given by

$$\begin{gathered} \mathrm{md}= \frac{\mathrm{M} \mathrm{xZ}}{{\mathrm{a}}^3 \mathrm{xN}{}_{\mathrm{A}}} \\ \mathrm{d}= \frac{100 \mathrm{x}2}{6.4 \mathrm{x} {10}^{-23} \mathrm{x} 6.023 \mathrm{x} 10{ }^{23}} = 51.88 \mathrm{g}/{\mathrm{cm}}^3 \\ \mathrm{volume} \mathrm{of} 10 \mathrm{g} \mathrm{of} \mathrm{A} = \frac{\mathrm{mass}}{\mathrm{Density} } \\ = \frac{10\mathrm{g}}{5.188\mathrm{g} {\mathrm{cm}}^{-3}} \\ = 1.9275 {\mathrm{cm}}^3 \\ \mathrm{number} \mathrm{of} \mathrm{unit} \mathrm{cell} \mathrm{in} 1.9275 {\mathrm{cm}}^3 \mathrm{volume} =\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{substance} }{\mathrm{unit} \mathrm{cell} \mathrm{volume} } \\ = 3.0 \mathrm{x}{10}^{22} \mathrm{unit} \mathrm{cell} \end{gathered}$$

we have given,.
mass =108 g
edge (a) = 409pm  
a³ = 6.84 x10²³ cm³
N_a =6.023 x10^-233
Z =4

Apply formula

$$\begin{gathered} \mathrm{d}=\frac{ \mathrm{Z} \mathrm{xM}}{{\mathrm{a}}^3 \mathrm{x} {\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{d} = \frac{4 \mathrm{x} 108}{6.84 \mathrm{x} {10}^{23} \mathrm{x} 6.02 \mathrm{x}{10}^{-23}} = 10.98 \mathrm{g} {\mathrm{cm}}^{-3} \end{gathered}$$

density =  10.98 g cm^–3

we know that

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

as given that,
density= 2.75 g cm^–3
a = 654 pm or a³ = 2.79 x 10^-23
mass of KBr is 119 gram



$Z =\frac{2.75 x6.54 x{10}^{-23} x6.022 x{10}^{23}}{119} =4$