Question
$$CsBr$$ crystallises in a body centred cubic lattice. The unit cell length is $$436.6\,pm.$$ Given that the atomic mass of $$Cs = 133\,u$$ and that of $$Br = 80\,u$$ and Avogadro number being $$6.023 \times {10^{23}}mo{l^{ - 1}},$$ the density of $$CsBr$$ is
A.
$$42.5\,g/c{m^3}$$
B.
$$0.425\,g/c{m^3}$$
C.
$$8.25\,g/c{m^3}$$
D.
$$4.25\,g/c{m^3}$$
Answer :
$$4.25\,g/c{m^3}$$
Solution :
$${\text{Density of }}CsBr = \frac{{Z \times M}}{{{a^3} \times {N_0}}}$$
$$Z \to $$ number of atoms in the bec unit cell = 2
$$M \to $$ molar mass of $$CsBr$$ = 133 + 80 =213
$$a \to $$ edge length of unit cell = 436.6 $$pm$$
$$ = 436.6 \times {10^{ - 10}}cm$$
$$\therefore {\text{Density}} = $$ $$\frac{{2 \times 213}}{{{{\left( {436.6 \times {{10}^{ - 10}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}$$
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8.49 \times {10^{ - 7}} \times {10^7}g/c{m^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8.50\,g/c{m^3} \cr} $$
$${\text{For a unit cell = }}\frac{{8.50}}{2} = 4.25\,g/c{m^3}$$