A metal crystallizes in such a lattice in which only $$70$$% of the total space of the crystal is occupied by the atoms. If the atomic mass of the metal is $$32\pi$$ $$g/mol$$ and the atomic radius is $$0.2nm$$, the density of the metal is
C
$$7.0g/{ cm }^{ 3 }\quad $$
Correct option is B. $$3.5g/{ cm }^{ 3 }$$
$$M= 32\pi g/mol$$, Packing Efficiency $$=0.70, \, R=0.2 \times 10^{-7} cm$$
$$D=\cfrac {ZM}{VN_A}$$..........(1)
$$\text {Packing Efficiency} = \cfrac {Z \times (4/3) \pi R^3}{V}$$ ..(2)
Divide eqn. (1) by (2)-
$$D/ 0.7 = \cfrac {M}{N_A \times (4/3) \pi R^3} $$
$$D= 0.70 \times \cfrac {32 \pi}{N_A \times (4/3) \pi R^3}= 0.70 \times \cfrac {24}{6.022 \times 10^{23} \times (0.2 \times 10^{-7})^3} $$
$$= 3.487 gcm^{-3}$$
Option B is correct.