An element (atomic weight $=250 \mathrm{u}$ ) crystallises in a simple cubic lattice. If the density of the unit cell is $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$. What is the radius (in $\mathop {\rm{A}}\limits^{\rm{o}}$ ) of the atom of the element?

$$ \left(N_A=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right) $$

A compound is formed by two elements $A$ and $B$. Atoms of the element $B$ (as anion) make ccp lattice and those of element $A$ (as cation) occupy all tetrahedral voids. The formula of the compound is

An element crystallises in bcc lattice. The atomic radius of the element is $2.598 \mathop {\rm{A}}\limits^{\rm{o}}$. What is the volume (in $\mathrm{cm}^3$ ) of one unit cell?

Gold crystallises in fcc lattice. The edge length of the unit cell is $4 \mathop {\rm{A}}\limits^{\rm{o}}$. The closest distance between gold atoms is ' $x$ ' $\mathop {\rm{A}}\limits^{\rm{o}}$ and density of gold is ' $y$ ' $\mathrm{g} \mathrm{cm}^{-3}$. What are $x$ and $y$ respectively?

(Molar mass of gold $=197 \mathrm{~g} \mathrm{~mol}^{-1} ; N=6 \times 10^{23} \mathrm{~mol}^{-1}$ )