$A$ metal $(M)$ crystallizes in an $fcc$ lattice with an edge length of $4.242 \mathring{A}$. What is the radius of the $M$ atom (in $\mathring{A}$)?

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is :- ............ $g \, mol^{-1}$ ( $N_A$ Avogadro constant $= 6.02 \times 10^{23} \, mol^{-1}$ )

An element crystallises in an $fcc$ unit cell with a cell edge length of $3.608 \times 10^{-8} \text{ cm}$. The density of the element is $8.92 \text{ g cm}^{-3}$. Calculate the atomic mass of the element $(N_A = 6.022 \times 10^{23} \text{ mol}^{-1})$. (in $\text{ g/mol}$)

$CsBr$ crystallises in a body-centred cubic lattice. The unit cell length is $436.6 \, pm$. Given that the atomic mass of $Cs = 133$ and that of $Br = 80 \, amu$ and Avogadro number being $6.02 \times 10^{23} \, mol^{-1}$,the density of $CsBr$ is .............. $g/cm^{3}$.

Gold crystallises in $fcc$ lattice. The edge length of the unit cell is $4 \ \mathring{A}$. The closest distance between gold atoms is '$x$' $\mathring{A}$ and density of gold is '$y$' $g \ cm^{-3}$. What are $x$ and $y$ respectively?
$($ Molar mass of gold $= 197 \ g \ mol^{-1} ; N_A = 6 \times 10^{23} \ mol^{-1} )$

$A$ copper complex crystallising in a $CCP$ lattice with a cell edge of $0.4518 \ nm$ has been revealed by employing $X$-ray diffraction studies. The density of the copper complex is found to be $7.62 \ g \ cm^{-3}$. The molar mass of the copper complex is $..... \ g \ mol^{-1}$. (Nearest integer)
[Given : $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$]