Question

A solid consisting of only $X$-atoms has a close packed structure with $X-X$ distance of $160\,pm$ . Assuming it to be a closed packed structure of hard spheres with radius equal to half of the $X-X$ bond length, the number of atoms in $1 \,cm^{3}$ would be

• A. $6.023 \times 10^{27}$
• B. $3.45 \times 10^{23}$
• C. $6.02 \times 10^{21}$
• D. $3.8 \times 10^{21}$

Answer 1

Answer: (B)

$d =160\,pm$
$\frac{d}{2}=\frac{a}{2\sqrt{2}}=r$ (for ccp or fcc)
$a=d \sqrt{2}$
$V=a^{3}=\left(d \sqrt{2}\right)^{3}=2\sqrt{2}d^{3}$
$V=2 \times 1.414 \times (160 \times 10^{-10}cm)^{3}$
$V=11583488 \times 10^{-30}\,cm^{3}$
$V=1.1583488 \times 10^{-23}\,cm^{3}$
$1.158 \times 10^{-23}\,cm^{3}$ contain 4 atom
$1\,cm^{3}$ contain $\frac{4}{1.158 \times 10^{-23}}$ atoms
$=3.45 \times 10^{23}$ atoms