Question

A solid consisting of only $X$-atoms has a close packed structure with $X-X$ distance of $160pm$ . 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 $1c{m}^3$ 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)

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