Question

The density of argon (face centered cubic cell) is $1.83g/c{m}^3$ at ${20}^{o}C$. What is the length of an edge a unit cell?

• A. $0.599nm$
• B. $0.569nm$
• C. $0.525nm$
• D. $0.551nm$

Answer 1

The correct option is C $0.525nm$

Density $d=\frac{ZM}{N{a}^3}$Density $d=\frac{ZM}{N{a}^3}$

$Z=4=$ number of atoms of Ar in one fcc unit cell
$M=40g/mol=$ atomic weight of argon.

$N=6.023\times {10}^{23}atoms/mol=$ Avogadro's number.
$a=$ edge length

Density $d=\frac{ZM}{N{a}^3}$

$1.83g/c{m}^3=\frac{4atoms\times 40g/mol}{6.023\times {10}^{23}atoms/mol\times a^3}$

$a^3=\frac{4atoms\times 40g/mol}{6.023\times {10}^{23}atoms/mol\times 1.83g/c{m}^3}$

$a^3=1.45\times {10}^{-22}c{m}^3$

$a=5.25\times {10}^{-8}cm$

Convert unit from cm to nm

$a=5.25\times {10}^{-8}cm\times \frac{{10}^{7}nm}{1cm}$

$a=0.525nm$