# A metal has BCC structure and edge length of it's unit cell is 3.04A. the volume of the unit cell in cm will be

Dear Student,
GIVEN: a=3.04 angstroms =$2.04×{10}^{-8}cm$
BCC structure so Z=2 ,
$d=\frac{Z×M}{{a}^{3}×{N}_{A}}$
we know that d= $\frac{M}{V}$
putting this value in the first equation we get
$\frac{M}{V}=\frac{Z×M}{{a}^{3}×{N}_{A}}$
cancelling M on both sides we get
$V=\frac{{a}^{3}×{N}_{A}}{Z}$
putting the given values in the above equation we get
$V=\frac{\left(3.04×{10}^{-8}{\right)}^{3}×6.023×{10}^{23}}{2}$
V= 8.46 cm3

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