# 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

GIVEN: a=3.04 angstroms =$2.04\times {10}^{-8}cm$

BCC structure so Z=2 ,

$d=\frac{Z\times M}{{a}^{3}\times {N}_{A}}$

we know that d= $\frac{M}{V}$

putting this value in the first equation we get

$\frac{M}{V}=\frac{Z\times M}{{a}^{3}\times {N}_{A}}$

cancelling M on both sides we get

$V=\frac{{a}^{3}\times {N}_{A}}{Z}$

putting the given values in the above equation we get

$V=\frac{(3.04\times {10}^{-8}{)}^{3}\times 6.023\times {10}^{23}}{2}$

V= 8.46 cm

^{3}

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