Question

While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theoretical density (Equation 3.8) and atomic weight. What is the volume of the zirconium unit cell in cubic meters?

Answer 1

Answer:

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

$z=12x\frac{1}{6}+2x\frac{1}{2}+3=6$

z = 6 atoms per unit cell

M = 91.224 g/mol

N = $6.023x10^{23} atoms/mol$

d = $6.51g/cm^{3}$

$V_{c}=\frac{zxM}{dxN}$

$V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }$

$V_{c}=1.396x10^{-22} cm^{3} /unit.cell$

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$