Question

Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of metals A and B are 6.17 and 8.00 g/cm3 , respectively, and their respective atomic weights are 171.3 and 162.0 g/mol, determine whether the crystal structure for this alloy is simple cubic, facecentered cubic, or body-centered cubic. Assume a unit cell edge length of 0.332 nm

Answer 1

Answer:

Simple cubic

Explanation:

The density of metal A (ρa) = 6.17 g/cm³, The density of metal B (ρb) = 8 g/cm³, The atomic weight of metal A (Aa) = 171.3 g/mol, The atomic weight of metal B (Ab) = 162 g/mol, the unit cell edge length (a) = 0.332 nm, concentration of metal A (Ca) = 25%, concentration of metal B (Cb) = 75%

The average density is given by:

$\rho_{ave}=\frac{100}{\frac{C_a}{\rho_a} +\frac{C_b}{\rho_b} } \\\\\rho_{ave}=\frac{100}{\frac{25}{6.17} +\frac{75}{8} } =7.45\ g/cm^3\\\\The\ average\ atomic\ weight\ is:\\\\A_{ave}=\frac{100}{\frac{C_a}{A_a} +\frac{C_b}{A_b} } \\\\A_{ave}=\frac{100}{\frac{25}{171.3} +\frac{75}{162} } =164.23\ g/mol\\\\The\ number\ of\ atoms\ per\ unit(n)\ is:\\\\n=\frac{\rho_{ave}*a^3*N_A}{A_{ave}} \\\\N_A=Avogadro\ constant=6.02*10^{22} \ mol^{-1},a=0.332\ nm=3.32*10^{-8}cm\\\\Substituting:$

$n=\frac{\rho_{ave}*a^3*N_A}{A_{ave}} =\frac{7.45*(3.32*10^{-8})^3*6.02*10^{23}}{164.23} \\\\n=0.999$

n≅1

Since n≅1, the crystal structure for this alloy is simple cubic